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Piecewise constant profiles minimizing total variation energies of Kobayashi-Warren-Carter type with fidelity

Yoshikazu Giga Graduate School of Mathematical Sciences, The University of Tokyo, 3-8-1 Komaba, Meguro-ku, Tokyo 153-8914, Japan. [email protected]. Ayato Kubo Department of Mathematics, Faculty of Science, Hokkaido University, Kita 10, Nishi 8, Kita-Ku, Sapporo, Hokkaido, 060-0810, Japan. [email protected] Hirotoshi Kuroda Department of Mathematics, Faculty of Science, Hokkaido University, Kita 10, Nishi 8, Kita-Ku, Sapporo, Hokkaido, 060-0810, Japan. [email protected] Jun Okamoto Institute for the Advanced Study of Human Biology, Kyoto University, Yoshida-Konoe-Cho, Sakyo-ku, Kyoto 606-8501, Japan. [email protected]  and  Koya Sakakibara Faculty of Mathematics and Physics, Institute of Science and Engineering, Kanazawa University, Kakuma-machi, Kanazawa-shi, Ishikawa 920-1192, Japan; RIKEN iTHEMS, 2-1 Hirosawa, Wako-shi, Saitama 351-0198, Japan. [email protected]
Abstract.

We consider a total variation type energy which measures the jump discontinuities different from usual total variation energy. Such a type of energy is obtained as a singular limit of the Kobayashi-Warren-Carter energy with minimization with respect to the order parameter. We consider the Rudin-Osher-Fatemi type energy by replacing relaxation term by this type of total variation energy. We show that all minimizers are piecewise constant if the data is continuous in one-dimensional setting. Moreover, the number of jumps is bounded by an explicit constant involving a constant related to the fidelity. This is quite different from conventional Rudin-Osher-Fatemi energy where a minimizer must have no jump if the data has no jumps. The existence of a minimizer is guaranteed in multi-dimensional setting when the data is bounded.

1. Introduction

We consider a kind of total variation energy which measures jumps different from the conventional total variation energy. Let Ω\Omega be a bounded domain in n\mathbb{R}^{n} with n1n\geq 1. Let uu be in BV(Ω)BV(\Omega), i.e., its distributional derivative DuDu is a finite Radon measure in Ω\Omega and let |Du||Du| denote its total variation measure. The total variation energy can be written of the form

TV(u)=Ω\Ju|Du|+Ju|u+u|𝑑n1,TV(u)=\int_{\Omega\backslash J_{u}}|Du|+\int_{J_{u}}|u^{+}-u^{-}|\;d\mathcal{H}^{n-1},

where JuJ_{u} denotes the (approximate) jump set of uu and u±u^{\pm} is a trace of uu from each side of JuJ_{u}; n1\mathcal{H}^{n-1} denotes the n1n-1 dimensional Hausdorff measure. For a precise meaning of this formula, see Section 2 and [AFP]. Let K(ρ)K(\rho) be a strictly increasing continuous function for ρ0\rho\geq 0 with K(0)=0K(0)=0. We set

TVK(u)=Ω\Ju|Du|+JuK(|u+u|)𝑑n1.TV_{K}(u)=\int_{\Omega\backslash J_{u}}|Du|+\int_{J_{u}}K\left(|u^{+}-u^{-}|\right)d\mathcal{H}^{n-1}.

For a given function gL2(Ω)g\in L^{2}(\Omega), we are interested in a minimizer of

TVKg(u)=TVK(u)+(u),(u)=λ2Ω|ug|2𝑑x,TV_{Kg}(u)=TV_{K}(u)+\mathcal{F}(u),\quad\mathcal{F}(u)=\frac{\lambda}{2}\int_{\Omega}|u-g|^{2}\;dx,

where λ>0\lambda>0 is a constant. The term \mathcal{F} is often called a fidelity term. If TVK=TVTV_{K}=TV, the functional TVg(u)=TV(u)+(u)TV_{g}(u)=TV(u)+\mathcal{F}(u) is often called Rudin-Osher-Fatemi functional since it is proposed by [ROF] to denoise the original image whose grey-level values equal gg. For TVgTV_{g}, there always exists a unique minimizer since the problem is strictly convex and lower semicontinuous in L2(Ω)L^{2}(\Omega). We are interested in regularity of a minimizer of TVKgTV_{Kg} assuming some regularity of gg. This problem is well studied for TVgTV_{g} started by [CCN]. Let uu_{*} be the minimizer of TVgTV_{g} for gBV(Ω)g\in BV(\Omega). In [CCN], it is shown that JuJgJ_{u_{*}}\subset J_{g} and u+(x)u(x)g+(x)g(x)u_{*}^{+}(x)-u_{*}^{-}(x)\leq g^{+}(x)-g^{-}(x) for n1\mathcal{H}^{n-1}-a.e. xJux\in J_{u}. In particular, if gg has no jumps, so does uu_{*}. This type of results is extended in various setting; see a reviewer paper [GKL].

In this paper, we shall show that a minimizer of TVKgTV_{Kg} may have jumps even if gg has no jumps for some class of subadditive function KK including

K(ρ)=ρ1+ρK(\rho)=\frac{\rho}{1+\rho}

as a particular example when Ω\Omega is an interval. This type of TVKTV_{K} appears as a kind of singular limit of the Kobayashi-Warren-Carter energy [GOU], [GOSU]. Actually, we shall prove a stronger result saying that a minimizer is a piecewise constant function with finitely many jumps for gC(Ω¯)g\in C(\overline{\Omega}) when Ω\Omega is a bounded interval. Here is a precise statement.

For a function measuring a jump, we assume that

  1. (K1)

    K:[0,)[0,)K:[0,\infty)\to[0,\infty) is a continuous, strictly increasing function with K(0)=0K(0)=0.

  2. (K2)

    For M>0M>0, there exists a positive constant CMC_{M} such that

    K(ρ1)+K(ρ2)K(ρ1+ρ2)+CMρ1ρ2K(\rho_{1})+K(\rho_{2})\geq K(\rho_{1}+\rho_{2})+C_{M}\rho_{1}\rho_{2}

    for all ρ1,ρ20\rho_{1},\rho_{2}\geq 0 with ρ1+ρ2M\rho_{1}+\rho_{2}\leq M. In particular, KK is subadditive.

  3. (K3)

    limρ0K(ρ)/ρ=1\lim_{\rho\to 0}K(\rho)/\rho=1.

If K(ρ)=ρK(\rho)=\rho so that TVK=TVTV_{K}=TV, KK satisfies (K1) and (K3) but does not satisfy (K2). If K(ρ)=ρ/(1+ρ)K(\rho)=\rho/(1+\rho), KK satisfies (K2) as well as (K1) and (K3). Indeed, a direct calculation shows that

ρ11+ρ1+ρ21+ρ2=ρ+2ρ1ρ21+ρ+ρ1ρ2=ρ1+ρ+(2+ρ)ρ1ρ2(1+ρ)(1+ρ+ρ1ρ2),ρ=ρ1+ρ2.\frac{\rho_{1}}{1+\rho_{1}}+\frac{\rho_{2}}{1+\rho_{2}}=\frac{\rho+2\rho_{1}\rho_{2}}{1+\rho+\rho_{1}\rho_{2}}=\frac{\rho}{1+\rho}+\frac{(2+\rho)\rho_{1}\rho_{2}}{(1+\rho)(1+\rho+\rho_{1}\rho_{2})},\quad\rho=\rho_{1}+\rho_{2}.

Thus, (K2) follows.

Theorem 1.1.

Assume that KK satisfies (K1), (K2) and (K3). Assume that gC[a,b]g\in C[a,b]. Let UBV(a,b)U\in BV(a,b) be a minimizer of TVKgTV_{Kg}. Then UU must be a piecewise constant function (with finite jumps) satisfying infgUsupg\inf g\leq U\leq\sup g on [a,b][a,b]. Let mm be the number of jumps of UU. Then

m[(ba)λ/AM]+1,m\leq\left[(b-a)\lambda/A_{M}\right]+1,

where AM=min{cM/M,2CM}A_{M}=\min\{c_{M}/M,2C_{M}\} with

osc[a,b]g:=max[a,b]gmin[a,b]gM.\operatorname{osc}_{[a,b]}g:=\max_{[a,b]}g-\min_{[a,b]}g\leq M.

Here cMc_{M} is a constant such that K(ρ)cMρK(\rho)\geq c_{M}\rho for ρ[0,M]\rho\in[0,M]. Here [r][r] denotes the integer part of r0r\geq 0.

We note that we do not assume that gBV(a,b)g\in BV(a,b). If gg is non-decreasing or non-increasing, we have a sharper estimate for mm.

Theorem 1.2.

Assume that KK satisfies (K1), (K2) and (K3). Assume that gC[a,b]g\in C[a,b] is non-decreasing (resp. non-increasing). Let UBV(a,b)U\in BV(a,b) be a minimizer of TVKgTV_{Kg}. Then UU must be a non-decreasing (non-increasing) piecewise constant function satisfying infgUsupg\inf g\leq U\leq\sup g on [a,b][a,b]. The number mm of jumps is estimated as

m[(ba)λ/(2CM)]+1m\leq\left[(b-a)\lambda/(2C_{M})\right]+1

for Mosc[a,b]g=|g(b)g(a)|M\geq\operatorname{osc}_{[a,b]}g=\left|g(b)-g(a)\right|.

To show Theorem 1.1 and also Theorem 1.2, we introduce the notion of a coincidence set CC, which is formally defined by

C={x[a,b]|U(x)=g(x)}.C=\left\{x\in[a,b]\bigm{|}U(x)=g(x)\right\}.

It turns out that a minimizer UU of TVKgTV_{Kg} is always continuous on CC and outside this set CC, UU is piecewise constant and

sup[a,b]Usup[a,b]g,inf[a,b]Uinf[a,b]g.\sup_{[a,b]}U\leq\sup_{[a,b]}g,\quad\inf_{[a,b]}U\geq\inf_{[a,b]}g. (1.1)

Moreover, we are able to prove that UU has at most one jump on (α,β)(\alpha,\beta) if (α,β)C=(\alpha,\beta)\cap C=\emptyset and α,βC\alpha,\beta\in C. We also prove that if α,βC\alpha,\beta\in C with α<β\alpha<\beta is too close, UU must be monotone on (α,β)(\alpha,\beta). Here, (α,β)(\alpha,\beta) may include some point of CC. To show these properties, it suffices to assume the subadditivity of KK instead of (K2). It includes the case TVKTV_{K}, where K(ρ)=ρK(\rho)=\rho so that TVKTV_{K} is the standard total variation.

We shall prove that if αC\alpha\in C and βC\beta\in C with α<β\alpha<\beta is too close, UU must be a constant on (α,β)(\alpha,\beta). A key step (Lemma 4.7) is to show that for a non-decreasing minimizer UU on (a,b)(a,b) and α,βC\alpha,\beta\in C with α<β\alpha<\beta, the estimate

βα>2CM/λ\beta-\alpha>2C_{M}/\lambda

holds provided that there is γC(α,β)\gamma\in C\cap(\alpha,\beta) and (γ,β)C=(\gamma,\beta)\cap C=\emptyset and that ρ2:=U(β)U(γ)ρ1:=U(γ)U(α)\rho_{2}:=U(\beta)-U(\gamma)\geq\rho_{1}:=U(\gamma)-U(\alpha), where CMC_{M} is a constant in (K2). Here is a rough strategy to prove the statement. Since (γ,β)C=(\gamma,\beta)\cap C=\emptyset, there is exactly one jump point x1x_{1} in (γ,β)(\gamma,\beta). We compare TVKg(U)TV_{Kg}(U) and TVKg(v)TV_{Kg}(v) in (α,β)(\alpha,\beta), where vv equals U(α)U(\alpha) on (α,x1)(\alpha,x_{1}) and equals UU on (x1,β)(x_{1},\beta). It is not difficult to prove

TVK(U)TVK(v)K(ρ1)+K(ρ2)K(ρ1+ρ2)CMρ1ρ2TV_{K}(U)-TV_{K}(v)\geq K(\rho_{1})+K(\rho_{2})-K(\rho_{1}+\rho_{2})\geq C_{M}\rho_{1}\rho_{2} (1.2)

since ρ1+ρ2M\rho_{1}+\rho_{2}\leq M by (1.1). The proof for

2λ((U)(v))ρ1ρ2(x1α)\frac{2}{\lambda}\left(\mathcal{F}(U)-\mathcal{F}(v)\right)\geq-\rho_{1}\rho_{2}(x_{1}-\alpha) (1.3)

is more involved. It is not difficult to show

γx1{(Ug)2(vg)2}𝑑xρ1(ρ1+ρ2)(x1γ).\int_{\gamma}^{x_{1}}\left\{(U-g)^{2}-(v-g)^{2}\right\}dx\geq-\rho_{1}(\rho_{1}+\rho_{2})(x_{1}-\gamma). (1.4)

The proof for

αγ{(Ug)2(vg)2}𝑑xρ12(γα)\int_{\alpha}^{\gamma}\left\{(U-g)^{2}-(v-g)^{2}\right\}dx\geq-\rho_{1}^{2}(\gamma-\alpha) (1.5)

is more difficult. If gg is non-decreasing, we are able to prove

g(x)UU(x2+0)U(x20)forxF,g(x)-U\leq U(x_{2}+0)-U(x_{2}-0)\quad\text{for}\quad x\in F,

where F=[x0,x2][a,b)F=[x_{0},x_{2}]\subset[a,b) is a maximal closed interval (called a facet) such that UU is a constant in the interior intF\operatorname{int}F of FF. For general FF, one only expects such an inequality just in the average sense, i.e.,

x0x2(g(x)U)𝑑x(U(x2+0)U(x20))(x2x0).\int_{x_{0}}^{x_{2}}\left(g(x)-U\right)dx\leq\left(U(x_{2}+0)-U(x_{2}-0)\right)(x_{2}-x_{0}). (1.6)

The estimate (1.5) follows from (1.6). Combining (1.4) and (1.5), we obtain (1.3) since ρ2ρ1\rho_{2}\geq\rho_{1}. The estimates (1.2) and (1.3) yield

TVKg(U)TVKg(v)ρ1ρ2(CM(x1α)λ/2).TV_{Kg}(U)-TV_{Kg}(v)\geq\rho_{1}\rho_{2}\left(C_{M}-(x_{1}-\alpha)\lambda\bigm{/}2\right).

If βα2CM/λ\beta-\alpha\leq 2C_{M}/\lambda, UU cannot be a minimizer. Similarly, we are able to prove that if UU is continuous on (α,β)(\alpha,\beta) with α,βC\alpha,\beta\in C, then UU is a constant on (α,β)(\alpha,\beta). These observations yield Theorem 1.1. Theorem 1.2 can be proved similarly to Theorem 1.1 by noting that a minimizer UU is a monotone function.

We also give a quantative estimate for TVKgTV_{Kg} for monotone data gC[a,b]g\in C[a,b]. It is easy to estimate TVKg(u)TV_{Kg}(u) for a piecewise constant uu from below. We approximate a general BVBV function uu by piecewise constant functions umu_{m} so that TVKg(um)TVKg(u)TV_{Kg}(u_{m})\to TV_{Kg}(u). Using such an approximation result, we establish an estimate of TVKgTV_{Kg} for a general BVBV function. We notice that such an estimate gives another way to prove Theorem 1.2.

It is not difficult to prove that TVKgTV_{Kg} is lower semicontinuous in the space of piecewise constant functions with at most kk jumps. Since the space is of finite dimension, its bounded closed set is compact. By Weierstrass’ theorem, TVKgTV_{Kg} admits a minimizer vv among piecewise constant functions with at most kk jumps with infgvsupg\inf g\leq v\leq\sup g. Thus, Theorem 1.1 guarantees the existence of a minimizer in BV(a,b)BV(a,b). The existence of minimizer of TVKgTV_{Kg} itself can be proved for general essential bounded measurable function gg, i.e., gL(Ω)g\in L^{\infty}(\Omega) for general Lipschitz domain in n\mathbb{R}^{n} since it is known [AFP] that TVKTV_{K} is lower semicontinuous in a suitable topology. We shall discuss this point in Section 2. Note that instead of (K2) subadditivity for KK is enough to have the existence of a minimizer.

We believe Theorem 1.1 extends for general gL(a,b)g\in L^{\infty}(a,b). In fact, we have a weaker version.

Theorem 1.3.

Assume that KK satisfies (K1), (K2) and (K3). For gL(a,b)g\in L^{\infty}(a,b), there exists a piecewise constant function UU having at most

[(ba)λ/AM]+1\left[(b-a)\lambda\bigm{/}A_{M}\right]+1

jumps and satisfying ess.infgUess.supg\mathrm{ess.}\inf g\leq U\leq\mathrm{ess.}\sup g on (a,b)(a,b), which minimizes TVKgTV_{Kg} on BV(a,b)BV(a,b). Here, Mosc[a,b]g=ess.supgess.infgM\geq\operatorname{osc}_{[a,b]}g=\mathrm{ess.}\sup g-\mathrm{ess.}\inf g.

This follows from above-mentioned approximation result and Theorem 1.1. Since the uniqueness of minimizer is not guaranteed in general, there might exist another minimizer which is not piecewise constant although it is unlikely.

