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Approximate ultrahomogeneity in LpLqL_{p}L_{q} lattices

Mary Angelica Tursi
Abstract.

We show that for 1p,q<1\leq p,q<\infty with p/qp/q\notin\mathbb{N}, the doubly atomless separable LpLqL_{p}L_{q} Banach lattice Lp(Lq)L_{p}(L_{q}) is approximately ultrahomogeneous (AUH) over the class of its finitely generated sublattices. The above is not true when p/qp/q\in\mathbb{N}. However, for any pqp\neq q, Lp(Lq)L_{p}(L_{q}) is AUH over the finitely generated lattices in the class BLpLqBL_{p}L_{q} of bands of LpLqL_{p}L_{q} lattices.

1. Introduction

In this paper, we explore the homogeneity properties (or lack thereof) of the class of LpLqL_{p}L_{q} lattices under various conditions.

The following is taken from [6]: A Banach lattice XX is an abstract LpLqL_{p}L_{q} lattice if there is a measure space (Ω,Σ,μ)(\Omega,\Sigma,\mu) such that XX can be equipped with an L(Ω)L_{\infty}(\Omega)-module and a map N:XLp(Ω)+N:X\rightarrow L_{p}(\Omega)_{+} such that

  • For all ϕL(Ω)+\phi\in L_{\infty}(\Omega)_{+} and xX+x\in X_{+}, ϕx0\phi\cdot x\geq 0,

  • For all ϕL(Ω)\phi\in L_{\infty}(\Omega) and xXx\in X, N[ϕx]=|ϕ|N[x]N[\phi\cdot x]=|\phi|N[x].

  • For all x,yXx,y\in X, N[x+y]N[x]+N[y]N[x+y]\leq N[x]+N[y]

  • If xx and yy are disjoint, N[x+y]q=N[x]q+N[y]qN[x+y]^{q}=N[x]^{q}+N[y]^{q}, and if |x||y||x|\leq|y|, then N[x]N[y]N[x]\leq N[y].

  • For all xXx\in X, x=N[x]Lp\|x\|=\|N[x]\|_{L_{p}}.

When the abstract LpLqL_{p}L_{q} space is separable, it has a concrete representation: Suppose (Ω,Σ,μ)(\Omega,\Sigma,\mu) and (Ω,Σ,μ)(\Omega^{\prime},\Sigma^{\prime},\mu^{\prime}) are measure spaces. Denote by Lp(Ω;Lq(Ω))L_{p}(\Omega;L_{q}(\Omega^{\prime})) the space of Bochner-measurable functions f:ΩLq(Ω)f:\Omega\rightarrow L_{q}(\Omega^{\prime}) such that the function N[f]N[f], with N[f](ω)=f(ω)qN[f](\omega)=\|f(\omega)\|_{q} for ωΩ\omega\in\Omega, is in Lp(Ω)L_{p}(\Omega). The class of bands in LpLqL_{p}L_{q} lattices, which we denote by BLpLqBL_{p}L_{q}, has certain analogous properties to those of LpL_{p} spaces, particularly with respect to its isometric theory.

LpLqL_{p}L_{q} lattices (and their sublattices) have been extensively studied for their model theoretic properties in [6] and [7]. It turns out that while abstract LpLqL_{p}L_{q} lattices themselves are not axiomatizable, the larger class BLpLqBL_{p}L_{q} is axiomatizable with certain properties corresponding to those of LpL_{p} spaces. For instance, it is known that the class of atomless LpL_{p} lattices is separably categorical, meaning that there exists one unique atomless separable LpL_{p} lattice up to lattice isometry. Correspondingly, the class of doubly atomless BLpLqBL_{p}L_{q} lattices is also separably categorical; in particular, up to lattice isometry, Lp([0,1];Lq[0,1])L_{p}([0,1];L_{q}[0,1]), which throughout will just be referred to as Lp(Lq)L_{p}(L_{q}), is the unique separable doubly atomless BLpLqBL_{p}L_{q} lattice (see [7, Proposition 2.6]).

