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The monopolist’s free boundary problem in the plane

Robert J. McCann and Cale Rankin and Kelvin Shuangjian Zhang Department of Mathematics, University of Toronto, Toronto Ontario M5S 2E4 Canada [email protected] School of Mathematics, Monash University, 9 Rainforest Walk Melbourne 3800, Australia [email protected] School of Mathematical Sciences and Center for Applied Mathematics, Fudan University, 220 Handan Road, Yangpu District, Shanghai 200433, P.R. China [email protected]
Abstract.

We study the Monopolist’s problem with a focus on the free boundary separating bunched from unbunched consumers, especially in the plane, and give a full description of its solution for the family of square domains {(a,a+1)2}a0\{(a,a+1)^{2}\}_{a\geq 0}. The Monopolist’s problem is fundamental in economics, yet widely considered analytically intractable when both consumers and products have more than one degree of heterogeneity. Mathematically, the problem is to minimize a smooth, uniformly convex Lagrangian over the space of nonnegative convex functions. What results is a free boundary problem between the regions of strict and nonstrict convexity. Our work is divided into three parts: a study of the structure of the free boundary problem on convex domains in 𝐑n{\mathbf{R}}^{n} showing that the product allocation map remains Lipschitz up to most of the fixed boundary and that each bunch extends to this boundary; a proof in 𝐑2{\mathbf{R}}^{2} that the interior free boundary can only fail to be smooth in one of four specific ways (cusp, high frequency oscillations, stray bunch, nontranversal bunch); and, finally, the first complete solution to Rochet and Choné’s example on the family of squares Ω=(a,a+1)2\Omega=(a,a+1)^{2}, where we discover bifurcations first to targeted and then to blunt bunching as the distance a0a\geq 0 to the origin is increased. We use techniques from the study of the Monge–Ampére equation, the obstacle problem, and localization for measures in convex-order.

MSC 2020: 35R35 49N10 91B41 Secondary: 35Q91 49Q22 90B50 91A65 91B43
Robert McCann’s work was supported in part by the Canada Research Chairs program CRC-2020-00289, Natural Sciences and Engineering Research Council of Canada Discovery Grants RGPIN–2020–04162, and a grant from the Simons Foundation (923125, McCann). Cale Rankin’s was supported in part by postdoctoral fellowships from the Fields Institute for Research in Mathematical Sciences and the University of Toronto. Kelvin Shuangjian Zhang’s was supported by a Start-Up Grant from Fudan University. The authors are grateful to Krzysztof Ciosmak and Ting-Kam Leonard Wong for fruitful conversations. ©

1. Introduction

The Monopolist’s problem is a principal-agent model for making decisions facing asymmetric information; it has fundamental importance in microeconomic theory. A simple form from [50] capturing multiple dimensions of heterogeneity that we rederive below is to

(1) {minimizeL[u]:=Ω(12|Dux|2+u)𝑑x,over 𝒰:={u:Ω¯𝐑;u is nonnegative and convex}.\begin{dcases}&\text{minimize}\quad L[u]:=\int_{\Omega}\left(\frac{1}{2}|Du-x|^{2}+u\right)\,dx,\\ &\text{over }\quad\mathcal{U}:=\{u:\overline{\Omega}\rightarrow\mathbf{R};u\text{ is nonnegative and convex}\}.\end{dcases}

We always take Ω𝐑n\Omega\subset\mathbf{R}^{n} to be open and convex with compact closure Ω¯\overline{\Omega}. Our goal in this paper is to elucidate the properties of solutions to this minimization problem, with an eventual focus on the two-dimensional setting. Because the minimization takes place over the set of convex functions the problem has a free boundary structure. The free boundary separates the region where the function is convex, but not strictly convex, from the region where the function is strictly convex. In this paper we refine our understanding of the free boundary structure in all dimensions by showing regions where uu is not strictly convex always extend to the fixed boundary Ω\partial\Omega. We show the regularity known for uu often extends from the interior to the fixed boundary. In two-dimensions, we show in a neighbourhood of a certain class of free boundary points (that we call tame), the minimizer solves the classical obstacle problem. From this we obtain the tame free boundary is locally piecewise Lipschitz except at accumulation points of its local maxima; it has Hausdorff dimension strictly less than two and is the graph of a function continuous 1\mathcal{H}^{1}-almost everywhere. We also establish a boot strapping procedure: if the free boundary is suitably Lipschitz, it is CC^{\infty}. As an application of our techniques we completely describe the solution on the square domains Ω=(a,a+1)2𝐑2\Omega=(a,a+1)^{2}\subset\mathbf{R}^{2} with a0a\geq 0. Despite significant numerical [7, 17, 18, 19, 25, 42, 44, 48], and analytic attempts [50, 35, 41] the description of the solution on the square has previously remained incomplete, and has come to be regarded as analytically intractable [7, 19, 34]. At least on the plane, we rebut this view by confirming for a722a\geq\frac{7}{2}-\sqrt{2} the solution recently hypothesized by McCann and Zhang [41]. We show how their solution can also be modified to accommodate smaller values of aa and other convex, planar domains. We show that the nature of the bunching undergoes unanticipated changes — from absent to targeted to blunt — as a0a\geq 0 is increased. We rigorously prove the support Du(Ω¯)Du(\overline{\Omega}) of the unknown distribution of products consumed has a lower boundary which is concave nondecreasing — as the above-mentioned numerics and stingray description suggest — and that all products selected by more than one type of consumer lie on this boundary or its reflection through the diagonal.

The problem (1) arises from the question of how a monopolist should price goods for optimal profit in the face of information asymmetry. Here is a simple derivation of (1). We assume a closed set of products Ω𝐑n\Omega^{*}\subset\mathbf{R}^{n} where each coordinate represents some attribute of the product, and an open set of consumers Ω𝐑n\Omega\subset\mathbf{R}^{n} where each coordinate represents some attribute of the consumer. Consumers are distributed according to a Borel probability measure μ𝒫(Ω)\mu\in\mathcal{P}(\Omega). The monopolist’s goal is to determine a price v(y)v(y) at which to sell product yy in a way which maximizes their profit. If consumer xΩx\in\Omega attains benefit b(x,y)b(x,y) from product yΩy\in\Omega^{*} then the consumer will choose the product yy which maximizes their utility

(2) u(x):=supyΩb(x,y)v(y).u(x):=\sup_{y\in\Omega^{*}}b(x,y)-v(y).

Provided it exists, we denote the yy which realizes this supremum by Yu(x)Yu(x). Assuming the monopolist pays cost c(y)c(y) for product yy, then the monopolist’s goal is to maximize their profit, the integral of price they sell for minus cost they pay,

Ω[v(Yu(x))c(Yu(x))]𝑑μ(x).\int_{\Omega}[v(Yu(x))-c(Yu(x))]\,d\mu(x).

The problem has been considered in this generality in e.g. [3] [12] [30] [38], and for even more general (non quasilinear) utility functions in [47], [39]. For this paper, to highlight the mathematical properties of most interest, we adopt several standard simplifying assumptions proposed by Rochet and Choné [50]: that products lie in the nonnegative orthant Ω=[0,)n\Omega^{*}=[0,\infty)^{n} and the monopolist’s direct cost to produce them is quadratic c(y)=|y|2/2c(y)=|y|^{2}/2, furthermore, that consumers are uniformly distributed on their domain Ω𝐑n\Omega\subset{\mathbf{R}}^{n} and their product preferences are bilinear b(x,y)=xyb(x,y)=x\cdot y. In this case (2) implies uu is the Legendre transform of vv (and thus a convex function), Yu(x)=u(x)Yu(x)=\partial u(x), and when Ω[0,)n\Omega\subset[0,\infty)^{n} the Monopolist’s goal becomes to maximize

Ω(xDu(x)u(x)|Du(x)|22)𝑑x,\int_{\Omega}\left(x\cdot Du(x)-u(x)-\frac{|Du(x)|^{2}}{2}\right)\,dx,

over nonnegative convex functions which, up to an irrelevant constant, is the problem (1). Since convex functions are differentiable almost everywhere the integrand is well-defined. The nonnegativity constraint on uu represents the additional requirement that v(0)=0v(0)=0, meaning consumers need not consume if the monopolist raises prices too high, or equivalently, are always free to pay nothing by choosing the zero product as an outside option.

This problem was first considered by Mussa and Rosen in the one-dimensional setting [46], (after related models of taxation [45], matching [4], and signaling [53] were introduced and analyzed by Mirrlees, Becker and Spence). The multidimensional problem was considered by Wilson [55] and Armstrong [2], while our formulation above is essentially that of Rochet and Choné [50]. Although this model is of significant importance to economists it presents serious mathematical difficulties. Indeed, were there only the nonnegativity constraint in (1) we would have a variant of the obstacle problem [8, 29]; with the nonnegativity and convexity constraint we have a free boundary problem for three different regions. In the region where the function is positive but not strictly convex, the fundamental tool of two-sided perturbation by an arbitrary test function is no longer applicable. As a result, until recent work of the first and third authors [41] it has not even been possible to write down the Euler–Lagrange equation in the region of nonstrict convexity. Despite this, other aspects of the problem have been studied, notably by Rochet and Choné [50], who derived a necessary and sufficient condition for optimality in terms of convex-ordering between the positive and negative parts of the variational derivative of the objective functional, Basov [3] who advanced a control theoretic approach to such problems, Carlier [12, 13] who considered existence and first-order conditions for the minimizer, and Carlier and Lachand-Robert [14, 20, 16] who studied regularity and gave a description of the polar cone.

In this paper we prove results of mathematical and economic interest. We invoke tools from diverse areas of mathematics: the theory of sweeping and convex orders of measures, Monge–Ampére equations, regularity theory for the obstacle problem, and the theory of optimal transport (which has deep links to the Monopolist’s problem). We also indicate a striking connection to the classical obstacle problem: Locally the minimizer uu solves an obstacle problem where the obstacle is the minimal convex extension of uu from its region of nonstrict convexity. We now outline our results.

If uu solves (1) and Ω\Omega is a convex open subset of 𝐑n\mathbf{R}^{n} it is known from the work of Rochet–Choné and Carlier–Lachand-Robert that uC1(Ω¯)u\in C^{1}(\overline{\Omega}) and from the work of Caffarelli and Lions [9] (see [38]) that uCloc1,1(Ω)u\in C^{1,1}_{\text{loc}}(\Omega). Any convex function uC1(Ω¯)u\in C^{1}(\overline{\Omega}) partitions Ω\Omega according to its sets of contact with supporting hyperplanes; these sets are convex. Namely for each x0Ω¯x_{0}\in\overline{\Omega} set

(3) px0(x)\displaystyle p_{x_{0}}(x) =u(x0)+Du(x0)(xx0),\displaystyle=u(x_{0})+Du(x_{0})\cdot(x-x_{0}),
andx0~\displaystyle\text{and}\quad\tilde{x_{0}} ={xΩ¯;u(x)=px0(x)}.\displaystyle=\{x\in\overline{\Omega};u(x)=p_{x_{0}}(x)\}.

Here x~\tilde{x} is the equivalence class of xx under the equivalence relation x0x1x_{0}\sim x_{1} if and only if Du(x0)=Du(x1)Du(x_{0})=Du(x_{1}). We call an equivalence class trivial if x~={x}\tilde{x}=\{x\}, in which case we say uu is strictly convex at xx. We call equivalence classes leaves, since they foliate the interior of Ωi\Omega_{i}. They are also called isochoice sets [23] or bunches if nontrivial [50]. We also call one-dimensional leaves rays. We set

(4) Ωi={xΩ¯;x~ is (ni)-dimensional}.\Omega_{i}=\{x\in\overline{\Omega};\tilde{x}\text{ is ($n-i$)-dimensional}\}.

Thus, for example, Ωn\Omega_{n} consists of all points at which uu is strictly convex and Ω0\Omega_{0} consists of all points xx lying in the closure of some open set on which uu is affine. These disjoint sets partition Ω¯\overline{\Omega} and our first result describes the qualitative behavior in each set.

Theorem 1.1 (Partition into foliations by leaves that extend to the boundary).

Let uu solve (1) where Ω𝐑n\Omega\subset{\mathbf{R}}^{n} is bounded, open and convex. Then

  1. (1)

    If Ω0\Omega_{0}\neq\emptyset then Ω0={xΩ¯;u(x)=0}\Omega_{0}=\{x\in\overline{\Omega};u(x)=0\} hence is closed and convex.

  2. (2)

    Ω1,,Ωn1\Omega_{1},\dots,\Omega_{n-1} are a union of equivalence classes on which uu is affine and each such equivalence class intersects the boundary Ω\partial\Omega.

  3. (3)

    ΩnΩ\Omega_{n}\cap\Omega is an open set on which uC(Ωn)u\in C^{\infty}(\Omega_{n}) solves Δu=n+1\Delta u=n+1.

It is immediate from the definition that Ωi\Omega_{i} for 1in11\leq i\leq n-1 are a union of nontrivial equivalence classes; the key conclusion is these extend to the boundary (i.e. if xΩix\in\Omega_{i} for 1in11\leq i\leq n-1 then x~Ω\tilde{x}\cap\partial\Omega\neq\emptyset). The PDE in point (3) has been considered in more generality by Rochet and Choné [50]. Note the economic interpretation of (1): no bunches of positive measure are sold apart from the null product; as we shall see during the proof of (2), any product sold to more than one consumer lies on the boundary of the set of products sold. Thus the entire interior of the set of products sold consists of individually customized products. Our proof of Theorem 1.1 and some subsequent results requires a new proposition asserting that a.e. on the boundary of a convex domain Ω\Omega, the minimizer of (1) satisfies the boundary condition (Du(x)x)𝐧0(Du(x)-x)\cdot\mathbf{n}\geq 0 where 𝐧\mathbf{n} is the outer unit normal to Ω\Omega. Established in Proposition 2.3, it can be interpreted to mean that the normal component of any boundary distortion in product selected can never be inward. Moreover, in convex polyhedral domains and certain other situations, we are able to extend the interior regularity uCloc1,1(Ω)u\in C^{1,1}_{\text{loc}}(\Omega) of Caffarelli and Lions [9] [38] to the smooth parts of the fixed boundary, Theorem 4.1.

The remainder of our results are restricted to the planar case Ω𝐑2\Omega\subset\mathbf{R}^{2}. Theorem 1.1 provides a complete description of the solution in Ω0\Omega_{0} and Ω2\Omega_{2}: it remains to better understand the behavior of the solution in Ω1\Omega_{1} as well as the properties of the domains Ω1\Omega_{1} and Ω2\Omega_{2}, (noting Ω0\Omega_{0} is, by Theorem 1.1, a closed convex set).

By Theorem 1.1, the free boundary Γ:=Ω1Ω2Ω\Gamma:=\partial\Omega_{1}\cap\partial\Omega_{2}\cap\Omega between Ω1\Omega_{1} and Ω2\Omega_{2} consists only of points in rays which foliate Ω1\Omega_{1} and extend to Ω\partial\Omega. In §5 we prove that the Neumann condition

(5) (Du(x0)x0)𝐧=0,(Du(x_{0})-x_{0})\cdot\mathbf{n}=0,

where 𝐧\mathbf{n} is the outer unit normal to the fixed boundary Ω{\partial}\Omega, can be used to characterize the presence of these rays. Namely if (5) is not satisfied at x0Ωx_{0}\in\partial\Omega then x0~{x0}\tilde{x_{0}}\neq\{x_{0}\}. Conversely, if xΩx\in\partial\Omega lies in a boundary neighbourhood Bε(x)ΩB_{\varepsilon}(x)\cap\partial\Omega on which the Neumann condition (5) is satisfied then xx is a point of strict convexity for uu. The remaining case — rays x~0{x0}\tilde{x}_{0}\neq\{x_{0}\} which satisfy (5) — is subtle: we call such rays stray and conjecture the union 𝒮\mathcal{S} of stray rays has zero area in general, but are able to show it is empty only for the squares Ω=(a,a+1)2\Omega=(a,a+1)^{2}.

Let x1x_{1} be a point in the free boundary Γ\Gamma which, necessarily, lies on the ray x1~\tilde{x_{1}}. Let x0=x1~Ωx_{0}=\tilde{x_{1}}\cap\partial\Omega be the boundary endpoint of x1~\tilde{x_{1}}. Note that provided Ω\partial\Omega is C1C^{1} in a neighbourhood of x0x_{0} and (Du(x0)x0)𝐧>0(Du(x_{0})-x_{0})\cdot\mathbf{n}>0, the same inequality holds for all xBε(x0)Ωx\in B_{\varepsilon}(x_{0})\cap\partial\Omega and thus such xx are also the boundary endpoints of nontrivial rays. In this case we call x1x_{1} a tame free boundary point (and x~1\tilde{x}_{1} a tame ray); we denote the set of tame free boundary points by 𝒯\mathcal{T}.

Theorem 1.2 (Regularity results for the free boundary).

Let uu solve (1) where Ω𝐑2\Omega\subset{\mathbf{R}}^{2} is a bounded, open, and convex with smooth boundary except possibly at finitely many points. For every x1𝒯x_{1}\in\mathcal{T} and x0x~1Ωx_{0}\in\tilde{x}_{1}\cap{\partial}\Omega there is ε>0\varepsilon>0 such that
(1) Γ:=Ω1Ω2Ω\Gamma:={\partial}\Omega_{1}\cap{\partial}\Omega_{2}\setminus{\partial}\Omega has Hausdorff dimension less than 22 in Bε(x1)B_{\varepsilon}(x_{1});
(2) the function D(x):=diam(x~)D(x):=\text{diam}(\tilde{x}) is continuous on Bε(x0)ΩAB_{\varepsilon}(x_{0})\cap{\partial}\Omega\setminus A, where AA denotes the accumulation points of DD’s local maxima;
(3) Γ{xx~:xBε(x0)ΩA}\Gamma\cap\{x^{\prime}\in\tilde{x}:x\in B_{\varepsilon}(x_{0})\cap{\partial}\Omega\setminus A\} is Lipschitz, except at those local maxima of DD where the Lebesgue density of Ω1\Omega_{1} happens to vanish;
(4) if DD is Lipschitz on Ω{\partial}\Omega near x0x_{0}, then a bootstrapping procedure yields ΓBε(x1)\Gamma\cap B_{\varepsilon}(x_{1}) is a CC^{\infty} curve and uC(Bε(x1)int Ω1)u\in C^{\infty}(B_{\varepsilon}(x_{1})\cap\text{int }\Omega_{1}).

Since we don’t know whether local maxima of D(x)=diam(x)D(x)=\text{diam}(x) can be dense, we augment (2) by proving the function ΩxD(x)\partial\Omega\ni x\mapsto D(x) is continuous 1\mathcal{H}^{1}-almost everywhere on the fixed boundary in Theorem 7.2. Establishing the Lipschitz regularity of DD, which permits the above-mentioned bootstrapping to a CC^{\infty} free boundary, remains an interesting open problem.

Remark 1.3 (Lipschitzianity, convexity, and smoothness).

Note the Lipschitz requirement on D|Bε(x0)ΩD|_{B_{\varepsilon}(x_{0})\cap{\partial}\Omega} from (4) is not necessarily satisfied even when the corresponding portion of Γ\Gamma lies in a Lipschitz submanifold given by (3). If {xx~:xBε(x0)Ω}\{x^{\prime}\in\tilde{x}:x^{\prime}\in B_{\varepsilon}(x_{0})\cap{\partial}\Omega\} happens to be convex this distinction disappears for every smaller value of ε\varepsilon; simulations [44] suggest this occurs in the square examples from Theorem 1.4. Thus if the region Ω1+\Omega_{1}^{+} depicted in Figure 1 is convex, as Mirebeau’s simulations lead us to conjecture, Theorem 1.2 guarantees the curved portion of its boundary is smooth (away from Ω{\partial}\Omega).

Theorem 1.2 is proved using new coordinates for the problem, new Euler–Lagrange equations, and a new observation: That in a neighbourhood of a tame free boundary point the difference between the minimizer uu and the minimal convex extension of u|Ω1u|_{\Omega_{1}} solves the classical obstacle problem. A priori, the obstacle is C1,1C^{1,1}, i.e. has a merely LL^{\infty} Laplacian, and thus, without first improving the regularity of u|Ω1u|_{\partial\Omega_{1}}, the above results are the best one can obtain from the theory of the obstacle problem.

For general convex domains it is difficult to study the structure of the stray set 𝒮=Γ𝒯\mathcal{S}=\Gamma\setminus\mathcal{T} which may include points in the relative interior of rays. We also have not ruled out, in general, that the relative boundary of

Ω:={xΩ;(Dux)𝐧0},\partial\Omega_{\neq}:=\{x\in\partial\Omega;(Du-x)\cdot\mathbf{n}\neq 0\},

in Ω{\partial}\Omega might have positive 1\mathcal{H}^{1}-measure and the corresponding free boundary be nonsmooth. However, in specific cases, a more complete description is possible. For example, on the squares Ω=(a,a+1)2\Omega=(a,a+1)^{2}, we show Ω\partial\Omega_{\neq} is a single connected component of Ω\partial\Omega. In fact, we are able to provide the first explicit and complete description of the solution on Ω=(a,a+1)2\Omega=(a,a+1)^{2}, including an unexpected trichotomy for a=0a=0, aa sufficiently small, and aa sufficiently large. To describe the solution we label the edges of Ω\Omega by their compass direction and set

ΩN=[a,a+1]×{a+1}\displaystyle\Omega_{N}=[a,a+1]\times\{a+1\} ΩE={a+1}×[a,a+1]\displaystyle\Omega_{E}=\{a+1\}\times[a,a+1]
ΩW={a}×[a,a+1]\displaystyle\Omega_{W}=\{a\}\times[a,a+1] ΩS=[a,a+1]×{a}.\displaystyle\Omega_{S}=[a,a+1]\times\{a\}.

The minimizer is described by the following bifurcation theorem (visualized in Figure 1).

Ω0\Omega_{0}Ω2\Omega_{2}
(a) a=0a=0
Ω0\Omega_{0}Ω1\Omega_{1}^{-}Ω1+\Omega_{1}^{+}Ω2\Omega_{2}
(b) a>0a>0 sufficiently small
Ω0\Omega_{0}Ω10\Omega_{1}^{0}Ω1\Omega_{1}^{-}Ω1+\Omega_{1}^{+}Ω2\Omega_{2}
(c) a722a\geq\frac{7}{2}-\sqrt{2}
Figure 1. Bifurcation to targeted then blunt bunching as distance a0a\geq 0 to zero is increased.
Theorem 1.4 (Blunt bunching is a symptom of a seller’s market).

Let uu solve (1) with Ω=(a,a+1)2\Omega=(a,a+1)^{2} where a0a\geq 0. Then

  1. (1)

    Ω0\Omega_{0} is a convex set which includes a neighbourhood of (a,a)(a,a) in Ω¯\overline{\Omega}.

  2. (2)

    The portion of Ω1\Omega_{1} consisting of rays having both endpoints on the boundary Ω{\partial}\Omega is connected and denoted by Ω10\Omega_{1}^{0}. It is nonempty when a7222.1a\geq\frac{7}{2}-\sqrt{2}\approx 2.1. All rays in Ω10\Omega_{1}^{0} are orthogonal to the diagonal with one endpoint on ΩW\Omega_{W} and the other on ΩS\Omega_{S}. On the other hand there is ε0>0\varepsilon_{0}>0 such that Ω01\Omega_{0}^{1} is empty when a<ε0a<\varepsilon_{0}.

  3. (3)

    For all a>0a>0 there are exactly two disjoint connected components of Ω1Ω10\Omega_{1}\setminus\Omega_{1}^{0}. In these regions, each ray has only one endpoint x0Ωx_{0}\in{\partial}\Omega on the boundary; it lies in ΩSΩW\Omega_{S}\cup\Omega_{W}, violates the Neumann condition (5), and the solution uu is described by the Euler–Lagrange equations of McCann and Zhang [41]; c.f. (7)–(14). For a=0a=0, Ω1\Omega_{1} is empty.

  4. (4)

    The set Ω2\Omega_{2} of strict convexity of uu contains ΩEΩN\Omega_{E}\cup\Omega_{N} and the Neumann condition (5) holds at each x0Ω2Ωx_{0}\in\Omega_{2}\cap{\partial}\Omega apart from the 3 vertices.

The following corollary may be of purely mathematical interest from the point of view of calculus of variations and partial differential equations; the smoothness asserted follows from the results of [10, 8].

Corollary 1.5 (Convexity of solution to, and contact set for, an obstacle problem).

For Ω=(1,1)2\Omega=(-1,1)^{2}, the minimizer of L(u)L(u) over non-negative functions 0uW1,2(Ω¯)0\leq u\in W^{1,2}(\overline{\Omega}) is convex. Its zero set is smooth, convex, has positive area, and is compactly contained in the centered square Ω\Omega.

Remark 1.6 (Concave nondecreasing profile of stingray’s tail).

Numerical simulations of the square example show the region Du(Ω)Du(\Omega) of products consumed to be shaped like a stingray, e.g. Figure 1 of [19]. Theorem 1.4 combines with Lemma 8.4 below to provide a rigorous proof that the lower edge of stingray is concave non-decreasing — as the simulations suggest — while Theorem 1.1 shows that every product selected by more than one type of consumer lies on this boundary or its mirror image across the diagonal.

Remark 1.7 (Absence and ordering of blunt vs targeted bunching).

The potential absence of blunt bunching from the square example — established on a nonempty interval a(0,ε0)a\in(0,\varepsilon_{0}) by the preceding theorem — has been overlooked in all previous investigations that we are aware of. It can be understood as a symptom of a buyer’s market, in which a lack of enthusiasm on the part of qualified buyers incentivizes the monopolist to sell to fewer buyers but cater more to the tastes of those who do buy. The persistence of targeted bunching Ω1±\Omega_{1}^{\pm}\neq\emptyset for all a>0a>0 reflects the need to transition continuously from vanishing Neumann condition (5) — satisfied on the customization region Ω2(ΩSΩW)\Omega_{2}\cap(\Omega_{S}\cup\Omega_{W}) where uu is strictly convex — to the uniformly positive Neumann condition (Dux)𝐧=a(Du-x)\cdot\mathbf{n}=a on the exclusion region Ω0\Omega_{0} where uu vanishes, in light of the known regularity uC1(Ω¯)u\in C^{1}(\overline{\Omega}) [15][50]. Such bunching is neither needed nor present when a=0a=0: in this case x0𝐧=ax_{0}\cdot\mathbf{n}=a on ΩSΩW\Omega_{S}\cup\Omega_{W} shows the Neumann conditions in Ω0\Omega_{0} and Ω2\Omega_{2} coincide. When the blunt bunching region Ω10\Omega_{1}^{0} is present, our proof of Theorem 1.4 shows it separates Ω0\Omega_{0} from Ω1±\Omega_{1}^{\pm}, which in turn separate all but one point of Ω10\Omega_{1}^{0} from Ω2\Omega_{2}. In particular, blunt bunching implies Ω0\Omega_{0} is a triangle, which is exceedingly rare in its absence.

Assuming Ω2\Omega_{2} is Lipschitz (or at least has finite perimeter), we arrive at a characterization of the solution to (1) on Ω=(a,a+1)2\Omega=(a,a+1)^{2} for every value of a0a\geq 0. Namely u𝒰u\in\mathcal{U} minimizes (1) if and only if (A) bunching is absent (Ω1=\Omega_{1}=\emptyset, as for a=0a=0), in which case uu solves 13Δu=1{u>0}\frac{1}{3}\Delta u=1_{\{u>0\}}, i.e the classical obstacle problem [49, 29, 27] and (Dux)𝐧=0(Du-x)\cdot\mathbf{n}=0 on Ω\partial\Omega (Figure 1(a)), or (B) bunching is present but blunt bunching is absent, (Ω10=Ω1\Omega_{1}^{0}=\emptyset\neq\Omega_{1}, as for a1a\ll 1) in which case we derive below necessary conditions, whose sufficiency can be confirmed as in [41] (Figure 1(b)), or (C) blunt bunching is present, (Ω10\Omega_{1}^{0}\neq\emptyset, as for a722a\geq\frac{7}{2}-\sqrt{2}) (Figure 1(c)), in which case the sufficient conditions for a minimum established by two of us [41] are also shown to be necessary below. (The only gap separating the necessary from the sufficient condition is the question of whether or not Ω2\Omega_{2} must have finite perimeter.)

If instead (B) blunt bunching is absent but bunching is present, Ω10=Ω1\Omega_{1}^{0}=\emptyset\neq\Omega_{1}, Theorem 1.4 asserts that Ω1=Ω1+Ω1\Omega_{1}=\Omega_{1}^{+}\cap\Omega_{1}^{-} splits into two connected components

(6) Ω1±\displaystyle\Omega_{1}^{\pm} :={(x1,x2)Ω1Ω10:±(x1x2)>0},\displaystyle:=\{(x_{1},x_{2})\in\Omega_{1}\setminus\Omega_{1}^{0}:\pm(x_{1}-x_{2})>0\},

placed symmetrically below and above the diagonal. The region Ω1\Omega_{1}^{-} and its reflection Ω1+\Omega_{1}^{+} below the diagonal are foliated by isochoice segments making continuously varying angles θ\theta with the horizontal. The limit of these segments is a segment of length R0>0R_{0}>0 lying on the boundary of the convex set Ω0\Omega_{0}, having endpoint (a,h0)(a,h_{0}) and making angle θ0[π/4,0)\theta_{0}\in[-\pi/4,0) with the horizontal.