As shown in [GOU], [GOSU], TVKgTV_{Kg} is obtained as a singular limit (ε0\varepsilon\downarrow 0 limit) of the Kobayashi-Warren-Carter type energy

EKWCgε(u,v)\displaystyle E_{\mathrm{KWC}g}^{\varepsilon}(u,v) :=EKWCε(u,v)+(u)\displaystyle:=E_{\mathrm{KWC}}^{\varepsilon}(u,v)+\mathcal{F}(u)
EKWCε(u,v)\displaystyle E_{\mathrm{KWC}}^{\varepsilon}(u,v) :=Ωsv2|Du|+EsMMε(v),s>0\displaystyle:=\int_{\Omega}sv^{2}|Du|+E_{\mathrm{sMM}}^{\varepsilon}(v),\quad s>0
EsMMε(v)\displaystyle E_{\mathrm{sMM}}^{\varepsilon}(v) :=ε2Ω|v|2𝑑x+12εΩF(v)𝑑x\displaystyle:=\frac{\varepsilon}{2}\int_{\Omega}|\nabla v|^{2}\;dx+\frac{1}{2\varepsilon}\int_{\Omega}F(v)\;dx

by minimizing order parameter vv; here F(v)F(v) is a single-well potential typically of the form F(v)=(v1)2F(v)=(v-1)^{2} and ss is a positive parameter. In fact, in one-dimensional setting [GOU], the Gamma limit of EsMMε(v)E_{\mathrm{sMM}}^{\varepsilon}(v) under the graph convergence formally equals

EsMM0(Ξ)=i=12(G(ξi+G(ξi+)),G(s)=|1sF(ξ)dξ|E_{\mathrm{sMM}}^{0}(\Xi)=\sum_{i=1}^{\infty}2\left(G(\xi_{i}^{-}+G(\xi_{i}^{+})\right),\quad G(s)=\left|\int_{1}^{s}\sqrt{F(\xi)}\;d\xi\right|

if the limit of vv in Ω=(a,b)\Omega=(a,b) equals a set-valued function Ξ\Xi of the form

Ξ(x)={1,xΣ,[ξi,ξi+](1)forxΣ,\Xi(x)=\left\{\begin{array}[]{ll}&1,\quad x\notin\Sigma,\\ &[\xi_{i}^{-},\xi_{i}^{+}](\ni 1)\quad\text{for}\quad x\in\Sigma,\end{array}\right.

where Σ\Sigma is some (at most) countable set. The Gamma limit of EKWCεE_{\mathrm{KWC}}^{\varepsilon} equals

EKWC0(u,Ξ)=i=1(s(ξi,+)2|u+(xi)u(xi)|+2(G(ξi)+G(ξi+)))+Ω\Ju|Du|,E_{\mathrm{KWC}}^{0}(u,\Xi)=\sum_{i=1}^{\infty}\left(s(\xi_{i,+}^{-})^{2}\left|u^{+}(x_{i})-u^{-}(x_{i})\right|+2\left(G(\xi_{i}^{-})+G(\xi_{i}^{+})\right)\right)+\int_{\Omega\backslash J_{u}}|Du|,

where ξ+=max(ξ,0)\xi_{+}=\max(\xi,0) and xiΣx_{i}\in\Sigma. Here for uu, L1L^{1} type limit is considered. If we minimize EKWC0(u,Ξ)E_{\mathrm{KWC}}^{0}(u,\Xi) by fixing uu, ξ+\xi^{+} must be one since [ξ,ξ+]1[\xi^{-},\xi^{+}]\ni 1 and G(1)=0G(1)=0. Thus

infΞEKWC0(u,Ξ)\displaystyle\inf_{\Xi}E_{\mathrm{KWC}}^{0}(u,\Xi) =i=1minξ>0(s(ξ+)2|u+(xi)u(xi)|+2G(ξ))+Ω\Ju|Du|\displaystyle=\sum_{i=1}^{\infty}\min_{\xi>0}\left(s(\xi_{+})^{2}\left|u^{+}(x_{i})-u^{-}(x_{i})\right|+2G(\xi)\right)+\int_{\Omega\backslash J_{u}}|Du|
=TVK(u)+Ω\Ju|Du|,\displaystyle=TV_{K}(u)+\int_{\Omega\backslash J_{u}}|Du|,

where

K(ρ)=minξ(s(ξ+)2ρ+2G(ξ)).K(\rho)=\min_{\xi}\left(s(\xi_{+})^{2}\rho+2G(\xi)\right). (1.7)

In the case, s=1s=1 and F(v)=(v1)2F(v)=(v-1)^{2}, a direct calculation shows that

K(ρ)=minξ>0(ξ2ρ+(ξ1)2)=ρρ+1.K(\rho)=\min_{\xi>0}\left(\xi^{2}\rho+(\xi-1)^{2}\right)=\frac{\rho}{\rho+1}.

We shall prove that KK defined in (1.7) satisfies (K2) provided that

lim¯v1F(v)/(v1)<\varlimsup_{v\uparrow 1}F^{\prime}(v)\bigm{/}(v-1)<\infty

for FF if FC()F\in C(\mathbb{R}) is non-negative and F(v)=0F(v)=0 if and only if v=0v=0.

If we replace sv2|Du|\int sv^{2}|Du| by sv2|Du|2\int sv^{2}|Du|^{2}, the energy corresponding to EKWCgεE_{\mathrm{KWC}g}^{\varepsilon} is nothing but what is called the Ambrosio-Tortorelli energy [AT]. Its singular limit is a Mumford-Shah functional

EMS(u,K):=sΩ\K|u|2+n1(K)+(u),E_{\mathrm{MS}}(u,K):=s\int_{\Omega\backslash K}|\nabla u|^{2}+\mathcal{H}^{n-1}(K)+\mathcal{F}(u),

where KK is a closed set in Ω\Omega [AT], [AT2], [FL]. The existence of a minimizer is obtained in [GCL] by using the space of special functions with bounded variation, i.e., SBVSBV functions which is a subspace of BV(Ω)BV(\Omega).

A modified total variation energy TVKTV_{K} is not limited to a singular limit of the Kobayashi-Warren-Carter energy. In fact, TVK(u)TV_{K}(u) like energy is derived as the surface tension of grain boundaries in polycrystals by [LL], where uu is taken as a piecewise constant (vector-valued) function; see also [GaFSp] for more recent development. The function KK measuring jumps may not be isotropic but still concave. In [ELM], TVKTV_{K} type energy for a piecewise constant function is also considered to study motion of a grain boundary. However, in their analysis, the convexity of KK is assumed.

This paper is organized as follows. In Section 2, we give a rigorous formulation of TVKTV_{K} and prove the existence of a minimizer of TVKgTV_{Kg} for gL(Ω)g\in L^{\infty}(\Omega) for a general Lipschitz domain in n\mathbb{R}^{n}. In Section 3, we study a profile of a minimizer UU for TVKgTV_{Kg} including TVgTV_{g} outside the coincidence set. We also prove that UU is monotone in (α,β)(\alpha,\beta) with α,βC\alpha,\beta\in C provided that α\alpha and β\beta is close. In Section 4, we prove that α,βC\alpha,\beta\in C cannot be too close if KK satisfies (K2). We prove Theorem 1.1, Theorem 1.2 and Theorem 1.3. We also establish an estimate for TVKgTV_{Kg} for monotone gg. In Section 5, we shall discuss a sufficient condition that KK in (1.7) satisfies (K2).

2. Existence of a minimizer

In this section, after giving a precise definition of TVKTV_{K}, we give an existence result for its Rudin-Osher-Fatemi type energy. The proof is based on a standard compactness result for TVTV and a classical lower semicontinuity result for TVKTV_{K}.

We recall a standard notation as in [AFP]. Let Ω\Omega be a domain in n\mathbb{R}^{n}. For a locally integrable (real-valued) function uu, we consider its total variation TV(u)TV(u) in Ω\Omega, i.e.,

TV(u)=sup{Ωudivφdx|φCc(Ω,n),|φ(x)|1inΩ},TV(u)=\sup\left\{\int_{\Omega}-u\operatorname{div}\varphi\;dx\biggm{|}\varphi\in C_{c}^{\infty}(\Omega,\mathbb{R}^{n}),\ \left|\varphi(x)\right|\leq 1\ \text{in}\ \Omega\right\},

where Cc(Ω,n)C_{c}^{\infty}(\Omega,\mathbb{R}^{n}) denotes the space of all n\mathbb{R}^{n}-valued smooth functions with compact support in Ω\Omega. An integrable function uu, i.e., uL1(Ω)u\in L^{1}(\Omega), is called a function of bounded variation if TV(u)<TV(u)<\infty. The space of all such function is denoted by BV(Ω)BV(\Omega), i.e.,

BV(Ω)={uL1(Ω)|TV(u)<}.BV(\Omega)=\left\{u\in L^{1}(\Omega)\bigm{|}TV(u)<\infty\right\}.

By Riesz’s representation theory, one easily observe that TV(u)TV(u) is finite if and only if the distributional derivative DuDu of uu is a finite Radon measure on Ω\Omega and its total variation |Du|(Ω)|Du|(\Omega) in Ω\Omega equals TV(u)TV(u).

We next define a jump discontinuity of a locally integrable function. Let Br(x)B_{r}(x) denote an open ball of radius rr centered at xx in n\mathbb{R}^{n}. In other words,

Br(x)={yn||yx|<r}.B_{r}(x)=\left\{y\in\mathbb{R}^{n}\bigm{|}|y-x|<r\right\}.

For a unit vector νn\nu\in\mathbb{R}^{n}, we define a half ball of the form

Br±(x,ν)={yBr(x)|±ν(yx)0},B_{r}^{\pm}(x,\nu)=\left\{y\in B_{r}(x)\bigm{|}\pm\nu\cdot(y-x)\geq 0\right\},

where aba\cdot b for a,bna,b\in\mathbb{R}^{n} denotes the standard inner product in n\mathbb{R}^{n}. Let ww be a locally integrable function in Ω\Omega. We say that a point xΩx\in\Omega is a (approximate) jump point of ww if there exists a unit vector νwn\nu_{w}\in\mathbb{R}^{n}, w±w^{\pm}\in\mathbb{R}, w+ww^{+}\neq w^{-}, such that

limr01n(Br±(x,νw))Br±(x,νw)|w(y)w±|𝑑y=0.\lim_{r\downarrow 0}\frac{1}{\mathcal{L}^{n}\left(B_{r}^{\pm}(x,\nu_{w})\right)}\int_{B_{r}^{\pm}(x,\nu_{w})}\left|w(y)-w^{\pm}\right|dy=0.

Here n\mathcal{L}^{n} denotes the Lebesgue measure in n\mathbb{R}^{n} so this integral is the average of |w(y)w±|\left|w(y)-w^{\pm}\right| over Br±(x,νw)B_{r}^{\pm}(x,\nu_{w}). The set of all jump points of ww is denoted by JwJ_{w} and called the (approximate) jump (set) of ww. By definition, JwJ_{w} is contained in the set SwS_{w} of (approximate) discontinuity point of ww, i.e.,

Sw={xΩ|limr01n(Br(x))Br(x)|w(y)z|dt=0does not holdfor any choice ofz}.S_{w}=\Biggl{\{}x\in\Omega\biggm{|}\lim_{r\downarrow 0}\frac{1}{\mathcal{L}^{n}\left(B_{r}(x)\right)}\int_{B_{r}(x)}\left|w(y)-z\right|dt=0\ \text{does not hold}\\ \text{for any choice of}\ z\in\mathbb{R}\Biggr{\}}.

By the Federer-Vol’pert theorem [AFP, Theorem 3.78], we know n1\mathcal{H}^{n-1}
(Su\Ju)=0(S_{u}\backslash J_{u})=0, where m\mathcal{H}^{m} denotes mm-dimensional Hausdorff measure provided that uBV(Ω)u\in BV(\Omega). Moreover, SuS_{u} is countably n1\mathcal{H}^{n-1}-rectifiable, i.e., SuS_{u} can be covered by a countable union of Lipschitz graphs up to an n1\mathcal{H}^{n-1} measure zero set. In particular, JuJ_{u} is also countably n1\mathcal{H}^{n-1}-rectifiable. Quite recently, it is proved that JuJ_{u} is always countably n1\mathcal{H}^{n-1}-rectifiable if we only assume that ww is locally integrable [DN]. For uBV(Ω)u\in BV(\Omega), the value u±u^{\pm} can be viewed as a trace of uu on a countably n1\mathcal{H}^{n-1}-rectifiable set [AFP, Theorem 3.77, Remark 3.79] except n1\mathcal{H}^{n-1} negligible set (up to permutation of u+u^{+} and uu^{-}). We now recall a unique decomposition of the Radon measure DuDu for uBV(Ω)u\in BV(\Omega) of the form

Du=Dau+Dsu,Dsu=Dcu+(u+u)νun1Ju;Du=D^{a}u+D^{s}u,\quad D^{s}u=D^{c}u+(u^{+}-u^{-})\cdot\nu_{u}\mathcal{H}^{n-1}\lfloor J_{u};

see [AFP, Section 3.8]. The term DauD^{a}u denotes the absolutely continuous part and DsuD^{s}u denotes the singular part with respect to the Lebesgue measure. The term Dau=unD^{a}u=\nabla u\mathcal{L}^{n}, where u(L1(Ω))n\nabla u\in\left(L^{1}(\Omega)\right)^{n}. The singular part is decomposed into two parts; DcuD^{c}u vanishes on sets of finite n1\mathcal{H}^{n-1} measure. For a measure μ\mu on Ω\Omega and a set AΩA\subset\Omega, the associate measure μA\mu\lfloor A is defined as

(μA)(W)=μ(AW),WΩ.(\mu\lfloor A)(W)=\mu(A\cap W),\quad W\subset\Omega.

We now consider a total variation type energy measuring jumps in a different way. For uBV(Ω)u\in BV(\Omega), we set

TVK(u):=(TVK(u,Ω):=)Ω\Ju|Du|+JuK(|u+u|)dn1.TV_{K}(u):=\left(TV_{K}(u,\Omega):=\right)\int_{\Omega\backslash J_{u}}|Du|+\int_{J_{u}}K\left(|u^{+}-u^{-}|\right)d\mathcal{H}^{n-1}.

Here the density function is assumed to satisfy following conditions.

  1. (K1w)

    K:(0,)[0,)K:(0,\infty)\to[0,\infty) is non-decreasing, lower semicontinuous.

  2. (K2w)

    KK is subadditive, i.e., K(ρ1+ρ2)K(ρ1)+K(ρ2)K(\rho_{1}+\rho_{2})\leq K(\rho_{1})+K(\rho_{2}).

  3. (K3)

    limρ0K(ρ)/ρ=1\lim_{\rho\to 0}K(\rho)/\rho=1.

Theorem 2.1.

Assume that KK satisfies (K1w), (K2w) and (K3). Let Ω\Omega be a bounded domain with Lipschitz boundary in n\mathbb{R}^{n}. Let \mathcal{E} be a lower semicontinuous function in L1(Ω)L^{1}(\Omega) with values in [0,][0,\infty]. Then TVK+TV_{K}+\mathcal{E} has a minimizer on BV(Ω)BV(\Omega) provided that a coercivity condition

infuBV(Ω)(TVK+)(u)=infuBV(Ω)uM(TVK+)(u)\inf_{u\in BV(\Omega)}(TV_{K}+\mathcal{E})(u)=\inf_{\begin{subarray}{c}u\in BV(\Omega)\\ \lVert u\rVert_{\infty}\leq M\end{subarray}}(TV_{K}+\mathcal{E})(u)

for some M>0M>0, where \lVert\cdot\rVert_{\infty} denotes the LL^{\infty}-norm.

We consider the Rudin-Osher-Fatemi type energy for TVKTV_{K}, i.e.,

TVKg(u):=TVK(u)+(u),(u)=λ2Ω|ug|2𝑑xTV_{Kg}(u):=TV_{K}(u)+\mathcal{F}(u),\quad\mathcal{F}(u)=\frac{\lambda}{2}\int_{\Omega}|u-g|^{2}\;dx

for gL2(Ω)g\in L^{2}(\Omega).

Theorem 2.2.

Assume that KK satisfies (K1w), (K2w) and (K3). Let Ω\Omega be a bounded domain with Lipschitz boundary in n\mathbb{R}^{n}. Assume that gL(Ω)g\in L^{\infty}(\Omega). Then there is an element u0BV(Ω)u_{0}\in BV(\Omega) such that

TVKg(u0)=infuBV(Ω)TVKg(u).TV_{Kg}(u_{0})=\inf_{u\in BV(\Omega)}TV_{Kg}(u).

In other words, there is at least one minimizer of TVKgTV_{Kg}.

Proof of Theorem 2.2 admitting Threorem 2.1..

In the case =\mathcal{E}=\mathcal{F}, the lower semicontinuity of \mathcal{E} in L1(Ω)L^{1}(\Omega) is rather clear. If gL(Ω)g\in L^{\infty}(\Omega), then for a chopped function uM=max(min(u,M),M)u_{M}=\max\left(\min(u,M),-M\right) with M>gM>\lVert g\rVert_{\infty}, we see that

2λ((u)(uM))=uM|ug|2𝑑xuM|Mg|2𝑑x\displaystyle\frac{2}{\lambda}\left(\mathcal{F}(u)-\mathcal{F}(u_{M})\right)=\int_{u\geq M}|u-g|^{2}\;dx-\int_{u\geq M}|M-g|^{2}\;dx
+uM|ug|2𝑑xuM|Mg|2𝑑x0.\displaystyle+\int_{u\leq-M}|u-g|^{2}\;dx-\int_{u\leq-M}|-M-g|^{2}\;dx\geq 0.

Thus the coercivity condition is fulfilled. Theorem 2.2 now follows from Theorem 2.1. ∎

In the rest of this section, we shall prove Theorem 2.1 by a simple direct method. We begin with compactness.

Proposition 2.3.

Assume that (K1w) and (K3) are fulfilled. Assume that Ω\Omega is a bounded domain with Lipschitz boundary in n\mathbb{R}^{n}. Let {uk}k=1\{u_{k}\}_{k=1}^{\infty} be a sequence in BV(Ω)BV(\Omega) such that

supk1TVK(uk)<andsupk1uk<.\sup_{k\geq 1}TV_{K}(u_{k})<\infty\quad\text{and}\quad\sup_{k\geq 1}\lVert u_{k}\rVert_{\infty}<\infty.

Then there is a subsequence {uk}\{u_{k^{\prime}}\} and uBV(Ω)u\in BV(\Omega) such that ukuu_{k^{\prime}}\to u strongly in L1(Ω)L^{1}(\Omega) and DukDuDu_{k^{\prime}}\to Du weak* in the space of bounded measures. In other words, uku_{k^{\prime}} is sequentially weakly* converges to uu in BV(Ω)BV(\Omega).

Proof.

By (K1) and (K3), we see that for any MM, there is cM>0c^{\prime}_{M}>0 such that

K(ρ)cMρforρM.K(\rho)\geq c^{\prime}_{M}\rho\quad\text{for}\quad\rho\leq M. (2.1)

If MM is chosen such that ukM\lVert u_{k}\rVert_{\infty}\leq M, then

TV(uk)1cMTVK(uk).TV(u_{k})\leq\frac{1}{c^{\prime}_{M}}TV_{K}(u_{k}).

Thus {TV(uk)}\left\{TV(u_{k})\right\} is bounded. By the standard compactness for BV(Ω)BV(\Omega) function [AFP, Theorem 3.23], [Giu, Theorem 1.19] yields the desired results. ∎

For a lower semicontinuity, we have

Proposition 2.4.

Assume (K1w), (K2w) and (K3), then TVKTV_{K} is sequentially weakly* lower semicontinuous in BV(Ω)BV(\Omega).

This is a special form of the lower semicontinuity result [AFP, Theorem 5.4]. We restate it for the reader’s convenience. We consider

F(u)=Ω\Juφ(|u|)𝑑x+β|Dcu|(Ω)+JuK(|u+u|)𝑑n1.F(u)=\int_{\Omega\backslash J_{u}}\varphi\left(|\nabla u|\right)dx+\beta|D^{c}u|(\Omega)+\int_{J_{u}}K\left(|u^{+}-u^{-}|\right)d\mathcal{H}^{n-1}.
Proposition 2.5.

Let φ:[0,)[0,)\varphi:[0,\infty)\to[0,\infty) be a non-decreasing, lower semicontinuous and convex function. Assume that K:(0,)[0,)K:(0,\infty)\to[0,\infty) is a non-decreasing, lower semicontinuous and subadditive function and β[0,)\beta\in[0,\infty). Then FF is sequentially weakly* lower semicontinuous in BV(Ω)BV(\Omega) provided that

limtφ(t)t=β=limt0K(t)t.\lim_{t\uparrow\infty}\frac{\varphi(t)}{t}=\beta=\lim_{t\downarrow 0}\frac{K(t)}{t}.

See [AFP, Theorem 5.4]. Such a lower semicontinuity is originally due to Bouchitté and Buttazzo [BB]. In our setting φ(t)=t\varphi(t)=t, β=1\beta=1.