Additionally, when pqp\neq q, the lattice isometries of LpLqL_{p}L_{q} lattices can be characterized in a manner echoing those of linear isometries over LpL_{p} spaces (with p2p\neq 2). Recall from [1, Ch. 11 Theorem 5.1] that a map T:Lp(0,1)Lp(0,1)T:L_{p}(0,1)\rightarrow L_{p}(0,1) is a surjective linear isometry iff Tf(t)=h(t)f(ϕ(t))Tf(t)=h(t)f(\phi(t)), where ϕ\phi is a measure-preserving transformation and hh is related to ϕ\phi through Radon-Nikodym derivatives. If we want TT to be a lattice isometry as well, then we also have hh positive (and the above characterization will also work for p=2p=2). In [3] (for the case of q=2q=2) and [13], a corresponding characterization of linear isometries is found for spaces of the form Lp(X;Y)L_{p}(X;Y), for certain pp and Banach spaces YY. In particular, for LpLqL_{p}L_{q} lattices with pqp\neq q: given fLp(Ω;Lq(Ω))f\in L_{p}(\Omega;L_{q}(\Omega^{\prime})), where ff is understood as a map from Ω\Omega to LqL_{q}, any surjective linear isometry TT is of the form

Tf(x)=S(x)(e(x)ϕf(x)),Tf(x)=S(x)\big{(}e(x)\phi f(x)\big{)},

where ϕ\phi is a set isomorphism (see [3] and [13] for definitions) ee is a measurable function related to ϕ\phi via Radon-Nikodym derivatives, and SS is a Bochner-measurable function from Ω\Omega to the space of linear maps from LqL_{q} to itself such that for each xx, S(x)S(x) is a linear isometry over LqL_{q}.

In [11], Raynaud obtained results on linear subspaces of LpLqL_{p}L_{q} spaces, showing that for 1qp<1\leq q\leq p<\infty, some r\ell_{r} linearly isomorphically embeds into Lp(Lq)L_{p}(L_{q}) iff it embeds either to LpL_{p} or to LqL_{q}. However, when 1pq<1\leq p\leq q<\infty, for prqp\leq r\leq q, the space r\ell_{r} isometrically embeds as a lattice in Lp(Lq)L_{p}(L_{q}), and for any pp-convex and qq-concave Orlicz function ϕ\phi, the lattice LϕL_{\phi} embeds lattice isomorphically into Lp(Lq)L_{p}(L_{q}). Thus, unlike with LpL_{p} lattices whose infinite dimensional sublattices are determined up to lattice isometry by the number of atoms, the sublattices of LpLqL_{p}L_{q} are not so simply classifiable.

In fact, the lattice isometry classes behave more like the LpL_{p} linear isometries, at least along the positive cone, as is evident in certain equimeasurability results for LpLqL_{p}L_{q} lattices. In [11], Raynaud also obtained the following on uniqueness of measures, a variation of a result which will be relevant in this paper: let α>0,α\alpha>0,\alpha\notin\mathbb{N}, and suppose two probability measures ν1\nu_{1} and ν2\nu_{2} on +\mathbb{R}_{+} are such that for all s>0s>0,

0(t+s)α𝑑ν1(t)=0(t+s)α𝑑ν1(t).\int_{0}^{\infty}(t+s)^{\alpha}\ d\nu_{1}(t)=\int_{0}^{\infty}(t+s)^{\alpha}\ d\nu_{1}(t).

Then ν1=ν2\nu_{1}=\nu_{2}. Linde gives an alternate proof of this result in [8].

Various versions and expansions of the above result appear in reference to LpL_{p} spaces: for instance, an early result from Rudin generalizes the above to equality of integrals over n\mathbb{R}^{n}: ([12]). Assume that α>0\alpha>0 with α2\alpha\notin 2\mathbb{N}, and suppose that for all 𝐯Rn\mathbf{v}\in R^{n},

n(1+𝐯z)α𝑑ν1(z)=n(1+𝐯z)α𝑑ν2(z)\int_{\mathbb{R}^{n}}(1+\mathbf{v}\cdot z)^{\alpha}\ d\nu_{1}(z)=\int_{\mathbb{R}^{n}}(1+\mathbf{v}\cdot z)^{\alpha}\ d\nu_{2}(z)