Fix any closed convex neighbourhood Ω0\Omega_{0} of (a,a)(a,a) in Ω¯\overline{\Omega} which is reflection symmetric around the diagonal and contains such a segment in its boundary. We describe the solution u=u1u=u_{1}^{-} in Ω1\Omega_{1}^{-} using an Euler-Lagrange equation (9) from [41], rederived more simply in Section 8 below. Index each isochoice segment in Ω1\Omega_{1}^{-} by its angle θ(π4,0)\theta\in(-\frac{\pi}{4},0); (angles which are less than π/4-\pi/4 or non-negative are ruled out in the proof of Theorem 1.4). Let (a,h(θ))(a,{h(\theta)}) denote its left-hand endpoint and parameterize the segment by distance r[0,R(θ)]{r}\in[0,R(\theta)] to this boundary point (a,h(θ))(a,h(\theta)). Along the hypothesized length R(θ)R(\theta) of this segment assume uu increases linearly with slope m(θ)m(\theta) and offset b(θ)b(\theta):

(7) u1((a,h(θ))+r(cosθ,sinθ))=m(θ)r+b(θ).u_{1}^{-}\Big{(}(a,h(\theta))+{r}(\cos\theta,\sin\theta)\Big{)}={m(\theta)}{r}+{b(\theta)}.

Given the initial (angle, height) pair (θ0,h0)[π/4,0)×(a,a+1)(\theta_{0},h_{0})\in[-\pi/4,0)\times(a,a+1), and R:[θ0,0][0,2/2){R}:\left[\theta_{0},0\right]\to\left[0,\sqrt{2}/2\right) piecewise Lipschitz with R(θ0)=R0R(\theta_{0})=R_{0}, solve

(8) m(θ0)=0,m(θ0)=0such that\textstyle m(\theta_{0})=0,\qquad m^{\prime}(\theta_{0})=0\qquad\mbox{\rm such that}
(9) (m′′(θ)+m(θ)2R(θ))(m(θ)sinθm(θ)cosθ+a)=32R2(θ)cosθ;({m^{\prime\prime}(\theta)+m(\theta)}-{2R(\theta)})({m^{\prime}(\theta)}\sin\theta-{m(\theta)}\cos\theta+a)=\frac{3}{2}{R^{2}(\theta)\cos\theta};

then set

(10) h(θ)\displaystyle{h(\theta)} =\displaystyle= h0+13θ0θ(m′′(ϑ)+m(ϑ)2R(ϑ))dϑcosϑ,\displaystyle h_{0}+\frac{1}{3}\int_{\theta_{0}}^{\theta}(m^{\prime\prime}(\vartheta)+m(\vartheta)-2{R(\vartheta)})\frac{d\vartheta}{\cos\vartheta},
(11) b(θ)\displaystyle{b(\theta)} =\displaystyle= θ0θ(m(ϑ)cosϑ+m(ϑ)sinϑ)h(ϑ)𝑑ϑ.\displaystyle\int_{\theta_{0}}^{\theta}(m^{\prime}(\vartheta)\cos\vartheta+m(\vartheta)\sin\vartheta)h^{\prime}(\vartheta)d\vartheta.

Given (θ0,h0,R0)(\theta_{0},h_{0},R_{0}) and R()R(\cdot) as above, the triple (m,b,h)(m,b,h) satisfying (9)–(11) exists and is unique on the interval where R()>0R(\cdot)>0. Thus the shape of Ω1\Omega_{1}^{-} and the value of u1u_{1}^{-} on it will be uniquely determined by Ω0\Omega_{0} and R:[θ0,0][0,2/2)R:\left[\theta_{0},0\right]\to\left[0,\sqrt{2}/2\right). We henceforth restrict our attention to choices of Ω0\Omega_{0} and R()R(\cdot) for which the resulting set Ω1\Omega_{1}^{-} lies above the diagonal. In this case Ω1+\Omega_{1}^{+} and the value of u=u1+u=u_{1}^{+} on Ω1+\Omega_{1}^{+} are determined by reflection symmetry x1x2x_{1}\leftrightarrow x_{2} across the diagonal. This defines u=u1u=u_{1} on Ω1\Omega_{1} and provides the boundary data on Ω1Ω2{\partial}\Omega_{1}\cap{\partial}\Omega_{2} needed for the mixed Dirichlet / Neumann boundary value problem for Poisson’s equation,

(12) {Δu2=3, on Ω2,(Du2(x)x)𝐧=0, on Ω2Ω,u2u1=0, on Ω2Ω1,u2=0 on Ω2Ω0,\displaystyle\begin{cases}\Delta u_{2}=3,&\text{ on }\Omega_{2},\\ (Du_{2}(x)-x)\cdot\mathbf{n}=0,&\text{ on }{\partial}\Omega_{2}\cap{\partial}\Omega,\\ u_{2}-u_{1}=0,&\text{ on }{\partial}\Omega_{2}\cap{\partial}\Omega_{1},\\ u_{2}=0&\text{ on }{\partial}\Omega_{2}\cap{\partial}\Omega_{0},\end{cases}

which determines u=u2u=u_{2} on Ω2:=Ω(Ω0Ω1)\Omega_{2}:=\Omega\setminus(\Omega_{0}\cup\Omega_{1}). The duality discovered in [41], implies that for at most one choice of Ω0\Omega_{0} and R()R(\cdot) Lipschitz can convex uu (pieced together from u1,u2u_{1},u_{2} as above with u0:=0u_{0}:=0 on Ω0\Omega_{0}) satisfy the supplemental Neumann conditions

(13) D(u2u1)𝐧^=0,\displaystyle D(u_{2}-u_{1})\cdot\hat{\mathbf{n}}=0, on Ω2Ω1\displaystyle\text{ on }{\partial}\Omega_{2}\cap{\partial}\Omega_{1}
(14) Du2𝐧^=0,\displaystyle Du_{2}\cdot\hat{\mathbf{n}}=0, on Ω2Ω0\displaystyle\text{ on }{\partial}\Omega_{2}\cap{\partial}\Omega_{0}

required on the free boundaries (since uC1(Ω¯)u\in C^{1}(\overline{\Omega}) [15][50]); here 𝐧^\hat{\mathbf{n}} denotes the outer unit normal to Ω2\Omega_{2} at xΩ2x\in{\partial}\Omega_{2}. In the course of proving Theorem 1.4 in Section 8 below we complete this circle of ideas — apart from the piecewise Lipschitz hypothesis which Theorem 1.2 falls just short of proving — by showing at least one such choice exists; this choice uniquely solves (1) on the square in case (B). In case (C), Theorem 1.4 shows at least (and therefore exactly [41]) one choice exists satisfying the free boundary problem from [40] in the analogous sense.

We conclude this introduction by outlining the structure of the paper. Section 2 contains preliminaries: the variational inequality associated to (1), some background on Alexandrov second derivatives, and localization results of Rochet–Choné. In Section 3 we prove Theorem 1.1 using perturbation techniques previously used to study the Monge–Ampère equation. In Sections 4 and 5 and we prove some technical results which facilitate our later work in Sections 6 and 7. First, in Section 4, a boundary C1,1C^{1,1} result which is new for this problem and extends the interior regularity result of Caffarelli and Lions [9] [38]. Next, in Section 5, propositions quantifying how at points of nonstrict convexity the Neumann boundary condition fails to be satisfied. Section 6 and Section 7 establish Theorem 1.2 using techniques from the study of the obstacle problem. Here we indicate a new connection to the classical obstacle problem. Namely, that the Monopolist’s problem gives rise to an obstacle problem where the obstacle is the minimal convex extension of the function defined on Ω1\Omega_{1}. The proof of Theorem 1.4 is completed in Section 8 using a case by case analysis based on a careful choice of coordinates. It confirms the economic intuition that the degree to which product selection (hence bunching) is influenced by the market presence of competing consumers decreases as we move away from the exclusion region, i.e. from the lower left toward the upper right region of the square, while on the other hand, increasing as we move the entire square of consumer types away from the outside option by increasing a0a\geq 0. We conclude with appendices containing some relevant background results. Table 1 contains a list of notation.

Notation Meaning
Ω\Omega A bounded open convex subset of 𝐑n\mathbf{R}^{n}.
Ω¯\overline{\Omega} The closure of Ω\Omega.
int Ω¯\text{int }\overline{\Omega} The interior of Ω¯\overline{\Omega}.
Ωc\Omega^{c} Set complement Ωc:=𝐑nΩ\Omega^{c}:=\mathbf{R}^{n}\setminus\Omega of Ω\Omega.
𝐧\mathbf{n} Outer unit normal at a point where Ω{\partial}\Omega is differentiable.
x~\tilde{x} Bunch x~:={zΩ¯;Du(z)=Du(x)}\tilde{x}:=\{z\in\overline{\Omega};Du(z)=Du(x)\}.
r.i.(x~)\mathop{\rm r.i.}(\tilde{x}) The relative interior of the convex set x~\tilde{x}.
Ωi\Omega_{i} Subset (4) of Ω¯\overline{\Omega} foliated by (nin-i)-dimensional bunches.
\subset\subset Compact containment.
v+v_{+} The positive part of a function, v+(x):=max{v(x),0}v_{+}(x):=\max\{v(x),0\}.
μ+\mu_{+} The positive part of a measure μ\mu.
sptf\text{spt}{f} The support of ff, i.e. sptf=closure{x;f(x)0}\text{spt}{f}=\text{closure}\{x;f(x)\neq 0\}.
𝒫(Ω)\mathcal{P}(\Omega) The set of Borel probability measures on Ω\Omega.
 \!\mathbin{\vrule height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule height=0.55974pt,depth=0.0pt,width=3.44444pt}\! Measure restriction: (μ A)(B)=μ(AB)(\mu\!\mathbin{\vrule height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule height=0.55974pt,depth=0.0pt,width=3.44444pt}\!A)(B)=\mu(A\cap B).
d\mathcal{H}^{d} dd-dimensional Hausdorff measure.
dx:=dn(x)dx:=d\mathcal{H}^{n}(x) nn-dimensional Lebesgue measure, i.e. volume measure.
dS:=dn1dS:=d\mathcal{H}^{n-1} Surface area measure (or arclength in special case n=2n=2).
Table 1. Table of notation.

2. Variational inequalities and Alexandrov second derivatives

2.1. Variational inequalities

Our basic tools for studying the unique minimizer of the functional

(15) L[u]=Ω(12|Dux|2+u)𝑑x,L[u]=\int_{\Omega}\left(\frac{1}{2}|Du-x|^{2}+u\right)\,dx,

over

𝒰={u:Ω¯𝐑; u is nonnegative and convex},\mathcal{U}=\{u:\overline{\Omega}\rightarrow\mathbf{R};\text{ $u$ is nonnegative and convex}\},

are the variational inequalities stated in the following lemma.

Lemma 2.1 (Variational inequalities).

Let uu solve (1). Let u¯𝒰\bar{u}\in\mathcal{U} be Lipschitz and w=u¯uw=\bar{u}-u. Then each of the following inequalities hold:

(16) 0\displaystyle 0 Lu(w):=Ω(n+1Δu)w𝑑x+Ω(Dux)𝐧w𝑑S,\displaystyle\leq L^{\prime}_{u}(w):=\int_{\Omega}(n+1-\Delta u)\,wdx+\int_{\partial\Omega}(Du-x)\cdot\mathbf{n}\,wdS,
(17) 0\displaystyle 0 Lu(u¯)=Ω(n+1Δu)u¯𝑑x+Ω(Dux)𝐧u¯𝑑S.\displaystyle\leq L^{\prime}_{u}(\bar{u})=\int_{\Omega}(n+1-\Delta u)\bar{u}\,dx+\int_{\partial\Omega}(Du-x)\cdot\mathbf{n}\,\bar{u}dS.

Moreover if DuDu¯Du\not\equiv D\bar{u} on a set of positive n\mathcal{H}^{n} measure, then

(18) 0\displaystyle 0 <Lu¯(w)=Ω(n+1Δu¯)w𝑑x+Ω(Du¯x)𝐧w𝑑S,\displaystyle<L^{\prime}_{\bar{u}}(w)=\int_{\Omega}(n+1-\Delta\bar{u})\,wdx+\int_{\partial\Omega}(D\bar{u}-x)\cdot\mathbf{n}\,wdS,

where Δu¯\Delta\bar{u} is interpreted as a Radon measure [26, Ch. 6] and Du¯𝐧D\bar{u}\cdot\mathbf{n} as the one-sided derivative limt0(u¯(x)u¯(xt𝐧))/t\lim\limits_{t\downarrow 0}(\bar{u}(x)-\bar{u}(x-t\mathbf{n}))/t which exists by convexity of u¯\bar{u}.

Proof.

We begin with (16). Let uu be the minimizer and observe 𝒰\mathcal{U} is convex. Thus for any u¯𝒰\bar{u}\in\mathcal{U}, w=u¯uw=\bar{u}-u and t[0,1]t\in[0,1] we have

L[u]L[u+tw],L[u]\leq L[u+tw],

so, in particular,

0\displaystyle 0 ddt|t=0L[u+tw]\displaystyle\leq\frac{d}{dt}\Big{|}_{t=0}L[u+tw]
(19) =Ω(Dux)Dw+wdx.\displaystyle=\int_{\Omega}(Du-x)\cdot Dw+w\,dx.

Note uC1(Ω¯)Cloc1,1(Ω)u\in C^{1}(\overline{\Omega})\cap C^{1,1}_{\text{loc}}(\Omega) with ii2u0{\partial}^{2}_{ii}u\geq 0 so we may apply the divergence theorem and obtain

0\displaystyle 0 Ω(n+1Δu)w𝑑x+Ω(Dux)𝐧w𝑑S.\displaystyle\leq\int_{\Omega}(n+1-\Delta u)\,wdx+\int_{\partial\Omega}(Du-x)\cdot\mathbf{n}\,wdS.

where 𝐧\mathbf{n} is the outer unit normal to Ω\Omega which exists n1\mathcal{H}^{n-1}-a.e. for the convex domain Ω\Omega.

Inequality (17) follows by performing the same argument with u+u¯u+\bar{u} in place of u¯\bar{u}. For (18) we perform similar calculations but use that h(t):=L[u¯tw]h(t):=L[\bar{u}-tw] is strictly convex with a minimum at 11 so h(0)<0h^{\prime}(0)<0. For the divergence theorem we take the one-sided directional derivatives and use that DuDu is of bounded variation [26, Ch. 6]. ∎

Remark 2.2.

(1) It is straightforward to see, again by arguing using a perturbation, that inequality (17) holds not just for u¯𝒰\bar{u}\in\mathcal{U} but for any convex u¯\bar{u} with sptu¯\text{spt}\,\bar{u}_{-} (the support of the negative part of u¯\bar{u}) disjoint from the set {u=0}\{u=0\}. The key observation is that for sufficiently small tt, u+t(u¯u)𝒰u+t(\bar{u}-u)\in\mathcal{U}.
(2) In any neighbourhood where uu is C2C^{2} and uniformly convex, that is uu satisfies an estimate D2uλI>0D^{2}u\geq\lambda I>0, one may perturb — as is standard in the calculus of variations — by smooth compactly supported functions and obtain Δu=3\Delta u=3 in the interior and (Dux)𝐧=0(Du-x)\cdot\mathbf{n}=0 on the fixed boundary of Ω\Omega. Without a local uniform convexity estimate, even for smooth functions u¯\bar{u}, it may be that there is no t>0t>0 small enough to ensure u+tu¯u+t\bar{u} is convex.

Inequality (18) is useful when one chooses u¯\bar{u} as paraboloid with prescribed Laplacian. We give an example now — the result we prove is required in subsequent sections. It is interesting to contrast the following result with the one-dimensional case, in which minimizers on domains Ω[0,)\Omega\subset[0,\infty) satisfy u(x)xu^{\prime}(x)\leq x.

Proposition 2.3 (Normal distortion is not inward).

Let uu solve (1) where Ω𝐑n\Omega\subset\mathbf{R}^{n} is bounded, open, and convex. Then for any x0Ωx_{0}\in\partial\Omega where the outer normal is defined,

(Du(x0)x0)𝐧0.(Du(x_{0})-x_{0})\cdot\mathbf{n}\geq 0.
Proof.

By approximation it suffices to prove the result for smooth strictly convex domains. Indeed, [30, Corollary 4.7] and its proof imply if Ω(k)Ω\Omega^{(k)}\supset\Omega is a sequence of smooth strictly convex approximating domains and u(k)u^{(k)} is the solution of (1) on Ω(k)\Omega^{(k)} then Du(k)(x)Du(x)Du^{(k)}(x)\rightarrow Du(x) at every xx where uu and each u(k)u^{(k)} is differentiable. The C1(Ω¯)C^{1}(\overline{\Omega}) result of [50, 15] implies this is every xΩ¯x\in\overline{\Omega}.

Thus we take Ω\Omega to be smooth and strictly convex. Up to a choice of coordinates we assume x0=0x_{0}=0 and 𝐧=e1\mathbf{n}=e_{1} (see Figure 2). Recall px0(x):=u(x0)+Du(x0)(xx0)p_{x_{0}}(x):=u(x_{0})+Du(x_{0})\cdot(x-x_{0}) is the affine support at x0x_{0}. For t>0t>0 sufficiently small and x=(x1,,xn)𝐑nx=(x^{1},\dots,x^{n})\in\mathbf{R}^{n} we consider the family of admissible perturbations (see Figure 2)

u^t(x)\displaystyle\hat{u}_{t}(x) :=n+12([x1+t]+)2+px0(x),\displaystyle:=\frac{n+1}{2}\big{(}[x^{1}+t]_{+}\big{)}^{2}+p_{x_{0}}(x),
u¯t(x)\displaystyle\bar{u}_{t}(x) :=max{u(x),u^t(x)},\displaystyle:=\text{max}\{u(x),\hat{u}_{t}(x)\},
Ωt\displaystyle\Omega_{t} :={xΩ¯;u¯t(x)>u(x)}.\displaystyle:=\{x\in\overline{\Omega};\bar{u}_{t}(x)>u(x)\}.

Note Ωt\Omega_{t} has positive measure since at x0=0x_{0}=0, u^t(0)=u(0)+(n+1)t2/2>u(0)\hat{u}_{t}(0)=u(0)+{(n+1)}t^{2}/2>u(0). Moreover if x1tx^{1}\leq-t then u¯t(x)u(x)\bar{u}_{t}(x)\leq u(x) and thus Ωt{xΩ¯;x1t}\Omega_{t}\subset\{x\in\overline{\Omega};x^{1}\geq-t\}. Thus, strict convexity of Ω\Omega implies Ω¯t{0}\overline{\Omega}_{t}\rightarrow\{0\} in Hausdorff distance [52, §1.8] as t0t\rightarrow 0. Clearly Δu¯t=n+1\Delta\bar{u}_{t}=n+1 on Ωt\Omega_{t}. To derive a contradiction assume (Du(x0)x0)𝐧<0(Du(x_{0})-x_{0})\cdot\mathbf{n}<0. Then for tt sufficiently small we also have

(20) (Du¯t(x)x)𝐧<0onΩtΩ,(D\bar{u}_{t}(x)-x)\cdot\mathbf{n}<0\quad\text{on}\quad\Omega_{t}\cap\partial\Omega,

which holds by the C1(Ω¯)C^{1}(\overline{\Omega}) continuity of uu and because |Du¯t(x0)Du(x0)|=O(t)|D\bar{u}_{t}(x_{0})-Du(x_{0})|=O(t). But (18) with Δu¯=n+1\Delta\bar{u}=n+1 implies

0<ΩtΩ(u¯tu)(Du¯t(x)x)𝐧𝑑S0<\int_{\Omega_{t}\cap\partial\Omega}(\bar{u}_{t}-u)(D\bar{u}_{t}(x)-x)\cdot\mathbf{n}\,dS

which contradicts (20) to conclude the proof.

(A)uu (minimizer) px0p_{x_{0}} (support) u^t\hat{u}_{t} (perturbation) e1e_{1}x0=0x_{0}=0x1=tx^{1}=-t(B)Ω\partial\Omegae1(=𝐧)e_{1}\,(=\mathbf{n})x0=0x_{0}=0{x1=0}\{x^{1}=0\}{x1=t}\{x^{1}=-t\}Ωt\Omega_{t}

Figure 2. Illustrates the constructions in the proof of Proposition 2.3. Subfigure (A) shows a cross-section of the minimizer uu, its support px0p_{x_{0}}, and the perturbation u^t\hat{u}_{t}. Subfigure (B) illustrates that because Ωt{x1t}\Omega_{t}\subset\{x^{1}\geq-t\} and Ω\Omega is strictly convex with outer normal 𝐧=e1\mathbf{n}=e_{1} at 0 we have Ωt¯{0}\overline{\Omega_{t}}\rightarrow\{0\} in the Hausdorff distance.

An essential tool is that the variational inequality (17) holds not only on Ω\Omega but restricted to the contact set x~\tilde{x} — at least for n\mathcal{H}^{n} almost every xx. This novel and powerful technique was pioneered in this context by Rochet and Chonè [50] who exploited the sweeping theory of measures in convex order. In this section we recall the statement of their localization result; for completeness we include a proof in Appendix A.

We introduce the following notation for the variational derivative σ:=δL/δu\sigma:=\delta L/\delta u of our objective L(u)L(u):

(21) dσ(x)=(n+1Δu)d2  Ω+(Dux)𝐧d1  Ω,d\sigma(x)=(n+1-\Delta u)d\mathcal{H}^{2}\!\mathbin{\vrule height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule height=0.55974pt,depth=0.0pt,width=3.44444pt}\!\Omega+(Du-x)\cdot\mathbf{n}d\mathcal{H}^{1}\!\mathbin{\vrule height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule height=0.55974pt,depth=0.0pt,width=3.44444pt}\!\partial\Omega,

which turns out to be a measure with finite total variation. The equivalence relation induced by DuDu, namely x1x2x_{1}\sim x_{2} if and only if Du(x1)=Du(x2)Du(x_{1})=Du(x_{2}) yields the partitioning of each Ωi\Omega_{i} into leaves. We let x~\tilde{x} denote the equivalence class of xx and can disintegrate σ\sigma by conditioning on this equivalence relation. Let the conditional measures σx~=σx~+σx~\sigma_{\tilde{x}}=\sigma_{\tilde{x}}^{+}-\sigma_{\tilde{x}}^{-} be defined by disintegrating separately the positive and negative parts of σ\sigma with respect to the given equivalence relation, we recall how

0x~v(z)𝑑σx~(z)0\leq\int_{\tilde{x}}v(z)d\sigma_{\tilde{x}}(z)

for all convex functions vv in Corollary A.9 of Appendix A below.

2.2. Legendre transforms and Alexandrov second derivatives

Recall if u:Ω𝐑u:\Omega\rightarrow\mathbf{R} is a convex function, then its Legendre transform is defined by

(22) v(y)=supxΩxyu(x).v(y)=\sup_{x\in\Omega}x\cdot y-u(x).

A function is called Alexandrov second differentiable at x0x_{0}, with Alexandrov Hessian D2u(x0)D^{2}u(x_{0}) (an n×nn\times n matrix), provided as xx0x\rightarrow x_{0} that

u(x)=u(x0)+Du(x0)(xx0)+12(xx0)TD2u(x0)(xx0)+o(|xx0|2).u(x)=u(x_{0})+Du(x_{0})\cdot(x-x_{0})+\frac{1}{2}(x-x_{0})^{T}D^{2}u(x_{0})(x-x_{0})+o(|x-x_{0}|^{2}).

Alexandrov proved convex functions are twice differentiable in this sense n\mathcal{H}^{n} almost everywhere.

It’s well known that if a differentiable convex function uu is Alexandrov differentiable at x0x_{0} and its Legendre transform is Alexandrov second differentiable at y0:=Du(x0)y_{0}:=Du(x_{0}) then

D2v(y0)=[D2u(x0)]1.D^{2}v(y_{0})=\big{[}D^{2}u(x_{0})\big{]}^{-1}.

We have an analogous result even when Alexandrov second differentiability is not assumed.

Lemma 2.4 (Legendre transform of Hessian bounds).

Assume u:Ω𝐑u:\Omega\rightarrow\mathbf{R} is a convex function with Legendre transform vv, that x0Ωx_{0}\in\Omega and MM is an invertible symmetric positive definite matrix. Assume y0u(x0)y_{0}\in\partial u(x_{0}). Then

u(x0+δx)u(x0)+y0δx+δxTMδx/2+o(|δx|2)asδx0,u(x_{0}+\delta x)\geq u(x_{0})+y_{0}\cdot\delta x+\delta x^{T}M\delta x/2+o(|\delta x|^{2})\quad{\rm as}\ \delta x\rightarrow 0,

if and only if

v(y0+δy)v(y0)+x0δy+δyTM1δy/2+o(|δy|2)asδy0.v(y_{0}+\delta y)\leq v(y_{0})+x_{0}\cdot\delta y+\delta y^{T}M^{-1}\delta y/2+o(|\delta y|^{2})\quad{\rm as}\ \delta y\rightarrow 0.
Proof.

We prove the “only if” statement; the “if” statement is proved similarly. Up to a choice of coordinates and subtracting an affine support we may assume x0,y0=0x_{0},y_{0}=0 and u(x0)=0u(x_{0})=0. Whereby we’re assuming

(23) u(x)xTMx/2+o(|x|2).u(x)\geq x^{T}Mx/2+o(|x|^{2}).

It is straightforward to see that (23) holds if and only if for every ε>0\varepsilon>0 there is a neighbourhood 𝒩ε\mathcal{N}_{\varepsilon} of 0 on which

(24) u(x)(1ε)xTMx/2.u(x)\geq(1-\varepsilon)x^{T}Mx/2.

Now, for yu(𝒩ε)y\in\partial u(\mathcal{N}_{\varepsilon}) we have

v(y)\displaystyle v(y) =supxΩxyu(x)\displaystyle=\sup_{x\in\Omega}x\cdot y-u(x)
=supx𝒩εxyu(x)\displaystyle=\sup_{x\in\mathcal{N}_{\varepsilon}}x\cdot y-u(x)
supx𝒩εxy(1ε)xTMx/2.\displaystyle\leq\sup_{x\in\mathcal{N}_{\varepsilon}}x\cdot y-(1-\varepsilon)x^{T}Mx/2.

Provided yy lies in the possibly smaller neighbourhood Yε:=[(1ε)M𝒩ε]u(𝒩ε)Y_{\varepsilon}:=[(1-\varepsilon)M\mathcal{N}_{\varepsilon}]\cap\partial u(\mathcal{N}_{\varepsilon}) the supremum is obtained at x=[(1ε)M]1yx=[(1-\varepsilon)M]^{-1}y and

v(y)yT[(1ε)M]1y/2.v(y)\leq y^{T}[(1-\varepsilon)M]^{-1}y/2.

The aforementioned equivalence between (24) and (23) gives the desired result. ∎

3. Partition into foliations by leaves that intersect the boundary

In this section we present the proof of Theorem 1.1. We use localization (Corollary A.9) to obtain the vanishing Neumann condition throughout ΩnΩ\Omega_{n}\cap{\partial}\Omega, but apart from that the only technique we use is energy comparison, and the variational inequalities (16) – (17) coupled with careful choice of the comparison functions, many of which are inspired by those used to study the Monge–Ampère equation [28, 36]. We denote a subsection to each point of Theorem 1.1.

3.1. Point 1: Ω0={u=0}\Omega_{0}=\{u=0\} if nonempty.

We shall only prove Ω0{u=0}\Omega_{0}\subset\{u=0\}; equality follows easily if Ω0\Omega_{0} is nonempty, which is known to be true on strictly convex domains [2].

Proof of Theorem 1.1 (1).

For a contradiction we assume there is x0Ω0x_{0}\in\Omega_{0} with u(x0)>0u(x_{0})>0. Applying Rochet–Chonè’s localization with v=uv=u we have

(25) 0x0~(n+1Δu)u𝑑x+x0~Ωu(Dux)𝐧𝑑S.0\leq\int_{\tilde{x_{0}}}(n+1-\Delta u)u\,dx+\int_{\tilde{x_{0}}\cap\partial\Omega}u(Du-x)\cdot\mathbf{n}\,dS.