Proof of Theorem 2.1.

Let {uk}k=1\{u_{k}\}_{k=1}^{\infty} be a minimizing sequence of TVK+TV_{K}+\mathcal{E}, i.e.,

limk(TVK+)(uk)=infuBV(TVK+).\lim_{k\to\infty}(TV_{K}+\mathcal{E})(u_{k})=\inf_{u\in BV}(TV_{K}+\mathcal{E}).

By compactness (Proposition 2.3), {uk}\{u_{k}\} contains a convergent subsequence still denoted by {uk}\{u_{k}\} to some uBV(Ω)u\in BV(\Omega), sequentially weakly* in BV(Ω)BV(\Omega). By lower semicontinuity of TVKTV_{K} (Proposition 2.4), we conclude that

(TVK+)(u)lim¯k(TVK+)(uk).(TV_{K}+\mathcal{E})(u)\leq\varliminf_{k\to\infty}(TV_{K}+\mathcal{E})(u_{k}).

Thus, uu is a minimizer of TVK+TV_{K}+\mathcal{E} in BV(Ω)BV(\Omega). ∎

3. Coincidence set of a minimizer

In this section, we discuss one-dimensional setting and study properties of coincidence set

C={xΩ|U(x)=g(x)}C=\left\{x\in\Omega\bigm{|}U(x)=g(x)\right\}

of a minimizer UU.

Let Ω\Omega be a bounded open interval, i.e., Ω=(a,b)\Omega=(a,b). We consider TVKg(u)TV_{Kg}(u) for gC[a,b]g\in C[a,b] for uBV(a,b)u\in BV(a,b). Since uu can be written as a difference of two non-decreasing function, we may assume that uu has a representative that JuJ_{u} is at most a countable set and outside JuJ_{u}, uu is continuous. Moreover, we may assume that uu is right (resp. left) continuous at aa (resp. at bb). For xJux\in J_{u}, u(x±0)u(x\pm 0) is well-defined.

For KK, we assume

  1. (K1)

    K:[0,)[0,)K:[0,\infty)\to[0,\infty) is continuous, strictly increasing function with K(0)=0K(0)=0.

We first prove that a minimizer UU is piecewise constant in the place U>gU>g or U<gU<g.

Lemma 3.1.

Assume that KK satisfies (K1) and that gC[a,b]g\in C[a,b]. Let UBV(a,b)U\in BV(a,b) be a minimizer of TVKgTV_{Kg}. Let x0(a,b)x_{0}\in(a,b) be a continuous point of UU and assume U(x0)>g(x0)U(x_{0})>g(x_{0}) (resp. U(x0)<g(x0)U(x_{0})<g(x_{0})). Then UU is constant in some interval (α,β)(\alpha,\beta) including x0x_{0} and U(x)>g(x)U(x)>g(x) (resp. U(x)<g(x)U(x)<g(x)) for x(α,β)x\in(\alpha,\beta). Moreover, we can take (α,β)(\alpha,\beta) such that one of following three cases occurs exclusively.

  1. (i)

    U(α0)<g(α)<U(α+0)U(\alpha-0)<g(\alpha)<U(\alpha+0) (resp. U(α0)>g(α)>U(α+0)U(\alpha-0)>g(\alpha)>U(\alpha+0)) and U(β0)=g(β)U(\beta-0)=g(\beta),

  2. (ii)

    U(α+0)=g(α)U(\alpha+0)=g(\alpha) and U(β0)>g(β)>U(β+0)U(\beta-0)>g(\beta)>U(\beta+0) (resp. U(β0)<g(β)<U(β+0)U(\beta-0)<g(\beta)<U(\beta+0)),

  3. (iii)

    U(α+0)=g(α)U(\alpha+0)=g(\alpha) and U(β0)=g(β)U(\beta-0)=g(\beta).

In the case α=a\alpha=a, we do not impose the condition g(α)>U(α0)g(\alpha)>U(\alpha-0) (resp. g(α)<U(α0)g(\alpha)<U(\alpha-0)) and similarly, g(β)>U(β+0)g(\beta)>U(\beta+0) (resp. g(β)<U(β+0)g(\beta)<U(\beta+0)) is not imposed for β=b\beta=b since U(α0)U(\alpha-0), U(β+0)U(\beta+0) are undefined.

Proof.

We shall only give a proof for the case U(x0)>g(x0)U(x_{0})>g(x_{0}) since the argument for U(x0)<g(x0)U(x_{0})<g(x_{0}) is symmetric. We consider the case that UU is continuous on (α,β)(\alpha,\beta) with U>gU>g on (α,β)(\alpha,\beta) and U(α+0)=g(α)U(\alpha+0)=g(\alpha), U(β0)=g(β)U(\beta-0)=g(\beta). We shall claim that UU is a constant. If not, max[α,β]U>min[α,β]U\max_{[\alpha,\beta]}U>\min_{[\alpha,\beta]}U. There would exist two points x1x_{1} and x2x_{2} in [α,β][\alpha,\beta] such that U(x1)=min[α,β]UU(x_{1})=\min_{[\alpha,\beta]}U and U(x2)=max[α,β]UU(x_{2})=\max_{[\alpha,\beta]}U such that U(x)(minU,maxU)U(x)\in(\min U,\max U) for xx between x1x_{1} and x2x_{2}. We may assume x1<x2x_{1}<x_{2} since the proof for the other case is symmetric. We set

v(x)={U~(x),x(x1,x2),U(x),x(x1,x2),v(x)=\left\{\begin{array}[]{ll}\tilde{U}(x),&x\in(x_{1},x_{2}),\\ U(x),&x\notin(x_{1},x_{2}),\end{array}\right.

where U~\tilde{U} is a continuous non-decreasing function on [x1,x2][x_{1},x_{2}] such that U~(x)<U(x)\tilde{U}(x)<U(x) for x(x1,x2)x\in(x_{1},x_{2}) and U(x1)=U~(x1)U(x_{1})=\tilde{U}(x_{1}), U(x2)=U~(x2)U(x_{2})=\tilde{U}(x_{2}). Then (v)<(U)\mathcal{F}(v)<\mathcal{F}(U) and TVK(v)=TV(v)TV(U)=TVK(U)TV_{K}(v)=TV(v)\leq TV(U)=TV_{K}(U). This would contradict to the assumption that UU is a minimizer of TVKgTV_{Kg}. We conclude that UU is a constant and case (iii) occurs.

Assume that there is a jump point α0\alpha_{0} of UU with α0<x0\alpha_{0}<x_{0} with U(α0+0)>g(α0)U(\alpha_{0}+0)>g(\alpha_{0}). Then U(α0+0)>U(α00)U(\alpha_{0}+0)>U(\alpha_{0}-0). If not, d=U(α00)U(α0+0)>0d=U(\alpha_{0}-0)-U(\alpha_{0}+0)>0 and set

v(x)={U(x)d,x(α0δ,α0)U(x),x(α0δ,α0).v(x)=\left\{\begin{array}[]{lll}U(x)-d,&x\in(\alpha_{0}-\delta,\alpha_{0})\\ U(x),&x\notin(\alpha_{0}-\delta,\alpha_{0}).\end{array}\right.

For a sufficiently small δ>0\delta>0, v(x)>g(x)v(x)>g(x); see Figure 1.

Refer to caption
Figure 1. shift of a jump

By definition, TVK(v)TVK(U)TV_{K}(v)\leq TV_{K}(U) and (v)<(U)\mathcal{F}(v)<\mathcal{F}(U). Thus UU is not a minimizer of TVKgTV_{Kg}. Similarly, if there is a jump point β0\beta_{0} of UU with x0<β0x_{0}<\beta_{0}, then U(β00)>U(β0+0)U(\beta_{0}-0)>U(\beta_{0}+0).

We shall prove that UU is continuous on (x0,β)(x_{0},\beta) provided that U>gU>g on (α0,β)(\alpha_{0},\beta). If not, there would exist a jump point β0\beta_{0} of UU with α0<x0<β0<β\alpha_{0}<x_{0}<\beta_{0}<\beta satisfying U(x)>g(x)U(x)>g(x) for x(α0,β0)x\in(\alpha_{0},\beta_{0}), U(α0+0)>g(α0)U(\alpha_{0}+0)>g(\alpha_{0}), U(β00)>g(β0)U(\beta_{0}-0)>g(\beta_{0}). We shall prove that such configuration does not occur. We first claim that U(α0+0)=U(β00)U(\alpha_{0}+0)=U(\beta_{0}-0). By symmetry, we may assume that U(α0+0)<U(β00)U(\alpha_{0}+0)<U(\beta_{0}-0). Since we know U(β0+0)<U(β00)U(\beta_{0}+0)<U(\beta_{0}-0), we take MM such that

max(U(α0+0),U(β0+0))<M<U(β00).\max\left(U(\alpha_{0}+0),U(\beta_{0}+0)\right)<M<U(\beta_{0}-0).

We set uM=min(M,U)u_{M}=\min(M,U) on (α0,β0)(\alpha_{0},\beta_{0}). Outside (α0,β0)(\alpha_{0},\beta_{0}), we set uM=Uu_{M}=U. If MM is taken close to U(β00)U(\beta_{0}-0), then uM>gu_{M}>g on (α0,β0)(\alpha_{0},\beta_{0}) and uM(α0+0)>g(α0)u_{M}(\alpha_{0}+0)>g(\alpha_{0}), uM(β00)>g(β0)u_{M}(\beta_{0}-0)>g(\beta_{0}); see Figure 2.

Refer to caption
Figure 2. a way of truncation

Thus (uM)(U)\mathcal{F}(u_{M})\leq\mathcal{F}(U) and by truncation TVK(uM)<TVK(U)TV_{K}(u_{M})<TV_{K}(U). This contradicts to the assumption that UU is a minimizer. Thus U(α0+0)=U(β00)U(\alpha_{0}+0)=U(\beta_{0}-0). However, if so, take MM again close to U(α0+0)U(\alpha_{0}+0) and M<U(α0+0)M<U(\alpha_{0}+0), TVK(uM)<TVK(U)TV_{K}(u_{M})<TV_{K}(U) with (uM)(U)\mathcal{F}(u_{M})\leq\mathcal{F}(U). This is a contradiction so if α0\alpha_{0} is a jump of UU for α0<x0\alpha_{0}<x_{0}, then there is no jump on (x0,β)(x_{0},\beta) provided that U(x)>g(x)U(x)>g(x). In other words, UU is continuous and satisfies U(x)>g(x)U(x)>g(x) on [x0,β)[x_{0},\beta) and U(β0)=g(β)U(\beta-0)=g(\beta). We now conclude that UU is constant on [x0,β)[x_{0},\beta) by using a similar argument at the beginning of this proof.

Since there is a sequence of continuity point xjx_{j} of UU converging to α0\alpha_{0} as jj\to\infty keeping xj>α0x_{j}>\alpha_{0}, we conclude that UU is a constant on (xj,β)(x_{j},\beta). This says that UU is constant on (α0,β)(\alpha_{0},\beta). To say that this corresponds to (i), it remains to prove that U(α0)<g(α)U(\alpha-0)<g(\alpha). If not, U(α0)g(α)U(\alpha-0)\geq g(\alpha), then we take

v(x)={max(U(x)d,g~(x)),x(α0,α0+δ),U(x),x(α0,α0+δ),U(x0),x=α0,v(x)=\left\{\begin{array}[]{ll}\max\left(U(x)-d,\tilde{g}(x)\right),&x\in(\alpha_{0},\alpha_{0}+\delta),\\ U(x),&x\notin(\alpha_{0},\alpha_{0}+\delta),\\ U(x-0),&x=\alpha_{0},\end{array}\right.

where d=U(α0+0)U(α0)d=U(\alpha_{0}+0)-U(\alpha-0). Here,

g~(x)=sup{g(y)g(α0)|α0yx}+g(α0)\tilde{g}(x)=\sup\left\{g(y)-g(\alpha_{0})\bigm{|}\alpha_{0}\leq y\leq x\right\}+g(\alpha_{0})

which is continuous and non-decreasing. Moreover,

TVK(v)TVK(U)+m(g(δ)),(v)(U)δ(dg~(δ))2,TV_{K}(v)\leq TV_{K}(U)+m\left(g(\delta)\right),\quad\mathcal{F}(v)\leq\mathcal{F}(U)-\delta\left(d-\tilde{g}(\delta)\right)^{2},

where m(σ)=sup{K(d+τ)K(d)|0τσ}m(\sigma)=\sup\left\{K(d+\tau)-K(d)\bigm{|}0\leq\tau\leq\sigma\right\}; see Figure 3.

Refer to caption
Figure 3. modification of UU

Thus

TVKg(v)TVKg(U)+m(g~(δ))δ(dg~(δ))2.TV_{Kg}(v)\leq TV_{Kg}(U)+m\left(\tilde{g}(\delta)\right)-\delta\left(d-\tilde{g}(\delta)\right)^{2}.

Since g~(δ)0\tilde{g}(\delta)\to 0 as δ0\delta\to 0, and m(σ)0m(\sigma)\to 0 as σ0\sigma\to 0 by (K1), we conclude that for sufficiently small δ>0\delta>0, TVKg(v)<TVKg(U)TV_{Kg}(v)<TV_{Kg}(U). We now obtain (i). A symmetric argument yields (ii). The proof is now complete. ∎

For uBV(a,b)u\in BV(a,b) and gC[a,b]g\in C[a,b], we set

C±={x[a,b]|u(x±0)=g(x)}.C_{\pm}=\left\{x\in[a,b]\bigm{|}u(x\pm 0)=g(x)\right\}.

If C+=CC_{+}=C_{-} in (a,b)(a,b), we simply write

C=C+CC=C_{+}\cup C_{-}

and call CC the coincidence set of uu. By Lemma 3.1, we obtain a few properties of CC.

Lemma 3.2.

Assume the same hypotheses of Lemma 3.1 concerning KK, gg and UU. Let C±C_{\pm} be defined for u=Uu=U. Then C=C+C_{-}=C_{+} in (a,b)(a,b), i.e., UU is continuous on the coincidence set CC. If (x1,x2)C=(x_{1},x_{2})\cap C=\emptyset with x1,x2Cx_{1},x_{2}\in C. then UU is piecewise constant on (x1,x2)(x_{1},x_{2}) with at most one jump.

Proof.

By Lemma 3.1 (i) and (ii), we easily see that JUC±=J_{U}\cap C_{\pm}=\emptyset. Thus, UU is continuous on CC.

It remains to prove the second statement. By Lemma 3.1, the value of UU on (x1,x2)(x_{1},x_{2}) is either g(x1)g(x_{1}) or g(x2)g(x_{2}). Moreover, if there are more than two jumps, UU must take value of gg at some point x(x1,x2)x_{*}\in(x_{1},x_{2}). In other words, xCx_{*}\in C. This contradicts to C(x1,x2)=C\cap(x_{1},x_{2})=\emptyset. ∎

We conclude this section by showing that if two points α,βC\alpha,\beta\in C with α<β\alpha<\beta for a minimizer UU is too close, then UU must be monotone in (α,β)(\alpha,\beta) under the assumption that KK satisfies (K1), (K2w) and (K3).

We first note comparison with usual total variation TVTV and TVKTV_{K}.

Lemma 3.3.

Assume that KK satisfies (K1), (K2w) and (K3). If uBV(Ω)u\in BV(\Omega) with Ω=(a,b)\Omega=(a,b) is continuous at aa and bb, then

TVK(u)K(ρ)withρ=|u(b)u(a)|.TV_{K}(u)\geq K(\rho)\quad\text{with}\quad\rho=\left\lvert u(b)-u(a)\right\rvert.
Proof.

We may assume that u(a)<u(b)u(a)<u(b). By definition,

TVK(u)=Ω\Ju|Du|+xiJuK(ρi)TV_{K}(u)=\int_{\Omega\backslash J_{u}}|Du|+\sum_{x_{i}\in J_{u}}K(\rho_{i})

where JuJ_{u} denote the jump discontinuity of uu and ρi=|u(xi+0)u(xi0)|\rho_{i}=\left\lvert u(x_{i}+0)-u(x_{i}-0)\right\rvert for xiJux_{i}\in J_{u}. We note that

Ω\Ju|Du|(ρxiJu+ρi)+,\int_{\Omega\backslash J_{u}}|Du|\geq\left(\rho-\sum_{x_{i}\in J_{u}^{+}}\rho_{i}\right)_{+},

where Ju+={xJu|u(x+0)u(x0)>0}J_{u}^{+}=\left\{x\in J_{u}\bigm{|}u(x+0)-u(x-0)>0\right\}. By subadditivity (K2w), we see that

xiJu+K(ρi)K(xiJu+ρi)\sum_{x_{i}\in J_{u}^{+}}K(\rho_{i})\geq K\left(\sum_{x_{i}\in J_{u}^{+}}\rho_{i}\right)

since KK is continuous by (K1). By (K2w), we have

K(ρ)2K(ρ/2)2mK(ρ/2m)=(K(q)/q)ρK(\rho)\leq 2K(\rho/2)\leq\cdots\leq 2^{m}K(\rho/2^{m})=\left(K(q)/q\right)\rho

for q=ρ/2mq=\rho/2^{m}. Sending mm\to\infty, we obtain by (K3) that

K(ρ)ρ.K(\rho)\leq\rho.

We thus observe that

TVK(u)\displaystyle TV_{K}(u) Ω\Ju|Du|+xiJu+K(ρi)(ρxiJu+ρi)++K(xiJu+ρi)\displaystyle\geq\int_{\Omega\backslash J_{u}}|Du|+\sum_{x_{i}\in J_{u}^{+}}K(\rho_{i})\geq\left(\rho-\sum_{x_{i}\in J_{u}^{+}}\rho_{i}\right)_{+}+K\left(\sum_{x_{i}\in J_{u}^{+}}\rho_{i}\right)
K((ρxiJu+ρi)+)+K(xiJu+ρi)K(ρ).\displaystyle\geq K\left(\left(\rho-\sum_{x_{i}\in J_{u}^{+}}\rho_{i}\right)_{+}\right)+K\left(\sum_{x_{i}\in J_{u}^{+}}\rho_{i}\right)\geq K(\rho).

The last inequality follows from the subadditivity (K2w). ∎

If we assume (K1) and (K3), we have, by (2.1),

K(ρ)>cMρfor0<ρM,K(\rho)>c_{M}\rho\quad\text{for}\quad 0<\rho\leq M, (3.1)

with some positive constant cMc_{M} depending only on MM. We next give our monotonicity result.

Theorem 3.4.

Assume that KK satisfies (K1), (K2w) and (K3), and that gC[a,b]g\in C[a,b]. Let UU be a minimizer of TVKgTV_{Kg} in BV(a,b)BV(a,b). Let α\alpha and β\beta with α<β\alpha<\beta be in CC, where CC denotes the coincide set. If U(α)U(β)U(\alpha)\leq U(\beta) (resp. U(α)U(β)U(\alpha)\geq U(\beta)), then UU is non-decreasing (non-increasing) provided that βα<cM/(λM)\beta-\alpha<c_{M}/(\lambda M) if oscg=max[α,β]gmin[α,β]gM\operatorname{osc}g=\max_{[\alpha,\beta]}g-\min_{[\alpha,\beta]}g\leq M, where cMc_{M} is a constant (3.1).