Then ν1=ν2\nu_{1}=\nu_{2}. An application of this result is a similar condition by which one can show that one collection of measurable functions F:nF:\mathbb{R}^{n}\rightarrow\mathbb{R}, with 𝐟=(f1,,fn)\mathbf{f}=(f_{1},...,f_{n}) is equimeasurable with another collection 𝐠=(g1,,gn)\mathbf{g}=(g_{1},...,g_{n}) By defining ν1\nu_{1} and ν2\nu_{2} as pushforward measures of FF and GG. In the case of LpL_{p} spaces, if ff and gg are corresponding basic sequences whose pushforward measures satisfy the above for α=p\alpha=p, then they generate isometric Banach spaces. Raynaud’s result shows the converse is true for α4,6,8,\alpha\neq 4,6,8,.... A similar result inLp(Lq)L_{p}(L_{q}) from [7] holds for α=p/q\alpha=p/q\notin\mathbb{N} under certain conditions, except instead of equimeasurable 𝐟\mathbf{f} and 𝐠\mathbf{g}, when the fif_{i}’s and gisg_{i}^{\prime}s are mutually disjoint and positive and the map figif_{i}\mapsto g_{i} generates a lattice isometry, (N[f1],,N[fn])(N[f_{1}],...,N[f_{n}]) and (N[g1],,N[gn])(N[g_{1}],...,N[g_{n}]) are equimeasurable.

Recall that a space XX is approximately ultrahomogeneous (AUH) over a class 𝒢\mathcal{G} of finitely generated spaces if for all appropriate embeddings fi;EXf_{i};E\hookrightarrow X with i=1,2i=1,2, for all E𝒢E\in\mathcal{G} generated by e1,,enEe_{1},...,e_{n}\in E, and for all ε>0\varepsilon>0, there exists an automorphism ϕ:XX\phi:X\rightarrow X such that for each 1jn1\leq j\leq n, ϕf1(ej)f2(ej)<ε\|\phi\circ f_{1}(e_{j})-f_{2}(e_{j})\|<\varepsilon.

X{X}   X{X}E{E}   ϕ\scriptstyle{\exists\phi}f1\scriptstyle{f_{1}}f2\scriptstyle{f_{2}}

In the Banach space setting, the embeddings are linear embeddings and the class of finitely generated spaces are finite dimensional spaces. In the lattice setting, the appropriate maps are isometric lattice embeddings, and one can either choose finite dimensional or finitely generated lattices.

The equimeasurability results described above can be used to show an approximate ultrahomogeneity of Lp([0,1])L_{p}([0,1]) over its finite dimensional linear subspaces only so long as p2p\notin 2\mathbb{N} (see [10]). Conversely, the cases where p2p\in 2\mathbb{N} are not AUH over finite dimensional linear subspaces, with counterexamples showing linearly isometric spaces whose corresponding basis elements are not equimeasurability. Alternate methods using continuous Fraïssé Theory have since then been used to give alternate proofs of linear approximate ultrahomogeneity of LpL_{p} for p2p\notin 2\mathbb{N} (see [5]) as well as lattice homogeneity of LpL_{p} for all 1p<1\leq p<\infty (see [2], [5]).

This paper is structured as follows: in section 2, we first establish basic notation and give a characterization of finite dimensional BLpLqBL_{p}L_{q} lattices. This characterization is used in subsequent sections for establishing both equimeasurability and ultrahomogeneity results later on.

In section 3 we show that when pqp\neq q, Lp(Lq):=Lp(Lq)L_{p}(L_{q}):=L_{p}(L_{q}) is AUH over the larger class of finite dimensional (and finitely generated) BLpLqBL_{p}L_{q} spaces. This is done by characterizing representations of BLpLqBL_{p}L_{q} sublattices Lp(Lq)L_{p}(L_{q}) in such a way that induces automorphisms over Lp(Lq)L_{p}(L_{q}) making the homogeneity diagram commute. The results here play a role in subsequent sections as well.

In section 4, we prove that if in addition p/qp/q\notin\mathbb{N}, Lp(Lq)L_{p}(L_{q}) is also AUH over the class of its finitely generated sublattices. First, we determine the isometric structure of finite dimensional sublattices of Lp(Lq)L_{p}(L_{q}) lattices by giving an alternate proof of [7, Proposition 3.2] showing that two sublattices EE and FF of Lp(Lq)L_{p}(L_{q}), with the eie_{i}’s and fif_{i}’s each forming the basis of atoms, are lattice isometric iff (N[e1],,N[en])(N[e_{1}],...,N[e_{n}]) and (N[f1),,N[fn])(N[f_{1}),...,N[f_{n}]) are equimeasurable. The equimeasurability result allows us to reduce a homogeneity diagram involving a finite dimensional sublattice of Lp(Lq)L_{p}(L_{q}) to one with a finite dimensional BLpLqBL_{p}L_{q} lattice, from which, in combination with the results in section 3, the main result follows.