We plan to show equality holds. By assumption u(x)=px0(x)=u(x0)+Du(x0)(xx0)u(x)=p_{x_{0}}(x)=u(x_{0})+Du(x_{0})\cdot(x-x_{0}) on x0~\tilde{x_{0}}. Since px0(x)-p_{x_{0}}(x) is convex and x0Ω{u=0}x_{0}\in\Omega\setminus\{u=0\} we may apply Corollary A.9 with v=px0v=p_{x_{0}} and obtain

(26) 0x0~(n+1Δu)u𝑑x+x0~Ωu(Dux)𝐧𝑑S.0\geq\int_{\tilde{x_{0}}}(n+1-\Delta u)u\,dx+\int_{\tilde{x_{0}}\cap\partial\Omega}u(Du-x)\cdot\mathbf{n}\,dS.

Using Δu=0\Delta u=0 on x0~\tilde{x_{0}}, (25) and (26) imply

(27) 0=(n+1)x0~u𝑑x+x0~Ωu(Dux)𝐧𝑑S.0=(n+1)\int_{\tilde{x_{0}}}u\,dx+\int_{\tilde{x_{0}}\cap\partial\Omega}u(Du-x)\cdot\mathbf{n}\,dS.

We know u0u\geq 0 and (Dux)𝐧0(Du-x)\cdot\mathbf{n}\geq 0 on Ω\partial\Omega (Proposition 2.3) and n(x~0)>0\mathcal{H}^{n}(\tilde{x}_{0})>0. Therefore (27) implies u=0u=0 on x0~\tilde{x_{0}} and this contradiction completes the proof. ∎

3.2. Point 3 of Theorem 1.1

Now we present point 3: that Δu=n+1\Delta u=n+1 in Ωn\Omega_{n} and that ΩnΩ\Omega_{n}\cap\Omega is open. One can immediately obtain Δu=n+1\Delta u=n+1 a.e. in Ωn\Omega_{n} via Rochet–Choné’s localization (Corollary A.9). However for point 3 of Theorem 1.1 we require in addition, an inequality for Δu(x0)\Delta u(x_{0}) at all points where Du(x0)intDu(Ω)Du(x_{0})\in\text{int}Du(\Omega). Thus we prove the following lemma directly using perturbations.

Lemma 3.1 (Sub- and super-Poisson for interior vs. customized consumption).

Assume u:Ω𝐑u:\Omega\rightarrow\mathbf{R} solves (1) and x0Ωx_{0}\in\Omega is a point of Alexandrov second differentiability satisfying Du(x0)int(Du(Ω))Du(x_{0})\in\mathop{\rm int}(Du(\Omega)). Then

  1. (1)

    There holds Δu(x0)n+1\Delta u(x_{0})\geq n+1.

  2. (2)

    If, in addition uu is strictly convex at x0x_{0}, then Δu(x0)n+1\Delta u(x_{0})\leq n+1.

Proof.

(1) We take x0Ωx_{0}\in\Omega with Du(x0)intDu(Ω)Du(x_{0})\in\mathop{\rm int}Du(\Omega) assumed to be a point of Alexandrov second differentiability. For convenience translate and subtract the affine support at x0x_{0} so that x0,u(x0)x_{0},u(x_{0}) and Du(x0)Du(x_{0}) all vanish.

For a contradiction assume Δu(x0)<n+1\Delta u(x_{0})<n+1 and take ε>0\varepsilon>0 satisfying Δu(x0)+nε<n+1\Delta u(x_{0})+n\varepsilon<n+1. There is a neighborhood of x0=0x_{0}=0 on which u(x)<xT(D2u(x0)+εI)x/2u(x)<x^{T}(D^{2}u(x_{0})+\varepsilon I)x/2.

Let vv denote the Legendre transform (22) of uu and set v~=yT[D2u(0)+εI]1y/2\tilde{v}=y^{T}[D^{2}u(0)+\varepsilon I]^{-1}y/2. Lemma 2.4 implies v~<v\tilde{v}<v in a punctured neighborhood of the origin. Thus as h0h\rightarrow 0 the connected component of {x;v<v~+h}\{x;v<\tilde{v}+h\} containing the origin, which we denote Ωh\Omega^{*}_{h}, converges to {0}\{0\} in the Hausdorff distance. Set

(28) vh(y)={v~(y)+hyΩhv(y)yΩh.v_{h}(y)=\begin{cases}\tilde{v}(y)+h&y\in\Omega^{*}_{h}\\ v(y)&y\not\in\Omega^{*}_{h}.\end{cases}

Let uhu_{h} be the Legendre transform of vhv_{h}. Note uhuu_{h}\leq u and this inequality is strict at x0x_{0}. Moreover because D2vh(D2u(0)+εI)1D^{2}v_{h}\geq(D^{2}u(0)+\varepsilon I)^{-1} at each yΩhy\in\Omega^{*}_{h} Lemma 2.4 implies ΔuhΔu(x0)+nε<n+1\Delta u_{h}\leq\Delta u(x_{0})+n\varepsilon<n+1 on the set {uh<u}\{u_{h}<u\}. This contradicts inequality (18) to establish (1), where Ω{uh=u}\partial\Omega\subset\{u_{h}=u\} for hh small enough has been used.

(2) Now suppose, in addition, x0x_{0} is a point of strict convexity for uu and, for a contradiction, that Δu(x0)>n+1\Delta u(x_{0})>n+1. Set u¯(x)=(1ε)xTD2u(0)x/2\bar{u}(x)=(1-\varepsilon)x^{T}D^{2}u(0)x/2 with ε>0\varepsilon>0 chosen so small that Δu¯=(1ε)Δu(x0)>n+1\Delta\bar{u}=(1-\varepsilon)\Delta u(x_{0})>n+1. Note that u¯<u\bar{u}<u in a punctured neighborhood of 0; (this relies on the strict convexity of uu at 0 in the case D2u(0)D^{2}u(0) has a zero eigenvalue). Thus for sufficiently small h>0h>0 the connected component of {x;u(x)<u¯(x)+h}\{x;u(x)<\bar{u}(x)+h\} containing x0=0x_{0}=0, which we call Ω(h)\Omega^{(h)}, converges to {0}\{0\} in the Hausdorff distance. Set

u¯h={u¯(x)+hxΩ(h),u(x)xΩ(h).\bar{u}_{h}=\begin{cases}\bar{u}(x)+h&x\in\Omega^{(h)},\\ u(x)&x\not\in\Omega^{(h)}.\end{cases}

Then u¯h\bar{u}_{h} is an admissible interior perturbation of uu with Δu¯h>n+1\Delta\bar{u}_{h}>n+1 on {u¯h>u}\{\bar{u}_{h}>u\}. Once again we contradict inequality (18). ∎

At any interior point of strict convexity, xΩnΩx\in\Omega_{n}\cap\Omega, we have Du(x)intDu(Ω)Du(x)\in\mathop{\rm int}Du(\Omega) since u{\partial}u is closed. It’s now immediate that Δu=n+1\Delta u=n+1 at each point of Alexandrov second differentiability in Ωn\Omega_{n}. It remains to show ΩnΩ\Omega_{n}\cap\Omega is open. We prove in the next subsection that Du(i=0n1Ωi)Du(Ω)Du(\bigcup^{n-1}_{i=0}\Omega_{i})\subset\partial Du(\Omega), that is if xi=0n1Ωix\in\bigcup^{n-1}_{i=0}\Omega_{i} then Du(x)Du(x) is in the boundary of the set of gradients. Combined with the Cloc1,1C^{1,1}_{\text{loc}} regularity of uu we obtain if xΩnΩx\in\Omega_{n}\cap\Omega the same is true for all sufficiently close x¯\bar{x} (this is because Du(x¯)Du(\bar{x}) is also in intDu(Ω)\mathop{\rm int}Du(\Omega)) and this is a sufficient condition for x¯ΩnΩ\bar{x}\in\Omega_{n}\cap\Omega. We conclude ΩnΩ\Omega_{n}\cap\Omega is open.

Since we now know uu is a Wloc2,W^{2,\infty}_{\text{loc}} (equivalently, Cloc1,1C^{1,1}_{\text{loc}}) solution of Δu=n+1\Delta u=n+1 almost everywhere on the open set ΩnΩ\Omega_{n}\cap\Omega the elliptic regularity [31, Theorem 9.19] implies uC(Ωn)u\in C^{\infty}(\Omega_{n}).

3.3. Point 2 of Theorem 1.1

To conclude the proof of Theorem 1.1 we show if xΩix\in\Omega_{i} for i=0,,n1i=0,\dots,{n-1} then x~\tilde{x} extends to the boundary.

For a contradiction assume otherwise. Because uu is convex, then there exists x0x_{0} with {x0}x0~Ω\{x_{0}\}\neq\tilde{x_{0}}\subset\subset\Omega and y0:=Du(x0)intDu(Ω)y_{0}:=Du(x_{0})\in\mathop{\rm int}Du(\Omega) since u{\partial}u is closed. Because uu is C1,1C^{1,1} and the sections

Ω(h):={xΩ¯;u(x)uh(x):=u(x0)+Du(x0)(xx0)+h}\Omega^{(h)}:=\{x\in\overline{\Omega};u(x)\leq u_{h}(x):=u(x_{0})+Du(x_{0})\cdot(x-x_{0})+h\}

converge to x0~\tilde{x_{0}} in the Hausdorff distance as h0h\rightarrow 0, we obtain Du(Ω(h))Du(Ω)Du(\Omega^{(h)})\subset\subset Du(\Omega) for hh sufficiently small. In particular by Lemma 3.1 we have Δun+1\Delta u\geq n+1 in Ω(h)\Omega^{(h)}. Using u^=max{u,uh}\hat{u}=\max\{u,u_{h}\} as a perturbation function in inequality (16) we see Δu=n+1\Delta u=n+1 almost everywhere in Ω(h)\Omega^{(h)} (if Δu>n+1\Delta u>n+1 on a subset of {uh>u}\{u_{h}>u\} with positive measure, inequality (16) is violated). As in the previous subsection the elliptic regularity implies uC(Ω(h))u\in C^{\infty}(\Omega^{(h)}) is a classical solution of Δu=n+1\Delta u=n+1 in Ω(h)\Omega^{(h)}. Differentiating this PDE twice implies the second derivatives of uu are harmonic (and nonnegative by convexity). The strong maximum principle for harmonic functions says in fact jj2u>0{\partial}^{2}_{jj}u>0 in Ω(h)\Omega^{(h)} for all j=1,2,,nj=1,2,...,n, so uu cannot be affine on x~\tilde{x}. This contradiction completes the proof. Note our use of the strong maximum principle requires jj2u>0{\partial}^{2}_{jj}u>0 at some point in Ω(h)\Omega^{(h)}. This, however, follows by considering v=uuhv=u-u_{h}. If jj2u=0{\partial}^{2}_{jj}u=0 throughout Ωh\Omega^{h} then Dv(x0)=0Dv(x_{0})=0, v(x0)=hv(x_{0})=-h and vv is independent of xjx_{j}, hence ΩhΩ\Omega^{h}\cap{\partial}\Omega\neq\emptyset, which would contradict Ωhx~0Ω\Omega^{h}\to\tilde{x}_{0}\subset\subset\Omega as h0h\downarrow 0.

In the course of the above proof we’ve proved the following lemma which we record here since we require it again and again.

Lemma 3.2 (Interior regularity and strong maximum principle).

Assume u:Ω𝐑u:\Omega\rightarrow\mathbf{R} is a Cloc1,1C^{1,1}_{\text{loc}} convex function. Let C𝐑C\in\mathbf{R}. Assume, in the sense of Alexandrov second derivatives, Δu=C\Delta u=C almost everywhere in Bε(x0)ΩB_{\varepsilon}(x_{0})\subset\Omega. Then uC(Bε(x0))u\in C^{\infty}(B_{\varepsilon}(x_{0})) and satisfies that for each unit vector ξ\xi either uξξ0u_{\xi\xi}\equiv 0 throughout Bε(x0)B_{\varepsilon}(x_{0}) or uξξ>0u_{\xi\xi}>0 throughout Bε(x0)B_{\varepsilon}(x_{0}).

4. Product selection remains Lipschitz up to the boundary

One of our key techniques for studying the planar Monopolist’s problem is the introduction of a new coordinate system defined in terms of the rays which foliate Ωn1\Omega_{n-1}. These new coordinates are a powerful tool for studying the behaviour of the minimizer uu but their justification requires two significant technicalities. The first is a boundary regularity result in arbitrary dimensions, namely that in convex polyhedral domains uu is C1,1C^{1,1} up to the boundary (away from the nondifferentiabilities of the boundary); furthermore, in smooth convex domains uu is C1,1C^{1,1} on the set of rays having only one end on the boundary (Theorem 4.1). The second required technicality, proved in Section 5, is an equivalence between the Neumann condition and strict convexity stated more precisely in Propositions 5.3 and 5.4. Readers who are interested primarily in the consequences of these technicalities rather than their proof may proceed directly to Section 6.

Boundary regularity beyond uC1(Ω¯)u\in C^{1}(\overline{\Omega}) [15][50] for the Monopolist’s problem is new. Previously only an interior C1,1C^{1,1} result was known [9, 38] and C1,1C^{1,1} regularity is known to be sharp.

Theorem 4.1 (Boundary C1,1C^{1,1} regularity on convex polyhedral domains).

Let uu minimize (1) where Ω𝐑n\Omega\subset\subset\mathbf{R}^{n} is open, bounded, and convex.

  1. (1)

    There is CC depending only on Ω\Omega such that if x0Ωn1Ωnx_{0}\in\Omega_{n-1}\cup\Omega_{n} is a point of Alexandrov second differentiability and x0~Ω\tilde{x_{0}}\cap\partial\Omega a singleton or empty then

    Δu(x0)C.\Delta u(x_{0})\leq C.
  2. (2)

    Assume, in addition, Ω\Omega is a convex polyhedron (i.e. an intersection of finitely many half spaces). Let Ωε\Omega_{\varepsilon} be Ω\Omega excluding an ε\varepsilon-ball about each point where Ω\partial\Omega is not smooth. Then there is CC depending only on ε\varepsilon and Ω\Omega such that

    uC1,1(Ωε)C.\|u\|_{C^{1,1}(\Omega_{\varepsilon})}\leq C.
Proof.

The key energy comparison ideas are inspired by Caffarelli and Lions’s proof of interior regularity [9] and its generalization [38]. However new ideas are required for perturbation near the boundary. We prove there exists CC depending only on ε\varepsilon and Ω\Omega such that for all x0Ωεx_{0}\in\Omega_{\varepsilon} there holds

(29) lim supr0supBr(x0)|upx0|r2C.\limsup_{r\rightarrow 0}\frac{\sup_{B_{r}(x_{0})}|u-p_{x_{0}}|}{r^{2}}\leq C.

Equivalently, there is some r0r_{0} such that for all r<r0r<r_{0} there holds supBr(upx0)Cr2\sup_{B_{r}}(u-p_{x_{0}})\leq Cr^{2}. We emphasize that r0r_{0} will be chosen small depending on quantities which the constant CC in (29) is not permitted to depend on, however this does not affect the C1,1C^{1,1} estimate111This is analogous to an estimate |f(x+h)f(x)h|C\left|\frac{f(x+h)-f(x)}{h}\right|\leq C for all sufficiently small hh implying |f(x)|C|f^{\prime}(x)|\leq C regardless of what dictates our small choice of hh. It is interesting to note such an approach would not work for boundary Hölder or C1,αC^{1,\alpha} estimates with 0<α<10<\alpha<1. .

We begin by explaining the proof for part (2), that is, when Ω\Omega is a polyhedron and then explain the changes required for part (1) namely, points on rays having an end in the interior of Ω\Omega. Note the result for xΩnx\in\Omega_{n} is a straightforward consequence of the convexity of uu and Δu=n+1\Delta u=n+1 in Ωn\Omega_{n}.

Step 1. (Construction of section and comparison function on polyhedrons) We fix x0Ωεx_{0}\in\Omega_{\varepsilon} and translate and subtract a support plane after which we may assume x0,u(x0)x_{0},u(x_{0}) and Du(x0)Du(x_{0}) all vanish and thus, u0u\geq 0222It is worth noting that LL is not translation invariant, so after this transformation we should work with L¯[u]=Ω12|Du|2+u(x+x0)Dudx\bar{L}[u]=\int_{\Omega}\frac{1}{2}|Du|^{2}+u-(x+x_{0})\cdot Du\,dx. Inspection of the proof reveals such a change is inconsequential.. Now, after a rotation we may assume the face closest to x0x_{0} is

Pd=Ω{x=(x1,,xn);x1=d}.P_{-d}=\partial\Omega\cap\{x=(x^{1},\dots,x^{n});x^{1}=-d\}.

We assume d<εd<\varepsilon. (If d>εd>\varepsilon then we already have a C1,1C^{1,1} estimate; Caffarelli and Lions’s estimate is |D2u(x)|C(n)dist(x,Ω)1sup|Du||D^{2}u(x)|\leq C(n)\text{dist}(x,\partial\Omega)^{-1}\sup|Du|.) Note that x0x_{0} may be close to a single face of the polyhedron but, because we work in Ωε\Omega_{\varepsilon}, satisfies

(30) dist(x0,ΩPd)=:δC(ε,Ω)\text{dist}(x_{0},\partial\Omega\setminus P_{-d})=:\delta\geq C(\varepsilon,\Omega)

for a positive constant C(ε,Ω)C(\varepsilon,\Omega) depending only on ε\varepsilon and Ω\Omega.

For r>0r>0 to be chosen sufficiently small, but initially r<dr<d, set

h=supBr(0)u=u(rξ),h=\sup_{B_{r}(0)}u=u(r\xi),

where the latter equality defines the unit vector ξ\xi as the direction in which the supremum is obtained. The section

S\displaystyle S :={xΩ;u(x)<p(x)},\displaystyle:=\left\{x\in\Omega;u(x)<p(x)\right\},
wherep(x)\displaystyle\text{where}\quad p(x) :=h2r(xξ+r),\displaystyle:=\frac{h}{2r}(x\cdot\xi+r),

satisfies the slab containment condition

(31) S{xΩ;r<xξ<r}=:Sξ,r.S\subset\left\{x\in\Omega;-r<x\cdot\xi<r\right\}=:S_{\xi,r}.

The lower estimate is because p(x)<0p(x)<0 when xξ<rx\cdot\xi<-r and u0u\geq 0. For the upper estimate note

Du(rξ)Dp(rξ),Du(r\xi)-Dp(r\xi),

is the outer unit normal to SS at rξr\xi. However, because uu attains its maximum over the boundary of the ball at rξr\xi, DuDu has zero tangential component and so, by convexity, Du(rξ)=κξDu(r\xi)=\kappa\xi for some κh/r\kappa\geq h/r meaning the outer normal is

Du(rξ)Dph(rξ)=κξh2rξ.Du(r\xi)-Dp_{h}(r\xi)=\kappa\xi-\frac{h}{2r}\xi.

Step 2. (Tilting and shifting at the boundary on polyhedrons) The possibility that SS intersects Ω\partial\Omega complicates the boundary estimate. The existing interior estimates use a bound n1(SΩ)Cdist(x0,SΩ)n(S)\mathcal{H}^{n-1}(\partial S\cap\partial\Omega)\leq\frac{C}{\text{dist}(x_{0},\partial S\cap\partial\Omega)}\mathcal{H}^{n}(S) which does not suffice near the boundary. Thus we must tilt the affine support to ensure points where S\partial S intersects Ω\partial\Omega lie sufficiently far (distance greater than C(ε,Ω)C(\varepsilon,\Omega)) from x0x_{0}.

We consider the modified plane and section (see Figure 3)

S¯\displaystyle\bar{S} ={xΩ;u(x)<p¯(x)},\displaystyle=\left\{x\in\Omega;u(x)<\bar{p}(x)\right\},
(32) wherep¯(x)\displaystyle\text{where}\quad\bar{p}(x) =h2r(xξ+r)2hd(𝐧x)+s,\displaystyle=\frac{h}{2r}(x\cdot\xi+r)-2\frac{h}{d}(\mathbf{n}\cdot x)+s,

where 𝐧=e1\mathbf{n}=-e_{1} is the outer unit normal to Ω\Omega along PdP_{-d} and ss is a small positive or negative shift to be specified. The key idea is that provided rdr\ll d, h/2r2h/dh/2r\gg 2h/d so Dp¯D\bar{p} is a small perturbation of DpDp (in both direction and magnitude). Observe that slab containment, (31), implies pu<hp-u<h and on Pd{x;x1=d}P_{-d}\subset\{x;x^{1}=-d\}, we have p¯=p2h+s\bar{p}=p-2h+s, so provided s<hs<h (which we enforce below), S¯\partial\bar{S} is disjoint from PdP_{-d}. We claim if rr is initially chosen sufficiently small, then

(33) S¯{xΩ;2r<xξ<2r}=Sξ,2r\bar{S}\subset\left\{x\in\Omega;-2r<x\cdot\xi<2r\right\}=S_{\xi,2r}

where the fact that Ω\Omega is bounded has been used.

First we prove the lower bound S¯{xΩ;2r<xξ}\bar{S}\subset\left\{x\in\Omega;-2r<x\cdot\xi\right\}. This follows because a choice of rr sufficiently small ensures the plane {p¯=0}\{\bar{p}=0\} makes an arbitrarily small angle with the original plane {p=0}={x;xξ=r}\{p=0\}=\{x;x\cdot\xi=-r\}. Moreover we can ensure p¯(tξ)=0\bar{p}(t\xi)=0 for some 0>t>3r/20>t>-3r/2. Indeed, the plane {p=0}\{p=0\}, which we used as our original lower bound for the slab SS, is necessarily orthogonal to Dp(x)=h2rξDp(x)=\frac{h}{2r}\xi. Similarly, the plane {p¯=0}\{\bar{p}=0\}, which we use as our lower bound for S¯\bar{S}, is orthogonal to

Dp¯(x)=h2rξ2hd𝐧,D\bar{p}(x)=\frac{h}{2r}\xi-\frac{2h}{d}\mathbf{n},

provided rdr\ll d the vectors Dp¯D\bar{p} and DpDp make arbitrarily small angle. Thus the planes {p¯=0}\{\bar{p}=0\} and {p=0}\{p=0\} make arbitrarily small angle. Since p¯(0)=h2+s\bar{p}(0)=\frac{h}{2}+s and p¯(3rξ/2)=h4+s+3hrd𝐧ξ\bar{p}(-3r\xi/2)=-\frac{h}{4}+s+3\frac{hr}{d}{\mathbf{n}}\cdot\xi, provided |s|<h/8|s|<h/8 and r<d/24r<d/24 there is a point on {tξ;0>t>3r/2}\{t\xi;0>t>-3r/2\} where p¯=0\bar{p}=0.

Next we prove the upper bound S{xΩ;xξ<2r}S\subset\left\{x\in\Omega;x\cdot\xi<2r\right\}. This is where we choose our vertical shift. Recall

u(rξ)=handDu(rξ)=κξ,u(r\xi)=h\quad\text{and}\quad Du(r\xi)=\kappa\xi,

where κhr\kappa\geq\frac{h}{r}. Note that

p¯(ξr)=h2hd𝐧(rξ)+s.\bar{p}(\xi r)=h-2\frac{h}{d}\mathbf{n}\cdot(r\xi)+s.

Because r<d/24r<d/24 the choice s=2hr𝐧ξ/ds=2hr\mathbf{n}\cdot\xi/d gives |s|<h/12|s|<h/12 and we have equality of p¯\bar{p} and uu at rξr\xi, i.e. p¯(rξ)=u(rξ)=h\bar{p}(r\xi)=u(r\xi)=h. Then, as before, Du(rξ)Dp¯(rξ)Du(r\xi)-D\bar{p}(r\xi) is a normal to a support of the convex set S¯\bar{S}. However

Du(rξ)Dp¯(rξ)=κξh2rξ+2hd𝐧.Du(r\xi)-D\bar{p}(r\xi)=\kappa\xi-\frac{h}{2r}\xi+\frac{2h}{d}\mathbf{n}.

Recalling κhr\kappa\geq\frac{h}{r} we see again for rr sufficiently small this vector makes arbitrarily small angle with ξ\xi. This yields the upper containment in (33).

PdP_{-d}x0x_{0}rξr\xiSSSξ,rS_{\xi,r}PdP_{-d}x0x_{0}rξr\xiS¯\bar{S}
Figure 3. An example of the original section and the tilted section. The tilted section is now disjoint from boundary portion PdP_{-d}. The trade off is it may leave the slab Sξ,rS_{\xi,r}. Nevertheless S¯\bar{S} is contained in the slightly larger slab Sξ,2rS_{\xi,2r}.

Our choice of tilted support p¯\bar{p} implies S¯\bar{S} is disjoint from PdP_{-d}. Moreover we have

(34) p¯(0)u(0)h/4andsupS¯p¯u4h,\bar{p}(0)-u(0)\geq h/4\quad\text{and}\quad\sup_{\bar{S}}\bar{p}-u\leq 4h,

where the second inequality is because each point in the containment slab Sξ,4rS_{\xi,4r} is of distance less than 4r4r from the plane {p¯=0}\{\bar{p}=0\} and |Dp¯|h/r|D\bar{p}|\leq h/r. These properties are enough for us to employ the dilation argument used by the authors in [38]. For completeness we include full details, but first explain how to obtain the section S¯\bar{S} in the other setting of the theorem: in arbitrary convex domains at rays with one endpoint on the boundary.

Step 3. (Rays with one endpoint on the boundary in convex domains) Now we explain the choice of a suitable perturbation for case (1) of Theorem 4.1, namely when Ω\Omega is merely open, bounded, and convex and x0Ωn1x_{0}\in\Omega_{n-1} satisfies that x0~\tilde{x_{0}} has only one point on the boundary. In this case a similar tilting procedure yields a section S¯\bar{S} which is in fact strictly contained in Ω\Omega. Assume, x0~={x0+tζ;atb}\tilde{x_{0}}=\{x_{0}+t\zeta;-a\leq t\leq b\} where x0aζΩx_{0}-a\zeta\in\partial\Omega. Note, if there is c0>0c_{0}>0 such that for all rr sufficiently small, |ξζ|>c0|\xi\cdot\zeta|>c_{0} (i.e. the angle between ξ\xi and ζ\zeta is bounded away from π/2\pi/2) then the slab containment (31) yields SΩS\subset\subset\Omega. On the other hand, if ξζ0\xi\cdot\zeta\rightarrow 0 (i.e. ξ\xi approaches the orthogonal direction to ζ\zeta) we consider the new perturbation

p¯(x)=h2r(xξ+r)+4haxζ+s.\bar{p}(x)=\frac{h}{2r}(x\cdot\xi+r)+\frac{4h}{a}x\cdot\zeta+s.

Provided |s|<h|s|<h, which we can enforce as above, we have for rr sufficiently small S¯{x;xζ3a/4}\bar{S}\subset\{x;x\cdot\zeta\geq-3a/4\}. Since, in addition, as h0h\rightarrow 0, supxS¯dist(x,x0~)0\sup_{x\in\bar{S}}\text{dist}(x,\tilde{x_{0}})\rightarrow 0 we have that for hh sufficiently small (obtained by an initial choice of rr sufficiently small) S¯\bar{S} will be strictly contained in Ω\Omega and satisfy both the slab containment condition (33) and the height estimates (34). Note in the steps that follow the argument is simpler for this case. Indeed, S¯Ω=\bar{S}\cap\partial\Omega=\emptyset implies we do not need to consider the boundary terms in what follows or the dilation argument in Step 5.

Step 4. Initial estimates. Now we use the minimality of the function uu for the functional LL (defined in (1)) to derive the desired inequality hCr2h\leq Cr^{2}. Set u¯=max{p¯,u}\bar{u}=\text{max}\{\bar{p},u\} where p¯\bar{p} is defined in (32). Minimality implies

0\displaystyle 0 L[u¯]L[u]\displaystyle\leq L[\bar{u}]-L[u]
=S¯12(|Du¯|2|Du|2)+(u¯u)(xDu¯xDu)dx\displaystyle=\int_{\bar{S}}\frac{1}{2}(|D\bar{u}|^{2}-|Du|^{2})+(\bar{u}-u)-(x\cdot D\bar{u}-x\cdot Du)\,dx
(35) S¯12(|Du¯|2|Du|2)x(Du¯Du)dx+4hn(S¯).\displaystyle\leq\int_{\bar{S}}\frac{1}{2}(|D\bar{u}|^{2}-|Du|^{2})-x\cdot(D\bar{u}-Du)\,dx+4h\mathcal{H}^{n}(\bar{S}).

The divergence theorem implies

S¯x(Du¯Du)𝑑x\displaystyle-\int_{\bar{S}}x\cdot(D\bar{u}-Du)\,dx =S¯Ω(u¯u)x𝐧𝑑n1+nS¯(u¯u)𝑑x\displaystyle=-\int_{\partial\bar{S}\cap\partial\Omega}(\bar{u}-u)x\cdot\mathbf{n}\,d\mathcal{H}^{n-1}+n\int_{\bar{S}}(\bar{u}-u)\,dx
(36) Chn1(S¯Ω)+4nhn(S¯).\displaystyle\leq Ch\mathcal{H}^{n-1}(\partial\bar{S}\cap\partial\Omega)+4nh\mathcal{H}^{n}(\bar{S}).