Proof.

We first note that UU is continuous at α,β\alpha,\beta by Lemma 3.2. Moreover, if there is no point of CC in (α,β)(\alpha,\beta), by Lemma 3.2, UU is piecewise constant with at most one jump in (α,β)(\alpha,\beta). Thus, UU is monotone.

We next consider the case that C(α,β)C\cap(\alpha,\beta)\neq\emptyset. We may assume that U(α)U(β)U(\alpha)\leq U(\beta) since the proof for the other case is symmetric. We shall prove that

U(α)U(x0)U(β)U(\alpha)\leq U(x_{0})\leq U(\beta)

for any x0C(α,β)x_{0}\in C\cap(\alpha,\beta). Suppose that there were a point x0C(α,β)x_{0}\in C\cap(\alpha,\beta) such that U(α)>U(x0)U(\alpha)>U(x_{0}) or U(β)<U(x0)U(\beta)<U(x_{0}). We may assume that U(x0)>U(β)U(x_{0})>U(\beta) since the proof for the other case is parallel. By Lemma 3.2, we see that there is x0C(α,β)x_{0}\in C\cap(\alpha,\beta) such that

U(x0)=max[α,β]U.U(x_{0})=\max_{[\alpha,\beta]}U.

We take x1C(x0,β]x_{1}\in C\cap(x_{0},\beta] such that

U(x1)=min[x0,β]U,U(x_{1})=\min_{[x_{0},\beta]}U,

see Figure 4.

Refer to caption
Figure 4.

We now define a truncated function

v(x)={min(U(x),U(x1)),x[α,x1]U(x),x(x1,β].v(x)=\left\{\begin{array}[]{ll}\min\left(U(x),U(x_{1})\right),&x\in[\alpha,x_{1}]\\ U(x),&x\in(x_{1},\beta].\end{array}\right.

We shall prove that TVKg(v)<TVKg(U)TV_{Kg}(v)<TV_{Kg}(U) if ρ=U(x0)U(x1)>0\rho=U(x_{0})-U(x_{1})>0. Since UU and vv are continuous at x=x0x=x_{0} and x1x_{1},

TVK(U)=TVK(U,(α,x0))+TVK(U,(x0,x1))+TVK(U,(x1,β)),\displaystyle TV_{K}(U)=TV_{K}\left(U,(\alpha,x_{0})\right)+TV_{K}\left(U,(x_{0},x_{1})\right)+TV_{K}\left(U,(x_{1},\beta)\right),
TVK(v)=TVK(v,(α,x0))+TVK(v,(x0,x1))+TVK(v,(x,β)).\displaystyle TV_{K}(v)=TV_{K}\left(v,(\alpha,x_{0})\right)+TV_{K}\left(v,(x_{0},x_{1})\right)+TV_{K}\left(v,(x,\beta)\right).

Since TVKTV_{K} does not increase by truncation, we see

TVK(U,(α,x0))TVK(v,(α,x0)).TV_{K}\left(U,(\alpha,x_{0})\right)\geq TV_{K}\left(v,(\alpha,x_{0})\right).

Since v=Uv=U on (x1,β)(x_{1},\beta), we proceed

TVK(U)TVK(v)\displaystyle TV_{K}(U)-TV_{K}(v) TVK(U,(x0,x1))TVK(v,(x0,x1))\displaystyle\geq TV_{K}\left(U,(x_{0},x_{1})\right)-TV_{K}\left(v,(x_{0},x_{1})\right)
=TVK(U,(x0,x1)).\displaystyle=TV_{K}\left(U,(x_{0},x_{1})\right).

By Lemma 3.3, we observe that

TVK(U,(x0,x1))K(ρ)withρ=U(x0)U(x1).TV_{K}\left(U,(x_{0},x_{1})\right)\geq K(\rho)\quad\text{with}\quad\rho=U(x_{0})-U(x_{1}).

Thus by (3.1), we now obtain

TVK(U)TVK(v)>cMρ.TV_{K}(U)-TV_{K}(v)>c_{M}\rho.

We next compare the values (v)\mathcal{F}(v) and (U)\mathcal{F}(U). By definition, we observe

2λ((v)(U))\displaystyle\frac{2}{\lambda}\left(\mathcal{F}(v)-\mathcal{F}(U)\right) =αx1(gv)2𝑑xαx1(gU)2𝑑x\displaystyle=\int_{\alpha}^{x_{1}}(g-v)^{2}\;dx-\int_{\alpha}^{x_{1}}(g-U)^{2}\;dx
αx1ρ(|gv|+|gU|)𝑑x\displaystyle\leq\int_{\alpha}^{x_{1}}\rho\left(|g-v|+|g-U|\right)dx
(βα)ρ2M.\displaystyle\leq(\beta-\alpha)\rho 2M.

The last inequality follows from the fact that the value of UU must be between min[α,β]g\min_{[\alpha,\beta]}g and max[α,β]g\max_{[\alpha,\beta]}g. We thus observe that

TVKg(U)TVKg(v)\displaystyle TV_{Kg}(U)-TV_{Kg}(v) >cMρλ(βα)ρM\displaystyle>c_{M}\rho-\lambda(\beta-\alpha)\rho M
=(cMρλ(βα)M)ρ.\displaystyle=\left(c_{M}\rho-\lambda(\beta-\alpha)M\right)\rho.

If βα\beta-\alpha satisfies βαcM/(λM)\beta-\alpha\leq c_{M}/(\lambda M), UU is not a minimizer provided that ρ>0\rho>0, i.e., U(x0)>U(x1)U(x_{0})>U(x_{1}). Thus, we conclude that U(x0)U(β)U(x_{0})\leq U(\beta).

So far we have proved that UU is non-decreasing on CC. By Lemma 3.2, we conclude that UU itself is non-decreasing in [α,β][\alpha,\beta]. ∎

4. Minimiziers for general one-dimensional data

In this section, we shall prove that a minimizer UU is piecewise linear if KK satisfies (K1), (K3) and (K2) instead of (K2w). In other words, we shall prove our main result.

If we assume (K2), then merging jumps decrease the value TVKTV_{K}. However, \mathcal{F} may increase. We have to estimate an increase of \mathcal{F}.

4.1. Bound for an increase of fidelity

We shall estimate an increase of fidelity \mathcal{F}. We begin with a simple setting. We set

U0γ(x)={g(α),x[α,γ),g(β),x[γ,β]U_{0}^{\gamma}(x)=\left\{\begin{array}[]{ll}g(\alpha),&x\in[\alpha,\gamma),\\ g(\beta),&x\in[\gamma,\beta]\end{array}\right.

for γ(α,β)\gamma\in(\alpha,\beta); see Figure 5.

Refer to caption
Figure 5. profile of U0γU_{0}^{\gamma} and gg

The fidelity of U0γU_{0}^{\gamma} on (α,β)(\alpha,\beta) is still denoted by (U0γ)\mathcal{F}(U_{0}^{\gamma}), i.e.,

F(γ)=2λ(U0γ)=αβ|U0γg|2𝑑xF(\gamma)=\frac{2}{\lambda}\mathcal{F}\left(U_{0}^{\gamma}\right)=\int_{\alpha}^{\beta}\left|U_{0}^{\gamma}-g\right|^{2}dx

for γ[α,β]\gamma\in[\alpha,\beta]. Since we do not assume that gg is non-decreasing, gg may be very large on (α,γ)(\alpha,\gamma). Fortunately, we observe that gg cannot be too large on (α0,γ)(\alpha_{0},\gamma) in the average if F(γ)F(\gamma) is smaller than F(α+0)F(\alpha+0).

Proposition 4.1.

Assume that gC[α,β]g\in C[\alpha,\beta] and U0γ(α)=g(α)U_{0}^{\gamma}(\alpha)=g(\alpha), U0γ(β)=g(β)U_{0}^{\gamma}(\beta)=g(\beta) with ρ=U0γ(β)U0γ(α)>0\rho=U_{0}^{\gamma}(\beta)-U_{0}^{\gamma}(\alpha)>0. If γ[α,β]\gamma\in[\alpha,\beta] satisfies F(α+0)F(γ)F(\alpha+0)\geq F(\gamma), then

αγg(x)𝑑x12(U0γ(α)+U0γ(β))(γα)\int_{\alpha}^{\gamma}g(x)\;dx\leq\frac{1}{2}\left(U_{0}^{\gamma}(\alpha)+U_{0}^{\gamma}(\beta)\right)(\gamma-\alpha)

or

αγ(g(x)U0γ(α))𝑑xρ(γα)/2.\int_{\alpha}^{\gamma}\left(g(x)-U_{0}^{\gamma}(\alpha)\right)dx\leq\rho(\gamma-\alpha)/2.
Proof.

We observe that

F(α+0)F(γ)\displaystyle F(\alpha+0)-F(\gamma) =αγ{(g(β)g(x))2(g(α)g(x))2}𝑑x\displaystyle=\int_{\alpha}^{\gamma}\left\{\left(g(\beta)-g(x)\right)^{2}-\left(g(\alpha)-g(x)\right)^{2}\right\}dx
=αγρ{2g(g(α)+g(β))}𝑑x.\displaystyle=-\int_{\alpha}^{\gamma}\rho\left\{2g-\left(g(\alpha)+g(\beta)\right)\right\}dx.

Since ρ>0\rho>0, F(α+0)F(γ)0F(\alpha+0)-F(\gamma)\geq 0 implies that

αγ2g𝑑xαγ(g(α)+g(β))𝑑x=(U0γ(α)+U0γ(β))(γα).\int_{\alpha}^{\gamma}2g\;dx\leq\int_{\alpha}^{\gamma}\left(g(\alpha)+g(\beta)\right)dx=\left(U_{0}^{\gamma}(\alpha)+U_{0}^{\gamma}(\beta)\right)(\gamma-\alpha).

We give a simple application. See Figure 5.

Lemma 4.2.

Assume the same hypotheses of Proposition 4.1. Then

αγ(U0γρ1g)2𝑑xαγ(U0γg)2𝑑xρ1(ρ1+ρ)(γα)\int_{\alpha}^{\gamma}\left(U_{0}^{\gamma}-\rho_{1}-g\right)^{2}dx-\int_{\alpha}^{\gamma}\left(U_{0}^{\gamma}-g\right)^{2}dx\leq\rho_{1}(\rho_{1}+\rho)(\gamma-\alpha)

for ρ1>0\rho_{1}>0.

Proof.

We may assume that U0γ(α)=0U_{0}^{\gamma}(\alpha)=0 by adding a constant to both U0γU_{0}^{\gamma} amd gg. The left-hand side equals

I=αγ(ρ1+g)2𝑑xαγg2𝑑x=ρ1αγ{ρ1+2g}𝑑x.I=\int_{\alpha}^{\gamma}(\rho_{1}+g)^{2}\;dx-\int_{\alpha}^{\gamma}g^{2}\;dx=\rho_{1}\int_{\alpha}^{\gamma}\{\rho_{1}+2g\}\;dx.

Since

2αγg(x)𝑑xρ(γα)2\int_{\alpha}^{\gamma}g(x)\;dx\leq\rho(\gamma-\alpha)

by Proposition 4.1, we end up with I(ρ12+ρ1ρ)(γα)I\leq(\rho_{1}^{2}+\rho_{1}\rho)(\gamma-\alpha). ∎

Remark 4.3.

In Proposition 4.1 and Lemma 4.2, we do not assume that g<U0γ(β)g<U_{0}^{\gamma}(\beta) on (γ,β)(\gamma,\beta) nor gU0γ(α)g\geq U_{0}^{\gamma}(\alpha) on (α,γ)(\alpha,\gamma).

We next consider behavior of gg between two points of the coincidence set where UU is a constant.

Proposition 4.4.

Assume that KK satisfies (K1), (K2w) and (K3). Assume that gC[a,b]g\in C[a,b]. Let UBV(a,b)U\in BV(a,b) by a minimizer of TVKgTV_{Kg}. Let α,β[a,b)\alpha,\beta\in[a,b) be α<β\alpha<\beta and α,βC\alpha,\beta\in C. Assume that UU is non-decreasing and there is γ[α,β)C\gamma\in[\alpha,\beta)\cap C such that U(γ)=U(α)=g(α)U(\gamma)=U(\alpha)=g(\alpha) and U(x)>U(γ)U(x)>U(\gamma) for x>γx>\gamma. Assume that there is pjU(C)p_{j}\in U(C) such that pj<U(β)p_{j}<U(\beta), pjU(γ)p_{j}\downarrow U(\gamma) as jj\to\infty. Then, αγ(g(x)U(α))𝑑x0\int_{\alpha}^{\gamma}\left(g(x)-U(\alpha)\right)dx\leq 0.

Proof.

We may assume that U(α)=g(α)=0U(\alpha)=g(\alpha)=0 so that limjpj=0\lim_{j\to\infty}p_{j}=0. For j2j\geq 2, we set

vj(x)={max(pj,U(x)),αj:=α+(γα)/j<xU(x),xαj;v_{j}(x)=\left\{\begin{array}[]{ll}\max\left(p_{j},U(x)\right),&\alpha_{j}:=\alpha+(\gamma-\alpha)/j<x\\ U(x),&x\leq\alpha_{j};\end{array}\right.

see Figure 6.

Refer to caption
Figure 6. vjv_{j} and UU

Since UU is a minimizer, by definition,

TVKg(vj)TVKg(U).TV_{Kg}(v_{j})\geq TV_{Kg}(U).

By Lemma 3.3,

TVK(U,(α,β))\displaystyle TV_{K}\left(U,(\alpha,\beta)\right) TVK(U,(α,γ))\displaystyle\geq TV_{K}\left(U,(\alpha,\gamma)\right)
+TVK(U,(γ,qj))+TVK(U,(qj,β))\displaystyle+TV_{K}\left(U,(\gamma,q_{j})\right)+TV_{K}\left(U,(q_{j},\beta)\right)
0+K(pj)+TVK(U,(qj,β))\displaystyle\geq 0+K(p_{j})+TV_{K}\left(U,(q_{j},\beta)\right)

where qjU1(pj)q_{j}\in U^{-1}(p_{j}). Since

TVK(vj,(α,β))=K(pj)+TVK(U,(qj,b)),TV_{K}\left(v_{j},(\alpha,\beta)\right)=K(p_{j})+TV_{K}\left(U,(q_{j},b)\right),

TVKg(vj)TVKg(U)TV_{Kg}(v_{j})\geq TV_{Kg}(U) implies that (vj)(U)\mathcal{F}(v_{j})\geq\mathcal{F}(U). In other words,

αqj{(vjg)2(Ug)2}𝑑x0.\int_{\alpha}^{q_{j}}\left\{(v_{j}-g)^{2}-(U-g)^{2}\right\}dx\geq 0.

Dividing the region of integration (α,qj)(\alpha,q_{j}) by (α,γ)(\alpha,\gamma) and (γ,qj)(\gamma,q_{j}), we obtain

αjγ{g2(pjg)2}𝑑xγqj{(pjg)2(Ug)2}𝑑x,\int_{\alpha_{j}}^{\gamma}\left\{g^{2}-(p_{j}-g)^{2}\right\}dx\leq\int_{\gamma}^{q_{j}}\left\{(p_{j}-g)^{2}-(U-g)^{2}\right\}dx,

or

pjαjγ(2gpj)𝑑xγqj(pjU)(pj+U2g)𝑑x.p_{j}\int_{\alpha_{j}}^{\gamma}(2g-p_{j})\;dx\leq\int_{\gamma}^{q_{j}}(p_{j}-U)(p_{j}+U-2g)\;dx.

Since UpjU\leq p_{j} on (γ,qj)(\gamma,q_{j}), the right-hand side is dominated by

pjγqj|pj+U2g|𝑑xpj|qjγ|(2pj+2g).p_{j}\int_{\gamma}^{q_{j}}|p_{j}+U-2g|\;dx\leq p_{j}|q_{j}-\gamma|\left(2p_{j}+2\|g\|_{\infty}\right).

Thus

αjγ(2gpj)𝑑x|qjγ|(2pj+2g).\int_{\alpha_{j}}^{\gamma}(2g-p_{j})\;dx\leq|q_{j}-\gamma|\left(2p_{j}+2\|g\|_{\infty}\right).

Sending jj\to\infty yields that

αγg𝑑x0,\int_{\alpha}^{\gamma}g\;dx\leq 0,

since qjγq_{j}\to\gamma by our assumption that U(x)>0U(x)>0 for x>γx>\gamma and UU is non-decreasing. The proof is now complete. ∎

We say that a closed interval FF is a facet of UU if FF is a maximal nontrivial closed interval such that UU is a constant on the interior intF\operatorname{int}F of FF. Let |F||F| denote its length. We are able to claim a similar statement for each facet of a minimizer UU.

Lemma 4.5.

Assume that KK satisfies (K1), (K2w) and (K3). Assume that gC[a,b]g\in C[a,b]. Let UBV(a,b)U\in BV(a,b) be a minimizer of TVKgTV_{Kg}. Assume that UU is non-decreasing. Let F=[x0,x1]F=[x_{0},x_{1}] with x1<bx_{1}<b be a facet of UU. Then

F(g(x)U)𝑑x(U(x1+0)U(x10))|F|/2.\int_{F}\left(g(x)-U\right)dx\leq\left(U(x_{1}+0)-U(x_{1}-0)\right)|F|/2. (4.1)
Proof.

We may assume U0U\equiv 0 on FF. By Lemma 3.1, FCF\cap C\neq\emptyset. We set

α=inf(FC)\alpha=\inf(F\cap C)

which is still in CC since CC is closed. By Lemma 3.1,

g(x)<U(x)=0on[x0,α)g(x)<U(x)=0\quad\text{on}\quad[x_{0},\alpha)

since UU is non-increasing. If U(x1+0)=U(x10)U(x_{1}+0)=U(x_{1}-0), then x1Cx_{1}\in C and UU is continuous by Lemma 3.2. Thus by Proposition 4.4,

αx1g(x)𝑑x0.\int_{\alpha}^{x_{1}}g(x)\;dx\leq 0.

Thus, we obtain (4.1) when UU does not jump at x1x_{1} since we know g<0g<0 on [x0,α)[x_{0},\alpha).

If U(x1+0)U(x10)>0U(x_{1}+0)-U(x_{1}-0)>0, we may apply Proposition 4.1 and conclude that

αx1g𝑑xU(x1+0)(x1α)/2.\int_{\alpha}^{x_{1}}g\;dx\leq U(x_{1}+0)(x_{1}-\alpha)/2.

Again by g<0g<0 on [x0,α)[x_{0},\alpha), we conclude that

αx1g𝑑xU(x1+0)(x1α)/2U(x1+0)|F|/2.\int_{\alpha}^{x_{1}}g\;dx\leq U(x_{1}+0)(x_{1}-\alpha)/2\leq U(x_{1}+0)|F|/2.

Since we know that g<0g<0 on [x0,α)[x_{0},\alpha), the proof of Lemma 4.5 is now complete. ∎

Theorem 4.6.