Section 5 considers the case of p/qp/q\in\mathbb{N}. Here, we provide a counterexample to equimeasurability in the case that p/qp/q\in\mathbb{N} and use this counterexample to show that in such cases, Lp(Lq)L_{p}(L_{q}) is not AUH over the class of its finite dimensional lattices.

2. Preliminaries

We begin with some basic notation and definitions. Given a measurable set AnA\subseteq\mathbb{R}^{n}, we let 𝟏A\mathbf{1}_{A} refer to the characteristic function over AA. For a lattice XX, let B(X)B(X) be the unit ball, and S(X)S(X) be the unit sphere.

For elements e1,,ene_{1},...,e_{n} in some lattice XX, use bracket notation <e1,,en>L<e_{1},...,e_{n}>_{L} to refer to the Banach lattice generated by the elements e1,,ene_{1},...,e_{n}. In addition, we write <e1,,en><e_{1},...,e_{n}> without the LL subscript to denote that the generating elements eie_{i} are also mutually disjoint positive elements in the unit sphere. Throughout, we will also use boldface notation to designate a finite sequence of elements: for instance, for x1,,xnx_{1},...,x_{n}\in\mathbb{R} or x1,,xnXx_{1},...,x_{n}\in X for some lattice xx, let 𝐱=(x1,,xn)\mathbf{x}=(x_{1},...,x_{n}). Use the same notation to denote a sequence of functions over corresponding elements: for example, let (f1,,fn)=𝐟(f_{1},...,f_{n})=\mathbf{f}, or (f1(x1),fn(xn))=𝐟(𝐱)(f_{1}(x_{1}),...f_{n}(x_{n}))=\mathbf{f}(\mathbf{x}), or (f(x1),,f(xn))=f(𝐱)(f(x_{1}),...,f(x_{n}))=f(\mathbf{x}). Finally, for any element ee or tuple 𝐞\mathbf{e} of elements in some lattice XX, let 𝜷(e)\boldsymbol{\beta}(e) and 𝜷(𝐞)\boldsymbol{\beta}(\mathbf{e}) be the band generated by ee and 𝐞\mathbf{e} in XX, respectively.

Recall that Bochner integrable functions are the norm limits of simple functions f:ΩLq(Ω)f:\Omega\rightarrow L_{q}(\Omega^{\prime}), with f(ω)=1nri𝟏Ai(ω)𝟏Bif(\omega)=\sum_{1}^{n}r_{i}\mathbf{1}_{A_{i}}(\omega)\mathbf{1}_{B_{i}}, where 𝟏Ai\mathbf{1}_{A_{i}} and 𝟏Bi\mathbf{1}_{B_{i}} are the characteristic functions for AiΣA_{i}\in\Sigma and BiΣB_{i}\in\Sigma^{\prime}, respectively. One can also consider fLp(Ω;Lq(Ω))f\in L_{p}(\Omega;L_{q}(\Omega^{\prime})) as a ΣΣ\Sigma\otimes\Sigma^{\prime}-measurable function such that

f=(Ωf(ω)qp𝑑ω)1/p=(Ω(Ω|f(ω,ω)|q𝑑ω)p/q𝑑ω)1/p\|f\|=\bigg{(}\int_{\Omega}\|f(\omega)\|_{q}^{p}\ d\omega\bigg{)}^{1/p}=\bigg{(}\int_{\Omega}\bigg{(}\int_{\Omega^{\prime}}|f(\omega,\omega^{\prime})|^{q}\ d\omega^{\prime}\bigg{)}^{p/q}\ d\omega\bigg{)}^{1/p}

Unlike the more familiar LpL_{p} lattices, the class of abstract LpLqL_{p}L_{q} lattices is not itself axiomatizable; however, the slightly more general class BLpLqBL_{p}L_{q} of bands in Lp(Lq)L_{p}(L_{q}) lattices is axiomatizable. Additionally, if XX is a separable BLpLqBL_{p}L_{q} lattice, it is lattice isometric to a lattice of the form

(pLp(Ωn;qn))pLp(Ω;q)\displaystyle\bigg{(}\bigoplus_{p}L_{p}(\Omega_{n};\ell_{q}^{n})\bigg{)}\oplus_{p}L_{p}(\Omega_{\infty};\ell_{q})
p\displaystyle\oplus_{p} (pLp(Ωn;Lq(0,1)qqn))\displaystyle\bigg{(}\bigoplus_{p}L_{p}(\Omega^{\prime}_{n};L_{q}(0,1)\oplus_{q}\ell_{q}^{n})\bigg{)}
p\displaystyle\oplus_{p} Lp(Ω;Lq(0,1)qq).\displaystyle\ L_{p}(\Omega^{\prime}_{\infty};L_{q}(0,1)\oplus_{q}\ell_{q}).