Next, using that u¯\bar{u} is linear in S¯\bar{S} (in particular, Δu¯=0\Delta\bar{u}=0), we compute

S¯|Du¯|2|Du|2dx\displaystyle\int_{\bar{S}}|D\bar{u}|^{2}-|Du|^{2}\,dx =S¯Du¯+Du,Du¯Du𝑑x\displaystyle=\int_{\bar{S}}\langle D\bar{u}+Du,D\bar{u}-Du\rangle\,dx
=S¯div((Du¯+Du)(u¯u))Δu(u¯u)dx\displaystyle=\int_{\bar{S}}\text{div}\big{(}(D\bar{u}+Du)(\bar{u}-u)\big{)}-\Delta u(\bar{u}-u)\,dx
=S¯div((Du¯+Du)(u¯u))\displaystyle=\int_{\bar{S}}\text{div}\big{(}(D\bar{u}+Du)(\bar{u}-u)\big{)}
+div((Du¯Du)(u¯u))|Du¯Du|2dx\displaystyle\quad\quad+\text{div}\big{(}(D\bar{u}-Du)(\bar{u}-u)\big{)}-|D\bar{u}-Du|^{2}\,dx
(37) S¯|Du¯Du|2𝑑x+2S¯Ω(u¯u)Du¯n𝑑n1.\displaystyle\leq-\int_{\bar{S}}|D\bar{u}-Du|^{2}\,dx+2\int_{\partial\bar{S}\cap\partial\Omega}(\bar{u}-u)D\bar{u}\cdot\textbf{n}\,d\mathcal{H}^{n-1}.

Substituting (36) and (37) into (35) we have for CC depending on sup|Du¯|\sup|D\bar{u}|

(38) S¯|Du¯Du|2\displaystyle\int_{\bar{S}}|D\bar{u}-Du|^{2} C(hn(S¯)+hn1(S¯Ω)).\displaystyle\leq C(h\mathcal{H}^{n}(\bar{S})+h\mathcal{H}^{n-1}(\partial\bar{S}\cap\partial\Omega)).

Step 5. (Final estimates) To complete the proof we prove for C>0C>0, which in the case of polyhedra depends in particular on ε\varepsilon, there holds

(39) n1(S¯Ω)\displaystyle\mathcal{H}^{n-1}(\partial\bar{S}\cap\partial\Omega) Cn(S¯)\displaystyle\leq C\mathcal{H}^{n}(\bar{S})
(40) andS¯|Du¯Du|2\displaystyle\text{and}\quad\int_{\bar{S}}|D\bar{u}-Du|^{2} Ch2r2n(S¯).\displaystyle\geq C\frac{h^{2}}{r^{2}}\mathcal{H}^{n}(\bar{S}).

For the first, in the case of polyhedral domains, recall S¯Pd\partial\bar{S}\cap P_{-d} is empty so S¯Ω\partial\bar{S}\cap\partial\Omega is of distance C(ε,Ω)C(\varepsilon,\Omega) from x0=0x_{0}=0. Thus the estimate (39), for CC depending on ε\varepsilon, is standard in convex geometry and may be proved either as in the work of Chen [22], or the authors [38]. In case (1) of the theorem the estimate is trivial because S¯Ω=.\partial\bar{S}\cap\partial\Omega=\emptyset.

Now we obtain (40). We let S¯/K\bar{S}/K denote the dilation of S¯\bar{S} by a factor of 1/K1/K with respect to x0x_{0}. What is again crucial is that S¯Pd=\partial\bar{S}\cap P_{-d}=\emptyset so for Dx0,ΩPdD_{x_{0},\Omega\setminus P_{-d}} defined as in (30), S¯Bδ/2(x0)\partial\bar{S}\cap B_{\delta/2}(x_{0}) consists of interior points of Ω\Omega on which u¯u=0\bar{u}-u=0. It is helpful now to choose coordinates such that ξ=e1\xi=e_{1}. For x=(x1,x)x=(x^{1},x^{\prime}), let P(x):=(0,x)P(x):=(0,x^{\prime}) be the projection onto {x;x1=0}\{x;x^{1}=0\}. For each (0,x)P(S¯/K)(0,x^{\prime})\in P(\bar{S}/K) the set (P1(0,x)S¯)(S¯/K)(P^{-1}(0,x^{\prime})\cap\bar{S})\setminus(\bar{S}/K) is two disjoint line segments. We let lxl_{x^{\prime}} be the line segment with greater x1x_{1} component and write lx=[ax,bx]×{x}l_{x^{\prime}}=[a_{x^{\prime}},b_{x^{\prime}}]\times\{x^{\prime}\} where bx>axb_{x^{\prime}}>a_{x^{\prime}}.

Choose K=max{2diam(Ω)/δ,2}K=\max\{2\text{diam}(\Omega)/\delta,2\} in case (2) which is bounded below by a positive constant depending on ε\varepsilon and Ω\Omega. Case (1) is simpler as this dilation is not required. Note that each line segment lxl_{x^{\prime}} for (0,x)P(S¯/K)(0,x^{\prime})\in P(\bar{S}/K) has u¯u=0\bar{u}-u=0 at the upper endpoint. This is because from the slab containment condition the upper endpoint lies distance less than 4rδ4r\ll\delta from Bε/2(0){x1=0}B_{\varepsilon/2}(0)\cap\{x^{1}=0\} whereas S¯Ω\partial\bar{S}\cap\partial\Omega lies distance at least δ\delta from x0=0x_{0}=0. Clearly on S¯Ω\partial\bar{S}\cap\Omega we have u¯u=0\bar{u}-u=0.

We claim each of the following

(41) u((bx,x))u¯((bx,x))\displaystyle u((b_{x^{\prime}},x^{\prime}))-\bar{u}((b_{x^{\prime}},x^{\prime})) =0,\displaystyle=0,
(42) u((ax,x))u¯((ax,x))\displaystyle u((a_{x^{\prime}},x^{\prime}))-\bar{u}((a_{x^{\prime}},x^{\prime})) K1Kh4,\displaystyle\leq-\frac{K-1}{K}\frac{h}{4},
(43) dx:=bxax\displaystyle d_{x^{\prime}}:=b_{x^{\prime}}-a_{x^{\prime}} 4r.\displaystyle\leq 4r.

As noted above (41) is because u¯u=0\bar{u}-u=0 on S¯Bε(x0)\partial\bar{S}\cap B_{\varepsilon}(x_{0}). Then (42) is by convexity of uu¯u-\bar{u} along a line segment joining the origin, where uu¯h/4u-\bar{u}\leq-h/4, to (Kax,Kx)S¯(Ka_{x^{\prime}},Kx^{\prime})\in\partial\bar{S}, where uu¯0u-\bar{u}\leq 0. Finally, (43) is by the modified slab containment condition (33).

Thus, by an application of Jensen’s inequality we have

axbx\displaystyle\int_{a_{x^{\prime}}}^{b_{x^{\prime}}} [Dx1u¯((t,x))Dx1u((t,x))]2dt\displaystyle[D_{x^{1}}\bar{u}((t,x^{\prime}))-D_{x^{1}}u((t,x^{\prime}))]^{2}dt
1dx(axbxDx1u¯((t,x))Dx1u((t,x))dt)2\displaystyle\geq\frac{1}{d_{x^{\prime}}}\left(\int_{a_{x^{\prime}}}^{b_{x^{\prime}}}D_{x^{1}}\bar{u}((t,x^{\prime}))-D_{x^{1}}u((t,x^{\prime}))dt\right)^{2}
(44) 1dx(K1K)2h216Ch2/r.\displaystyle\geq\frac{1}{d_{x^{\prime}}}\left(\frac{K-1}{K}\right)^{2}\frac{h^{2}}{16}\geq Ch^{2}/r.

To conclude we integrate along all lines lxl_{x^{\prime}} for xP(S¯/K)x^{\prime}\in P(\bar{S}/K). Indeed

S¯|Du¯Du|2𝑑x\displaystyle\int_{\bar{S}}|D\bar{u}-Du|^{2}\ dx P(S¯/K)axbx|Dx1u¯((t,x))Dx1u((t,x))|2𝑑t𝑑x\displaystyle\geq\int_{P(\bar{S}/K)}\int_{a_{x^{\prime}}}^{b_{x^{\prime}}}|D_{x^{1}}\bar{u}((t,x^{\prime}))-D_{x^{1}}u((t,x^{\prime}))|^{2}\ dt\ dx^{\prime}
P(S¯/K)Ch2/r𝑑x\displaystyle\geq\int_{P(\bar{S}/K)}Ch^{2}/r\ dx^{\prime}
=Ch2r2(r|P(S¯/K)|).\displaystyle=C\frac{h^{2}}{r^{2}}(r|P(\bar{S}/K)|).

Finally, the convexity of S¯\bar{S} and slab containment (33) implies

(45) S¯|Du¯Du|2𝑑x\displaystyle\int_{\bar{S}}|D\bar{u}-Du|^{2}\ dx Ch2r2|S¯|.\displaystyle\geq C\frac{h^{2}}{r^{2}}|\bar{S}|.

Having obtained (40) , substituting inequalities (39) and (40) yields (38) and completes the proof. ∎

5. Strict convexity implies the Neumann condition

In this section we continue establishing technical conditions required for the coordinates introduced in Section 6. For planar domains we prove the equivalence between the Neumann condition and strict convexity stated precisely in Propositions 5.3 and 5.4. We begin with two lemmas concerning convex functions in 𝐑n\mathbf{R}^{n}. The first states the upper semicontinuity of the function xdiam(x~)x\mapsto\text{diam}(\tilde{x}) and the second yields the convexity of ii2u\partial^{2}_{ii}u when restricted to a contact set x~\tilde{x}.

Lemma 5.1 (Upper semicontinuity of leaf diameter).

Let Ω\Omega be a bounded open convex subset of 𝐑n\mathbf{R}^{n} and uC1(Ω¯)u\in C^{1}(\overline{\Omega}) a convex function. Then the function xdiam(x)x\mapsto\text{diam}(x) is upper semicontinuous.

Proof.

We fix a sequence (xk)k1(x_{k})_{k\geq 1} converging to some xΩx_{\infty}\in\Omega and note it suffices to prove that

lim supkdiam(xk~)diam(x~).\limsup_{k\rightarrow\infty}\text{diam}(\tilde{x_{k}})\leq\text{diam}(\tilde{x_{\infty}}).

To this end, let pk=Du(xk)p_{k}=Du(x_{k}) and take xk(1)x^{(1)}_{k}, xk(2)x^{(2)}_{k} in Ω¯\bar{\Omega} realizing

|xk(1)xk(2)|\displaystyle|x^{(1)}_{k}-x^{(2)}_{k}| =diam(xk~)\displaystyle=\text{diam}(\tilde{x_{k}})
andlimk|xk(1)xk(2)|\displaystyle\text{and}\quad\lim_{k\rightarrow\infty}|x^{(1)}_{k}-x^{(2)}_{k}| =lim supkdiam(xk~).\displaystyle=\limsup_{k\rightarrow\infty}\text{diam}(\tilde{x_{k}}).

The convergence properties of the subdifferential of a convex function imply Du(xk)Du(x)Du(x_{k})\rightarrow Du(x_{\infty}) and we may assume that, up to a subsequence, xk(i)x(i)Ω¯x^{(i)}_{k}\rightarrow x^{(i)}_{\infty}\in\overline{\Omega} for i=1,2i=1,2. Thus we may send kk\rightarrow\infty in the identity

(46) u(xk(i))=u(xk)+Du(xk)(xk(i)xk)\displaystyle u(x_{k}^{(i)})=u(x_{k})+Du(x_{k})\cdot(x_{k}^{(i)}-x_{k})

to obtain that for i=1,2i=1,2 we have x(i)x~x^{(i)}_{\infty}\in\tilde{x_{\infty}} and thus x~\tilde{x} has diameter greater than or equal to

|x(1)x(2)|=limk|xk(1)xk(2)|=lim supkdiam(xk~).|x^{(1)}_{\infty}-x^{(2)}_{\infty}|=\lim_{k\rightarrow\infty}|x^{(1)}_{k}-x^{(2)}_{k}|=\limsup_{k\rightarrow\infty}\text{diam}(\tilde{x_{k}}).

Let r.i.(x~)\mathop{\rm r.i.}(\tilde{x}) denote the relative interior of the convex set x~\tilde{x}.

Lemma 5.2 (Existence a.e. of D2uD^{2}u on a leaf implies convexity of ii2u\partial^{2}_{ii}u).

Let u:Ω𝐑u:\Omega\rightarrow{\mathbf{R}} be a differentiable convex function defined on an open convex subset Ω𝐑n\Omega\subset\mathbf{R}^{n}. Fix any xΩx\in\Omega. If dimx~(x~domD2u)=0\mathcal{H}^{\dim\tilde{x}}(\tilde{x}\setminus\mathop{\rm dom}D^{2}u)=0, where domD2uΩ\mathop{\rm dom}D^{2}u\subset\Omega denotes the set of second differentiability of uu, then u|r.i.(x~)Cloc2(r.i.(x~))u|_{\mathop{\rm r.i.}(\tilde{x})}\in C^{2}_{\text{loc}}(\mathop{\rm r.i.}(\tilde{x})) and ii2u|x~\partial^{2}_{ii}u|_{\tilde{x}} is a convex function for each i=1,,ni=1,\dots,n.

Proof.

Fix xx satisfying dimx~(x~domD2u)=0\mathcal{H}^{\dim\tilde{x}}(\tilde{x}\setminus\mathop{\rm dom}D^{2}u)=0. After subtracting the support at xx, we may assume u(x),Du(x)=0u(x),Du(x)=0 and that x~\tilde{x} is not a singleton (since otherwise the result holds trivially). We will show ii2u\partial^{2}_{ii}u is convex along any of those line segments contained in x~\tilde{x} for which Alexandrov second differentiability holds a.e. . To this end fix x0,x1x~domD2ux_{0},x_{1}\in\tilde{x}\cap\mathop{\rm dom}D^{2}u along such a segment. Choose orthonormal coordinates such that x1=x0+Te1x_{1}=x_{0}+Te_{1} for some T>0T>0. Then since uu is affine on x~\tilde{x} and {x0+te1;0tT}x~\{x_{0}+te_{1};0\leq t\leq T\}\subset\tilde{x} we have 112u=0\partial^{2}_{11}u=0 a.e. on {x0+te1;0tT}\{x_{0}+te_{1};0\leq t\leq T\}.

Next, for i=2,,ni=2,\dots,n and any t(0,1)t\in(0,1), convexity of uu implies for r>0r>0 sufficiently small

(47) u((1t)x0+tx1+rei)(1t)u(x0+rei)+tu(x1+rei).u((1-t)x_{0}+tx_{1}+re_{i})\leq(1-t)u(x_{0}+re_{i})+tu(x_{1}+re_{i}).

Here, by rr sufficiently small we mean small enough to ensure the above arguments of uu are contained in Ω\Omega. The definition of Alexandrov second differentiability along with u,Du=0u,Du=0 on x~\tilde{x} implies

u((1t)x0+tx1+rei)=r2ii2u((1t)x0+tx1)/2+o(r2),u((1-t)x_{0}+tx_{1}+re_{i})=r^{2}\partial^{2}_{ii}u((1-t)x_{0}+tx_{1})/2+o(r^{2}),

for a.e. t(0,1)t\in(0,1) and similarly at x0+re1x_{0}+re_{1} and x1+re1x_{1}+re_{1}. Thus (47) becomes

r2ii2u((1t)x0+tx1)/2+o(r2)(1t)r2ii2u(x0)/2+tr2ii2u(x1)/2+o(r2).r^{2}\partial^{2}_{ii}u((1-t)x_{0}+tx_{1})/2+o(r^{2})\leq(1-t)r^{2}\partial^{2}_{ii}u(x_{0})/2+tr^{2}\partial^{2}_{ii}u(x_{1})/2+o(r^{2}).

Dividing by r2r^{2} and sending r0r\rightarrow 0 yields that ii2u\partial^{2}_{ii}u is the restriction of a convex function to the segment [x0,x1][x_{0},x_{1}] — hence continuous on [x0,x1][x_{0},x_{1}]. The polarization identity implies the continuity of mixed second order partial derivatives. It follows that u|r.i.(x~)Cloc2u|_{\mathop{\rm r.i.}(\tilde{x})}\in C^{2}_{\text{loc}}. ∎

Proposition 5.3 (No normal distortion nearby implies strict convexity).

Let uu minimize (1) where Ω𝐑2\Omega\subset\subset\mathbf{R}^{2} is open and convex. Let x0Ωx_{0}\in\partial\Omega be a point where u(x0)>0u(x_{0})>0 and x0~Ω={x0}\tilde{x_{0}}\cap\partial\Omega=\{x_{0}\}. Assume there is ε>0\varepsilon>0 with

(Du(x)x)𝐧=0 on Bε(x0)Ω.(Du(x)-x)\cdot\mathbf{n}=0\text{ on }B_{\varepsilon}(x_{0})\cap\partial\Omega.

Then x0~={x0}\tilde{x_{0}}=\{x_{0}\}, that is uu is strictly convex at x0x_{0}.

Proof.

Because uC1(Ω¯)u\in C^{1}(\overline{\Omega}) and Lemma 5.1 implies the upper semicontinuity of RR, we may find a possibly smaller ε>0\varepsilon>0 such that u(x)>0u(x)>0 and x~Ω={x}\tilde{x}\cap\partial\Omega=\{x\} for each x𝒩:=Bε(x0)Ωx\in\mathcal{N}:=B_{\varepsilon}(x_{0})\cap\partial\Omega. Rochet and Choné’s localization (Corollary A.9) with 0 boundary term implies Δu=3\Delta u=3 almost everywhere on

𝒩~:={xx~;x𝒩}.\tilde{\mathcal{N}}:=\{x^{\prime}\in\ \tilde{x};x\in\mathcal{N}\}.

To see this, note, by Lemma 5.2, Δu\Delta u restricted to any x~\tilde{x} for which dimx~(x~domD2u)=0\mathcal{H}^{\dim\tilde{x}}(\tilde{x}\setminus\mathop{\rm dom}D^{2}u)=0 is a convex function. Thus v=(3Δu)v=-(3-\Delta u) is a permissible test function in the localization Corollary A.9 from which we obtain

(48) 0x~(3Δu)2(dx)x~,0\leq-\int_{\tilde{x}}(3-\Delta u)^{2}\,(dx)_{\tilde{x}},

where (dx)x~(dx)_{\tilde{x}} denotes the disintegration of the Lebesgue measure with respect to the contact sets (an explanation of disintegration is provided in Appendix A). Inequality (48) implies that dimx~\mathcal{H}^{\text{dim}\tilde{x}} almost everywhere on x~\tilde{x} there holds (3Δu)2=0(3-\Delta u)^{2}=0 and thus the same equality holds on 𝒩~\tilde{\mathcal{N}}.

Let x1x_{1} be the interior endpoint of x0~\tilde{x_{0}} (if x0~={x0}\tilde{x_{0}}=\{x_{0}\} then we are already done). In a sufficiently small ball Bδ(x1)B_{\delta}(x_{1}), we have just shown Δu=3\Delta u=3 a.e. in Bδ(x1)𝒩~B_{\delta}(x_{1})\cap\tilde{\mathcal{N}}. Moreover, Theorem 1.1 implies Δu=3\Delta u=3 in Bδ(x1)𝒩~=Bδ(x1)Ω2B_{\delta}(x_{1})\setminus\tilde{\mathcal{N}}=B_{\delta}(x_{1})\cap\Omega_{2}, which is nonempty. Our usual maximum principle argument, Lemma 3.2, implies uu is strictly convex inside Bδ(x1)B_{\delta}(x_{1}), contradicting that x1x_{1} is the endpoint of a ray. ∎

The next proof requires Lemma A.6 which gives the pushforward Du#(σ)=δ0Du_{\#}(\sigma)=\delta_{0} of the variational derivative — and is proved in Appendix A by combining the neutrality implied by localization away from the excluded region {u=0}\{u=0\} with the fact that our objective responds proportionately to a uniform increase in indirect utility. We use these to estimate the following:

Proposition 5.4 (One-ended ray lengths bound normal distortion).

Let uu solve (1) where Ω𝐑2\Omega\subset\subset\mathbf{R}^{2} is open and convex. Let {x0}=x0~Ω\{x_{0}\}=\tilde{x_{0}}\cap\partial\Omega with Ω\partial\Omega smooth in a neighbourhood of x0x_{0} and u(x0)>0u(x_{0})>0. Set R(x0)=diam(x0~)R(x_{0})=\text{diam}(\tilde{x_{0}}). Then

(49) 0(Du(x0)x0)nCR(x0),0\leq(Du(x_{0})-x_{0})\cdot\textbf{n}\leq CR(x_{0}),

where CC depends only on a C1,1C^{1,1} bound for uu in a neighbourhood of x0~\tilde{x_{0}}.

Proof.

The lower bound (49) was established in Proposition 2.3. We first prove (49) assuming x0~\tilde{x_{0}} nontrivial and at the conclusion of the proof explain why it holds for all x0x_{0} in the theorem. Let x0Ωx_{0}\in\partial\Omega and let Ω\partial\Omega be locally represented by a smooth curve with an arc length parametrization γ:(ε,ε)𝐑2\gamma:(-\varepsilon,\varepsilon)\rightarrow\mathbf{R}^{2} traversing Ω\partial\Omega in the anticlockwise direction with x0=γ(0)x_{0}=\gamma(0) and without loss of generality γ˙(0)=e2\dot{\gamma}(0)=e_{2}.

The upper semicontinuity of RR from Lemma 5.1 implies

lim supxx0R(x)R(x0).\limsup_{x\rightarrow x_{0}}R(x)\leq R(x_{0}).

On the other hand we know from Lemma 3.2 that no subinterval of x0~\tilde{x_{0}} can be exposed to Ω2\Omega_{2} by which we mean there is no xx0~x\in\tilde{x_{0}} and δ>0\delta>0 with Bδ(x)x0~=Ω2B_{\delta}(x)\setminus\tilde{x_{0}}=\Omega_{2}. Note in two-dimensions, if (xn)n1Ω(x_{n})_{n\geq 1}\subset\partial\Omega satisfies xnx0x_{n}\rightarrow x_{0} and R(xn)R(x0)R(x_{n})\rightarrow R(x_{0}) then xn~x0~\tilde{x_{n}}\rightarrow\tilde{x_{0}} in the Hausdorff distance (and such sequences can be found).

As a result there exists sufficiently small α,β>0\alpha,\beta>0 such that

  1. (1)

    The leaves γ(α)~\widetilde{\gamma(-\alpha)} and γ(β)~\widetilde{\gamma(\beta)} have length at least 3R(x0)/43R(x_{0})/4.

  2. (2)

    The leaves γ(α)~\widetilde{\gamma(-\alpha)} and γ(β)~\widetilde{\gamma(\beta)} can be chosen to make fixed but arbitrarily small angle with x0~\tilde{x_{0}}, by e.g. [11, Lemma 16].

  3. (3)

    All leaves intersecting the boundary in γ([2α,2β])\gamma\big{(}[-2\alpha,2\beta]\big{)} have length less than 9R(x0)/89R(x_{0})/8 (this holds by the upper semicontinuity of RR).

Moreover α,β\alpha,\beta can be taken as close to 0 as desired. With the smoothness of γ\gamma, our two-dimensional setting, and the fact that leaves cannot intersect, this significantly constrains the geometry of

A:=(Du)1(Du(γ([α,β]))=t[α,β]γ(t)~.A:=(Du)^{-1}\big{(}Du(\gamma\big{(}[-\alpha,\beta]\big{)}\big{)}=\bigcup_{t\in[-\alpha,\beta]}\widetilde{\gamma(t)}.

The set AA is strictly contained in a set with left edge γ([2α,2β])\gamma\big{(}[-2\alpha,2\beta]\big{)} and a vertical right edge of lengths bounded by 2(β+α)2(\beta+\alpha), and top and bottom side lengths bounded by 5R(x0)/45R(x_{0})/4 (see Figure 4). Finally we note we can choose sequences αk,βk\alpha_{k},\beta_{k} satisfying the above requirements and αk,βk0\alpha_{k},\beta_{k}\rightarrow 0.

Boundary portion γ([2α,2β])Ω\gamma([-2\alpha,2\beta])\subset\partial\Omega x0~\tilde{x_{0}}γ(β)~\widetilde{\gamma(\beta)}γ(α)~\widetilde{\gamma(-\alpha)}xb~\tilde{x_{b}} vertical height— 2(α+β)2(\alpha+\beta) length 5R(x0)/45R(x_{0})/4 Containment set for Du1(Du(γ[α,β]))Du^{-1}\big{(}Du(\gamma[-\alpha,\beta])\big{)}
Figure 4. Geometry of the constructed set AA. Note apriori (though not expected) there may be errant leaves such as xb~\tilde{x_{b}} or those between γ(α)~\widetilde{\gamma(-\alpha)} and x0~\tilde{x_{0}}. However we have constrained the length of such leaves as less than 9R(x0)/89R(x_{0})/8 and, when long, their angles are constrained by the outer leaves γ(α)~\widetilde{\gamma(-\alpha)} and γ(β)~\widetilde{\gamma(\beta)} (which other leaves may not intersect). Thus we obtain the (crude) containment estimate indicated by dotted lines.

Now, by Lemma A.6, σ(A)=((Du)#σ)(Du(γ([α,β])))=0\sigma(A)=((Du)_{\#}\sigma)\big{(}Du(\gamma([-\alpha,\beta]))\big{)}=0. That is,

(50) 0=A(n+1Δu)𝑑x+γ(α,β)(Dux)n𝑑1.0=\int_{A}(n+1-\Delta u)\,dx+\int_{\gamma(-\alpha,\beta)}(Du-x)\cdot\textbf{n}\,d\mathcal{H}^{1}.

Using the boundary C1,1C^{1,1} estimate from Theorem 4.1 (proved in Section 4) near x0~\tilde{x_{0}}333Because x0~\tilde{x_{0}} doesn’t intersect Ω\partial\Omega at both endpoints and RR is upper semicontinuous the same is true for all sufficiently close leaves., the constrained geometry of AA, and nonnegativity of (Dux)𝐧(Du-x)\cdot\mathbf{n} already established, we see (50) implies

0\displaystyle 0 γ(α,β)(Dux)n𝑑1\displaystyle\leq\int_{\gamma(-\alpha,\beta)}(Du-x)\cdot\textbf{n}\,d\mathcal{H}^{1}
supA|n+1Δu|2(A)\displaystyle\leq\sup_{A}|n+1-\Delta u|\mathcal{H}^{2}(A)
CR(x0)(α+β)\displaystyle\leq CR(x_{0})(\alpha+\beta)

which is precisely the desired estimate. Indeed, employing this estimate with αk,βk\alpha_{k},\beta_{k} in place of α,β\alpha,\beta, dividing by αk+βk\alpha_{k}+\beta_{k}, and sending α,β0\alpha,\beta\rightarrow 0 we obtain (49) (after dividing we have an average and (Du(x0)x0)𝐧(Du(x_{0})-x_{0})\cdot\mathbf{n} is continuous).

Now, we explain how to obtain the estimate when x~\tilde{x} is trivial. If xΩx\in\partial\Omega is such that x~\tilde{x} is trivial and there is a sequence of nontrivial leaves Ωxkx\partial\Omega\ni x_{k}\rightarrow x then the estimate follows by the upper semicontinuity of RR. If there is no such sequence then xΩx\in\partial\Omega lies in a relatively open subset of the boundary 𝒩=Bε0(x)Ω\mathcal{N}=B_{\varepsilon_{0}}(x)\cap\partial\Omega on which uu is strictly convex. Using the nonnegativity of (Dux)𝐧(Du-x)\cdot\mathbf{n} and Corollary A.9 with v=±1v=\pm 1 applied to 𝒩\mathcal{N} yields (Dux)𝐧=0(Du-x)\cdot\mathbf{n}=0 on 𝒩\mathcal{N}. ∎

6. Leafwise coordinates parameterizing bunches in the plane

In this section and the next we study the behavior of the minimizer on Ω1\Omega_{1} and the free boundary Γ=Ω1Ω2Ω\Gamma=\partial\Omega_{1}\cap\partial\Omega_{2}\cap\Omega in two-dimensions. We introduce one of our main tools: a coordinate system to study the problem on Ω1\Omega_{1} which is flexible enough to include the coordinates proposed earlier by the first and third authors [41], and for which we are finally able to provide a rigorous foundation by proving biLipschitz equivalence to Cartesian coordinates. Moreover, by combining these coordinates with Rochet and Choné’s localization technique (Corollary A.9) we are able to provide a radically simpler derivation of the Euler-Lagrange equations (9)–(10) first expressed in [41]; c.f. (67)–(68) and (80)–(81) below,

Let γ:[a,b]𝐑2\gamma:[-a,b]\rightarrow\mathbf{R}^{2} be a curve parameterizing Ω\partial\Omega in the clockwise direction with γ˙(t)0\dot{\gamma}(t)\neq 0 and write γ(t)=(γ1(t),γ2(t))\gamma(t)=(\gamma^{1}(t),\gamma^{2}(t)).