Assume that KK satisfies (K1), (K2w) and (K3). Assume that gC[a,b]g\in C[a,b]. Let UBV(a,b)U\in BV(a,b) be a minimizer of TVKgTV_{Kg}. Assume that UU is non-decreasing. Let α,βC\alpha,\beta\in C with α<β<b\alpha<\beta<b. Then

αβ(U(α)g)2𝑑xαβ(Ug)2𝑑xρ2(βα)\int_{\alpha}^{\beta}\left(U(\alpha)-g\right)^{2}dx-\int_{\alpha}^{\beta}(U-g)^{2}\;dx\leq\rho^{2}(\beta-\alpha)

when ρ=U(β)U(α)0\rho=U(\beta)-U(\alpha)\geq 0.

Proof.

If ρ=0\rho=0, a minimizer must be constant U(α)U(\alpha) so the above inequality is trivially fulfilled. We may assume ρ>0\rho>0.

As before, we may assume U(α)=0U(\alpha)=0 so that U(x)0U(x)\geq 0. We proceed

αβg2𝑑xαβ(Ug)2𝑑x=αβU(2gU)𝑑x.\int_{\alpha}^{\beta}g^{2}\;dx-\int_{\alpha}^{\beta}(U-g)^{2}\;dx=\int_{\alpha}^{\beta}U(2g-U)\;dx.

On the coincidence set CC,

CU(2gU)𝑑x=CU2\int_{C}U(2g-U)\;dx=\int_{C}U^{2}

since g=Ug=U on CC. On a facet F=[x0,x1]F=[x_{0},x_{1}], by Lemma 4.5,

FU(2gU)𝑑x\displaystyle\int_{F}U(2g-U)\;dx =U(x10)F{(2g2U(x10))+U(x10)}𝑑x\displaystyle=U(x_{1}-0)\int_{F}\left\{\left(2g-2U(x_{1}-0)\right)+U(x_{1}-0)\right\}dx
U(x10)((U(x1+0)U(x10))+U(x10))|F|\displaystyle\leq U(x_{1}-0)\left(\left(U(x_{1}+0)-U(x_{1}-0)\right)+U(x_{1}-0)\right)|F|
=U(x10)(U(x1+0)U(x10))|F|\displaystyle=U(x_{1}-0)\left(U(x_{1}+0)-U(x_{1}-0)\right)|F|
ρ2|F|\displaystyle\leq\rho^{2}|F|

since x1β<bx_{1}\leq\beta<b. We now conclude that

αβU(2gU)𝑑x\displaystyle\int_{\alpha}^{\beta}U(2g-U)\;dx i=1FiU(2gU)𝑑x+CU(2gU)𝑑x\displaystyle\leq\sum_{i=1}^{\infty}\int_{F_{i}}U(2g-U)\;dx+\int_{C}U(2g-U)\;dx
ρ2i=1|Fi|+CU2𝑑x\displaystyle\leq\rho^{2}\sum_{i=1}^{\infty}|F_{i}|+\int_{C}U^{2}\;dx
ρ2i=1|Fi|+ρ2|C|\displaystyle\leq\rho^{2}\sum_{i=1}^{\infty}|F_{i}|+\rho^{2}|C|

Since (α,β)=i=1FiC(\alpha,\beta)=\bigcup_{i=1}^{\infty}F_{i}\cup C with at most countably many facets {Fi}\{F_{i}\}, this implies that

αβg2𝑑xαβ(Ug)2𝑑xρ2(i=1|Fi|+|C|)=ρ2(βα).\int_{\alpha}^{\beta}g^{2}\;dx-\int_{\alpha}^{\beta}(U-g)^{2}\;dx\leq\rho^{2}\left(\sum_{i=1}^{\infty}|F_{i}|+|C|\right)=\rho^{2}(\beta-\alpha).

4.2. No possibility of fine structure

Our goal in this subsection is to prove our main theorem. In other words, we shall prove that a minimizer does not allow to have a “fine” structure under the assumption (K2). At the end of this subsolution, we prove our main theorem Theorem 1.1.

Lemma 4.7.

Assume that KK satisfies (K1), (K2) and (K3). Assume that gC[a,b]g\in C[a,b]. Let UBV(a,b)U\in BV(a,b) be a minimizer of TVKgTV_{Kg}. Assume that UU is non-decreasing. Let α,β,γC\alpha,\beta,\gamma\in C satisfy aα<γ<βba\leq\alpha<\gamma<\beta\leq b and U(β)U(γ)U(γ)U(α)>0U(\beta)-U(\gamma)\geq U(\gamma)-U(\alpha)>0. Assume that (γ,β)C=(\gamma,\beta)\cap C=\emptyset. Then

βα>2CM/λ\beta-\alpha>2C_{M}/\lambda

for Mosc[α,β]gM\geq\operatorname{osc}_{[\alpha,\beta]}g, where CMC_{M} is the constant in (K2).

Proof.

We set ρ1=U(γ)U(α)\rho_{1}=U(\gamma)-U(\alpha), ρ2=U(β)U(γ)\rho_{2}=U(\beta)-U(\gamma). Since (γ,β)C=(\gamma,\beta)\cap C=\emptyset, by Lemma 3.2, UU has only one jump at x1(γ,β)x_{1}\in(\gamma,\beta) and U=U(β)U=U(\beta) for x(x1,β)x\in(x_{1},\beta), U=U(γ)U=U(\gamma) for x(γ,x1)x\in(\gamma,x_{1}).

We set

v(x)={U(α),x(α,x1),U(x),x(α,x1)v(x)=\left\{\begin{array}[]{ll}U(\alpha),&x\in(\alpha,x_{1}),\\ U(x),&x\notin(\alpha,x_{1})\end{array}\right.

and compare TVK(v)TV_{K}(v) with TVK(U)TV_{K}(U) on (α,β)(\alpha,\beta); see Figure 7.

Refer to caption
Figure 7. the graph of UU, gg and vv

Then

TVK(U,(α,β))\displaystyle TV_{K}\left(U,(\alpha,\beta)\right) =K(ρ2)+TVK(U,(α,γ))\displaystyle=K(\rho_{2})+TV_{K}\left(U,(\alpha,\gamma)\right)
K(ρ2)+K(ρ1)\displaystyle\geq K(\rho_{2})+K(\rho_{1})

by Lemma 3.3. Clearly,

TVK(v,(α,β))=K(ρ1+ρ2).TV_{K}\left(v,(\alpha,\beta)\right)=K(\rho_{1}+\rho_{2}).

By (K2), we see that

TVK(U,(α,β))TVK(v,(α,β))\displaystyle TV_{K}\left(U,(\alpha,\beta)\right)-TV_{K}\left(v,(\alpha,\beta)\right) K(ρ1)+K(ρ2)K(ρ1+ρ2)\displaystyle\geq K(\rho_{1})+K(\rho_{2})-K(\rho_{1}+\rho_{2})
CMρ1ρ2.\displaystyle\geq C_{M}\rho_{1}\rho_{2}.

Since UU minimizes γβ|Ug|2𝑑x\int_{\gamma}^{\beta}|U-g|^{2}\;dx, we have, by Lemma 4.2,

γx1{(Ug)2(vg)2}𝑑xρ1(ρ1+ρ2)(x1γ).\int_{\gamma}^{x_{1}}\left\{(U-g)^{2}-(v-g)^{2}\right\}dx\geq-\rho_{1}(\rho_{1}+\rho_{2})(x_{1}-\gamma).

Since γ<x1<b\gamma<x_{1}<b and UU is a minimizer of TVKgTV_{Kg} on (α,x1)(\alpha,x_{1}), we have, by Theorem 4.6,

αγ{(Ug)2(vg)2}𝑑xρ12(γα).\int_{\alpha}^{\gamma}\left\{(U-g)^{2}-(v-g)^{2}\right\}dx\geq-\rho_{1}^{2}(\gamma-\alpha).

We now conclude that

TVKg(U)TVKg(v)CMρ1ρ2(ρ1ρ2(x1γ)+ρ12(γα))λ/2.TV_{Kg}(U)-TV_{Kg}(v)\geq C_{M}\rho_{1}\rho_{2}-\left(\rho_{1}\rho_{2}(x_{1}-\gamma)+\rho_{1}^{2}(\gamma-\alpha)\right)\lambda\bigm{/}2.

Since we assume that ρ1ρ2\rho_{1}\leq\rho_{2}, this implies that

TVKg(U)TVKg(v)ρ1ρ2(CM(x1α)λ/2).TV_{Kg}(U)-TV_{Kg}(v)\geq\rho_{1}\rho_{2}\left(C_{M}-(x_{1}-\alpha)\lambda/2\right).

Since UU is a minimizer, CM(x1α)λ/20C_{M}-(x_{1}-\alpha)\lambda/2\leq 0, we conclude that

βαx1α2CM/λ.\beta-\alpha\geq x_{1}-\alpha\geq 2C_{M}/\lambda.

From the proof of Lemma 4.7, we have a rather general estimate for U0γU_{0}^{\gamma} defined at the beginning of Section 4.1.

Lemma 4.8.

Assume that KK satisfies (K1), (K2) and (K3). Assume that gC[α,β]g\in C[\alpha,\beta] and Mosg[α,β]gM\geq\operatorname{osg}_{[\alpha,\beta]}g. Assume that ρ=g(β)g(α)>0\rho=g(\beta)-g(\alpha)>0 and x1(α,β)x_{1}\in(\alpha,\beta). Let UU be a non-decreasing function with U(α)=g(α)U(\alpha)=g(\alpha), U(β)=g(β)U(\beta)=g(\beta) which is continuous at α\alpha and β\beta. Assume that (α,β)=i=1FiC(\alpha,\beta)=\bigcup_{i=1}^{\infty}F_{i}\cup C where FiF_{i} is a facet and CC is a coincidence set with F1=[x1,β]F_{1}=[x_{1},\beta] and F2=[x0,x1]F_{2}=[x_{0},x_{1}], x0(α,x1)x_{0}\in(\alpha,x_{1}). (The set FiF_{i} for i3i\geq 3 could be empty.) Assume that U(β0)=g(β)U(\beta-0)=g(\beta) and g(β)U(x10):=ρ2>ρ/2g(\beta)-U(x_{1}-0):=\rho_{2}>\rho/2. Assume further that UU satisfies (4.1) on each FiF_{i} for i3i\geq 3. Assume that F(x1)F(α+0)F(x_{1})\leq F(\alpha+0) for U0x1U_{0}^{x_{1}}. If CC contains an interior point of F2F_{2}, then

TVKg(U)TVKg(U0x1)ρ1ρ2(CM(x1α)λ/2),TV_{Kg}(U)-TV_{Kg}(U_{0}^{x_{1}})\geq\rho_{1}\rho_{2}\left(C_{M}-(x_{1}-\alpha)\lambda/2\right),

with ρ1=ρρ2\rho_{1}=\rho-\rho_{2}.

We are interested in the case that there are no jumps. We begin with an elementary property of KK.

Lemma 4.9.

Assume that KK satisfies (K2) and (K3). Then, ρK(ρ)CMρ2/2\rho-K(\rho)\geq C_{M}\rho^{2}/2.

Proof.

An iterative use of (K2) yields

K(ρ)\displaystyle K(\rho) 2K(ρ2)CM(ρ2)2\displaystyle\leq 2K\left(\frac{\rho}{2}\right)-C_{M}\left(\frac{\rho}{2}\right)^{2}
2(2K(ρ4)CM(ρ4)2)CM(ρ2)2\displaystyle\leq 2\left(2K\left(\frac{\rho}{4}\right)-C_{M}\left(\frac{\rho}{4}\right)^{2}\right)-C_{M}\left(\frac{\rho}{2}\right)^{2}
2mK(ρ2m)CMρ2j=1m2j1(12j)2\displaystyle\leq 2^{m}K\left(\frac{\rho}{2^{m}}\right)-C_{M}\rho^{2}\sum_{j=1}^{m}2^{j-1}\left(\frac{1}{2^{j}}\right)^{2}

for m=1,2,m=1,2,\ldots. Since j=1m2j11/2\sum_{j=1}^{m}2^{-j-1}\to 1/2 as mm\to\infty, sending mm\to\infty yields

K(ρ)ρCMρ2/2K(\rho)\leq\rho-C_{M}\rho^{2}/2

by (K3). The proof is now complete. ∎

Lemma 4.10.

Assume the same hypotheses of Lemma 4.7 concerning KK and gg. Let UBV(a,b)U\in BV(a,b) be a minimizer of TVKgTV_{Kg}. Assume that UU is non-decreasing and continuous on [α,β][a,b][\alpha,\beta]\subset[a,b] with α,βC\alpha,\beta\in C. Then U(α)=U(β)U(\alpha)=U(\beta).

Proof.

Suppose that U(α)U(β)U(\alpha)\neq U(\beta) so that U(α)<U(β)U(\alpha)<U(\beta), there would exist at least one p(U(α),U(β))p\in\left(U(\alpha),U(\beta)\right) such that U1(p)U^{-1}(p) is a singleton {x0}\{x_{0}\} since UU is continuous. By Lemma 3.1, x0Cx_{0}\in C. Moreover, there is a sequence pjpp_{j}\downarrow p (jj\to\infty) such that U1(pj)U^{-1}(p_{j}) is a singleton {xj}\{x_{j}\}. This is possible since the set of values qq where U1(q)U^{-1}(q) is not a singleton is at most a countable set. Since U1(p)U^{-1}(p) is a singleton, xjx0x_{j}\downarrow x_{0}. Again by Lemma 3.1, xjCx_{j}\in C. We set

vj(x)={p,x(x0,xj),U(x),x(x0,xj).v_{j}(x)=\left\{\begin{array}[]{ll}p,&x\in(x_{0},x_{j}),\\ U(x),&x\notin(x_{0},x_{j}).\end{array}\right.

By Theorem 4.6, we obtain

2λ((vj)(U))ρj2(xjx0)\frac{2}{\lambda}\left(\mathcal{F}(v_{j})-\mathcal{F}(U)\right)\leq\rho_{j}^{2}(x_{j}-x_{0})

with ρj=pjp\rho_{j}=p_{j}-p.

Since TV=TVKTV=TV_{K} for a continuous function, we see that

TVK(U,(x0,xj))=ρj.TV_{K}\left(U,(x_{0},x_{j})\right)=\rho_{j}.

By Lemma 4.9, we observe that

TVK(U)TVK(vj)=ρjK(ρj)CMρj2/2,M>osc[α,β]g.TV_{K}(U)-TV_{K}(v_{j})=\rho_{j}-K(\rho_{j})\geq C_{M}\rho_{j}^{2}/2,\quad M>\operatorname{osc}_{[\alpha,\beta]}g.

We thus conclude that

TVKg(U)TVKg(vj)\displaystyle TV_{Kg}(U)-TV_{Kg}(v_{j}) (CMρj2λρj2(xjx0))/2\displaystyle\geq\left(C_{M}\rho_{j}^{2}-\lambda\rho_{j}^{2}(x_{j}-x_{0})\right)/2
=ρj2(CMλ(xjx0))/2.\displaystyle=\rho_{j}^{2}\left(C_{M}-\lambda(x_{j}-x_{0})\right)/2.

For a sufficiently large jj, CMλ(xjx0)>0C_{M}-\lambda(x_{j}-x_{0})>0 since xjx0x_{j}\downarrow x_{0}. This would contradict to our assumption that UU is a minimizer. Thus U(α)=U(β)U(\alpha)=U(\beta). ∎

Theorem 4.11.

Assume that KK satisfies (K1), (K2) and (K3). Assume that gC[a,b]g\in C[a,b]. Let UBV(a,b)U\in BV(a,b) be a minimizer of TVKgTV_{Kg}. Let α,βC\alpha,\beta\in C with aα<βba\leq\alpha<\beta\leq b. Assume that βαAM/λ\beta-\alpha\leq A_{M}/\lambda with AM=min{cM/M,2CM}A_{M}=\min\{c_{M}/M,2C_{M}\} and osc[α,β]gM\operatorname{osc}_{[\alpha,\beta]}g\leq M, where cMc_{M} is in (3.1) and CMC_{M} is in (K2). Then UU takes either U(α)U(\alpha) or U(β)U(\beta) on [α,β][\alpha,\beta] and it has at most one jump point in (α,β)(\alpha,\beta).

Proof.

By Theorem 3.4, UU is non-decreasing in [α,β][\alpha,\beta]. We may assume U(α)<U(β)U(\alpha)<U(\beta). If there is no jump, i.e., UC[α,β]U\in C[\alpha,\beta], by Lemma 4.10, UU is not a minimizer so UU must have at least one jump point in (α,β)(\alpha,\beta). We take a jump point x0x_{0} such that jump size

U(x0+0)U(x00)(>0)U(x_{0}+0)-U(x_{0}-0)\ (>0)

is maximum among all jump size of UU in (α,β)(\alpha,\beta). If U(x00)=U(α)U(x_{0}-0)=U(\alpha), U(x0+0)=U(β)U(x_{0}+0)=U(\beta), we get the conclusion. Suppose that U(x00)>U(α)U(x_{0}-0)>U(\alpha). We set

β=inf{x(x0,β]|xC},\displaystyle\beta^{\prime}=\inf\left\{x\in(x_{0},\beta]\bigm{|}x\in C\right\},
γ=sup{x(α,x0]|xC}.\displaystyle\gamma=\sup\left\{x\in(\alpha,x_{0}]\bigm{|}x\in C\right\}.

By Lemma 3.1, β>x0\beta^{\prime}>x_{0} and γ<x0\gamma<x_{0}. Since U(x00)>U(α)U(x_{0}-0)>U(\alpha), we see γ>α\gamma>\alpha. Since the jump at x0x_{0} is a maximal jump, we apply Lemma 4.7 on (α,β)(\alpha,\beta^{\prime}) to get

βα>2CM/λ,\beta^{\prime}-\alpha>2C_{M}/\lambda,

which would contradict the assumption βα<AM/λ\beta-\alpha<A_{M}/\lambda. We thus conclude that U(x00)=U(α)U(x_{0}-0)=U(\alpha). If U(x0+0)<U(β)U(x_{0}+0)<U(\beta), we consider U(x)-U(-x) instead of UU. We argue in the same way. We apply Lemma 4.7 and get a contradiction. We thus conclude that U(x0+0)=U(β)U(x_{0}+0)=U(\beta). The proof is now complete. ∎

Proof of Theorem 1.1.

By Lemma 3.1, infgUsupg\inf g\leq U\leq\sup g on [a,b][a,b] and CC is non-empty. We take an integer mm so that

m>(ba)λ/AM.m>(b-a)\lambda/A_{M}.