First we give conditions to ensure a neighbourhood of a ray is foliated by rays.

Lemma 6.1 (Local foliation around each tame ray).

Let Ω\partial\Omega be smooth in a neighbourhood of x0𝐑2x_{0}\in{\mathbf{R}}^{2} satisfying (Du(x0)x0)𝐧0(Du(x_{0})-x_{0})\cdot\mathbf{n}\neq 0, u(x0)0u(x_{0})\neq 0 and {x0}=x0~Ω\{x_{0}\}=\tilde{x_{0}}\cap\partial\Omega. Then there exist ε,r0>0\varepsilon,r_{0}>0 such that diam(x~)r0\text{diam}(\tilde{x})\geq r_{0} and x~Ω={x}\tilde{x}\cap\partial\Omega=\{x\} for all xΩBε(x0)x\in\partial\Omega\cap B_{\varepsilon}(x_{0}).

Proof.

We assume without loss of generality that γ(0)=x0\gamma(0)=x_{0} and thus there is ε>0\varepsilon>0 such that γ:(ε,ε)Ω\gamma:(-\varepsilon,\varepsilon)\rightarrow\partial\Omega is a smooth curve. Lemma 5.4 implies x0~\tilde{x_{0}} is nontrivial. For a possibly smaller ε>0\varepsilon>0 and all t(ε,ε)t\in(-\varepsilon,\varepsilon), the C1(Ω¯)C^{1}(\overline{\Omega}) regularity of uu implies γ(t)~\widetilde{\gamma(t)} is nontrivial, with length bounded below by some r0r_{0} determined by (49). Furthermore, the upper semicontinuity of xdiam(x~)x\mapsto\text{diam}(\tilde{x}) (Lemma 5.1) implies for a possibly smaller ε\varepsilon, γ(t)~Ω={γ(t)}\widetilde{\gamma(t)}\cap\partial\Omega=\{\gamma(t)\} for each t(ε,ε)t\in(-\varepsilon,\varepsilon). ∎

Recall x1Γx_{1}\in\Gamma is a tame point of the free boundary Γ:=Ω1Ω2Ω\Gamma:={\partial}\Omega_{1}\cap{\partial}\Omega_{2}\cap\Omega provided x1~Ω={x0}\tilde{x_{1}}\cap\partial\Omega=\{x_{0}\} for an x0x_{0} satisfying the hypothesis of Lemma 6.1.

We define ξ(t)=(ξ1(t),ξ2(t))\xi(t)=(\xi^{1}(t),\xi^{2}(t)) as the unit direction vector of the leaf γ(t)~\widetilde{\gamma(t)} pointing into Ω\Omega. This means, with R(t):=diam(γ(t)~)R(t):=\text{diam}(\widetilde{\gamma(t)}),

γ(t)~={γ(t)+rξ(t);0rR(t)},\widetilde{\gamma(t)}=\{\gamma(t)+r\xi(t);0\leq r\leq R(t)\},

and subsequently we can write a subset of the connected component of Ω1\Omega_{1} containing x0x_{0} as

(51) 𝒩=𝒩Ω1\displaystyle\mathcal{N}=\mathcal{N}\cap\Omega_{1} =t(ε,ε)γ(t)~\displaystyle=\bigcup_{t\in(-\varepsilon,\varepsilon)}\widetilde{\gamma(t)}
(52) ={x(r,t)=γ(t)+rξ(t);ε<t<ε,0rR(t)},\displaystyle=\{x(r,t)=\gamma(t)+r\xi(t);-\varepsilon<t<\varepsilon,0\leq r\leq R(t)\},

and we take (r,t)(r,t) as new coordinates for 𝒩\mathcal{N}. Because each ray is a contact set along which uu is affine there exists functions b,m:(ε,ε)𝐑b,m:(-\varepsilon,\varepsilon)\rightarrow\mathbf{R} such that

(53) u(x(r,t))=b(t)+rm(t),u(x(r,t))=b(t)+rm(t),

and Du(x(r,t))Du(x(r,t)) is independent of tt.

Our goal is to derive the Euler–Lagrange equations of Lemma 6.3 below which describe the equations the minimizer satisfies in terms of RR, ξ\xi, mm and bb . First, we record the key structural equalities for the new coordinates in the following lemma, which holds under the biLipschitz hypothesis we eventually establish in Corollary 6.6. The quantities (55)–(57) from this lemma also yield a formula for the Laplacian of uC1,1(𝒩)u\in C^{1,1}(\mathcal{N}):

(54) Δu=ξ×wJ(r,t)=:δ(t)J(r,t).\Delta u=\frac{\xi\times w^{\prime}}{J(r,t)}=:\frac{\delta(t)}{J(r,t)}.
Lemma 6.2 (Gradient and Hessian of uu in coordinates along tame rays).

Suppose uu solves (1) where Ω𝐑n\Omega\subset{\mathbf{R}}^{n} is bounded, open and convex. Let γ,ξ,R,m,b\gamma,\xi,R,m,b on 𝒩\mathcal{N} be as above (52). If the transformation x(r,t)=γ(t)+rξ(t)x(r,t)=\gamma(t)+r\xi(t) is biLipschitz on 𝒩\mathcal{N}, then its Jacobian determinant is positive and given by

(55) 0<J(r,t)=det((x1,x2)(r,t))=ξ×γ˙+rξ×ξ˙=:j(t)+rfξ(t)0<J(r,t)=\det\left(\frac{\partial(x^{1},x^{2})}{\partial(r,t)}\right)=\xi\times\dot{\gamma}+r\xi\times\dot{\xi}=:j(t)+rf_{\xi}(t)

where ξ×γ˙=ξ1γ˙2ξ2γ˙1\xi\times\dot{\gamma}=\xi^{1}\dot{\gamma}^{2}-\xi^{2}\dot{\gamma}^{1} and similarly ξ×ξ˙\xi\times\dot{\xi} are evaluated at tt. In addition the following formulas for the gradient and entries of the Hessian of (53) hold 2\mathcal{H}^{2}-a.e.:

(56) Du(x(r,t))\displaystyle Du(x(r,t)) =(D1uD2u)=1ξ×γ˙(γ˙2ξ2γ˙1ξ1)(m(t)b(t))=:(w1(t)w2(t)),\displaystyle=\begin{pmatrix}D_{1}u\\ D_{2}u\end{pmatrix}=\frac{1}{\xi\times\dot{\gamma}}\begin{pmatrix}\dot{\gamma}^{2}&-\xi^{2}\\ -\dot{\gamma}^{1}&\xi^{1}\end{pmatrix}\begin{pmatrix}m(t)\\ b^{\prime}(t)\end{pmatrix}=:\begin{pmatrix}w_{1}(t)\\ w_{2}(t)\end{pmatrix},
(57) D2u(x(r,t))\displaystyle D^{2}u(x(r,t)) =(112u122u212u222u)=1J(r,t)(ξ2(t)w1(t)ξ1(t)w1(t)ξ2(t)w2(t)ξ1(t)w2(t)).\displaystyle=\begin{pmatrix}{\partial}^{2}_{11}u&{\partial}^{2}_{12}u\\[4.30554pt] {\partial}^{2}_{21}u&{\partial}^{2}_{22}u\end{pmatrix}=\frac{1}{J(r,t)}\begin{pmatrix}-\xi^{2}(t)w_{1}^{\prime}(t)&\xi^{1}(t)w_{1}^{\prime}(t)\\ -\xi^{2}(t)w_{2}^{\prime}(t)&\xi^{1}(t)w_{2}^{\prime}(t)\end{pmatrix}.
Proof.

Where the transformation x(r,t)=γ(t)+rξ(t)x(r,t)=\gamma(t)+r\xi(t) and its inverse are Lipschitz, their Jacobian derivatives are easily compute to be:

(58) (x1,x2)(r,t)\displaystyle\frac{\partial(x^{1},x^{2})}{\partial(r,t)} =(ξ1(t)γ˙1(t)+rξ˙1(t)ξ2(t)γ˙2(t)+rξ˙2(t))\displaystyle=\begin{pmatrix}\xi^{1}(t)&\dot{\gamma}^{1}(t)+r\dot{\xi}^{1}(t)\\ \xi^{2}(t)&\dot{\gamma}^{2}(t)+r\dot{\xi}^{2}(t)\end{pmatrix}
(59) (r,t)(x1,x2)\displaystyle\frac{\partial(r,t)}{\partial(x^{1},x^{2})} =1J(r,t)(γ˙2(t)+rξ˙2(t)γ˙1(t)rξ˙1(t)ξ2(t)ξ1(t)),\displaystyle=\frac{1}{J(r,t)}\begin{pmatrix}\dot{\gamma}^{2}(t)+r\dot{\xi}^{2}(t)&-\dot{\gamma}^{1}(t)-r\dot{\xi}^{1}(t)\\ -\xi^{2}(t)&\xi^{1}(t)\end{pmatrix},

with J(r,t)J(r,t) from (55). Next, to obtain the gradient expressions we differentiate equation (53) with respect to rr to obtain

(60) m(t)\displaystyle m(t) =ξ1(t)u1(x(r,t))+ξ2(t)u2(x(r,t))=ξ,Du.\displaystyle=\xi^{1}(t)u_{1}(x(r,t))+\xi^{2}(t)u_{2}(x(r,t))=\langle\xi,Du\rangle.

Similarly, differentiating (53) with respect to tt and equating coefficients of rr yields

(61) m(t)\displaystyle m^{\prime}(t) =ξ˙1(t)D1u(x(r,t))+ξ˙2(t)u2(x(r,t))=ξ˙,Du,\displaystyle=\dot{\xi}^{1}(t)D_{1}u(x(r,t))+\dot{\xi}^{2}(t)u_{2}(x(r,t))=\langle\dot{\xi},Du\rangle,
(62) b(t)\displaystyle b^{\prime}(t) =γ˙1(t)D1u(x(r,t))+γ˙2(t)D2u(x(r,t))=γ˙,Du,\displaystyle=\dot{\gamma}^{1}(t)D_{1}u(x(r,t))+\dot{\gamma}^{2}(t)D_{2}u(x(r,t))=\langle\dot{\gamma},Du\rangle,

where Diu=uxiD_{i}u=\frac{\partial u}{\partial x^{i}} and that Du(x(r,t))Du(x(r,t)) is independent from rr has been used. We solve (60) and (62) for D1uD_{1}u, D2uD_{2}u and obtain (56). Note the functions w1w_{1} and w2w_{2} in (56) are Lipschitz because uC1,1(𝒩)u\in C^{1,1}(\mathcal{N}). Thus, differentiating the expressions for D1uD_{1}u and D2uD_{2}u given by (56) with respect to, respectively, x1x_{1} and x2x_{2} and using the Jacobian (59) gives the formula (57).

We note two facts about the functions jj and fξf_{\xi} which determine the Jacobian determinant. First

(63) j(t):=J(0,t)=ξ×γ˙>0,j(t):=J(0,t)=\xi\times\dot{\gamma}>0,

where the nonnegativity is by our chosen orientation: γ\gamma traverses Ω\partial\Omega in a clockwise direction and ξ\xi points into the convex domain Ω\Omega. Inequality (63) is strict because γ(t)~\widetilde{\gamma(t)} is nontrivial and has only one endpoint on Ω\partial\Omega for each t(ε,ε)t\in(-\varepsilon,\varepsilon). Next, because ξ\xi is a unit vector, whence ξ˙\dot{\xi} is orthogonal to ξ\xi, we have

(64) fξ(t)=ξ(t)×ξ˙(t)=±|ξ˙|,\displaystyle f_{\xi}(t)=\xi(t)\times\dot{\xi}(t)=\pm|\dot{\xi}|,

where the value of ±\pm is determined by the sign of fξ(t)f_{\xi}(t). From our biLipschitz hypothesis (or nonnegativity of the Laplacian (54)) the sign of J(r,t)J(r,t) is independent of rr. Combined with (63) we obtain J(r,t)>0J(r,t)>0 for t(ε,ε)t\in(-\varepsilon,\varepsilon) and 0rR(t)0\leq r\leq R(t). ∎

Now we combine our raywise coordinates (r,t)(r,t) with Rochet–Choné’s localization (Corollary A.9). We obtain the following Euler–Lagrange equations which are central to the remainder of our work.

Lemma 6.3 (Poisson data along tame rays).

Let uu solve (1) where Ω𝐑n\Omega\subset{\mathbf{R}}^{n} is bounded, open and convex. Let the coordinates r,tr,t and functions γ,ξ,R,m,b\gamma,\xi,R,m,b be as in Lemma 6.2. Then the minimality of uu, more precisely Rochet–Chone’s localization, implies the Euler–Lagrange equation relating the fixed and free boundaries

(65) R2(t)|ξ˙(t)|=2|γ˙(t)|(Dux)𝐧>0,\displaystyle R^{2}(t)|\dot{\xi}(t)|=2|\dot{\gamma}(t)|(Du-x)\cdot\mathbf{n}>0,

and the Euler–Lagrange equation for uu

(66) 3Δu=3j(t)+3r|ξ˙|δ(t)J(r,t)=3r2R(t)r+j(t)/|ξ˙(t)|.\displaystyle 3-\Delta u=\frac{3j(t)+3r|\dot{\xi}|-\delta(t)}{J(r,t)}=\frac{3r-2R(t)}{r+j(t)/|\dot{\xi}(t)|}.
Proof.

We compute the disintegration (110) and obtain for 1\mathcal{H}^{1}-a.e. t(ε,ε)t\in(-\varepsilon,\varepsilon) that

0v(Dux)𝐧|γ˙(t)|+0R(t)(3δ(t)J(r,t))J(r,t)v𝑑r.0\leq v(Du-x)\cdot\mathbf{n}|\dot{\gamma}(t)|+\int_{0}^{R(t)}\left(3-\frac{\delta(t)}{J(r,t)}\right)J(r,t)\,vdr.

The terms outside the integral are evaluated at x=x(0,t)x=x(0,t). We consider four choices of test functions, v|γ(t)~=±1v|_{\widetilde{\gamma(t)}}=\pm 1 and v|γ(t)~=±rv|_{\widetilde{\gamma(t)}}=\pm r. Using these in the localization formula we obtain the equalities

(67) ((3j(t)δ(t))R(t)+32R2(t)fξ(t))+|γ˙(t)|(Dux)𝐧=0\displaystyle\left((3j(t)-\delta(t))R(t)+\frac{3}{2}R^{2}(t)f_{\xi}(t)\right)+|\dot{\gamma}(t)|(Du-x)\cdot\mathbf{n}=0
(68) (3j(t)δ(t))R(t)22+R3(t)fξ(t)=0.\displaystyle(3j(t)-\delta(t))\frac{R(t)^{2}}{2}+R^{3}(t)f_{\xi}(t)=0.

These combine to imply

(69) R2(t)fξ(t)=2|γ˙(t)|(Dux)𝐧>0.\displaystyle R^{2}(t)f_{\xi}(t)=2|\dot{\gamma}(t)|(Du-x)\cdot\mathbf{n}>0.

We recall (64) and note (69) determines the sign fξ(t)=|ξ˙|f_{\xi}(t)=|\dot{\xi}|. Thus the Euler–Lagrange equation (65) for the free boundary holds. Equation (65) implies |ξ˙||\dot{\xi}| is bounded away from zero and from above depending only on estimates for the continuous function (Dux)𝐧(Du-x)\cdot\mathbf{n} and our previously given estimate R(t)>r0R(t)>r_{0} (Lemma 6.1).

Combining (54) with (68) we obtain (66). ∎

Remark 6.4 (Tame rays must spread as they leave the boundary).

Comparing (55) to (66) we recover fξ(t)=ξ×ξ˙>0f_{\xi}(t)=\xi\times\dot{\xi}>0, which asserts that the Jacobian J(r,t)J(r,t) is an increasing function of rr: that is, tame rays spread out as they move away from the boundary. This implies all the rays may be extended slightly further into the domain interior without intersecting each other.

Recalling that Δu=3\Delta u=3 in Ω2\Omega_{2}, we see (66) quantifies how Poisson’s equation fails to be satisfied along leaves (with the equation only satisfied at the point r=2R(t)/3r=2R(t)/3). It also implies that when we move from Ω1\Omega_{1} into Ω2\Omega_{2} the Laplacian jumps discontinuously across their common boundary.

In Section 7 we show this discontinuity yields quadratic separation of uu from its contact sets and exploit this to obtain estimates on the Hausdorff dimension of Γ𝒩\Gamma\cap\mathcal{N}.

Let us conclude by justifying the aforementioned Lipschitz continuity of ξ\xi. Note both (66) and (65) yield Lipschitz estimates. However, since their derivation assumed ξ\xi was Lipschitz we need to redo these calculations with a perturbed, Lipschitz, ξδ\xi_{\delta} and obtain uniform estimates as the perturbation parameter δ\delta approaches 0. Notice the Lipschitz constant from the following lemma does not depend on the Neumann values log((Dux)𝐧)L(𝒩Ω)\|\log((Du-x)\cdot\mathbf{n})\|_{L^{\infty}(\mathcal{N}\cap{\partial}\Omega)}.

Lemma 6.5 (Tame ray directions are Lipschitz on the fixed boundary).

With ξ\xi defined as above, the function tξ(t)t\mapsto\xi(t) is Lipschitz on the interval (ε,ε)(-\varepsilon,\varepsilon) provided by Lemma 6.1, with Lipschitz constant depending only on an upper bound for sup𝒩Δu\sup_{\mathcal{N}}\Delta u and a lower bound on R(t)R(t).

Proof.

Assume a collection of non-intersecting rays foliate an open set in 𝐑2\mathbf{R}^{2} and pierce a smooth curve γ(t)\gamma(t). Provided the intersection of each ray with the curve occurs some fixed distance dd from either endpoint of the ray, the assertion of Caffarelli, Feldman, and McCann [11, Lemma 16] says the directions ξ(t)\xi(t) (of the ray passing through γ(t)\gamma(t)) is a locally Lipschitz function with Lipschitz constant depending on γ\gamma and dd.

Thus, in our setting, if we could extend each ray by length δ\delta outside the domain, ξ(t)\xi(t) would be locally Lipschitz with a constant depending on δ\delta. Then, once we’ve obtained (66) the Lipschitz constant of ξ\xi is independent of δ\delta. Apriori such an extension outside the domain may not be possible. Thus our strategy below will be to translate the boundary distance δ\delta and use the translated boundary to redefine the r=0r=0 axis of the (r,t)(r,t) coordinates. In this setting the corresponding direction vector ξδ\xi_{\delta} is locally Lipschitz, the above calculations are justified, and sending δ0\delta\rightarrow 0 gives the Lipschitz estimate on ξ\xi.

Thus, with γ\gamma as above let 𝐧(0)\mathbf{n}(0) be the outer unit normal at γ(0)\gamma(0) and set

η(t)=γ(t)δ𝐧(0).\eta(t)=\gamma(t)-\delta\mathbf{n}(0).

Because the length of rays intersecting γ(ε,ε)\gamma(-\varepsilon,\varepsilon) is bounded below by r0r_{0}, up to a smaller choice of ε,δ\varepsilon,\delta we may assume for each t(ε,ε)t\in(-\varepsilon,\varepsilon) that η(t)Ω1\eta(t)\in\Omega_{1} and lies distance at least δ/2\delta/2 from each endpoint of η(t)~\widetilde{\eta(t)}. Let ξδ(t)\xi_{\delta}(t) denote the unit vector parallel to η(t)~\widetilde{\eta(t)}, where by [11, Lemma 16] ξδ\xi_{\delta} is a locally Lipschitz function of tt. We may redefine the (r,t)(r,t) coordinates and write a connected component of Ω1\Omega_{1} containing η(0)\eta(0) as

𝒩={x(r,t)=η(t)+rξδ(t);t(ε,ε) and R0(t)rR1(t)},\mathcal{N}=\{x(r,t)=\eta(t)+r\xi_{\delta}(t);t\in(-\varepsilon,\varepsilon)\text{ and }-R_{0}(t)\leq r\leq R_{1}(t)\},

where R0(t)R_{0}(t) is defined so that η(t)+R0(t)ξδ(t)\eta(t)+R_{0}(t)\xi_{\delta}(t) is the point where the ray η(t)~\widetilde{\eta(t)} intersects Ω\partial\Omega. The function R0(t)R_{0}(t) is locally Lipschitz because the curves γ,η\gamma,\eta are smooth and ξδ\xi_{\delta} is Lipschitz (though we don’t assert that the Lipschitz constant of R0R_{0} is independent of δ\delta). Thus all our earlier computations may be repeated in these new coordinates. The computations leading to (66) now yield the equation

(70) 3Δu=|ξ˙δ(t)|3r2R1+R0ξδ×η˙+r|ξ˙δ|,3-\Delta u=|\dot{\xi}_{\delta}(t)|\frac{3r-2R_{1}+R_{0}}{\xi_{\delta}\times\dot{\eta}+r|\dot{\xi}_{\delta}|},

which we note satisfies 3=Δu3=\Delta u when r=(2R1R0)/3r=(2R_{1}-R_{0})/3 and thus agrees with our earlier coordinate system (in our modified coordinates r=(2R1R0)/3r=(2R_{1}-R_{0})/3 is the point of distance 2(R1+R0)/32(R_{1}+R_{0})/3 from the endpoint of the ray).

Evaluating (70) at r=0r=0 we obtain a Lipschitz estimate on ξδ\xi_{\delta} which is independent of δ\delta (using, crucially, the Laplacian estimates of Theorem 4.1). The pointwise convergence of ξδ\xi_{\delta} to ξ\xi ensures ξ\xi is Lipschitz. ∎

Corollary 6.6 (Raywise coordinates (52) are biLipschitz).

Let 𝒩\mathcal{N} denote the subset of Ω1\Omega_{1} defined in (51), (x1,x2)(x^{1},x^{2}) Euclidean coordinates, and (r,t)(r,t) the coordinates defined in (52). Then the change of variables from (x1,x2)(x^{1},x^{2}) to (r,t)(r,t) is biLipschitz with Lipschitz constant depending only on sup|Du|\sup|Du|, supt(ε,ε)|γ˙|\sup_{t\in(-\varepsilon,\varepsilon)}|\dot{\gamma}|, inft(ε,ε)R(t)\inf_{t\in(-\varepsilon,\varepsilon)}R(t), and inft(ε,ε)ξ×γ˙\inf_{t\in(-\varepsilon,\varepsilon)}\xi\times\dot{\gamma} (i.e. the transversality of the intersections of rays with the fixed boundary {γ(t)}t(ε,ε)\{\gamma(t)\}_{t\in(-\varepsilon,\varepsilon)}).

Proof.

It suffices to estimate each of the entries in the Jacobians (x1,x2)(r,t)\frac{\partial(x^{1},x^{2})}{\partial(r,t)} and (r,t)(x1,x2)\frac{\partial(r,t)}{\partial(x^{1},x^{2})} computed earlier in (58) and (59). From (58) it’s clear that the entries of this Jacobian permit an estimate from above in terms of supt(ε,ε)|γ˙|\sup_{t\in(-\varepsilon,\varepsilon)}|\dot{\gamma}| and supt(ε,ε)|ξ˙|\sup_{t\in(-\varepsilon,\varepsilon)}|\dot{\xi}|, where the latter may, in turn, be estimated in terms of sup|Du|\sup|Du| and inft(ε,ε)R(t)\inf_{t\in(-\varepsilon,\varepsilon)}R(t) thanks to (65). The only additional term we must estimate for the second Jacobian, that is (59)\eqref{eq:jac-2}, is 1/J(r,t)1/J(r,t). Since 1/J(r,t)1/J(r,t) is decreasing in rr by Remark 6.4, it suffices to estimate J(0,t)J(0,t) and this is an immediate consequence of (55) which we recall says j(t)=ξ×γ˙>0j(t)=\xi\times\dot{\gamma}>0. ∎

7. On the regularity of the free boundary

In this section we study local properties of the free boundary by transformation to an obstacle problem and prove Theorem 1.2. If u1u_{1} denotes the minimal convex extension of u|𝒩u|_{\mathcal{N}} we show that v:=uu1v:=u-u_{1} solves an obstacle problem with the same free boundary as our original problem. Standard results for the obstacle problem then imply the free boundary Γ𝒩\Gamma\cap\mathcal{N} has Lebesgue measure 0 and, the stronger result, Γ𝒩\Gamma\cap\mathcal{N} has Hausdorff dimension strictly less than nn.

We use these estimates to establish that the function tR(t)t\mapsto R(t) from Section 6 is continuous almost everywhere on (ε,ε)(-\varepsilon,\varepsilon), Theorem 7.2. In Proposition 7.3 we describe a bootstrapping procedure which shows that if the free boundary — or rather the function tR(t)t\mapsto R(t) — is Lipschitz, then it is C.C^{\infty}. In Theorem 7.5 and its corollary, we show RR has a Lipschitz graph away from its local maxima.

7.1. Transformation to the classical obstacle problem

Let x1Ωx_{1}\in\Omega denote the endpoint of a tame ray and x1~Ω={x0}\tilde{x_{1}}\cap\partial\Omega=\{x_{0}\} as in Lemma 6.1. Using the coordinates from the previous section we consider a subset of Ω1\Omega_{1},

𝒩=𝒩Ω1\displaystyle\mathcal{N}=\mathcal{N}\cap\Omega_{1} ={γ(t)~;t(ε,ε)},\displaystyle=\{\widetilde{\gamma(t)};t\in(-\varepsilon,\varepsilon)\},
={x(r,t)=γ(t)+rξ(t);t(ε,ε) and 0rR(t)}.\displaystyle=\{x(r,t)=\gamma(t)+r\xi(t);t\in(-\varepsilon,\varepsilon)\text{ and }0\leq r\leq R(t)\}.

In Remark 6.4 we observed that rays spread out as they leave the boundary. Thus the coordinates (r,t)(r,t) remain well-defined on an extension of 𝒩\mathcal{N}. We denote this extension by 𝒩ext\mathcal{N}_{\text{ext}}:

𝒩ext={x(r,t)=γ(t)+rξ(t);t(ε,ε) and 0r<+}.\mathcal{N}_{\text{ext}}=\{x(r,t)=\gamma(t)+r\xi(t);t\in(-\varepsilon,\varepsilon)\text{ and }0\leq r<+\infty\}.

On 𝒩ext\mathcal{N}_{\text{ext}} we define the minimal convex extension of u|𝒩u|_{\mathcal{N}} by the formula (53)

u1(x)=b(t)+rm(t) for x=x(r,t)𝒩ext.u_{1}(x)=b(t)+rm(t)\text{ for $x=x(r,t)\in\mathcal{N}_{\text{ext}}$}.

Note there is some α>0\alpha>0 such that Bα(x1)𝒩extΩB_{\alpha}(x_{1})\subset\subset\mathcal{N}_{\text{ext}}\cap\Omega. Moreover on 𝒩extΩ1\mathcal{N}_{\text{ext}}\setminus\Omega_{1} we have, by (66), Δu13c0\Delta u_{1}\leq 3-c_{0} for c0>0c_{0}>0 depending only on a lower bound for R(t)R(t) and (Dux)𝐧(Du-x)\cdot\mathbf{n}. Let v=uu1v=u-u_{1} and let 1{v>0}1_{\{v>0\}} denote the characteristic function of {v>0}\{v>0\}. Then Δu=3\Delta u=3 in Ω2\Omega_{2} implies that

(71) Δv\displaystyle\Delta v =f(x)1{v>0}c01{v>0} in Bα(x1),\displaystyle=f(x)1_{\{v>0\}}\geq c_{0}1_{\{v>0\}}\text{ in }B_{\alpha}(x_{1}),
(72) v\displaystyle v 0 in Bα(x1),\displaystyle\geq 0\text{ in }B_{\alpha}(x_{1}),
(73) v\displaystyle v =0 in Bα(x1)Ω1 and v>0 in Bα(x1)Ω2\displaystyle=0\text{ in }B_{\alpha}(x_{1})\cap\Omega_{1}\text{ and }v>0\text{ in }B_{\alpha}(x_{1})\cap\Omega_{2}

where f(x)=3Δu1f(x)=3-\Delta u_{1}. Thus vv solves the classical obstacle problem in Bα(x1)B_{\alpha}(x_{1}) with the same free boundary as our original problem. The regularity theory for the obstacle problem yields estimates for the measure of the free boundary. What prevents us from using higher regularity theory for the obstacle problem is that on Bα(x1){v>0}B_{\alpha}(x_{1})\cap\{v>0\},

Δv=f(x)=3r2R(t)r+ξ×γ˙/|ξ˙|,\Delta v=f(x)=\frac{3r-2R(t)}{r+\xi\times\dot{\gamma}/|\dot{\xi}|},

may not be continuous — the minimum regularity required to apply typical regularity results for the obstacle problem is ff Hölder continuous. If, RR — i.e. the free boundary — were Lipschitz, Δv\Delta v would also be Lipschitz in which case one can bootstrap to CC^{\infty} regularity of RR; see Proposition 7.3. As a partial result in this direction we prove that tR(t)t\mapsto R(t) is continuous almost everywhere. We begin with Hausdorff dimension estimates for the free boundary.

Lemma 7.1 (Hausdorff dimension estimate for the free boundary).

Let x0,x1,αx_{0},x_{1},\alpha be as given above, so that x1Ωx_{1}\in\Omega is the endpoint of a tame ray; c.f. Lemma 6.1. Then the Hausdorff dimension of Bα/2(x1)ΓB_{\alpha/2}(x_{1})\cap\Gamma equals 2δ12-\delta_{1} for some δ1\delta_{1} depending only on α\alpha, infBα(x0)Ω(Dux)𝐧\inf_{B_{\alpha}(x_{0})\cap\partial\Omega}(Du-x)\cdot\mathbf{n} and inft(ε,ε)R(t)\inf_{t\in(-\varepsilon,\varepsilon)}R(t).

Proof.

This is a standard result for the obstacle problem once one notes that ff in (71) satisfies 0<c0f=Δv30<c_{0}\leq f=\Delta v\leq 3 on {v>0}\{v>0\} for c0c_{0} depending only on infBα(x0)(Dux)𝐧\inf_{B_{\alpha}(x_{0})}(Du-x)\cdot\mathbf{n} and inft(ε,ε)R(t)\inf_{t\in(-\varepsilon,\varepsilon)}R(t). We follow the clear exposition of Petrosyan, Shahgholian, and Uraltseva [49, §3.1, 3.2] to establish first quadratic detachment, then porosity.

Step 1. (Quadratic detachment at free boundary points) We claim if x2Bα/2(x1)Γx_{2}\in B_{\alpha/2}(x_{1})\cap\Gamma then

(74) supBρ(x2)vc02ρ2.\sup_{B_{\rho}(x_{2})}v\geq\frac{c_{0}}{2}\rho^{2}.

Fix such an x2x_{2} and x¯Ω2Bα/2(x1)\overline{x}\in\Omega_{2}\cap B_{\alpha/2}(x_{1}); we will eventually take x¯x2\overline{x}\rightarrow x_{2}. Set w(x)=v(x)c0|xx¯|2/2.w(x)=v(x)-c_{0}|x-\overline{x}|^{2}/2. On the set {v>0}\{v>0\} we have Δw>0\Delta w>0. Thus, the maximum principle implies

0v(x¯)=w(x¯)supBρ(x¯){v>0}w=sup[Bρ(x¯){v>0}]w.0\leq v(\overline{x})=w(\overline{x})\leq\sup_{B_{\rho}(\overline{x})\cap\{v>0\}}w=\sup_{\partial[B_{\rho}(\overline{x})\cap\{v>0\}]}w.

Because w<0w<0 on {v>0}\partial\{v>0\} the supremum is attained at some xx on (Bρ(x¯)){v>0}\partial(B_{\rho}(\overline{x}))\cap\{v>0\}. Because |xx¯|=ρ|x-\overline{x}|=\rho we obtain

0v(x)c02ρ2,0\leq v(x)-\frac{c_{0}}{2}\rho^{2},

for some xBρ(x¯)x\in\partial B_{\rho}(\overline{x}). We send x¯x2\overline{x}\rightarrow x_{2} to establish (74).

Step 2. (Nondegeneracy implies porosity) We recall a measurable set E𝐑nE\subset\mathbf{R}^{n} is called porous with porosity constant δ\delta if for all x𝐑nx\in\mathbf{R}^{n} and ρ>0\rho>0 there is yBρ(x)y\in B_{\rho}(x) with

Bδρ(y)Bρ(x)Ec.B_{\delta\rho}(y)\subset B_{\rho}(x)\cap E^{c}.

We prove that nondegeneracy, i.e. (74), and Caffarelli–Lions’s Cloc1,1C^{1,1}_{\text{loc}} implies ΓBα/2(x1)\Gamma\cap B_{\alpha/2}(x_{1}) is porous. Take x2ΓBα/2(x1)x_{2}\in\Gamma\cap B_{\alpha/2}(x_{1}). Note (74) implies supBρ(x2)|Dv|c0ρ/2.\sup_{B_{\rho}(x_{2})}|Dv|\geq c_{0}\rho/2. Indeed, with x¯Bρ(x2)\overline{x}\in B_{\rho}(x_{2}) such that v(x¯)c0ρ2/2v(\overline{x})\geq c_{0}\rho^{2}/2 we have

c0ρ2/2v(x¯)v(x2)=01Dv(x2+τ(x¯x2))(x¯x2)𝑑τρsupBr(x2)|Dv|.c_{0}\rho^{2}/2\leq v(\overline{x})-v(x_{2})=\int_{0}^{1}Dv(x_{2}+\tau(\overline{x}-x_{2}))\cdot(\overline{x}-x_{2})\,d\tau\leq\rho\sup_{B_{r}(x_{2})}|Dv|.

Now, redefine x¯\overline{x} as a point in Bρ/2(x2)B_{\rho/2}(x_{2}) where |Dv(x¯)|c0ρ/4|Dv(\overline{x})|\geq c_{0}\rho/4. Using that vCloc1,1M\|v\|_{C^{1,1}_{\text{loc}}}\leq M (where Δu=3\Delta u=3, Δu13\Delta u_{1}\leq 3 gives the obvious estimate M=6M=6), we have if xBδρ(x¯)x\in B_{\delta\rho}(\overline{x}) for δ=c0/8M\delta=c_{0}/8M then

|Dv(x)||Dv(x¯)||Dv(x)Dv(x¯)|c0ρ/4Mδρ=c0ρ/8.|Dv(x)|\geq|Dv(\overline{x})|-|Dv(x)-Dv(\overline{x})|\geq c_{0}\rho/4-M\delta\rho=c_{0}\rho/8.

Since Dv0Dv\equiv 0 along Γ\Gamma this proves Bδρ(x¯)B_{\delta\rho}(\overline{x}) lies in Γc\Gamma^{c}. Thus we’ve established the porosity condition for balls centered on ΓBα/2(x1)\Gamma\cap B_{\alpha/2}(x_{1}). To establish the porosity condition for any ball in 𝐑n\mathbf{R}^{n} we argue as follows. Let Bρ(x)𝐑nB_{\rho}(x)\subset\mathbf{R}^{n}. We take x¯Bρ/2(x)ΓBα/2(x1)\overline{x}\in B_{\rho/2}(x)\cap\Gamma\cap B_{\alpha/2}(x_{1}), noting if no such x¯\overline{x} exists we’re done. Our porosity result applied on Bρ/2(x¯)Bρ(x)B_{\rho/2}(\overline{x})\subset B_{\rho}(x) gives porosity of Bρ(x)B_{\rho}(x).

Step 3. (Conclusion) As noted in [49, §3.2.2] by the work of [51] a porous set in 𝐑n\mathbf{R}^{n} has Hausdorff dimension less than nn. ∎

Theorem 7.2 (Continuity of the free boundary a.e.).

Taking ε\varepsilon as in Lemma 6.1, the function R:(ε,ε)𝐑R:(-\varepsilon,\varepsilon)\rightarrow\mathbf{R} defined by R(t)=diam(γ(t)~)R(t)=\text{diam}(\widetilde{\gamma(t)}) is continuous for 1\mathcal{H}^{1} almost every t(ε,ε)t\in(-\varepsilon,\varepsilon).

Proof.

Assume RR is not continuous at some t(ε,ε)t_{\infty}\in(-\varepsilon,\varepsilon). Because RR is upper semicontinuous this implies there exists a sequence tktt_{k}\rightarrow t_{\infty} with R¯:=limR(tk)<R(t)=:R¯\underline{R}:=\lim R(t_{k})<R(t_{\infty})=:\overline{R}. We show each x=x(r,t)x=x(r,t_{\infty}) for r[R¯,R¯]r\in[\underline{R},\overline{R}] lies in the free boundary Γ𝒩.\Gamma\cap\mathcal{N}. Then, because Lemma 7.1 implies 2(Ω1Ω2)=0\mathcal{H}^{2}(\partial\Omega_{1}\cap\partial\Omega_{2})=0, Fubini’s theorem and Corollary 6.6 imply the union of all such tt_{\infty} has 1\mathcal{H}^{1}-measure 0.

To show for each r[R¯,R¯]r\in[\underline{R},\overline{R}], x(r,t)Ω1x(r,t_{\infty})\in\partial\Omega_{1} we suppose otherwise. Then since no such point x(r,t)x(r,t_{\infty}) can be interior to Ω2\Omega_{2} without violating Lemma 3.2, there is r(R¯,R¯)r\in(\underline{R},\overline{R}) such that x(r,t)x(r,t_{\infty}) is an interior point of Ω1\Omega_{1}.

Upper semicontinuity of RR implies rays sufficiently close to x(r,t)x(r,t_{\infty}) have intersection with the boundary close to x(0,t)=Ωx(r,t)~x(0,t_{\infty})=\partial\Omega\cap\widetilde{x(r,t_{\infty})}. More precisely for every ε>0\varepsilon>0 there is δ>0\delta>0 such that

{x~;xBδ(x(r,t))}ΩBε(x(0,t)).\{\tilde{x};x\in B_{\delta}(x(r,t_{\infty}))\}\cap\partial\Omega\subset B_{\varepsilon}(x(0,t_{\infty})).

Thus, our planar foliation implies that because x(r,t)x(r,t_{\infty}) is an interior point of Ω1\Omega_{1} then x(ρ,t)x(\rho,t_{\infty}) is also an interior point of Ω1\Omega_{1} for each ρr\rho\leq r which contradicts that x(R¯,t)Ω1x(\underline{R},t_{\infty})\in\partial\Omega_{1} and completes the proof. ∎

Next we show prove the fourth point of Theorem 1.2: that if this continuity could be strengthened to Lipschitz continuity then one can bootstrap to a smooth free boundary and minimizer on Ω1\Omega_{1}.

Proposition 7.3 (Tame part of free boundary is smooth where Lipschitz).

Let uu solve (1). Let x1Γ𝐑2x_{1}\in\Gamma\subset{\mathbf{R}}^{2} be a tame point of the free boundary and x0=Ωx1~x_{0}=\partial\Omega\cap\tilde{x_{1}}. Assume xdiam(x~)x\mapsto\text{diam}(\tilde{x}) is Lipschitz on some Bε(x0)ΩB_{\varepsilon}(x_{0})\cap\partial\Omega. Then there is δ>0\delta>0 such that Bδ(x1)ΓB_{\delta}(x_{1})\cap\Gamma is a smooth curve. On the portion 𝒩\mathcal{N} of Ω1\Omega_{1} consisting of rays which intersect Bδ(x1)B_{\delta}(x_{1}), the transformation (r,t)x(r,t)(r,t)\to x(r,t) of (52) is a smooth diffeomorphism and uC(𝒩intΩ)u\in C^{\infty}(\mathcal{N}\cap\mathop{\rm int}\Omega).

Proof.

We prove by induction on k=0,1,2,k=0,1,2,\dots that there is some neighbourhood on which the curve Bδ(x0)Ω1B_{\delta}(x_{0})\cap\partial\Omega_{1} and the function RR are both Ck,αC^{k,\alpha} for some 0<α<10<\alpha<1, while the coordinate transformations and uu are Ck+1,αC^{k+1,\alpha} in Bδ(x1)Ω1B_{\delta}(x_{1})\cap\Omega_{1}. For k=0k=0 our assumption is that tR(t)t\mapsto R(t) is Lipschitz, and from Corollary 6.6 the coordinate transformations are biLipschitz. From the formula (65), reproduced here for the readers convenience

(75) R2(t)|ξ˙(t)|=2|γ˙(t)|(Dux)𝐧,\displaystyle R^{2}(t)|\dot{\xi}(t)|=2|\dot{\gamma}(t)|(Du-x)\cdot\mathbf{n},

and Caffarelli and Lions uCloc1,1u\in C^{1,1}_{\text{loc}} (or Theorem 4.1) it follows that tξ˙(t)t\mapsto\dot{\xi}(t) is also Lipschitz, hence the coordinate transformations improve to C1,1C^{1,1} by the Jacobian expressions (58)–(59). From (66), i.e.

(76) 3Δu=|ξ˙(t)J(r,t)|(3r2R),3-\Delta u=\left|\frac{\dot{\xi}(t)}{J(r,t)}\right|(3r-2R),

we see ΔuC0,α\Delta u\in C^{0,\alpha}, hence the regularity theory for Poisson’s equation implies uCk+2,αu\in C^{k+2,\alpha} when k=0k=0 (one derivative more than needed).

Now assume the inductive hypothesis for some fixed kk. From (75)–(76) we again deduce uu has Ck,αC^{k,\alpha} Laplacian in Ω1\Omega_{1}. Thus the regularity theory for Poisson’s equation implies uCk+2,αu\in C^{k+2,\alpha}. The regularity theory for the obstacle problem (where the obstacle has a Ck,αC^{k,\alpha} Laplacian; (due to Caffarelli [10, 8], and Kinderlehrer [33] with Nirenberg [32], though the clearest statement we’ve found is by Blank [6, 5]) implies the free boundary is Ck+1,αC^{k+1,\alpha}. Note to apply the classical regularity theory for the obstacle problem we are using that the Lipschitz regularity of RR implies the set Ω¯1={v=0}\overline{\Omega}_{1}=\{v=0\} has positive density at each boundary point; here v=uu1v=u-u_{1} as in (71) . Now that RR is Ck+1,αC^{k+1,\alpha} the same is again true for ξ˙\dot{\xi} by equation (75) since the smoothness DuCk+1,αDu\in C^{k+1,\alpha} established in Ω1Bδ(x1)\Omega_{1}\cap B_{\delta}(x_{1}) propagates down the rays from the free to the fixed boundary using the Ck+1,αC^{k+1,\alpha} coordinate transformations; these transformations then improve to Ck+2,αC^{k+2,\alpha} by equations (58)–(59) so the induction is established and the proof is complete. ∎

Theorem 1.2(1) and (4) are obtained by combining Lemmas 7.1 and 7.3; parts (2) and (3) will be established in Theorem 7.5 of the next section.

7.2. Criteria for the tame part of the free boundary to be Lipschitz

Having deduced regularity of the free boundary when it is Lipschitz we now turn our attention to the question of characterising the set on which the free boundary is Lipschitz. We will rely on the well known Caffarelli dichotomy for the blow-up of solutions to the obstacle problem. We recall that blowing-up at the edge of the contact region in the classical obstacle problem (without convexity constraints) led Caffarelli to formulate his celebrated alternative [10, 8]: If wCloc1,1(𝐑n)w\in C^{1,1}_{\text{loc}}({\mathbf{R}}^{n}) satisfies

(77) Δw(x)=1{w>0}(x)a.e.on𝐑n{\Delta w(x)=1_{\{w>0\}}(x)\quad a.e.\ {\rm on}\ {\mathbf{R}}^{n}}

then ww is convex and either a quadratic polynomial or a rotated translate of the half-parabola solution

w(x1,,xn)={12x12ifx1>00else.{w(x_{1},\ldots,x_{n})=\begin{cases}\frac{1}{2}x_{1}^{2}&{\rm if}\ x_{1}>0\\ 0&{\rm else}.\end{cases}}

At each point in the free boundary, the density of the contact region is therefore either 0 (called singular) or 12\frac{1}{2} (called regular); it cannot equal 11 because of quadratic detachment (as in e.g. the proof of Lemma 7.1). Furthermore, the dichotomy holds for blowups of solution to equations of the form Δu(x)=f(x)1{u>0}(x)\Delta u(x)=f(x)1_{\{u>0\}}(x) in a domain Ω\Omega where ff is continuous in the following sense: Take x0Ω{u=0}x_{0}\in\Omega\cap\partial\{u=0\} and a sequence rk0r_{k}\rightarrow 0. Note that up to taking a sequence the limit

u0(x):=limku(rk(xx0))rk2,u_{0}(x):=\lim_{k\rightarrow\infty}\frac{u\big{(}r_{k}(x-x_{0})\big{)}}{r_{k}^{2}},

exists and is a globally defined solution of Δu(x)=f(x0)1{u=0}(x)\Delta u(x)=f(x_{0})1_{\{u=0\}}(x) so that Caffarelli’s dichotomy applies to the function u0u_{0}. Unfortunately, in our setting we only know fLlocf\in L^{\infty}_{\text{loc}} and not the Hölder continuity required for higher regularity [33].

A real-valued function SS on an interval J𝐑J\subset{\mathbf{R}} is called unimodal if it is monotone, or else if it attains its maximum on a (possibly degenerate interval) IJI\subset J, with SS being non-decreasing throughout the connected component of JIJ\setminus I to the left of II, and non-increasing through the connected component to the right of II. The following lemma shows lower semicontinuous functions are unimodal away from their local minima.

Lemma 7.4 (Lower semicontinuous unimodality away from local minima).

Let S:E[,)S:E\longrightarrow[-\infty,\infty) be lower semicontinuous on an interval E𝐑E\subset{\mathbf{R}}. Let TT denote the subset of EE consisting of local minima for SS, and T¯\overline{T} its closure. Then SS is unimodal on each connected component JJ of ET¯E\setminus\overline{T}.

Proof.

Fix any open interval JET¯J\subset E\setminus\overline{T}. We claim SS is unimodal on JJ. Since SS is lower semicontinuous but has no local minima on JJ, for each c𝐑c\in{\mathbf{R}} it follows that J(c):={tJS(t)>c}=i(ai,bi)J(c):=\{t\in J\mid S(t)>c\}=\cup_{i}(a_{i},b_{i}) is a countable union of open intervals on which S>cS>c with ScS\leq c on JJ(c)J\setminus J(c). If there were more than one open interval in this union, say (a1,b1)(a_{1},b_{1}) and (a2,b2)(a_{2},b_{2}) with b1a2b_{1}\leq a_{2}, then SS would attain a local minimum on the compact set [b1,a2]J[b_{1},a_{2}]\subset J, contradicting the fact JET¯J\subset E\setminus\overline{T}. Thus the set J(c)J(c) consists of at most one open interval, which is monotone nonincreasing with c𝐑c\in{\mathbf{R}}. Let c0c_{0} denote the infimum of c𝐑c\in{\mathbf{R}} for which J(c)J(c) is empty, and set I=c<c0J(c)I=\cap_{c<c_{0}}J(c). Then SS is non-decreasing to the left of II, non-increasing to the right of II, and — if II is nonempty — attains its maximum value on II. ∎

We apply this lemma to the diameter R=SR=-S of the rays along the tame part of the free boundary to deduce the free boundary is Lipschitz away from its local maxima.

Theorem 7.5 (Tame free boundary is Lipschitz away from local maxima).

Let γ:EΩ\gamma:E\longrightarrow{\partial}\Omega with γ˙(t)0\dot{\gamma}(t)\neq 0 for tE:=(ε,ε)t\in E:=(-\varepsilon,\varepsilon) smoothly parameterize a fixed boundary interval throughout which the Neumann condition (5) is violated. Let TT denote the subset of EE consisting of local maxima for R(t):=diam(γ(t)~)R(t):=\text{\rm diam}(\widetilde{\gamma(t)}), and JJ any connected component of ET¯E\setminus\overline{T}, where T¯\overline{T} is the closure of TT. Then RR extends continuously to J¯\overline{J} and its graph is a Lipschitz submanifold of J¯×𝐑\overline{J}\times{\mathbf{R}}. Similarly, the graph of F(t):=γ(t)+R(t)ξ(t)F(t):=\gamma(t)+R(t)\xi(t) over J¯\overline{J} is a Lipschitz submanifold of Ω¯\overline{\Omega} (except perhaps at t=±εt=\pm\varepsilon), and FF is continuous on J¯\overline{J}.

Proof.

Corollary 6.6 shows the coordinates x(t,r)=γ(t)+rξ(t)x(t,r)=\gamma(t)+r\xi(t) are locally biLipschitz on E×(0,)E\times(0,\infty), so the final sentence follows from showing RR has a continuous extension to J¯\overline{J} whose graph is a Lipschitz submanifold.

Proposition 5.4 asserts RR is upper semicontinuous on J¯\overline{J}. Unless RR is monotone on JJ, Lemma 7.4 shows JJ decomposes into two subintervals on which R-R is monotone and they overlap at least at one point tt^{\prime}. Although a monotone function need not be Lipschitz — or even continuous — its graph has Lipschitz constant at most 11. A discontinuity in RR on the closure of either of these subintervals would correspond to a line segment {x(t,r):r[R(t)δ,R(t)]}\{x(t,r):{r\in[R(t)-\delta,R(t)]}\} of length δ>0\delta>0 in the free boundary along which uu is affine. This contradicts the strong maximum principle (Lemma 3.2) after constructing the appropriate reflection of u2:=u|Ω2u_{2}:=u|_{\Omega_{2}} across this segment. Thus RR is continuous on J¯\overline{J}. The graph of RR on J¯\bar{J} is obviously Lipschitz, except perhaps when the minimum value of RR is uniquely attained at some tJt^{\prime}\in J. Since tt^{\prime} is a local minimum, RR is continuous at tt^{\prime} hence Caffarelli’s alternative holds for the blow-up at F(t)F(t^{\prime}): the Lebesgue density of Ω1\Omega_{1} at F(t)F(t^{\prime}) cannot be zero since R(t)R(t^{\prime}) is a local minimum, so it must be exactly 1/21/2 [8, 10]. The blow-up limits of u2u1u_{2}-u_{1} at F(t)F(t^{\prime}) all coincide with the same half-parabola, and RR is differentiable at tt^{\prime}. The Lipschitz graph of RR to the left of tt^{\prime} shares the same tangent at tt^{\prime} as the Lipschitz graph of RR to the right of tt^{\prime}, which completes the proof. ∎

Our next corollary shows that the tame part of the free boundary can only fail to be locally Lipschitz when oscillations with unbounded frequency cause local maxima of RR to accumulate, or when an isolated local maximum forms a cusp. In the latter case, the tame free boundary is locally piecewise Lipschitz and the perimeter of Ω1\Omega_{1} is locally finite in this region.

Corollary 7.6 (Is the tame free boundary piecewise Lipschitz?).

Assume TT has only finitely many connected components in Theorem 7.5 and RR is constant on each of them — as when RR has only finitely many local maxima on EE. Then the graph of F(t):=γ(t)+R(t)ξ(t)F(t):=\gamma(t)+R(t)\xi(t) is a piecewise Lipschitz submanifold of (ε+δ,εδ)×(0,)(-\varepsilon+\delta,\varepsilon-\delta)\times(0,\infty) for each δ>0\delta>0. Moreover, if the graph of FF fails to be Lipschitz at F(t)F(t^{\prime}) for some tEt^{\prime}\in E, then RR has an isolated local maximum at tt^{\prime} and Ω1\Omega_{1} has Lebesgue density zero at F(t)F(t^{\prime}).

Proof.

Under the hypotheses of Theorem 7.5, assume TT has only finitely many connected components and RR is constant on each of them. Then these components must be intervals which are relatively closed in EE: otherwise the upper semicontinuous function RR has a jump increase at the end of one of them, which leads to a segment in the graph of u2u_{2} — producing the same contradiction to Lemma 3.2 as in the proof of Theorem 7.5. Thus T¯=T\overline{T}=T. For each of the open intervals JJ comprising ET¯E\setminus\overline{T}, Theorem 7.5 already asserts that RR is continuous and has Lipschitz graph on J¯\overline{J}; the only question is whether the graph extends past each endpoint of J¯\overline{J} in EE in a Lipschitz fashion. If the endpoint of J¯\overline{J} belongs to a nondegenerate interval in TT this is obvious. When the endpoint of J¯\overline{J} is an isolated point tt^{\prime} in TT, then Lemma 7.4 shows RR nearby is monotone on either side hence must be continuous at tt^{\prime} to avoid an affine segment in the graph of u2u_{2} as before. Now Caffarelli’s alternative applies, so the density of Ω1\Omega_{1} at F(t)F(t^{\prime}) must be either 0 or 1/21/2. In the latter case RR has a Lipschitz graph in a neighbourhood of tt^{\prime}, as in the proof of Theorem 7.5, hence the corollary is established. ∎

Remark 7.7 (A partial converse).

If Ω1\Omega_{1} fails to have Lebesgue density 1/21/2 at some tame point x=F(t)ΩΩ1x^{\prime}=F(t^{\prime})\in\Omega\cap{\partial}\Omega_{1} where F(t)=γ(t)+R(t)ξ(t)F(t)=\gamma(t)+R(t)\xi(t) is continuous, then there is no neigbourhood of xx^{\prime} whose intersection with Ω1\Omega_{1} is a Lipschitz domain. This follows from the Caffarelli alternative, which requires Ω1\Omega_{1} to have Lebesgue density 0 at xx^{\prime} [8, 10].

It remains to see whether accumulation points of local maxima of R(t)R(t) and/or cusps might be ruled out by combining quadratic detachment shown in Lemma 7.1 with estimates in the spirit of the following lemma.

Lemma 7.8 (A variant on Clarke’s implicit function theorem).

Let EE be a Lipschitz manifold whose topology is metrized by dEd_{E}. Fix a Lipschitz function f:E×[a,b]𝐑f:E\times[a,b]\longrightarrow{\mathbf{R}} such that for each r[a,b]r\in[a,b], tf(t,r)t\mapsto f(t,r) has Lipschitz constant LL on EE. Assume there exists a set TET\subset E, and nonempty interval (α,β)𝐑(\alpha,\beta)\subset{\mathbf{R}} such that for each tTt\in T there exists R(t)[a,b]R(t)\in[a,b] satisfying

(78) f(t,r)f(t,R(t))(rR(t))β\displaystyle f(t,r)-f(t,R(t))\geq(r-R(t))\beta  for allr(R(t),b)\displaystyle\mbox{\rm\ for all}\ r\in(R(t),b)\
(79) andf(t,R(t))f(t,r)(R(t)r)α\displaystyle\mbox{\rm and}\ f(t,R(t))-f(t,r)\leq(R(t)-r)\alpha  for allr(a,R(t)).\displaystyle\mbox{\rm\ for all}\ r\in(a,R(t)).

Then the restriction of RR to TT has Lipschitz constant 2L/(βα)2L/(\beta-\alpha).

Proof.

Let t0,t1Tt_{0},t_{1}\in T and set ri:=R(ti)r_{i}:=R(t_{i}) for i=0,1i=0,1. Relabel if necessary so that δR:=r1r00{\delta}R:=r_{1}-r_{0}\geq 0. Conditions (78)–(79) and the Lipschitz continuity of ff give

f(t0,r1)f(t0,r0)\displaystyle f(t_{0},r_{1})-f(t_{0},r_{0}) βδR\displaystyle\geq\beta{\delta}R
andf(t1,r1)f(t1,r0)\displaystyle{\rm and}\quad f(t_{1},r_{1})-f(t_{1},r_{0}) αδR.\displaystyle\leq\alpha\delta R.

Subtracting yields

2LdE(t0,t1)(βα)δR.2Ld_{E}(t_{0},t_{1})\geq(\beta-\alpha)\delta R.

This shows RR has the asserted Lipschitz constant on TT. ∎

8. Bifurcations to bunching in the family of square examples

In this section we apply our results and techniques to a concrete example and completely describe the solution on the domain Ω=(a,a+1)2\Omega=(a,a+1)^{2}. We prove Theorem 1.4. For a722a\geq\frac{7}{2}-\sqrt{2} the solution is as hypothesized in the earlier work of McCann and Zhang [41]. Recall Ω10\Omega_{1}^{0} denotes the set of leaves with one endpoint on ΩW\Omega_{W} and the other on ΩS\Omega_{S} and here the solution is given explicitly in [50, 41]. We let Ω1\Omega_{1}^{-} denote the set of leaves with one endpoint in Ω\Omega and the other on ΩW\Omega_{W}, and, finally, let Ω1+\Omega_{1}^{+} denote the set of leaves with one endpoint in Ω\Omega and the other on ΩS\Omega_{S}. Because, in the course of our proof, we prove Ω1\Omega_{1} does not intersect ΩN\Omega_{N} or ΩE\Omega_{E} we have Ω1=Ω10Ω1Ω1+\Omega_{1}=\Omega_{1}^{0}\cup\Omega_{1}^{-}\cup\Omega_{1}^{+}.