We divide [a,b][a,b] into mm intervals so that the length of each interval is less than AM/λA_{M}/\lambda and the boundaries of each interval [x0,x1][x_{0},x_{1}] does not contain a jump point of UU. (This is possible if we shift x0,x1x_{0},x_{1} a little bit unless x0=ax_{0}=a^{\prime}, or x1=bx_{1}=b^{\prime} since JUJ_{U} is at most a countable set. If x0=ax_{0}=a^{\prime} (resp. xbx-b^{\prime}), UU must be continuous at x0=ax_{0}=a^{\prime} (x1=bx_{1}=b^{\prime}) since UU is continuous on the coincidence set CC by Lemma 3.2.) If C[x0,x1]C\cap[x_{0},x_{1}] is empty or singleton, then UU must be a constant on [x0,x1][x_{0},x_{1}] by Lemma 3.1. We next consider the case that C[x0,x1]C\cap[x_{0},x_{1}] has at least two points. We set

α=inf(C[x0,x1]),β=sup(C[x0,x1])\alpha^{\prime}=\inf\left(C\cap[x_{0},x_{1}]\right),\quad\beta^{\prime}=\sup\left(C\cap[x_{0},x_{1}]\right)

and may assume α<β\alpha^{\prime}<\beta^{\prime}. Since βα<AM/λ\beta^{\prime}-\alpha^{\prime}<A_{M}/\lambda, Theorem 4.11 implies that UU only takes two values U(x0)U(x_{0}) and U(x1)U(x_{1}) and has at most one jump on (α,β)(\alpha^{\prime},\beta^{\prime}). By Lemma 3.1, UU is constant on [x0,α][x_{0},\alpha^{\prime}] and [β,x1][\beta^{\prime},x_{1}].

We now observe that on each [x0,x1][x_{0},x_{1}], UU has at most one jump. We thus conclude that UU is a piecewise constant function with at most mm jumps on (a,b)(a,b). ∎

4.3. Minimizers for monotone data

We shall prove that the bound for number of jumps is improved when gg is monotone. In other words, we shall prove Theorem 1.2.

We first observe the monotonicity of a minimizer for TVKgTV_{Kg} when gg is monotone.

Lemma 4.12.

Assume that KK satifies (K1) and that gC[a,b]g\in C[a,b] is non-decreasing. Then a minimizer of TVKgTV_{Kg} (in BV(Ω)BV(\Omega)) is non-decreasing.

Proof.

Let UU be a minimizer. We take its right continuous representation. Suppose that U(x0)>U(y0)g(y0)U(x_{0})>U(y_{0})\geq g(y_{0}) with some x0<y0x_{0}<y_{0}. Then a chopped function

v(x)=min(U(x0),U(x))v(x)=\min\left(U(x_{0}),U(x)\right)

decreases both TVKTV_{K} and the fidelity term so that

TVKg(v)<TVKg(U);TV_{Kg}(v)<TV_{Kg}(U);

see Figure 8.

Refer to caption
Figure 8. chopped function

Thus for UgU\geq g, UU is always non-decreasing. A symmetric argument implies that UU is always non-decreasing for UgU\leq g. If UU is not non-decreasing in (a,b)(a,b), then it must jump at some x0x_{0} with

U(x0+0)g(x0)U(x00),U(x00)U(x0+0).U(x_{0}+0)\leq g(x_{0})\leq U(x_{0}-0),\quad U(x_{0}-0)\neq U(x_{0}+0).

Our competitor can be taken as

v(x)={min{U(x),g(x0)}forx<x0max{U(x),g(x0)}forx>x0.v(x)=\left\{\begin{array}[]{lll}\min\left\{U(x),g(x_{0})\right\}&\text{for}&x<x_{0}\\ \max\left\{U(x),g(x_{0})\right\}&\text{for}&x>x_{0}.\end{array}\right.

By definition, TVKg(v)<TVKg(U)TV_{Kg}(v)<TV_{Kg}(U) so such jump cannot occur for a minimizer. Thus UU is non-decreasing. ∎

Proof of Theorem 1.2.

If gg is monotone, a minimizer of TVKgTV_{Kg} is automatically monotone by Lemma 4.12. We don’t need to invoke Theorem 3.4 so the bound cM/Mc_{M}/M is unnecessary. Thus Theorem 1.2 follows from Theorem 1.1. ∎

It is not difficult to get a minimizer when gg is strictly increasing for TVgTV_{g}, i.e.,

TVg(u)=TV(u)+(u).TV_{g}(u)=TV(u)+\mathcal{F}(u).

By Lemma 4.12, a minimizer UU must be non-decreasing. (In this problem, TVgTV_{g} is strictly convex and lower semicontinuous in L2(Ω)L^{2}(\Omega), so there exists a unique minimizer.) We note that

TV(u)=u(b)u(a)TV(u)=u(b)-u(a)

provided that uu is non-decreasing. We set d1=u(a)g(a)d_{1}=u(a)-g(a), d2=g(b)u(b)d_{2}=g(b)-u(b) and a1=g1(u(a))a_{1}=g^{-1}\left(u(a)\right), a2=g1(u(b))a_{2}=g^{-1}\left(u(b)\right). To minimize (u)\mathcal{F}(u), we take d1d_{1} and d2d_{2} such that

d1=λ2aa1|u(a)g|2𝑑x\displaystyle d_{1}=\frac{\lambda}{2}\int_{a}^{a_{1}}\left|u(a)-g\right|^{2}dx
d2=λ2a2a|u(b)g|2𝑑x.\displaystyle d_{2}=\frac{\lambda}{2}\int_{a_{2}}^{a}\left|u(b)-g\right|^{2}dx.

See Figure 9.

Refer to caption
Figure 9. near the boundary

Then the minimizer UU must be

U(x)=min{max(g(a)+d1,u(x)),g(b)d2}U(x)=\min\left\{\max\left(g(a)+d_{1},u(x)\right),g(b)-d_{2}\right\}

provided that g(a)+d1g(b)d2g(a)+d_{1}\leq g(b)-d_{2}. This formation of a flat part near the boundary occurs by the natural boundary condition. (In general, the minimizer has no jumps if gg is continuous for TVgTV_{g} (cf. [CL], [GKL]).)

In the case when gg is monotone, we are able to prove a conclusion stronger than Proposition 4.1.

Proposition 4.13.

Assume that gC[α,β]g\in C[\alpha,\beta] and U0γ(α)=g(α)U_{0}^{\gamma}(\alpha)=g(\alpha), U0γ(β)=g(β)U_{0}^{\gamma}(\beta)=g(\beta) with ρ=U(β)U(α)>0\rho=U(\beta)-U(\alpha)>0. If γ[α,β]\gamma\in[\alpha,\beta] satisfies F(α+0)F(γ)F(\alpha+0)\geq F(\gamma), then g(x)U0γ(α)ρ/2g(x)-U_{0}^{\gamma}(\alpha)\leq\rho/2 for x[α,γ]x\in[\alpha,\gamma].

This easily follows from the next observation.

Proposition 4.14.

Assume that gC[α,β]g\in C[\alpha,\beta] is non-decreasing. Then F(γ)F(\gamma) is minimized if and only if g(γ)=(g(α)+g(β))/2g(\gamma)=\left(g(\alpha)+g(\beta)\right)/2.

Proof.

A direct calculation shows that

F(γ)\displaystyle F(\gamma) =αβ|U0γg|2𝑑x\displaystyle=\int_{\alpha}^{\beta}|U_{0}^{\gamma}-g|^{2}\;dx
=αγ|g(x)g(α)|2𝑑x+γβ|g(β)g(x)|2𝑑x.\displaystyle=\int_{\alpha}^{\gamma}\left|g(x)-g(\alpha)\right|^{2}dx+\int_{\gamma}^{\beta}\left|g(\beta)-g(x)\right|^{2}dx.

Then

F(γ)\displaystyle F^{\prime}(\gamma) =|g(γ)g(α)|2|g(β)g(γ)|2\displaystyle=\left|g(\gamma)-g(\alpha)\right|^{2}-\left|g(\beta)-g(\gamma)\right|^{2}
=(g(α)+g(α)2g(γ))(g(β)g(α)).\displaystyle=\left(g(\alpha)+g(\alpha)-2g(\gamma)\right)\left(g(\beta)-g(\alpha)\right).

Thus, FF takes its only minimum at (g(α)+g(β))/2\left(g(\alpha)+g(\beta)\right)/2. ∎

We give a few explicit estimate of TVKgTV_{Kg} for monotone function gg to give an alternate proof for Theorem 1.2.

We recall U0γU_{0}^{\gamma}. We shall simply write U0γU_{0}^{\gamma} by U0U_{0} if g(γ)=(g(α)+g(β))/2g(\gamma)=\left(g(\alpha)+g(\beta)\right)/2. Note that such γ\gamma always exists since gg is continuous. It is unique if gg is increasing. In general, it is not unique and we choose one of such γ\gamma. We next consider a non-decreasing piecewise constant function with two jumps. We set ρ=g(β)g(α)\rho=g(\beta)-g(\alpha) and δ(0,1)\delta\in(0,1) and define

Uδ(x)={g(α),x[α,x1)g(α)+δρ,x[x1,x2)g(β),x[x2,β)U_{\delta}(x)=\left\{\begin{array}[]{ll}g(\alpha),&x\in[\alpha,x_{1})\\ g(\alpha)+\delta\rho,&x\in[x_{1},x_{2})\\ g(\beta),&x\in[x_{2},\beta)\end{array}\right.

for x1<x2x_{1}<x_{2} with x1,x2(α,β)x_{1},x_{2}\in(\alpha,\beta). If (Uδ)\mathcal{F}(U_{\delta}) is minimized by moving x1x_{1} and x2x_{2} by Lemma 3.1, there is a point y0(x1,x2)y_{0}\in(x_{1},x_{2}) such that Uδ(y0)=g(α)+δρU_{\delta}(y_{0})=g(\alpha)+\delta\rho. Moreover, by Proposition 4.14, (Uδ)\mathcal{F}(U_{\delta}) is minimized by taking x1=x1x_{1}=x_{1}^{*}, x2=x2x_{2}=x_{2}^{*}, where x1x_{1}^{*} and x2x_{2}^{*} are defined by

g(x1)=g(α)+δρ/2,g(x2)=(g(β)+g(α)+δρ)/2=g(β)(1δ)ρ/2,g(x_{1}^{*})=g(\alpha)+\delta\rho/2,\quad g(x_{2}^{*})=\left(g(\beta)+g(\alpha)+\delta\rho\right)/2=g(\beta)-(1-\delta)\rho/2,

see Figure 10.

Refer to caption
Figure 10. U0U_{0} and UδU_{\delta}

Again x1x_{1}^{*}, x2x_{2}^{*} may not be unique. We take one of them.

We shall estimate

TVKg(Uδ)TVKg(U0).TV_{Kg}(U_{\delta})-TV_{Kg}(U_{0}).

This can be considered as a special case of Lemma 4.8 since (4.1) is automatically fulfilled by Proposition 4.13. However, the proof is clearer than that of general gg, we give a full proof.

Lemma 4.15.

Assume that gC[α,β]g\in C[\alpha,\beta] is increasing. Let vv be a non-decreasing piecewise constant function with three values g(α)g(\alpha), g(α)+δρg(\alpha)+\delta\rho, g(β)g(\beta) for δ(0,1)\delta\in(0,1), where ρ=g(β)g(α)\rho=g(\beta)-g(\alpha). Then

αβ(U0g)2𝑑xαβ(vg)2𝑑xδ(1δ)ρ2(βα).\int_{\alpha}^{\beta}(U_{0}-g)^{2}\;dx-\int_{\alpha}^{\beta}(v-g)^{2}\;dx\leq\delta(1-\delta)\rho^{2}(\beta-\alpha).
Proof.

We may assume δ1/2\delta\leq 1/2 by symmetry. For UδU_{\delta}, we fix x1=x1x_{1}=x_{1}^{*} and x2=x2x_{2}=x_{2}^{*}. We first observe that x1<x0<x2x_{1}^{*}<x_{0}<x_{2}^{*}. We calculate

αβ|Uδg|2𝑑x=\displaystyle\int_{\alpha}^{\beta}|U_{\delta}-g|^{2}dx= αx1|g(x)g(α)|2𝑑x+x1x2|g(x)g(α)δρ|2𝑑x\displaystyle\int_{\alpha}^{x_{1}^{*}}\left|g(x)-g(\alpha)\right|^{2}dx+\int_{x_{1}^{*}}^{x_{2}^{*}}\left|g(x)-g(\alpha)-\delta\rho\right|^{2}dx
+x2β|g(β)g(x)|2𝑑x.\displaystyle+\int_{x_{2}^{*}}^{\beta}\left|g(\beta)-g(x)\right|^{2}dx.

We shall estimate the second term of the right-hand side by dividing

x1x2=x1x0+x0x2=I1+I2.\int_{x_{1}^{*}}^{x_{2}^{*}}=\int_{x_{1}^{*}}^{x_{0}}+\int_{x_{0}}^{x_{2}^{*}}=I_{1}+I_{2}.

For I1I_{1}, we proceed

I1x1x0|g(x)g(α)|2𝑑x\displaystyle I_{1}-\int_{x_{1}^{*}}^{x_{0}}\left|g(x)-g(\alpha)\right|^{2}dx =x1x0δρ(2g(x)2g(α)δρ)dx\displaystyle=\int_{x_{1}^{*}}^{x_{0}}-\delta\rho\left(2g(x)-2g(\alpha)-\delta\rho\right)dx
δ(1δ)ρ2(x0x1)\displaystyle\geq-\delta(1-\delta)\rho^{2}(x_{0}-x_{1}^{*})

since g(x)g(α)ρ/2g(x)-g(\alpha)\leq\rho/2 by δ<1/2\delta<1/2 on (x1,x0)(x_{1}^{*},x_{0}) so that 2g(x)2g(α)δρ(1δ)ρ2g(x)-2g(\alpha)-\delta\rho\leq(1-\delta)\rho. Since

I2=x0x2|g(x)g(β)(1δ)ρ|2𝑑x,I_{2}=\int_{x_{0}}^{x_{2}^{*}}\left|g(x)-g(\beta)-(1-\delta)\rho\right|^{2}dx,

as for I1I_{1}, we have

I2x0x2|g(β)g(x)|2𝑑xδ(1δ)ρ2(x2x0).I_{2}-\int_{x_{0}}^{x_{2}^{*}}\left|g(\beta)-g(x)\right|^{2}dx\geq-\delta(1-\delta)\rho^{2}(x_{2}^{*}-x_{0}).

We thus conclude that

αβ|Uδg|2𝑑xαβ|U0g|2𝑑xδ(1δ)ρ2(x2x1).\int_{\alpha}^{\beta}|U_{\delta}-g|^{2}dx-\int_{\alpha}^{\beta}|U_{0}-g|^{2}dx\geq-\delta(1-\delta)\rho^{2}(x_{2}^{*}-x_{1}^{*}).

This yields the desired inequality for vv since (v)(Uδ)\mathcal{F}(v)\geq\mathcal{F}(U_{\delta}). ∎

We set ρ=g(β)g(α)\rho=g(\beta)-g(\alpha) and

Xδ={v|vis a non-decreasing piecewise constant function with three facets andv(α)=g(α),v(β)=g(β).Moreover, the value on the middle facet equalsg(α)+δρ}.X_{\delta}=\Bigl{\{}\;v\bigm{|}v\ \text{is a non-decreasing piecewise constant function }\\ \text{with three facets and}\ v(\alpha)=g(\alpha),v(\beta)=g(\beta).\ \text{Moreover, }\\ \text{the value on the middle facet equals}\ g(\alpha)+\delta\rho\;\Bigr{\}}.
Lemma 4.16.

Assume that KK satisfies (K1) and (K2). Assume that gC[α,β]g\in C[\alpha,\beta] is non-decreasing. If g(β)g(α)Mg(\beta)-g(\alpha)\leq M and C:=CM(βα)λ/2>0C_{*}:=C_{M}-(\beta-\alpha)\lambda/2>0, then

infvXδTVKg(v)TVKg(U0)+Cδ(1δ)ρ2\inf_{v\in X_{\delta}}TV_{Kg}(v)\geq TV_{Kg}(U_{0})+C_{*}\delta(1-\delta)\rho^{2}

for δ(0,1)\delta\in(0,1), where ρ=g(β)g(α)\rho=g(\beta)-g(\alpha).

Proof.

Since TVK(Uδ)TV_{K}(U_{\delta}) is independent of the choice of x1,x2x_{1},x_{2}, we observe that

infvXδTVKg(v)TVKg(Uδ)\inf_{v\in X_{\delta}}TV_{Kg}(v)\geq TV_{Kg}(U_{\delta})

with x1=x1x_{1}=x_{1}^{*}, x2=x2x_{2}=x_{2}^{*}. For ρ=g(β)g(α)\rho=g(\beta)-g(\alpha), we set ρ1=δρ\rho_{1}=\delta\rho, ρ2=(1δ)ρ\rho_{2}=(1-\delta)\rho so that ρ1+ρ2=ρ\rho_{1}+\rho_{2}=\rho. We observe that

TVK(Uδ)=K(ρ1)+K(ρ2),TVK(U0)=K(ρ).TV_{K}(U_{\delta})=K(\rho_{1})+K(\rho_{2}),\quad TV_{K}(U_{0})=K(\rho).

By (K2), we observe that

TVK(Uδ)TVK(U0)+CMρ1ρ2TV_{K}(U_{\delta})\geq TV_{K}(U_{0})+C_{M}\rho_{1}\rho_{2}

for ρM\rho\leq M. By Lemma 4.15, we conclude that

TVKg(Uδ)\displaystyle TV_{Kg}(U_{\delta}) =TVK(Uδ)+(Uδ)\displaystyle=TV_{K}(U_{\delta})+\mathcal{F}(U_{\delta})
TVK(U0)+CMρ1ρ2+(U0)λ2δ(1δ)ρ2(βα)\displaystyle\geq TV_{K}(U_{0})+C_{M}\rho_{1}\rho_{2}+\mathcal{F}(U_{0})-\frac{\lambda}{2}\delta(1-\delta)\rho^{2}(\beta-\alpha)
=TVKg(U0)+δ(1δ)ρ2(CM(βα)λ/2).\displaystyle=TV_{Kg}(U_{0})+\delta(1-\delta)\rho^{2}\left(C_{M}-(\beta-\alpha)\lambda/2\right).

Thus

TVKg(Uδ)TVKg(U0)Cδ(1δ)ρ2TV_{Kg}(U_{\delta})-TV_{Kg}(U_{0})\geq C_{*}\delta(1-\delta)\rho^{2}

provided that (βα)λ/2<CM(\beta-\alpha)\lambda/2<C_{M}, i.e., C>0C_{*}>0. The proof is now complete. ∎

To prove Theorem 1.2 directly, we approximate a general function by a piecewise constant function such that TVKTV_{K} is also approximated.

Lemma 4.17.

Assume that KK satisfies (K1), (K3) and that Ω=(a,b)\Omega=(a,b). For any u(LpBV)(Ω)u\in(L^{p}\cap BV)(\Omega) with p1p\geq 1, there is a sequence of piecewise constant functions {um}\{u_{m}\} (with finitely many jumps) such that umuu_{m}\to u in Lp(Ω)L^{p}(\Omega) and TVK(um)TVK(u)TV_{K}(u_{m})\to TV_{K}(u) as mm\to\infty.

Since umuu_{m}\to u in L2(Ω)L^{2}(\Omega) implies (um)(u)\mathcal{F}(u_{m})\to\mathcal{F}(u) when gL2(Ω)g\in L^{2}(\Omega), this lemma yields

Lemma 4.18.