Our main tool to study the minimizer is the coordinates introduced in Section 6. Let us consider a component of Ω1\Omega_{1}^{-} consisting of leaves with one endpoint on the boundary ΩW={a}×[a,a+1]\Omega_{W}=\{a\}\times[a,a+1] and the other interior to Ω\Omega. The argument is similar on each side. We may take the angle θ\theta made by leaves with the horizontal (that is, with the vector (1,0)(1,0)), as the parametrization coordinate of our boundary (i.e. t=θt=\theta and ξ(t)=(cos(θ),sin(θ))\xi(t)=(\cos(\theta),\sin(\theta))). Then γ(θ)=(a,h(θ))\gamma(\theta)=(a,h(\theta)) and (r,θ)(r,\theta) satisfy

x(r,θ)=(a+rcosθ,h(θ)+rsinθ),x(r,\theta)=(a+r\cos\theta,h(\theta)+r\sin\theta),

where h(θ)h(\theta) is the height at which the leaf that makes angle θ\theta with the horizontal intersects ΩW\Omega_{W}. We work in a connected subset of Ω1\Omega_{1}

𝒩=𝒩Ω1={(r,θ);θ¯θθ¯ and 0rR(θ)}.\mathcal{N}=\mathcal{N}\cap\Omega_{1}^{-}=\{(r,\theta);\underline{\theta}\leq\theta\leq\overline{\theta}\text{ and }0\leq r\leq R(\theta)\}.

In this setting (67) and (68) become

(80) 0\displaystyle 0 =(3hcosθm′′m)R(θ)+32R2(θ)+h(θ)(Du(x)x)𝐧\displaystyle=(3h^{\prime}\cos\theta-m^{\prime\prime}-m)R(\theta)+\frac{3}{2}R^{2}(\theta)+h^{\prime}(\theta)(Du({x})-{x})\cdot\mathbf{n}
(81) 0\displaystyle 0 =(3hcosθm′′m)R2(θ)2+R3(θ)\displaystyle=(3h^{\prime}\cos\theta-m^{\prime\prime}-m)\frac{R^{2}(\theta)}{2}+R^{3}(\theta)

where we use the prime notation for derivatives of roman characters as opposed to the dot notation for derivatives of greek characters, and equation (65) becomes

(82) R2(θ)=2h(θ)(Dux)𝐧.R^{2}(\theta)=2h^{\prime}(\theta)(Du-x)\cdot\mathbf{n}.

Note when we parameterize with respect to θ\theta, |ξ˙|=1|\dot{\xi}|=1 so mm^{\prime} is Lipschitz by (61). We’ve used that h(θ)>0h^{\prime}(\theta)>0 as is easily seen by first working with the parametrization γ(t)=(a,t)\gamma(t)=(a,t) and the angles ξ(t)=(cosθ(t),sinθ(t))\xi(t)=(\cos\theta(t),\sin\theta(t)), for which the identity ξ×ξ˙>0\xi\times\dot{\xi}>0 derived in Section 6 implies θ(t)>0\theta^{\prime}(t)>0. Equations (80) – (82) yield a new, expedited, proof of the Euler–Lagrange equations in Ω1±\Omega_{1}^{\pm} originally derived by the first and third author [41] via a complicated perturbation argument. Solving (60) and (61) gives

(83) u1(x)\displaystyle u_{1}(x) =m(θ)cosθm(θ)sinθ\displaystyle=m(\theta)\cos\theta-m^{\prime}(\theta)\sin\theta
(84) andu2(x)\displaystyle\text{and}\quad u_{2}(x) =m(θ)sinθ+m(θ)cosθ.\displaystyle=m(\theta)\sin\theta+m^{\prime}(\theta)\cos\theta.

Thus, along ΩW\Omega_{W}

(Du(x0)x0)𝐧=au1(x0)=a+m(θ)sin(θ)m(θ)cosθ(Du(x_{0})-x_{0})\cdot\mathbf{n}=a-u_{1}(x_{0})=a+m^{\prime}(\theta)\sin(\theta)-m(\theta)\cos\theta

Substituting into (82) we obtain

R2(θ)=2h(θ)(a+msin(θ)m(θ)cosθ).R^{2}(\theta)=2h^{\prime}(\theta)(a+m^{\prime}\sin(\theta)-m(\theta)\cos\theta).

After multiplying by cosθ\cos\theta and solving (81) for hcosθh^{\prime}\cos\theta we obtain (9), which coincides precisely with equation (4.22) of [41].

We obtain Theorem 1.4 as a combination of Lemmas. As required by the theorem, we henceforth make the tacit assumption a0a\geq 0.

Lemma 8.1 (Exclusion includes right-angled triangle in lower left corner).

Let uu minimize (1) with Ω=(a,a+1)2\Omega=(a,a+1)^{2}. Then 2(Ω0)>0\mathcal{H}^{2}(\Omega_{0})>0 and Ω0[a,c]2\Omega_{0}\subset[a,c]^{2} for some c>ac>a satisfying Ω0Ω=[a,c]2Ω\Omega_{0}\cap\partial\Omega=[a,c]^{2}\cap\partial\Omega.

Proof.

Whenever Ω\Omega is a subset of the first quadrant, symmetry shows the minimizer satisfies Diu0D_{i}u\geq 0. Since the inclusion Ω0{u=0}\Omega_{0}\subset\{u=0\} of Theorem 1.1 becomes an equality if Ω0\Omega_{0} is nonempty, monotonicity of convex gradients implies Ω0={u=0}[a,c]2\Omega_{0}=\{u=0\}\subset[a,c]^{2} for some c>ac>a such that Ω0Ω\Omega_{0}\cap\partial\Omega = [a,c]2Ω[a,c]^{2}\cap\partial\Omega in this case; here symmetry across the diagonal and the fact that {u=0}\{u=0\} is closed have been used. Armstrong has proved that Ω0\Omega_{0} has positive measure whenever Ω\Omega is strictly convex [2] and this result has been extended to general benefit functions by Figalli, Kim, and McCann [30]. It is straightforward to adapt their proof to our setting. Indeed, convexity of Ω0\Omega_{0} and symmetry across the diagonal means 1(Ω0Ω)>0\mathcal{H}^{1}(\partial\Omega_{0}\cap\partial\Omega)>0 implies 2(Ω0)>0\mathcal{H}^{2}(\Omega_{0})>0 and this implication is the only place strict convexity is used in [30, Theorem 4.8]. ∎

Proposition 8.2 (No normal distortion along top right boundaries).

Let uu solve (1) with Ω=(a,a+1)2\Omega=(a,a+1)^{2}. Then (Dux)𝐧=0(Du-x)\cdot\mathbf{n}=0 throughout ΩN\Omega_{N} and ΩE\Omega_{E}. Consequently, ΩEΩNΩ2\Omega_{E}\cup\Omega_{N}\subset\Omega_{2}.

Proof.

For a contradiction we assume (without loss of generality, by Proposition 2.3) there is x0=(a+1,t0)ΩEx_{0}=(a+1,t_{0})\in\Omega_{E}, at which (Du(x0)x0)𝐧>0(Du(x_{0})-x_{0})\cdot\mathbf{n}>0; when x0ΩEx_{0}\in\Omega_{E} is a vertex of Ω\Omega we interpret 𝐧=(1,0)\mathbf{n}=(1,0). With this interpretation the continuity of DuDu implies we may in fact assume, without loss of generality, that x0x_{0} lies in the relative interior of ΩE\Omega_{E}. Thus x0~\tilde{x_{0}} has positive diameter and the same is true in a relatively open portion of the boundary, by Proposition 5.4. Working on ΩE\Omega_{E}, it is convenient to let θ\theta denote the clockwise angle a ray makes with the inward normal (1,0)(-1,0). Thus θ>0\theta>0 corresponds to a ray with nonpositive slope. Parametrizing the boundary as γ(θ)=(a+1,h(θ))\gamma(\theta)=(a+1,h(\theta)) using this θ\theta, the derivation of the equation (82) is unchanged along ΩE\Omega_{E}. In particular h(θ)>0h^{\prime}(\theta)>0, which is most easily seen by beginning with the clockwise oriented parametrization γ(t)=(a+1,a+1t)\gamma(t)=(a+1,a+1-t) and ξ(t)=(cosθ(t),sinθ(t))\xi(t)=(-\cos\theta(t),\sin\theta(t)) in the coordinate arguments of Section 6 and recalling ξ×ξ˙>0\xi\times\dot{\xi}>0.

Clearly θ|x0~0\theta|_{\tilde{x_{0}}}\geq 0 or θ|x0~<0\theta|_{\tilde{x_{0}}}<0; we will derive a contradiction in either case.

Case 1. (Nonpositively sloped leaf). First let’s assume the leaf has nonpositive slope (i.e. θ0\theta\geq 0). The inequality h(θ)>0h^{\prime}(\theta)>0 implies leaves above, but in the same connected component of Ω1\Omega_{1}, as x0x_{0} with one endpoint on {a+1}×[t0,a+1]\{a+1\}\times[t_{0},a+1] are also nonpositively sloped.

At the endpoint of each leaf the Neumann condition is not satisfied, that is (Du(x)x)𝐧>0(Du(x)-x)\cdot\mathbf{n}>0, equivalently D1u(x)>a+1D_{1}u(x)>a+1 (by the sign condition on the Neumann value). On the boundary portion where leaves have nonnegative slope, tu1(a+1,t)t\mapsto u_{1}(a+1,t) is a nondecreasing function (see Figure 5(A)). Thus D1u(a+1,t)>a+1D_{1}u(a+1,t)>a+1 for all t[t0,a+1]t\in[t_{0},a+1] and, by Proposition 5.4, each x{a+1}×[t0,a+1]x\in\{a+1\}\times[t_{0},a+1] is the endpoint of a nontrivial leaf of nonpositive slope. We consider the following dichotomy and derive a contradiction in either case: (1) there is a sequence of leaves approaching (a+1,a+1)(a+1,a+1) with one endpoint on ΩE\Omega_{E} and the other in Ω\Omega or else there is not, in which case all sequences of leaves approaching (a+1,a+1)(a+1,a+1) have one end on ΩE\Omega_{E} and the other on ΩN\Omega_{N}.

ΩE\Omega_{E} Leaf containing x0,x1x_{0},x_{1} x0x_{0}x1x_{1}x2x_{2}
(a)
Triangle TTx1~\tilde{x_{1}}x1x_{1}
(b)
Figure 5. (A) Explanation of why tu1(a+1,t)t\mapsto u_{1}(a+1,t) is monotone nondecreasing when leaves make positive angle with the horizontal (i.e. have nonpositive slope). Because x1x0~x_{1}\in\tilde{x_{0}}, Du(x0)=Du(x1)Du(x_{0})=Du(x_{1}). Then monotonicity of the gradient of a convex function implies D1u(x2)D1u(x1)=D1u(x0)D_{1}u(x_{2})\geq D_{1}u(x_{1})=D_{1}u(x_{0}). Thus tD1u(a+1,t)t\mapsto D_{1}u(a+1,t) is nondecreasing.
(B) Since DuDu is constant along x1~\tilde{x_{1}}, monotonicity of the gradient implies D1u(x)D1u(x1)D_{1}u(x)\geq D_{1}u(x_{1}) and D2u(x)D2u(x1)D_{2}u(x)\geq D_{2}u(x_{1}) for all xTx\in T.

Case 1a. (All leaves approaching the vertex have one end in the interior). In the first case take a sequence (xk)k1ΩE(x_{k})_{k\geq 1}\subset\Omega_{E} with xk=(xk1,xk2)=(a+1,xk2)x_{k}=(x_{k}^{1},x_{k}^{2})=(a+1,x^{2}_{k}) satisfying xk(a+1,a+1)x_{k}\rightarrow(a+1,a+1) and that xk2x^{2}_{k} is increasing in kk. We can take x~k\tilde{x}_{k} to contain points of Alexandrov second differentiability of uu since the leaves occupy positive area by Corollary 6.6 and Fubini’s theorem. Let the corresponding angles be θk\theta_{k}. Because the leaves don’t intersect other sides of the square, symmetry and the sign of the angle yield R(θk)0R(\theta_{k})\rightarrow 0. Theorem 4.1(1) provides a C1,1C^{1,1} estimate along leaves with one endpoint on the boundary. Thus from (66),

Δu3=2R3rhcosθ+r,\Delta u-3=\frac{2R-3r}{h^{\prime}\cos\theta+r},

evaluated at r=0r=0 we obtain an estimate

R(θk)h(θk)C.\frac{R(\theta_{k})}{h^{\prime}(\theta_{k})}\leq C.

Combined with (82), i.e. R2(θ)=2h(θ)(Dux)𝐧R^{2}(\theta)=2h^{\prime}(\theta)(Du-x)\cdot\mathbf{n}, we contradict that R(θ)0R(\theta)\rightarrow 0 but (Du(x)x)𝐧(Du(x)-x)\cdot\mathbf{n} is positive and increasing.

Case 1b. (There exists a leaf crossing the domain). In the second case we pick any leaf with one endpoint (call it x1x_{1}) on ΩE\Omega_{E} and the other on ΩN\Omega_{N}. Then |θ|=π/4|\theta|=\pi/4 by symmetry. Note D1u(x1),D2u(x1)>a+1D_{1}u(x_{1}),D_{2}u(x_{1})>a+1 (by the Neumann inequality on ΩE\Omega_{E} and ΩN\Omega_{N}). Also this leaf bounds a right triangle TT with sides x1~\tilde{x_{1}}, and segments of ΩN,ΩE\Omega_{N},\Omega_{E} (Figure 5(B)). Define

(85) u¯(x):={u(x1)+Du(x1)(xx1)for xTu(x)xΩT.\bar{u}(x):=\begin{cases}u(x_{1})+Du(x_{1})\cdot(x-x_{1})&\text{for }x\in T\\ u(x)&x\in\Omega\setminus T\end{cases}.

Because u¯\bar{u} is defined by extension of an affine support for uu, for all xintTx\in\text{int}\,T, u¯(x)<u(x)\bar{u}(x)<u(x). Moreover for xTx\in T we have

|Du¯(x)x||Du(x)x|,|D\bar{u}(x)-x|\leq|Du(x)-x|,

this is because monotonicity of the gradient and the Neumann condition implies for xTx\in T, Diu(x)Diu(x1)>a+1D_{i}u(x)\geq D_{i}u(x_{1})>a+1, whereas xia+1x^{i}\leq a+1. Thus u¯\bar{u} is admissible for (1) and strictly decreases L[u]=Ω|Dux|2/2+udxL[u]=\int_{\Omega}|Du-x|^{2}/2+u\,dx, a contradiction, given that uu minimizes LL.

Case 2. (Positively sloped leaf). If our originally chosen leaf has positive slope (i.e. θ<0\theta<0) the proof is similar, with slight modifications in the lower right corner. Indeed, h(θ)>0h^{\prime}(\theta)>0 implies all leaves below our chosen leaf also have positive slope and on such leaves tD1u(a+1,t)t\mapsto D_{1}u(a+1,t) is a decreasing function (by monotonicity of DuDu, as in Case 1). Thus the Neumann value (Dux)𝐧=D1u(a+1,t)a1(Du-x)\cdot\mathbf{n}=D_{1}u(a+1,t)-a-1 increases as we move towards the lower right corner. Proposition 5.4 then implies each x{a+1}×[a,t0]x\in\{a+1\}\times[a,t_{0}] is the endpoint of a nontrivial leaf with positive slope. Consider the same two alternatives as above: there is a sequence of leaves whose endpoints on ΩE\Omega_{E} converge to (a+1,a)(a+1,a) and whose other endpoint is interior to Ω\Omega, or there is not.

In the first case the contradiction is the same as in Case 1a above. In the second case choose a leaf with endpoint x1x_{1} on ΩE\Omega_{E} and other endpoint on ΩS\Omega_{S}. By the Neumann inequality, Proposition 2.3, D1u(x1)>a+1D_{1}u(x_{1})>a+1 while D2u(x1)aD_{2}u(x_{1})\leq a. For xx in the interior of the right triangle TT formed by x1~\tilde{x_{1}} and segments of ΩE\Omega_{E}, ΩS\Omega_{S} monotonicity of the gradient implies

(86) D1u(x)>D1u(x1)>a+1,\displaystyle D_{1}u(x)>D_{1}u(x_{1})>a+1,
(87) D2u(x)D2u(x1)a.\displaystyle D_{2}u(x)\leq D_{2}u(x_{1})\leq a.

Thus the affine extension as in (85) once again satisfies |Du¯x|<|Dux||D\bar{u}-x|<|Du-x| (because x[a,a+1]2x\in[a,a+1]^{2}) and u¯<u\bar{u}<u in TT. Thus L[u¯]<L[u]L[\bar{u}]<L[u] — the same contradiction as in Case 1b above.

Conclusion: ΩEΩNΩ2\Omega_{E}\cup\Omega_{N}\subset\Omega_{2}. It remains to be shown that the corners (a,a+1),(a+1,a),(a,a+1),(a+1,a), and (a+1,a+1)(a+1,a+1) are not the endpoints of rays. This follows by the maximum principle, Lemma 3.2, combined with a reflection argument. For example, suppose a ray x0=(a,a+1)ΩNx_{0}=(a,a+1)\in\Omega_{N} is the endpoint of a nontrivial ray x0~\tilde{x_{0}} which has negative slope and thus enters Ω\Omega. Fix any point x1x_{1} in the relative interior of x0~\tilde{x_{0}} and let ν\nu be the normal to x0~\tilde{x_{0}} that has positive components. Then for ε\varepsilon sufficiently small Bε(x)ΩB_{\varepsilon}(x)\subset\Omega and the half ball

Bε+(x)=Bε(x){x;(xx1)ν>0},B_{\varepsilon}^{+}(x)=B_{\varepsilon}(x)\cap\{x;(x-x_{1})\cdot\nu>0\},

is contained in Ω2\Omega_{2} (because no rays intersect the relative interior of Ωn\Omega_{n}). After subtracting from uu its support at xx and extending the resulting function to

Bε(x)=Bε(x){x;(xx1)ν<0},B_{\varepsilon}^{-}(x)=B_{\varepsilon}(x)\cap\{x;(x-x_{1})\cdot\nu<0\},

via reflection from Bε+(x)B_{\varepsilon}^{+}(x) we obtain a function which violates Lemma 3.2. We conclude no rays intersect (a,a+1)(a,a+1) and, via an identical argument, no rays intersect (a+1,a)(a+1,a) and (a+1,a+1)(a+1,a+1). ∎

Remark 8.3 (No ray has positive slope).

A similar argument to the above implies no leaf intersecting ΩW\Omega_{W} or ΩS\Omega_{S} has positive slope. Indeed, if a leaf on ΩW\Omega_{W} has positive slope its other endpoint is interior to Ω\Omega (the leaf cannot intersect ΩN\Omega_{N} or ΩE\Omega_{E}). The same argument as Case 1 above implies as one moves vertically up ΩW\Omega_{W} each boundary point remains the endpoint of a nontrivial leaf of nonpositive slope. Leaves must have length shrinking to 0 as they approach (a,a+1)(a,a+1) and thus we obtain the same contradiction as in Case 1a above. As a result along ΩWΩ1\Omega_{W}\cap\Omega_{1} the function θu1(x(0,θ))\theta\mapsto u_{1}(x(0,\theta)) is nondecreasing (equivalently θ(Dux)𝐧|x=x(0,θ)\theta\mapsto(Du-x)\cdot\mathbf{n}\big{|}_{x=x(0,\theta)} is nonincreasing; see again Figure 5(A)).

Another key point that it will be helpful to have at our disposal is the following.

Lemma 8.4 (Concave nondecreasing profile of stingray’s tail).

Let ω\omega be some connected subset of Ω1\Omega_{1}^{-}. Then Du(ω)={Du(x)=(y1,y2);xω}Du(\omega)=\{Du(x)=(y^{1},y^{2});x\in\omega\} is such that y2y^{2} is a strictly convex increasing function of y1y^{1} lying above the diagonal whose monotonicity (88) and convexity (89) are easily quantified below in terms of the parameters θ(t)=t\theta(t)=t and m(t)=m(θ)m(t)=m(\theta) from (53).

Proof.

Connectivity of ωΩ1\omega\subset\Omega_{1}^{-} combines with Theorem 1.1(2) and the definition of Ω1\Omega_{1}^{-} to imply the set ω^={x~Ω;xω}\hat{\omega}=\{\tilde{x}\cap\partial\Omega;x\in\omega\} of fixed boundary endpoints of rays intersecting ω\omega is also connected — hence forms an interval on the left boundary ΩW\Omega_{W} of the square; it cannot intersect the relative interior of ΩN\Omega_{N} according to Proposition 8.2. On the set of rays whose endpoints lie in the relative interior of this interval ω^\hat{\omega}, the Lipschitz regularity of mm^{\prime} from Theorem 4.1 and Corollary 6.6 justifies the following computations.

We wish to consider the convexity of the curve y(θ)=(y1(θ),y2(θ))y(\theta)=(y^{1}(\theta),y^{2}(\theta)) where by (83) and (84)

y1(θ)\displaystyle y^{1}(\theta) =m(θ)cosθm(θ)sinθ\displaystyle=m(\theta)\cos\theta-m^{\prime}(\theta)\sin\theta
y2(θ)\displaystyle y^{2}(\theta) =m(θ)sinθ+m(θ)cosθ.\displaystyle=m(\theta)\sin\theta+m^{\prime}(\theta)\cos\theta.

Using Lipschitz regularity of mm we have for 1\mathcal{H}^{1} almost every θ\theta

(88) dy2dy1=y˙2(θ)y˙1(θ)=cosθsinθ>0.\frac{dy_{2}}{dy_{1}}=\frac{\dot{y}^{2}(\theta)}{\dot{y}^{1}(\theta)}=-\frac{\cos\theta}{\sin\theta}>0.

Here the sign condition comes from Remark 8.3. Note this implies the curve y(θ)y(\theta) is such that y2y^{2} is an increasing function of y1y_{1}. We see dy2dy1\frac{dy_{2}}{dy_{1}} is an increasing function of θ\theta, namely

(89) ddθdy2dy1=1sin2θ\frac{d}{d\theta}\frac{dy_{2}}{dy_{1}}=\frac{1}{\sin^{2}\theta}

and subsequently, by Remark 8.3 which implies θ\theta is an increasing function of y1y^{1}, dy2dy1\frac{dy_{2}}{dy_{1}} is an increasing function y1=u1y_{1}=u_{1}.Thus we have the required monotonicity of the derivative to conclude y2y_{2} is a convex function of y1y_{1}. ∎

Symmetry implies connected components of Du(Ω1+)Du(\Omega_{1}^{+}) are curves with y2y^{2} a concave function of y1y^{1}.

Now we can combine all the Lemmas we’ve just proved and complete the proof of Theorem 1.4.

Proof.

(Theorem 1.4). By symmetry about the diagonal and Ω1ΩE=\Omega_{1}\cap\Omega_{E}=\emptyset we can prove each point of the theorem by an analysis of the function t(Dux)𝐧|x=(a,t)t\mapsto(Du-x)\cdot\mathbf{n}\big{|}_{x=(a,t)}. Lemma 8.1 asserts (a,a)(a,a) is in Ω0\Omega_{0} as is {(a,a+t);0tα}\{(a,a+t);0\leq t\leq\alpha\} for some α=ca(0,1]\alpha=c-a\in(0,1]. On Ω0ΩW\Omega_{0}\cap\Omega_{W} we have (Dux)𝐧=a(Du-x)\cdot\mathbf{n}=a. whereas on Ω2ΩW\Omega_{2}\cap\Omega_{W} we have (Dux)𝐧=0(Du-x)\cdot\mathbf{n}=0. Thus, for a>0a>0, uC1(Ω¯)u\in C^{1}(\overline{\Omega}) [15, 50] implies some portion of Ω1\Omega_{1} must abut Ω0\Omega_{0} as one moves up ΩW\Omega_{W}.

Now we consider the configuration of Ω1\Omega_{1}. Since leaves in Ω10\Omega_{1}^{0} reach the diagonal, by symmetry they are orthogonal to it and Du(x)=(b,b)Du(x)=(b,b) on such leaves, i.e. the product Du(x)Du(x) selected lies on the diagonal.

Step 1. (Configuration of domains) We claim as one moves vertically up ΩW\Omega_{W} there is, in order, (i) a closed interval of Ω0\Omega_{0} with positive length, (ii) a half-open interval of Ω10\Omega_{1}^{0} which is empty for aa sufficiently small and nonempty for aa sufficiently large, (iii) a nonempty open interval of Ω1\Omega_{1}^{-}, and finally an interval of Ω2\Omega_{2}. All we must show is there is at most a single component of Ω10\Omega_{1}^{0}, and it is followed by Ω1\Omega_{1}^{-}. This is because, if Ω10\Omega_{1}^{0} and Ω1\Omega_{1}^{-} exist their ordering follows from Lemma 8.4. Indeed if a portion of Ω1\Omega_{1}^{-} is preceded by Ω0\Omega_{0} or Ω10\Omega_{1}^{0} then followed by Ω10\Omega_{1}^{0} we have the contradiction of a strictly convex curve lying above the diagonal with a start and endpoint on the diagonal in the stingray’s profile.

Step 2. (Blunt bunching (i.e. Ω10\Omega_{1}^{0}\neq\emptyset) for a722a\geq\frac{7}{2}-\sqrt{2}). Recall uu cannot be affine on any segment in the closure of Ω2\Omega_{2}, by a reflection argument combined with the strong maximum principle of Lemma 3.2. It follows that Ω1\Omega_{1}^{-} is nonempty whenever Ω10\Omega_{1}^{0} is nonempty; this was previously established by a different approach in [40]. Next we assume Ω10\Omega_{1}^{0} is empty and show a<722a<\frac{7}{2}-\sqrt{2}. Let (a,x¯2)(a,\underline{x}_{2}) be the upper endpoint of Ω0ΩW\Omega_{0}\cap\Omega_{W} and let (a,x¯2)(a,\overline{x}_{2}) be the lower endpoint of Ω2ΩW\Omega_{2}\cap\Omega_{W}. The segment {a}×(x¯2,x¯2)\{a\}\times(\underline{x}_{2},\overline{x}_{2}) consists of endpoints of leaves in Ω1\Omega_{1}^{-}. By the Neumann condition we have D1u(a,x¯2)=0D_{1}u(a,\underline{x}_{2})=0 and D1u(a,x¯2)=aD_{1}u(a,\overline{x}_{2})=a. Thus,

x¯2x¯2122u(a,x2)dx2=a.\int_{\underline{x}_{2}}^{\overline{x}_{2}}{\partial}^{2}_{12}u(a,x_{2})\,dx_{2}=a.

As in (4.17) of [41], using the (r,θ)(r,\theta) coordinates we compute

122u(a,x2)=sin(θ)m′′(θ)+m(θ)h(θ).{\partial}^{2}_{12}u(a,x_{2})=-\sin(\theta)\frac{m^{\prime\prime}(\theta)+m(\theta)}{h^{\prime}(\theta)}.

From (81), which reads m′′(θ)+m(θ)3h(θ)cosθ=2R(θ),m^{\prime\prime}(\theta)+m(\theta)-3h^{\prime}(\theta)\cos\theta=2R(\theta), we have

122u(a,x2)\displaystyle{\partial}^{2}_{12}u(a,x_{2}) =sin(θ)2R(θ)h(θ)3sin(θ)cos(θ)\displaystyle=-\sin(\theta)\frac{2R(\theta)}{h^{\prime}(\theta)}-3\sin(\theta)\cos(\theta)
=2sin(θ)R(θ)θ(x2)32sin(2θ).\displaystyle=-2\sin(\theta)R(\theta)\theta^{\prime}(x_{2})-\frac{3}{2}\sin(2\theta).

We’ve used the inverse function theorem to rewrite 1h(θ)=θ(x2)\frac{1}{h^{\prime}(\theta)}=\theta^{\prime}(x_{2}). Using sin(θ)0-\sin(\theta)\geq 0 from Remark 8.3 and R1R\leq 1 and θπ4\theta\geq-\frac{\pi}{4} for convexity of Ω0\Omega_{0} we conclude

a\displaystyle a =x¯2x¯2122u(a,x2)dx2\displaystyle=\int_{\underline{x}_{2}}^{\overline{x}_{2}}{\partial}^{2}_{12}u(a,x_{2})\,dx_{2}
=x¯2x¯2[2sin(θ)R(θ)θ(x2)+32sin(2θ)]𝑑x2\displaystyle=-\int_{\underline{x}_{2}}^{\overline{x}_{2}}[2\sin(\theta)R(\theta)\theta^{\prime}(x_{2})+\frac{3}{2}\sin(2\theta)]\,dx_{2}
<2R[cos(0)cos(π4)]+32[x¯2x¯2]\displaystyle<2\|R\|_{\infty}[\cos(0)-\cos(-\frac{\pi}{4})]+\frac{3}{2}[\bar{x}^{2}-\underline{x}^{2}]
722.\displaystyle\leq\frac{7}{2}-\sqrt{2}.