Assume that KK satisfies (K1), (K3) and Ω=(a,b)\Omega=(a,b). For any u(L2BV)(Ω)u\in(L^{2}\cap BV)(\Omega), there is a sequence of piecewise constant function {um}\{u_{m}\} such that umuu_{m}\to u in L2(Ω)L^{2}(\Omega) and TVKg(um)TVKg(u)TV_{Kg}(u_{m})\to TV_{Kg}(u) as mm\to\infty.

We shall prove Lemma 4.17 by reducing the problem when uu is continuous, i.e., uC[a,b]u\in C[a,b].

Lemma 4.19.

Assume that KK satisfies (K3) and Ω=(a,b)\Omega=(a,b). For any uC[a,b]BV(Ω)u\in C[a,b]\cap BV(\Omega), there is a sequence of piecewise constant functions {um}\{u_{m}\} such that umuu_{m}\to u in C[a,b]C[a,b] and TVK(um)TVK(u)TV_{K}(u_{m})\to TV_{K}(u) as mm\to\infty.

Proof.

For a continuous function uu, TVK(u)TV_{K}(u) agrees with usual TV(u)TV(u). We extend uu continuously in some neighborhood if Ω¯\overline{\Omega} and denote it by u¯\overline{u}. We mollify u¯\overline{u} by a symmetric mollifier ρε\rho_{\varepsilon}. It is well known that uε=u¯ρεu_{\varepsilon}=\overline{u}*\rho_{\varepsilon} is CC^{\infty} in [a,b][a,b] and uεuu_{\varepsilon}\to u in C[a,b]C[a,b] as ε0\varepsilon\to 0. Moreover, TV(uε)TV(u)TV(u_{\varepsilon})\to TV(u) [Giu, Proposition 1.15]. Since ρε\rho_{\varepsilon} can be approximated (in C1C^{1} sense) by polynomials in a bounded set, we approximate uεu_{\varepsilon} by a polynomial with its derivative in C[a,b]C[a,b]. Thus, we may assume that uu is a polynomial.

For a given η>0\eta>0, we define a piecewise constant function

uη(x)=kηifkηu(x)<(k+1)η.u^{\eta}(x)=k\eta\quad\text{if}\quad k\eta\leq u(x)<(k+1)\eta.

We divide the interval (a,b)(a,b) into finitely many subintervals {(ai,ai+1)}i=0\left\{(a_{i},a_{i+1})\right\}_{i=0}^{\ell} with a=a0<a1<<a<a+1=ba=a_{0}<a_{1}<\cdots<a_{\ell}<a_{\ell+1}=b such that on each such an interval uu is either increasing or decreasing. This is possible since uu is a polynomial. Let TV(u,(ai,ai+1))TV\left(u,(a_{i},a_{i+1})\right) denote the total variation of uu in (ai,ai+1)(a_{i},a_{i+1}) for i=0,,i=0,\ldots,\ell. Then

0TV(u,(ai,ai+1))TV(uη,(ai,ai+1))2η.0\leq TV\left(u,(a_{i},a_{i+1})\right)-TV\left(u^{\eta},(a_{i},a_{i+1})\right)\leq 2\eta.

By the assumption (K3), we see that for any δ>0\delta>0 there is η0>0\eta_{0}>0 such that |ηK(η)|<δη\left|\eta-K(\eta)\right|<\delta\eta for η<η0\eta<\eta_{0}. Since uu is continuous, the size of jumps of uηu^{\eta} is always η\eta so

|TV(uη,(ai,ai+1))TVK(uη,(ai,ai+1))|δTV(uη,(ai,ai+1))forη<η0,\left\lvert TV\left(u^{\eta},(a_{i},a_{i+1})\right)-TV_{K}\left(u^{\eta},(a_{i},a_{i+1})\right)\right\rvert\leq\delta TV\left(u_{\eta},(a_{i},a_{i+1})\right)\quad\text{for}\quad\eta<\eta_{0},

where we use the same convention to TVKTV_{K}. We thus observe that

|TV(u)TVK(uη)|i=0(2η+δTV(uη,(ai,ai+1)))\displaystyle\left\lvert TV(u)-TV_{K}(u^{\eta})\right\rvert\leq\sum_{i=0}^{\ell}\left(2\eta+\delta TV\left(u^{\eta},(a_{i},a_{i+1})\right)\right)
=2η(+1)+δTV(uη)2η(+1)+δTV(u).\displaystyle=2\eta(\ell+1)+\delta TV(u^{\eta})\leq 2\eta(\ell+1)+\delta TV(u).

Sending η0\eta\to 0, we now conclude that

limη0|TV(u)TVK(uη)|δTV(u).\lim_{\eta\to 0}\left\lvert TV(u)-TV_{K}(u^{\eta})\right\rvert\leq\delta TV(u).

Since δ>0\delta>0 is arbitrary, the convergence TVK(uη)TV(u)TV_{K}(u^{\eta})\to TV(u) as η0\eta\to 0 follows. By definition, uηuu^{\eta}\to u in C[a,b]C[a,b]. The proof is now complete. ∎

Proof of Lemma 4.17.

Since uu is bounded by uBV(Ω)u\in BV(\Omega), for any δ>0\delta>0, the set JδJ_{\delta} of jump discontinuities of uu whose jump size greater than δ\delta is a finite set. For any ε>0\varepsilon>0, we take δ>0\delta>0 such that

xJu\JδK(|u+u|(x))<ε.\sum_{x\in J_{u}\backslash J_{\delta}}K\left(|u^{+}-u^{-}|(x)\right)<\varepsilon.

We may assume that Jδ={aj}j=1J_{\delta}=\{a_{j}\}_{j=1}^{\ell} with aj<aj+1a_{j}<a_{j+1} and a0=aa_{0}=a, a+1=ba_{\ell+1}=b. In each interval (aj,aj+1)(a_{j},a_{j+1}), we approximate uu in LpL^{p} with continuous function uεu_{\varepsilon}. We now apply Lemma 4.19 on each interval (aj,aj+1)(a_{j},a_{j+1}) to approximate uεu_{\varepsilon} by a piecewise constant function uεηu_{\varepsilon}^{\eta}. Although the jump at aja_{j} is not exactly equal to

|u+u|(aj),|u^{+}-u^{-}|(a_{j}),

it converges to this value as η0\eta\to 0 and ε0\varepsilon\to 0. We now obtain a desired sequence of piecewise constant functions. ∎

Remark 4.20.

If uu is non-decreasing, it is rather clear that umu_{m} in Lemmas 4.17, 4.18, 4.19 can be taken as a non-decreasing function by construction.

We give a sufficient condition (Lemma 4.16) that a larger jump costs less for TVKgTV_{Kg} not only TVKTV_{K} by giving a quantative estimate.

In the rest of this section, we give an alternate proof of Theorem 1.2 without using Theorem 1.1 by establishing a quantative estimate of TVKgTV_{Kg}. By approximation (Lemma 4.18), if there is a minimizer of TVKgTV_{Kg} among piecewise constant functions, this minimizer is also a minimizer of TVKgTV_{Kg} in BVBV. The existence of a minimizer among piecewise constant functions can be proved since the number of facets are restricted. We shall come back to this point at the end of this section. Unfortunately, this argument is not enough to prove Theorem 1.2 since we do not know the uniqueness of a minimizer. To show Theorem 1.2, we need a quantified version of Lemma 4.16. Let U0U_{0} be a minimizer with one jump.

Lemma 4.21.

Assume that KK satisfies (K1), (K2) and (K3). Assume that gC[α,β]g\in C[\alpha,\beta] is non-decreasing. Let MM be a constant such that ρ=g(β)g(α)M\rho=g(\beta)-g(\alpha)\leq M. Assume that C=CMλ(βα)/2>0C_{*}=C_{M}-\lambda(\beta-\alpha)/2>0. Let vv is a non-decreasing function on [α,β][\alpha,\beta] which is continuous at α\alpha and β\beta and v(α)=g(α)v(\alpha)=g(\alpha), v(β)=g(β)v(\beta)=g(\beta). Then

TVKg(v)C2((i=1ρi)2i=1ρi2+(ρi=1ρi)2)+TVKg(U0),TV_{Kg}(v)\geq\frac{C_{*}}{2}\left(\left(\sum_{i=1}^{\infty}\rho_{i}\right)^{2}-\sum_{i=1}^{\infty}\rho_{i}^{2}+\left(\rho-\sum_{i=1}^{\infty}\rho_{i}\right)^{2}\right)+TV_{Kg}(U_{0}),

where ρi=v(zi+0)v(zi0)>0\rho_{i}=v(z_{i}+0)-v(z_{i}-0)>0 and Jv={zi}i=1J_{v}=\{z_{i}\}_{i=1}^{\infty} ((α,β)\subset(\alpha,\beta)) is the set of jump discontinuities. The non-negative term

(i=1ρi)2i=1ρi2+(ρi=1ρi)2\left(\sum_{i=1}^{\infty}\rho_{i}\right)^{2}-\sum_{i=1}^{\infty}\rho_{i}^{2}+\left(\rho-\sum_{i=1}^{\infty}\rho_{i}\right)^{2}

vanishes if and only if ρ=ρi0\rho=\rho_{i_{0}} with some i0i_{0}.

We begin with an elementary lemma on a sequence.

Lemma 4.22.

Let {mk}k=1\{m_{k}\}_{k=1}^{\infty} be an increasing sequence of natural numbers and limkmk=\lim_{k\to\infty}m_{k}=\infty. Let {ρik}1imk\{\rho_{i}^{k}\}_{1\leq i\leq m_{k}} be a set of non-negative real numbers. Assume that

s:=limksk>0forsk=i=1mkρik.s:=\lim_{k\to\infty}s_{k}>0\quad\text{for}\quad s_{k}=\sum_{i=1}^{m_{k}}\rho_{i}^{k}.
  1. (i)
    limk1i,jmkρikρjk=s2/2\lim_{k\to\infty}\sum_{1\leq i,j\leq m_{k}}\rho_{i}^{k}\rho_{j}^{k}=s^{2}/2

    provided that limiρik=0\lim_{i\to\infty}\rho_{i}^{k}=0.

  2. (ii)
    lim¯k1i,jmkρikρjk1i,jmkρiρj\varliminf_{k\to\infty}\sum_{1\leq i,j\leq m_{k}}\rho_{i}^{k}\rho_{j}^{k}\geq\sum_{1\leq i,j\leq m_{k}}\rho_{i}\rho_{j}

    provided that limiρik=ρi0\lim_{i\to\infty}\rho_{i}^{k}=\rho_{i}\geq 0.

  3. (iii)

    Let II be the set of ii such that ρi=0\rho_{i}=0 and iIi\not\in I implies ρi>0\rho_{i}>0. Then

    lim¯k1i,jmkρikρjk\displaystyle\varliminf_{k\to\infty}\sum_{1\leq i,j\leq m_{k}}\rho_{i}^{k}\rho_{j}^{k} 1i,jmkρiρj+sIc2/2fors=sI+sIc,sI=i=1ρi\displaystyle\geq\sum_{1\leq i,j\leq m_{k}}\rho_{i}\rho_{j}+s_{I^{c}}^{2}/2\quad\text{for}\quad s=s_{I}+s_{I^{c}},\quad s_{I}=\sum_{i=1}^{\infty}\rho_{i}
    =(sI2i=1(ρi)2+sIc2)/2.\displaystyle=\left(s_{I}^{2}-\sum_{i=1}^{\infty}(\rho_{i})^{2}+s_{I^{c}}^{2}\right)\Biggm{/}2.
Proof.
  1. (i)

    We may assume that ρ1kρ2k\rho_{1}^{k}\geq\rho_{2}^{k}\geq\cdots. Since

    1i,jmkρikρjk=(sk2i=1mk(ρik)2)/2\sum_{1\leq i,j\leq m_{k}}\rho_{i}^{k}\rho_{j}^{k}=\left(s_{k}^{2}-\sum_{i=1}^{m_{k}}(\rho_{i}^{k})^{2}\right)\Biggm{/}2

    and

    i=1(ρik)2ρ1ki=1ρik=ρ1ksk0ask,\sum_{i=1}^{\infty}(\rho_{i}^{k})^{2}\leq\rho_{1}^{k}\sum_{i=1}^{\infty}\rho_{i}^{k}=\rho_{1}^{k}s_{k}\to 0\quad\text{as}\quad k\to\infty,

    we obtain (i).

  2. (ii)

    This follows from Fatou’s lemma.

  3. (iii)

    We divide the set of indices by II and IcI^{c}, the complement of II. Since

    1i,jmkρikρjk1i<jmki,jIρikρjk+1i,jmki,jIcρikρjk,\sum_{1\leq i,j\leq m_{k}}\rho_{i}^{k}\rho_{j}^{k}\geq\sum_{\begin{subarray}{c}1\leq i<j\leq m_{k}\\ i,j\in I\end{subarray}}\rho_{i}^{k}\rho_{j}^{k}+\sum_{\begin{subarray}{c}1\leq i,j\leq m_{k}\\ i,j\in I^{c}\end{subarray}}\rho_{i}^{k}\rho_{j}^{k},

    applying the results (i), (ii) yield (iii).

Proof of Lemma 4.21.

We may assume that g(β)>g(α)g(\beta)>g(\alpha). Let UU be a minimizer of TVKgTV_{Kg} on BVBV. Its existence is proved in Section 2. By Lemma 4.12, UU is non-decreasing. Moreover, by Lemma 3.2, UU is a piecewise constant outside the set

C={xΩ|U(x)=g(x)}C=\left\{x\in\Omega\bigm{|}U(x)=g(x)\right\}

and each facet contains a point zz such that U(z)=g(z)U(z)=g(z) and UU is continuous at zz. We set

a=infC,b=supC.a=\inf C,\quad b=\sup C.

By definition, aαa\geq\alpha and bβb\leq\beta. By Lemma 4.12 and Lemma 3.2, g(α)Ug(β)g(\alpha)\leq U\leq g(\beta) and UU is continuous at aa and bb. We approximate UU on [a,b][a,b] by a piecewise constant non-decreasing function uku_{k} with mk+1m_{k}+1 facets by Lemma 4.19. Since UU is continuous at aa and bb, we may assume that uk(x)=g(α)u_{k}(x)=g(\alpha) for x>ax>a close to aa and uk(x)=g(β)u_{k}(x)=g(\beta) for x<bx<b close to bb. We denote jumps by a1<a2<<am1a_{1}<a_{2}<\cdots<a_{m-1} with m=mk+1m=m_{k}+1 and set hi=uk(ai+0)h_{i}=u_{k}(a_{i}+0) with convention that a0=aa_{0}=a, am=ba_{m}=b. We set

ρi=hihi1fori=1,,mk\rho_{i}=h_{i}-h_{i-1}\quad\text{for}\quad i=1,\ldots,m_{k}

which denotes the jump at each aia_{i}. We fix ρi\rho_{i} and minimize TVKgTV_{Kg} on (a,b)(a,b). In other words, we minimize \mathcal{F} by moving aia_{i}’s. Let u¯k\bar{u}_{k} be its minimizer. By Proposition 4.14, facets of u¯k\bar{u}_{k} consist of

[a,x1],[x1,x2],,[xmk1,b][a,x_{1}],[x_{1},x_{2}],\ldots,[x_{m_{k}-1},b]

with g(ai)=(g(xi)+g(xi+1))/2g(a_{i})=\left(g(x_{i})+g(x_{i+1})\right)/2, i=1,,mk1i=1,\ldots,m_{k}-1; see Figure 11 with x0=a0,x_{0}=a_{0},\ldots, xm1=am=bx_{m-1}=a_{m}=b.

Refer to caption
Figure 11. graphs of gg and UU

By this choice,

TVKg(uk)TVKg(u¯k).TV_{Kg}(u_{k})\geq TV_{Kg}(\bar{u}_{k}).

We shall estimate TVKg(u¯k)TV_{Kg}(\bar{u}_{k}) from below as in Lemma 4.16. For (a,xi)(a,x_{i}), let ViV_{i} be a piecewise constant function on [a,xi][a,x_{i}] with one jump at the point yiy_{i} such that

g(yi)=(g(xi)+g(a))/2andVi(a)=g(α),Vi(xi)=uk(xi).g(y_{i})=\left(g(x_{i})+g(a)\right)\bigm{/}2\quad\text{and}\quad V_{i}(a)=g(\alpha),\quad V_{i}(x_{i})=u_{k}(x_{i}).

We set

Wi(x)={Vi(x),x[a,xi)u¯k(x),x[xi,b].W_{i}(x)=\left\{\begin{array}[]{ll}V_{i}(x),&x\in[a,x_{i})\\ \bar{u}_{k}(x),&x\in[x_{i},b].\end{array}\right.

We note that Wmk=Vmk=U0W_{m_{k}}=V_{m_{k}}=U_{0}. Since C>0C_{*}>0, we argue as in Lemma 4.16 and observe that

TVKg(u¯k)\displaystyle TV_{Kg}(\bar{u}_{k}) TVKg(W2)+Cρ1ρ2\displaystyle\geq TV_{Kg}(W_{2})+C_{*}\rho_{1}\rho_{2}
TVKg(W3)+Cρ3(ρ1+ρ2)+Cρ1ρ2\displaystyle\geq TV_{Kg}(W_{3})+C_{*}\rho_{3}(\rho_{1}+\rho_{2})+C_{*}\rho_{1}\rho_{2}
\displaystyle\cdots
TVKg(Vmk)+C1i,jmkρiρj=TVKg(U0)+C1i,jmkρiρj.\displaystyle\geq TV_{Kg}(V_{m_{k}})+C_{*}\sum_{1\leq i,j\leq m_{k}}\rho_{i}\rho_{j}=TV_{Kg}(U_{0})+C_{*}\sum_{1\leq i,j\leq m_{k}}\rho_{i}\rho_{j}.

From now on, we write jumps of uku_{k} by ρik\rho_{i}^{k} instead of ρi\rho_{i}. Our estimate for TVK(u¯k)TV_{K}(\bar{u}_{k}) yields

TVKg(uk)TVKg(U0)+C1i,jmkρikρjk.TV_{Kg}(u_{k})\geq TV_{Kg}(U_{0})+C_{*}\sum_{1\leq i,j\leq m_{k}}\rho_{i}^{k}\rho_{j}^{k}.