Step 3. (No blunt bunching (i.e. Ω10=\Omega_{1}^{0}=\emptyset) for a1a\ll 1 sufficiently small) Suppose for a contradiction that there is a sequence ak0a_{k}\downarrow 0 such that the minimizer on Ω(k)=(ak,ak+1)2\Omega^{(k)}=(a_{k},a_{k}+1)^{2} has Ω10\Omega_{1}^{0} nonempty. Let uku_{k} be the minimal convex extension to 𝐑n\mathbf{R}^{n} of the corresponding minimizer to (1). Let, for example (Ω01)(k)(\Omega^{1}_{0})^{(k)} denote Ω01\Omega^{1}_{0} for the problem on (ak,ak+1)2(a_{k},a_{k}+1)^{2}, and domains with no superscript denote the corresponding domain for a=0a=0. The convergence result [30, Corollary 4.7] implies uk|Bεu|Bεu_{k}|_{B_{\varepsilon}}\rightarrow u|_{B_{\varepsilon}} locally uniformly for any Bε(0,1)2B_{\varepsilon}\subset(0,1)^{2} where uu is the minimizer for a=0a=0.

It is clear that Ω1\Omega_{1} is empty when a=0a=0. Indeed, for a=0a=0 the solution on [0,1]2[0,1]^{2} is the restriction of the solution on [1,1]2[-1,1]^{2}. The solution on [1,1]2[-1,1]^{2} satisfies Ω1=\Omega_{1}=\emptyset: Theorem 1.1(2) asserts the rays all extend to the boundary but Proposition 8.2, which is valid also on [1,1]2[-1,1]^{2} asserts there can be no ray intersecting the boundaries {1}×[0,1]\{1\}\times[0,1] and [0,1]×{1}[0,1]\times\{1\}. By symmetry there are no rays intersecting anywhere on [1,1]2\partial[-1,1]^{2} and hence no rays whatsoever.

Recall (4.18) of [41] asserts {uk=0}\{u_{k}=0\} is the triangle (x1,x2)[ak,ak+1]2(x^{1},x^{2})\in[a_{k},a_{k}+1]^{2} defined by x1+x22a+2a3(1+32a21)x^{1}+x^{2}\leq 2a+\frac{2a}{3}(\sqrt{1+\frac{3}{2a^{2}}}-1). The limit Ω0:={u=0}\Omega_{0}:=\{u=0\} from Theorem 1.1 must therefore contain the triangle x1+x22/3x^{1}+x^{2}\leq\sqrt{2/3} of area 13\frac{1}{3} in [0,1]2[0,1]^{2}. Nor can Ω0\Omega_{0} be larger than this triangle, since Proposition 2.3 implies both integrands are non-negative in the identity

1=Ω03𝑑2+Ω0Ω(Dux)𝐧𝑑11=\int_{\Omega_{0}}3d\mathcal{H}^{2}+\int_{\Omega_{0}\cap{\partial}\Omega}(Du-x)\cdot\mathbf{n}d\mathcal{H}^{1}

asserted by Lemma A.6. But the previous paragraph implies Ω1=\Omega_{1}=\emptyset, so outside the triangle Ω0\Omega_{0} the minimizing uCloc1,1((0,1)2)u\in C^{1,1}_{\text{loc}}((0,1)^{2}) is a strictly convex solution to Poisson’s equation Δu=3\Delta u=3. Reflecting this solution across the line x1+x2=2/3x^{1}+x^{2}=\sqrt{2/3} where it vanishes contradicts the strong maximum principle (Lemma 3.2). This is the desired contradiction which establishes Step 3.

Step 4. (No further Ω1\Omega_{1} components) From Remark 8.3 any leaves intersecting ΩW\Omega_{W} have nonpositive slope. Thus, recalling Figure 5(A), the Neumann value (Dux)𝐧(Du-x)\cdot\mathbf{n} is a positive decreasing function of x2x_{2} along Ω1ΩW\Omega_{1}\cap\Omega_{W} and 0 along Ω2ΩW\Omega_{2}\cap\Omega_{W}. Thus there cannot exist an Ω1\Omega_{1} component above an Ω2\Omega_{2} component on ΩW\Omega_{W}. ∎

Remark 8.5 (Estimating the bifurcation point).

It is clear that a722a\geq\frac{7}{2}-\sqrt{2} is not sharp for the existence of Ω10\Omega_{1}^{0}. However the existence of a bifurcation reflects the radically different behavior we have shown the model to display for small and large aa. We expect there is a single bifurcation value a0a_{0} such that Ω10\Omega_{1}^{0} is nonempty for a>a0a>a_{0} while Ω10\Omega_{1}^{0} is empty for 0<aa00<a\leq a_{0}. It would be interesting to confirm this expectation, and to find or estimate a0a_{0} more precisely.

Appendix A Rochet and Choné’s sweeping and localization

We recall as in (17) the minimizer uu of (1) satisfies the variational inequality

(90) 0Ω(n+1Δu)v(x)𝑑x+Ωv(x)(Dux)n𝑑n1,0\leq\int_{\Omega}(n+1-\Delta u)v(x)\,dx+\int_{\partial\Omega}v(x)(Du-x)\cdot\textbf{n}\,d\mathcal{H}^{n-1},

for all convex vv with sptv\text{spt}\,v_{-} disjoint from Ω0\Omega_{0}. Since uC1(Ω¯)u\in C^{1}(\overline{\Omega}) [15], we know

(91) dσ:=(n+1Δu)dx+(Dux)ndn1  Ω,d\sigma:=(n+1-\Delta u)\,dx+(Du-x)\cdot\textbf{n}\,d\mathcal{H}^{n-1}\!\mathbin{\vrule height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule height=0.55974pt,depth=0.0pt,width=3.44444pt}\!\partial\Omega,

is a measure of finite total-variation and we can rewrite (90) as

(92) 0Ω¯v(x)𝑑σ(x).0\leq\int_{\overline{\Omega}}v(x)\,d\sigma(x).

The disintegration theorem, which we state in Section A.3 (see also [1, Theorem 5.3.1], [24, 78-III]) implies we may disintegrate the measure σ\sigma (by separately disintegrating its positive and negative parts σ+\sigma^{+} and σ\sigma^{-}) with respect to the map DuDu (equivalently with respect to the contact sets x~=Du1(Du(x))\tilde{x}=Du^{-1}(Du(x))). Our goal in this section is to prove Corollary A.9, namely to show that for n\mathcal{H}^{n} almost every xΩ¯x\in\overline{\Omega} (90) holds for the disintegration on x~\tilde{x}. In fact, provided xΩ0x\not\in\Omega_{0} we will prove the result for general convex vv, and for xΩ0x\in\Omega_{0} we will prove the result for vv satisfying u+v0u+v\geq 0. More precisely, the disintegration theorem implies there exists families of measures σ+,y\sigma_{+,y} and σ,y\sigma_{-,y} such that for all Borel ff

Ω¯f(x)𝑑σ+(x)=Du(Ω¯)Du1(y)f(x)𝑑σ+,y(x)d(Du#σ+)(y),\displaystyle\int_{\overline{\Omega}}f(x)\,d\sigma_{+}(x)=\int_{Du({\overline{\Omega}})}\int_{Du^{-1}(y)}f(x)d\sigma_{+,y}(x)\,d(Du_{\#}\sigma_{+})(y),
Ω¯f(x)𝑑σ(x)=Du(Ω¯)Du1(y)f(x)𝑑σ,y(x)d(Du#σ)(y).\displaystyle\int_{\overline{\Omega}}f(x)\,d\sigma_{-}(x)=\int_{Du({\overline{\Omega}})}\int_{Du^{-1}(y)}f(x)d\sigma_{-,y}(x)\,d(Du_{\#}\sigma_{-})(y).

We show for (Du)#σ+(Du)_{\#}\sigma_{+} almost every yy

(93) 0Du1(y)v(x)d(σ+,yσ,y)(x),0\leq\int_{Du^{-1}(y)}v(x)d(\sigma_{+,y}-\sigma_{-,y})(x),

for all convex vv with sptv\text{spt}\,v_{-} disjoint from {u=0}\{u=0\}. As we will prove the same result (with x~\tilde{x} replacing Du1(y)Du^{-1}(y)) then holds for n\mathcal{H}^{n} almost every xΩx\in\Omega and n1\mathcal{H}^{n-1} almost every xΩx\in\partial\Omega.

We emphasize that this appendix, whilst included for completeness, merely provides some more details on Rochet and Chonè’s proof of this localization property [50].

A.1. Measures in convex order and sweeping operators.

We begin with the following definition which is used in the theory of martingales and clearly related to (92).

Definition A.1 (Convex order).

Let Ω𝐑n\Omega\subset\mathbf{R}^{n} be a Borel set. Let σ1,σ2𝒫(Ω)\sigma_{1},\sigma_{2}\in\mathcal{P}(\Omega) be Borel probability measures. We write σ1σ2\sigma_{1}\preceq\sigma_{2} (read as σ1\sigma_{1} precedes σ2\sigma_{2} in convex order) provided for every continuous convex u:Ω𝐑u:\Omega\rightarrow\mathbf{R} there holds

Ωu𝑑σ1Ωu𝑑σ2.\int_{\Omega}u\,d\sigma_{1}\leq\int_{\Omega}u\,d\sigma_{2}.

The “Sweeping Theorem” characterizes measure in convex order and requires some more definitions. We take this Theorem from the work of Strassen [54, Theorem 2] where it’s attributed to “Hardy–Littlewood–Pólya–Blackwell–Stein–Sherman–Cartier–Fell–Meyer” (see also [21, Théorème 1], [43, T51, T53]).

Definition A.2 (Sweeping operators and Markov kernels).

(1) By a Markov kernel on Ω\Omega we mean a function T:Ω𝒫(Ω)T:\Omega\rightarrow\mathcal{P}(\Omega) where for each xΩx\in\Omega, Tx:=T(x)𝒫(Ω)T_{x}:=T(x)\in\mathcal{P}(\Omega) is a probability measure. As a technicality we require for each Borel EE that xTx(E)x\mapsto T_{x}(E) is Borel measurable.
(2) A Markov kernel TT is called a sweeping operator444The term sweeping is also sometimes known as a balayage or dilation. provided it satisfies that for each affine p:Ω𝐑p:\Omega\rightarrow\mathbf{R},

(94) p(x)=Ωp(ξ)dTx(ξ)=:(Tp)(x).p(x)=\int_{\Omega}p(\xi)\,dT_{x}(\xi)=:(Tp)(x).

(3) If σ𝒫(Ω)\sigma\in\mathcal{P}(\Omega) and TT is a Markov kernel on Ω\Omega, then we define Tσ𝒫(Ω)T\sigma\in\mathcal{P}(\Omega) by

(95) (Tσ)(A)=ΩTx(A)𝑑σ(x).(T\sigma)(A)=\int_{\Omega}T_{x}(A)\,d\sigma(x).

We note two points: First, (94) is equivalent to the requirement

Ωξ𝑑Tx(ξ)=x.\int_{\Omega}\xi\,dT_{x}(\xi)=x.

Second, for each integrable fL1(Tσ)f\in L^{1}(T\sigma), from (95) we have

(96) Ωf(x)d(Tσ)(x)=ΩΩf(ξ)𝑑Tx(ξ)𝑑σ(x).\int_{\Omega}f(x)\,d(T\sigma)(x)=\int_{\Omega}\int_{\Omega}f(\xi)\,dT_{x}(\xi)\,d\sigma(x).

The sweeping theorem gives a necessary and sufficient condition for σ1\sigma_{1} to precede σ2\sigma_{2} in convex order.

Theorem A.3 (Sweeping characterization of convex order; see [54]).

The measures σ1,σ2𝒫(Ω)\sigma_{1},\sigma_{2}\in\mathcal{P}(\Omega) satisfy σ1σ2\sigma_{1}\preceq\sigma_{2} if and only if there exists a sweeping operator TT such that σ2=Tσ1\sigma_{2}=T\sigma_{1}.

A.2. First characterization; Lagrange multiplier and sweeping

We aim to apply Theorem A.3 to the positive and negative parts of σ\sigma. However Theorem A.3 does not apply directly because we do not have the condition σσ+\sigma_{-}\preceq\sigma_{+} but only the weaker condition obtained by testing against nonnegative convex functions. Nevertheless, we obtain the following lemma.

Lemma A.4 (Restoring neutrality).

Let σ\sigma be as in (91). There exists a nonnegative measure λ\lambda supported on {u=0}\{u=0\} and a sweeping operator TT on Ω¯\overline{\Omega} such that

(97) σλ\displaystyle\sigma-\lambda =(σλ)+(σλ)\displaystyle=(\sigma-\lambda)_{+}-(\sigma-\lambda)_{-}
(98) =Tωω,\displaystyle=T\omega-\omega,

for ω:=(σλ).\omega:=(\sigma-\lambda)_{-}.

The representation (97) is, of course, trivial. The essential conclusion is that after subtracting the Lagrange multiplier λ\lambda (our reason for designating it so will be clear in the proof) (σλ)(σλ)+(\sigma-\lambda)_{-}\preceq(\sigma-\lambda)_{+}.

Proof of Lemma A.4.

First note for all t0t\geq 0, tutu is admissible for the minimization problem. Thus by minimality

(99) 0=ddt|t=1L[tu]=Ω¯u(x)𝑑σ.0=\frac{d}{dt}\Big{|}_{t=1}L[tu]=\int_{\overline{\Omega}}u(x)\,d\sigma.

Combined with the minimality condition (92) we have

0=inf{Ω¯v𝑑σ;v is convex with v0},0=\inf\left\{\int_{\overline{\Omega}}v\,d\sigma;v\text{ is convex with $v\geq 0$}\right\},

with v=uv=u realizing the infimum. A classical theorem in the calculus of variations says the constraint v0v\geq 0 may be realized by a Lagrange multiplier (see, for example, [37, Theorem 1,pg. 217]). Thus there is a nonnegative Radon measure λ\lambda such that

(100) 0=inf{Ω¯vd(σλ);v is convex},0=\inf\left\{\int_{\overline{\Omega}}v\,d(\sigma-\lambda);v\text{ is convex}\right\},

with v=uv=u still realizing the infimum. Using that uu attains the infimum along with (99) yields

u𝑑λ=0.\int u\,d\lambda=0.

Since λ\lambda is nonnegative we conclude sptλ{u=0}\text{spt}\lambda\subset\{u=0\}. Since (100) implies (σλ)(σλ)+(\sigma-\lambda)_{-}\preceq(\sigma-\lambda)_{+} we see (98) follows by Theorem A.3. Note that (uλ)(u-\lambda)_{-} and (uλ)+(u-\lambda)_{+} may not be probability measures but are of finite and equal mass (finiteness follows from uC1(Ω¯)u\in C^{1}(\overline{\Omega}) [15]), equality from testing against v=±1v=\pm 1) and this justifies our use of Theorem A.3. ∎

The following property of the sweeping operator TT is essential to our arguments.

Lemma A.5 (Localization to leaves).

Let TT be the sweeping operator given by Lemma A.4. Then for ω\omega almost every xx the measure TxT_{x} is supported on x~\tilde{x}.

Proof.

First, by (96)

(101) Ω¯u(x)d(Tω)(x)\displaystyle\int_{\overline{\Omega}}u(x)\,d(T\omega)(x) =Ω¯Ω¯u(ξ)𝑑Tx(ξ)𝑑ω(x).\displaystyle=\int_{\overline{\Omega}}\int_{\overline{\Omega}}u(\xi)\,dT_{x}(\xi)\,d\omega(x).

For each xΩ¯x\in\overline{\Omega} we have u(ξ)px(ξ)=u(x)+Du(x)(ξx)u(\xi)\geq p_{x}(\xi)=u(x)+Du(x)\cdot(\xi-x) with equality if and only if ξx~\xi\in\tilde{x}. Thus (101) and the definition of a sweeping operator (94) imply

Ω¯u(x)d(Tω)(x)\displaystyle\int_{\overline{\Omega}}u(x)\,d(T\omega)(x) Ω¯Ω¯px(ξ)𝑑Tx(ξ)𝑑ω(x)\displaystyle\geq\int_{\overline{\Omega}}\int_{\overline{\Omega}}p_{x}(\xi)\,dT_{x}(\xi)\,d\omega(x)
=Ω¯u(x)𝑑ω(x).\displaystyle=\int_{\overline{\Omega}}u(x)\,d\omega(x).

However (100) implies we must have equality. Subsequently we obtain

Ω¯u(ξ)𝑑Tx(ξ)=Ω¯px(ξ)𝑑Tx(ξ),\int_{\overline{\Omega}}u(\xi)\,dT_{x}(\xi)=\int_{\overline{\Omega}}p_{x}(\xi)\,dT_{x}(\xi),

for ω\omega a.e. xx, which can only occur if for a.e. xx, TxT_{x} is supported on x~\tilde{x}. ∎

At this point we have almost everything needed to obtain the localization property. Now we establish the following lemma which was was used earlier in the proof of Proposition 5.4.

Lemma A.6 (Objective responds proportionately to uniform utility increase).

Let σ\sigma denote the variational derivative (recall equation (21)). Then (Du)#σ=δ0,(Du)_{\#}\sigma=\delta_{0}, that is (Du)#σ(Du)_{\#}\sigma is a unit Dirac mass at the origin.

Proof.

That the variational derivative σ\sigma assigns measure 11 to {u=0}\{u=0\} was observed by Rochet and Chonè [50]: this follows from the leafwise neutrality outside {u=0}\{u=0\} implied by Lemmas A.4A.5, and from L(u+t)=t+L(u)L(u+t)=t+L(u) for all t𝐑t\in{\mathbf{R}}. Thus it suffices to prove (Du)#σ(A)=0(Du)_{\#}\sigma(A)=0 for any ADu(Ω¯)A\subset Du(\overline{\Omega}) not containing 0 or, equivalently,

(102) (Tω)(Du1(A))=ω(Du1(A)).\displaystyle(T\omega)(Du^{-1}(A))=\omega(Du^{-1}(A)).

Because TxT_{x} is supported on x~\tilde{x} for ω\omega a.e. xx we have

(Tω)(Du1(A))\displaystyle(T\omega)(Du^{-1}(A)) =ΩTx(Du1(A))𝑑ω(x)=Du1(A)Tx(Du1(A))𝑑ω(x),\displaystyle=\int_{\Omega}T_{x}(Du^{-1}(A))\,d\omega(x)=\int_{Du^{-1}(A)}T_{x}(Du^{-1}(A))\,d\omega(x),

where we’ve used Tx(Du1(A))=0T_{x}(Du^{-1}(A))=0 for xDu1(A)x\not\in Du^{-1}(A). Since for xDu1(A)x\in Du^{-1}(A) we have (up to normalization) Tx(Du1(A))=1T_{x}(Du^{-1}(A))=1 this proves (102). ∎

A.3. Disintegration and localization

In this section we complete the proof that the variational inequality (18) holds on almost every contact set. We use the Disintegration Theorem in the following form, which can be viewed as a continuum generalization of Bayes’ theorem (see [1, Theorem 5.3.1] and [24, 78-III]).

Theorem A.7 (Disintegration of measure).

Let X,YX,Y be Radon separable metric spaces, μ𝒫(X),\mu\in\mathcal{P}(X), and let F:XYF:X\rightarrow Y be a Borel map used to define the push forward ν=F#μ𝒫(Y)\nu=F_{\#}\mu\in\mathcal{P}(Y). Then there exists a ν\nu-a.e. uniquely determined Borel family of probability measures {μy}yY𝒫(X)\{\mu_{y}\}_{y\in Y}\subset\mathcal{P}(X) such that μy\mu_{y} vanishes outside F1(y)F^{-1}(y) for ν\nu-a.e. yy, and

Xf(x)𝑑μ(x)=YF1(y)f(x)𝑑μy(x)𝑑ν(y)\int_{X}f(x)\,d\mu(x)=\int_{Y}\int_{F^{-1}(y)}f(x)\,d\mu_{y}(x)\,d\nu(y)

for every Borel f:[0,+]f:\rightarrow[0,+\infty].

Theorem A.8 (Leafwise Euler-Lagrange condition).

Let σ+,y\sigma_{+,y} and σ,y\sigma_{-,y} be the conditional measures obtained from (91) by applying Theorem A.7 with the projection F=DuF=Du, x~=Du1(y)\tilde{x}=Du^{-1}(y), and μ\mu equal to σ+\sigma_{+} and σ\sigma_{-} respectively. Put σy=σ+,yσ,y\sigma_{y}=\sigma_{+,y}-\sigma_{-,y} and let vv be a convex function with sptv\text{spt}\,v_{-} disjoint from {u=0}\{u=0\}. Then for both n\mathcal{H}^{n} almost every xΩx\in\Omega and n1\mathcal{H}^{n-1} almost every xΩx\in\partial\Omega there holds for y=Du(x)y=Du(x)

(103) x~v(ξ)𝑑σy(ξ)0.\int_{\tilde{x}}v(\xi)\,d\sigma_{y}(\xi)\geq 0.

We work away from {u=0}\{u=0\} and use ω  {u>0}=σ  {u>0}\omega\!\mathbin{\vrule height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule height=0.55974pt,depth=0.0pt,width=3.44444pt}\!\{u>0\}=\sigma_{-}\!\mathbin{\vrule height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule height=0.55974pt,depth=0.0pt,width=3.44444pt}\!\{u>0\} without further reference to this restriction. The idea of the proof is that because TxT_{x} is supported on x~\tilde{x} the conditioning of TωT\omega, denoted (Tω)y(T\omega)_{y}, is obtained by sweeping the conditioning of ω\omega, denoted ωy\omega_{y}. More succinctly

(104) σ+,y=(Tω)y=T(ωy)=T(σ,y).\sigma_{+,y}=(T\omega)_{y}=T(\omega_{y})=T\big{(}\sigma_{-,y}\big{)}.

Then Theorem A.3 implies (103).

Proof.

We apply the Disintegration Theorem a number of times in this proof, each time with F(x)=Du(x)F(x)=Du(x). Applying the Disintegration Theorem to the measure ω=(σλ)\omega=(\sigma-\lambda)_{-} from Lemma A.4, we obtain a family of measures ωy\omega_{y} such that for any Borel f:Ω¯[0,+]f:{\overline{\Omega}}\rightarrow[0,+\infty]

(105) Ω¯f(x)𝑑ω(x)=Du(Ω¯)Du1(y)f(x)𝑑ωy(x)d((Du)#ω)(y).\int_{\overline{\Omega}}f(x)\,d\omega(x)=\int_{Du({\overline{\Omega}})}\int_{Du^{-1}(y)}f(x)\,d\omega_{y}(x)\,d\big{(}(Du)_{\#}\omega\big{)}(y).

We consider two ways of expressing the result of disintegrating TωT\omega. First, by a direct application of the disintegration theorem

(106) Ω¯f(x)d(Tω)(x)=Du(Ω¯)Du1(y)f(x)d(Tω)y(x)d((Du)#(Tω))(y).\displaystyle\int_{\overline{\Omega}}f(x)\,d(T\omega)(x)=\int_{Du({\overline{\Omega}})}\int_{Du^{-1}(y)}f(x)\,d(T\omega)_{y}(x)\,d\big{(}(Du)_{\#}(T\omega)\big{)}(y).

On the other hand using the sweeping operator and the disintegration of ω\omega, (105), we obtain

Ω¯f(x)d(Tω)(x)\displaystyle\int_{\overline{\Omega}}f(x)\,d(T\omega)(x) =Ω¯Ω¯f(ξ)𝑑Tx(ξ)𝑑ω\displaystyle=\int_{\overline{\Omega}}\int_{\overline{\Omega}}f(\xi)\,dT_{x}(\xi)\,d\omega
(107) =Du(Ω¯)Du1(y)Ω¯f(ξ)𝑑Tx(ξ)𝑑ωy(x)d((Du)#ω)(y).\displaystyle=\int_{Du({\overline{\Omega}})}\int_{Du^{-1}(y)}\int_{\overline{\Omega}}f(\xi)\,dT_{x}(\xi)\,d\omega_{y}(x)\,d\big{(}(Du)_{\#}\omega\big{)}(y).

Using that TxT_{x} is supported on Du1(Du(x))Du^{-1}(Du(x)) the inner two integrals in (107) become integration against T(ωy)T(\omega_{y}). Namely,

(108) Ω¯f(x)d(Tω)(x)=Du(Ω¯)Du1(y)f(x)d(T(ωy))(x)d((Du)#ω)(y).\int_{\overline{\Omega}}f(x)\,d(T\omega)(x)=\int_{Du({\overline{\Omega}})}\int_{Du^{-1}(y)}f(x)\,d(T(\omega_{y}))(x)\,d\big{(}(Du)_{\#}\omega\big{)}(y).

To conclude we recall from Lemma A.6 that (Du)#(Tω)=(Du)#ω(Du)_{\#}(T\omega)=(Du)_{\#}\omega. Thus comparing (108) and (106) and using the uniqueness a.e of the conditional measures we obtain

(Tω)y=T(ωy).(T\omega)_{y}=T(\omega_{y}).

which is (104) so by the sweeping characterization of convex order (Theorem A.3) we obtain (103) for (Du)#σ+(Du)_{\#}\sigma_{+} almost every yDu(Ω¯)y\in Du({\overline{\Omega}}).

To finish the proof we must translate back to stating the result in terms of n\mathcal{H}^{n} a.e. xΩ¯x\in\overline{\Omega}. Let BΩ¯B\subset\overline{\Omega} be the set of xx for which the leafwise localized inequality (103) does not hold. Then (Du)#σ+(B)=0(Du)_{\#}\sigma_{+}(B)=0 and by Lemma A.6 (Du)#σ(B)=0(Du)_{\#}\sigma_{-}(B)=0. Hence

(109) 0=Du1(Du(B))(n+1Δu)±𝑑x+Du1(Du(B))Ω((Dux)𝐧)±𝑑n1.0=\int_{Du^{-1}(Du(B))}(n+1-\Delta u)_{\pm}\,dx+\int_{Du^{-1}(Du(B))\cap\partial\Omega}((Du-x)\cdot\mathbf{n})_{\pm}\,d\mathcal{H}^{n-1}.

It follows that

0=Du1(Du(B))|n+1Δu|𝑑x+Du1(Du(B))Ω|(Dux)𝐧|𝑑n1.0=\int_{Du^{-1}(Du(B))}|n+1-\Delta u|\,dx+\int_{Du^{-1}(Du(B))\cap\partial\Omega}|(Du-x)\cdot\mathbf{n}|\,d\mathcal{H}^{n-1}.

Now on each x~\tilde{x} with xBx\in B because the leafwise inequality does not hold we have n+1Δu0n+1-\Delta u\not=0 on a positive dim x~\mathcal{H}^{\text{dim }\tilde{x}} measure subset of x~\tilde{x}. Indeed, since the leafwise inequality does not hold either n+1Δu0n+1-\Delta u\not\equiv 0 or (Dux)𝐧0(Du-x)\cdot\mathbf{n}\neq 0 and in this latter case mass balance, Lemma A.6, implies n+1Δu0n+1-\Delta u\not\equiv 0. Thus Du1(Du(B))Du^{-1}(Du(B)), which clearly contains BB, has measure 0. Thus n(B)=0\mathcal{H}^{n}(B)=0.

Finally to obtain the result for n1\mathcal{H}^{n-1} almost every xΩx\in\partial\Omega, note at boundary points of strict convexity the leafwise (now, pointwise) inequality is satisfied by Lemma A.6. Moreover if the set of xΩBx\in\partial\Omega\cap B which are contained in nontrivial contact sets and has n1(BΩ)>0\mathcal{H}^{n-1}(B\cap\partial\Omega)>0 then, by Fubini’s theorem n(B)>0\mathcal{H}^{n}(B)>0. We conclude n1(BΩ)=0.\mathcal{H}^{n-1}(B\cap\partial\Omega)=0.

Corollary A.9 (Rochet and Chonè’s leafwise localization).

Every convex v:𝐑n𝐑v:{\mathbf{R}}^{n}\longrightarrow{\mathbf{R}} satisfies

(110) 0x~v(z)𝑑σx~(z)0\leq\int_{\tilde{x}}v(z)d\sigma_{\tilde{x}}(z)

for n\mathcal{H}^{n} almost every xΩ¯{u=0}x\in\overline{\Omega}\setminus\{u=0\} and for n1\mathcal{H}^{n-1} almost every xΩ{u=0}x\in\partial\Omega\setminus\{u=0\}. (The same conclusions extend to x{u=0}x\in\{u=0\} if also u+v0u+v\geq 0.)

Proof.

To obtain Corollary A.9 from Theorem A.8: for xΩ¯{u=0}x\in\overline{\Omega}\setminus\{u=0\} apply Theorem A.8 with the convex function ξv(ξ)+Mdist(ξ,x~)\xi\mapsto v(\xi)+M\text{dist}(\xi,\tilde{x}) for MM chosen sufficiently large; if instead u(x)=0u(x)=0 and u+v0u+v\geq 0 apply Theorem A.8 to the convex function v+ε+Mdist(,x~)v+\varepsilon+M\text{dist}(\cdot,\tilde{x}) which becomes positive for MuC0,1(Ω)M\geq\|u\|_{C^{0,1}(\Omega)}, and then send ε0\varepsilon\downarrow 0. ∎

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