Since ukUu_{k}\to U in L2L^{2} and TVKg(uk)TVKg(U)TV_{Kg}(u_{k})\to TV_{Kg}(U), we observe, by changing numbering of ii in ρik\rho_{i}^{k} if necessary, that ρikρi>0\rho_{i}^{k}\to\rho_{i}>0 as k0k\to 0 for ii such that ρi\rho_{i} corresponds to jumps at ziz_{i} of UU and ρik0\rho_{i}^{k}\to 0 for other ii’s. The convergence of TVKgTV_{Kg} prevents that two jumps with positive length merges at the limit unless one of them tends to zero. Let II be the set such that ρikρi>0\rho_{i}^{k}\to\rho_{i}>0. We now apply Lemma 4.22 to conclude that

TVKg(v)\displaystyle TV_{Kg}(v) TVKg(U)TVKg(U0)+lim¯kC1i,jmkρikρjk\displaystyle\geq TV_{Kg}(U)\geq TV_{Kg}(U_{0})+\varliminf_{k\to\infty}C_{*}\sum_{1\leq i,j\leq m_{k}}\rho_{i}^{k}\rho_{j}^{k}
TVKg(U0)+C2(sI2i=1ρi2+sIc2),\displaystyle\geq TV_{Kg}(U_{0})+\frac{C_{*}}{2}\left(s_{I}^{2}-\sum_{i=1}^{\infty}\rho_{i}^{2}+s_{I^{c}}^{2}\right),

where sI=i=1ρig(β)g(α)s_{I}=\sum_{i=1}^{\infty}\rho_{i}\leq g(\beta)-g(\alpha). Since

sI2i=1ρi20andsIc20,s_{I}^{2}-\sum_{i=1}^{\infty}\rho_{i}^{2}\geq 0\quad\text{and}\quad s_{I^{c}}^{2}\geq 0,

the quantity

sI2i=1ρi2+sIc2=0s_{I}^{2}-\sum_{i=1}^{\infty}\rho_{i}^{2}+s_{I^{c}}^{2}=0

if and only if sIcc=0s_{I^{c}}^{c}=0 and (i=1ρi)2=i=1ρi2\left(\sum_{i=1}^{\infty}\rho_{i}\right)^{2}=\sum_{i=1}^{\infty}\rho_{i}^{2}, i.e., II is a singleton, i.e., I={i0}I=\{i_{0}\} with some i0Ii_{0}\in I and ρi0=ρ\rho_{i_{0}}=\rho. ∎

We conclude this subsection by proving Theorem 1.2 based on the quantative estimate.

Alternate proof of Theorem 1.2.

Again, we may assume that gg is not a constant. By Lemma 4.12, a minimizer UU is non-decreasing. By Lemma 3.2, at the place where C={U(x)=g(x)}C=\left\{U(x)=g(x)\right\}, the function UU is continuous. Let Q[g(a),g(b)]Q\subset\left[g(a),g(b)\right] be the set of qq such that

Iq={x|q(x)=q},Iq={xIq|x>infIq}I_{q}=\left\{x\bigm{|}q(x)=q\right\},\quad I_{q}^{*}=\left\{x\in I_{q}\bigm{|}x>\inf I_{q}\right\}

are not a singleton. We set

CΓ=C\qQIq.C_{\Gamma}=C\Bigm{\backslash}\bigcup_{q\in Q}I_{q}^{*}.

Since gg is continuous, CΓC_{\Gamma} is not empty by Lemma 3.1.

If we prove that CΓC_{\Gamma} is a discrete set so that it is a finite set. Then, by Lemma 3.1, UU is a piecewise constant function. We shall prove that CΓC_{\Gamma} is a discrete set. By definition, if x1<x2x_{1}<x_{2} for x1,x2CΓx_{1},x_{2}\in C_{\Gamma}, then g(x1)<g(x2)g(x_{1})<g(x_{2}). Suppose that CΓC_{\Gamma} were not discrete, then for any small ε\varepsilon, there would exist x1,x2CΓx_{1},x_{2}\in C_{\Gamma} with x1<x2<x1+εx_{1}<x_{2}<x_{1}+\varepsilon such that the interval (x1,x2)(x_{1},x_{2}) would contain infinitely many element of CΓC_{\Gamma}. Since UU minimizes TVKgTV_{Kg} on (x1x2)(x_{1}x_{2}) with U(x1)=g(x1)U(x_{1})=g(x_{1}), U(x2)=g(x2)U(x_{2})=g(x_{2}), applying Lemma 4.21 to α=x1\alpha=x_{1}, β=x2\beta=x_{2} to conclude that UU must have at most one jump provided that ε<2CM/λ\varepsilon<2C_{M}/\lambda. This yields a contradiction so we conclude that CΓC_{\Gamma} is discrete and UU is a non-decreasing piecewise constant function. The number of jumps can be estimated since the distance of two points in CC is at most 2CM/λ2C_{M}/\lambda. ∎

4.4. Application of approximation lemma

We shall prove Theorem 1.3. We approximate gL(a,b)g\in L^{\infty}(a,b) by gC[a,b]g_{\ell}\in C[a,b] such that ggg_{\ell}\to g in L2(a,b)L^{2}(a,b) as \ell\to\infty and

ess.infggess.supgon(a,b)\operatorname{ess.}\inf g\leq g_{\ell}\leq\operatorname{ess.}\sup g\quad\text{on}\quad(a,b)

for all 1\ell\geq 1. Let uBV(a,b)u\in BV(a,b) with TVKg(u)<TV_{Kg}(u)<\infty. By Lemma 4.17, there is a sequence {u}\{u_{\ell}\} of piecewise constant functions such that TVK(u)TVK(U)TV_{K}(u_{\ell})\to TV_{K}(U) with uuu_{\ell}\to u in L2(a,b)L^{2}(a,b) as \ell\to\infty. We thus observe that {u}\{u_{\ell}\} is a “recovery” sequence in the sense that TVKg(u)TVKg(u)TV_{Kg_{\ell}}(u_{\ell})\to TV_{Kg}(u). Let UU_{\ell} be a minimizer of TVKgTV_{Kg_{\ell}}. By Theorem 1.1, the number of jump mm_{\ell} is bounded by

m[(ba)λ/(2CM)]+1=mm_{\ell}\leq\left[(b-a)\lambda\bigm{/}(2C_{M})\right]+1=m_{*}

with M=osc[a,b]gosc[a,b]gM=\operatorname{osc}_{[a,b]}g\geq\operatorname{osc}_{[a,b]}g_{\ell}. By compactness of a bounded set in a finite dimensional space, UU_{\ell} has a convergent subsequence, i.e., UUU_{\ell}\to U with some UU in L2(a,b)L^{2}(a,b) (and also with respect to weak* topology of BVBV). Moreover, the limit UU is still a piecewise constant function with at most mm_{*} jumps. By lower semicontinuity of TVKTV_{K} (Proposition 2.4), we see that

TVKg(U)lim¯TVKg(U)lim¯TVKg(u)=TVKg(u).TV_{Kg}(U)\leq\varliminf_{\ell\to\infty}TV_{Kg_{\ell}}(U_{\ell})\leq\varliminf_{\ell\to\infty}TV_{Kg}(u_{\ell})=TV_{Kg}(u).

We thus conclude that UU is a desired minimizer of TVKgTV_{Kg} since uBV(a,b)u\in BV(a,b) is an arbitrary element.

5. A sufficient condition for (K2)

We give a sufficient condition for (K2) if KK is derived as a limit of the Kobayashi-Warren-Carter energy, i.e., KK is of the form (1.7).

We first consider a kind of Fenchel dual of a function ff. We set

H(ρ)=infx>0(ρx+f(x))H(\rho)=\inf_{x>0}\left(\rho x+f(x)\right) (5.1)

for a real-valued function ff on [0,)[0,\infty). If we use the Fenchel dual, it can be written as

H(ρ)=f(ρ)=supx((ρ)xf¯(x)),H(\rho)=-f^{*}(-\rho)=-\sup_{x}\left((-\rho)x-\bar{f}(x)\right),

where f¯(x)=f(x)\bar{f}(x)=f(x) for x0x\geq 0 and f¯(x)=\bar{f}(x)=\infty for x<0x<0. We assume

  1. (f1)

    fC1(0,1]C[0,1]f\in C^{1}(0,1]\cap C[0,1];

  2. (f2)

    ff takes the only minimum 0 at x=1x=1. In other words, f(x)0f(x)\geq 0 for all x0x\geq 0 and f(x)=0f(x)=0 if and only if x=1x=1;

  3. (f3)

    f(x)<0f^{\prime}(x)<0 for x(0,1)x\in(0,1).

Lemma 5.1.

Assume (f2). Then H(ρ)>0H(\rho)>0 for ρ>0\rho>0 and H(0)=0H(0)=0. Assume further (f2) and (f3). Then H(ρ)<f(0)H(\rho)<f(0) for all ρ>0\rho>0 and there is xρ(0,1)x_{\rho}\in(0,1) for ρ>0\rho>0 such that H(ρ)=ρxρ+f(xρ)H(\rho)=\rho x_{\rho}+f(x_{\rho}) and H(ρ)H(\rho) is strictly increasing in ρ\rho. Moreover, xρx_{\rho} is strictly increasing and xρ1x_{\rho}\uparrow 1 as ρ0\rho\downarrow 0. Furthermore, it satisfies (K2) with K=HK=H if ff satisfies

lim¯ρ0(f(f)1)(ρ)/ρ2>0.\varliminf_{\rho\downarrow 0}\left(f\circ(-f^{\prime})^{-1}\right)(\rho)\bigm{/}\rho^{2}>0. (5.2)

Here we define

(f)1(ρ)=min{x[0,1]|f(x)=ρ}.(-f^{\prime})^{-1}(\rho)=\min\left\{x\in[0,1]\bigm{|}f^{\prime}(x)=-\rho\right\}.
Proof.

The positivity for H(ρ)H(\rho) for ρ>0\rho>0 and H(0)=0H(0)=0 is clear by the definition (5.1). Also, the existence in [0,η][0,\eta] of a minimizer easily follows by (f1) and (f2). Although xρx_{\rho} may not be unique, xρ1x_{\rho}\uparrow 1 as well as monotonicity of xρx_{\rho} is guaranteed by (f3). The bound H(ρ)<f(0)H(\rho)<f(0) is rather clear.

It remains to prove that (5.2) yields (K2). Since xρx_{\rho} is the minimizer, it must satisfy

ρ=(f)(xρ).\rho=(-f^{\prime})(x_{\rho}).

We take xρ=(f)1(ρ)x_{\rho}=(-f^{\prime})^{-1}(\rho) in Lemma 5.1. Since

H(ρ)=ρxρ+f(xρ)H(\rho)=\rho x_{\rho}+f(x_{\rho})

by definition, we observe that for δ(0,1)\delta\in(0,1)

H(δρ)δH(ρ)ρ1xρ1+f(xρ1)(δρxρ1+δf(xρ1))withρ1=δρ=(1δ)f(xρ1).\displaystyle\begin{aligned} H(\delta\rho)-\delta H(\rho)&\geq\rho_{1}x_{\rho_{1}}+f(x_{\rho_{1}})-\left(\delta\rho x_{\rho_{1}}+\delta f(x_{\rho_{1}})\right)\ \text{with}\ \rho_{1}=\delta\rho\\ &=(1-\delta)f(x_{\rho_{1}}).\end{aligned} (5.3)

For ρ2=(1δ)ρ\rho_{2}=(1-\delta)\rho, we have

H((1δ)ρ)(1δ)H(ρ)δf(xρ2).H\left((1-\delta)\rho\right)-(1-\delta)H(\rho)\geq\delta f(x_{\rho_{2}}).

We thus observe that

H(ρ1)+H(ρ2)H(ρ)(1δ)f(xρ1)+δf(xρ2).H(\rho_{1})+H(\rho_{2})-H(\rho)\geq(1-\delta)f(x_{\rho_{1}})+\delta f(x_{\rho_{2}}).

By (5.2), we may assume that

(f(f)1)(ρ)CMρ2\left(f\circ(-f^{\prime})^{-1}\right)(\rho)\geq C_{M}\rho^{2}

with some CM>0C_{M}>0 provided that 0ρM0\leq\rho\leq M. Thus

(1δ)f(xρ1)\displaystyle(1-\delta)f(x_{\rho_{1}}) =(1δ)(f(f)1)(ρ1)CM(1δ)(δρ)2,\displaystyle=(1-\delta)\left(f\circ(-f^{\prime})^{-1}\right)(\rho_{1})\geq C_{M}(1-\delta)(\delta\rho)^{2},
δf(xρ1)\displaystyle\delta f(x_{\rho_{1}}) CMδ((1δ)ρ)2.\displaystyle\geq C_{M}\delta\left((1-\delta)\rho\right)^{2}.

We now observe that

H(ρ1)+H(ρ2)H(ρ)\displaystyle H(\rho_{1})+H(\rho_{2})-H(\rho) CM((1δ)δρ2δ+(1δ)δρ2(1δ))\displaystyle\geq C_{M}\left((1-\delta)\delta\rho^{2}\delta+(1-\delta)\delta\rho^{2}(1-\delta)\right)
=CM((1δ)δρ2)=CMρ1ρ2.\displaystyle=C_{M}\left((1-\delta)\delta\rho^{2}\right)=C_{M}\rho_{1}\rho_{2}.

We have proved (K2) for HH. ∎

Lemma 5.2.

Assume that (f1), (f2) and (f3). Then limρ0H(ρ)/ρ=0\lim_{\rho\downarrow 0}H(\rho)/\rho=0. In other words, HH satisfies (K3) with K=HK=H.

Proof.

Taking x=1x=1 in (5.1), we see that H(ρ)ρH(\rho)\leq\rho. We observe that

H(ρ)ρ\displaystyle H(\rho)-\rho =minx>0(ρ(x1)+f(x))\displaystyle=\min_{x>0}\left(\rho(x-1)+f(x)\right)
=ρ((xρ1)+f(xρ)/ρ)or\displaystyle=\rho\left((x_{\rho}-1)+f(x_{\rho})\bigm{/}\rho\right)\ \text{or}
H(ρ)ρ1\displaystyle\frac{H(\rho)}{\rho}-1 =xρ1+f(xρ)ρ.\displaystyle=x_{\rho}-1+\frac{f(x_{\rho})}{\rho}.

Since f0f\geq 0 and xρ1x_{\rho}\to 1 as ρ0\rho\downarrow 0, we conclude

lim¯ρ0(H(ρ)ρ1)0+0,\varliminf_{\rho\downarrow 0}\left(\frac{H(\rho)}{\rho}-1\right)\geq 0+0,

which now yields (K3). ∎

Remark 5.3.
  1. (i)

    Without (5.2) we only get (K2w) since f0f\geq 0 and the estimate (5.3).

  2. (ii)

    If f(x)=|x1|mf(x)=|x-1|^{m} for m>0m>0, then (5.2) holds if and only if m2m\geq 2. In fact, f(x)=m|x1|m2(x1)f^{\prime}(x)=m|x-1|^{m-2}(x-1) so that (f)1(ρ)=1(ρ/m)1/(m1)(-f^{\prime})^{-1}(\rho)=1-(\rho/m)^{1/(m-1)}. The function (f(f)1)(ρ)=(ρ/m)m/(m1)\left(f\circ(-f^{\prime})^{-1}\right)(\rho)=(\rho/m)^{m/(m-1)} so (5.2) holds if and only if m2m\geq 2.

  3. (iii)

    We may take other element of a preimage of f-f^{\prime} of ρ\rho but as a sufficient condition the present choice is the weakest assumption.

We come back to (1.7). In other words,

K(ρ)=minξ(ρ(ξ+)2+2G(ξ)),G(ξ)=|1ξF(τ)𝑑τ|.K(\rho)=\min_{\xi}\left(\rho(\xi_{+})^{2}+2G(\xi)\right),\quad G(\xi)=\left|\int_{1}^{\xi}\sqrt{F(\tau)}\;d\tau\right|.

We assume that

  1. (F1)

    FC[0,)F\in C[0,\infty) and FF takes the only minimum 0 at x=1x=1.

If we set f(x)=2G(x1/2)f(x)=2G(x^{1/2}), the property (F1) implies (f1), (f2) and (f3). Since (5.2) is property near x=1x=1, (5.2) for ff and GG are equivalent. We thus obtain an sufficient condition so that KK in (1.7) satisfies (K1), (K2) and (K3)

Proposition 5.4.

Assume (F1) and

lim¯ρ0(G(G)1)(ρ)/ρ2>0.\varliminf_{\rho\downarrow 0}\left(G\circ(-G^{\prime})^{-1}\right)(\rho)\bigm{/}\rho^{2}>0. (5.4)

Then KK in (1.7) satisfies (K1), (K2) and (K3).

Proof.

By Lemma 5.1, conditions (K1), (K2) are fulfilled. The property (K3) follows from Lemma 5.2. ∎

We conclude this section by examining the property (5.4). The left-hand side is equivalent to saying that

lim¯ρ0F1(ρ2)1F(τ)𝑑τ/ρ2>0,\varliminf_{\rho\downarrow 0}\int_{F^{-1}(\rho^{2})}^{1}\sqrt{F(\tau)}\;d\tau\bigm{/}\rho^{2}>0,

where F1(ρ2)=min{x[0,1]|F(x)=ρ2}F^{-1}(\rho^{2})=\min\left\{x\in[0,1]\bigm{|}F(x)=\rho^{2}\right\}. We set

F¯(x)=F(1x).\bar{F}(x)=F(1-x).

The condition (5.4) is now equivalent to

lim¯ρ00F¯1(ρ2)F¯(τ)𝑑τ/ρ2>0,\varliminf_{\rho\downarrow 0}\int_{0}^{\bar{F}^{-1}(\rho^{2})}\sqrt{\bar{F}(\tau)}\;d\tau\bigm{/}\rho^{2}>0, (5.5)

where F¯1(y)=max{x(0,1)|F¯(x)=y}\bar{F}^{-1}(y)=\max\left\{x\in(0,1)\bigm{|}\bar{F}(x)=y\right\}. To simplify the argument, we further assume that

  1. (F2)

    F<0F^{\prime}<0 in (0,1)(0,1) so that the inverse function F1F^{-1} in (0,F(0))\left(0,F(0)\right) is uniquely determined.

If we assume (F1) and (F2), by changing the variable of integration τ=F¯1(s2)\tau=\bar{F}^{-1}(s^{2}), we have

0F¯1(ρ2)F¯(τ)𝑑τ=0ρ2sdτds𝑑s=0ρ22s2(F¯1)(s2)𝑑s.\int_{0}^{\bar{F}^{-1}(\rho^{2})}\sqrt{\bar{F}(\tau)}\;d\tau=\int_{0}^{\rho^{2}}s\frac{d\tau}{ds}\;ds=\int_{0}^{\rho^{2}}2s^{2}(\bar{F}^{-1})^{\prime}(s^{2})\;ds.

The condition (5.5) is fulfilled if

lim¯σ0σ(F¯1)(σ)>0\varliminf_{\sigma\downarrow 0}\sigma(\bar{F}^{-1})^{\prime}(\sigma)>0

or equivalently

lim¯η0F¯(η)/η<.\varlimsup_{\eta\downarrow 0}\bar{F}^{\prime}(\eta)\Bigm{/}\eta<\infty.

We thus obtain a simple sufficient condition.

Theorem 5.5.

Assume that (F1) and (F2). Then KK in (1.7) satisfies (K1), (K2), (K3) provided that

lim¯x1F(x)/(x1)<.\varlimsup_{x\uparrow 1}F^{\prime}(x)/(x-1)<\infty.

Acknowledgements

The work of the first author was partly supported by JSPS KAKENHI Grant Numbers JP19H00639, JP20K20342, JP24K00531 and JP24H00183 and by Arithmer Inc., Daikin Industries, Ltd. and Ebara Corporation through collaborative grants. The work of the third author was partly supported by JSPS KAKENHI Grant Number JP20K20342. The work of the fifth author was partly supported by JSPS KAKENHI Grant Numbers JP22K03425, JP22K18677, 23H00086.

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