BLpLqBL_{p}L_{q} lattices may also contain what are called base disjoint elements. xx and yy are base disjoint if N[x]N[y]N[x]\perp N[y]. Based on this, we call xx a base atom if whenever 0y,zx0\leq y,z\leq x with yy and zz base disjoint, then either N[y]=0N[y]=0 or N[z]=0N[z]=0. Observe this implies that N[x]N[x] is an atom in LpL_{p}. Alternatively, we call xx a fiber atom if any disjoint 0y,zx0\leq y,z\leq x are also base disjoint. Finally, we say that XX is doubly atomless if it contains neither base atoms nor fiber atoms.

Another representation of BLpLqBL_{p}L_{q} involves its finite dimensional subspaces. We say that XX is an (pq)λ(\mathcal{L}_{p}\mathcal{L}_{q})_{\lambda} lattice, with λ1\lambda\geq 1 if for all disjoint x1,,xnXx_{1},...,x_{n}\in X and ε>0\varepsilon>0, there is a finite dimensional FF of XX that is (1+ε)(1+\varepsilon)-isometric to a finite dimensional BLpLqBL_{p}L_{q} space containing x1,,xnx^{\prime}_{1},...,x^{\prime}_{n} such that for each 1in1\leq i\leq n, xixi<ε\|x_{i}-x^{\prime}_{i}\|<\varepsilon. Henson and Raynaud proved that in fact, any lattice XX is a BLpLqBL_{p}L_{q} space iff XX is (pq)1(\mathcal{L}_{p}\mathcal{L}_{q})_{1} (see [6]). This equivalence can be used to show the following:

Proposition 2.1.

(Henson, Raynaud) If XX is a separable BLpLqBL_{p}L_{q} lattice, then it is the inductive limit of finite dimensional BLpLqBL_{p}L_{q} lattices.

The latter statement is not explicitly in the statement of Lemma 3.53.5 in [6], but the proof showing that any BLpLqBL_{p}L_{q} lattice is (pq)1(\mathcal{L}_{p}\mathcal{L}_{q})_{1} was demonstrated by proving the statement itself.

Throughout this paper, we refer to this class of finite dimensional BLpLqBL_{p}L_{q} lattices as B𝒦p,qB\mathcal{K}_{{p},{q}}. Observe that if EB𝒦p,qE\in B\mathcal{K}_{{p},{q}}, then it is of the form p(qmi)1N\oplus_{p}(\ell_{q}^{m_{i}})_{1}^{N} where for 1kN1\leq k\leq N, the atoms e(1,1),,e(k,mk)e(1,1),...,e(k,m_{k}) generate qmk\ell_{q}^{m_{k}}.

Proposition 2.2.

Let EE be a B𝒦p,qB\mathcal{K}_{{p},{q}} sublattice of Lp(Lq)L_{p}(L_{q}) with atoms e(k,j)e(k,j) as described above. Then the following are true:

  1. (1)

    There exist disjoint measurable A(k)[0,1]A(k)\subseteq[0,1] such that for all ii, supp(e(k,j))A(k)×[0,1]\mathrm{supp}(e(k,j))\subseteq A(k)\times[0,1],

  2. (2)

    For all kk and for all j,jj,j^{\prime}, N[e(k,j)]=N[e(k,j)]N[e(k,j)]=N[e(k,j^{\prime})].

Conversely, if EE is a finite dimensional sublattice of Lp(Lq)L_{p}(L_{q}) satisfying properties (1) and (2), then EE is in B𝒦p,qB\mathcal{K}_{{p},{q}}.

In order to prove this theorem, we first need the following lemma:

Lemma 2.3.

Let 0<r<0<r<\infty, with r1r\neq 1. suppose x1,,xnLr+x_{1},...,x_{n}\in L_{r}+ are such that

1nxkrr=xkrr\|\sum_{1}^{n}x_{k}\|_{r}^{r}=\sum\|x_{k}\|^{r}_{r}

Then the xix_{i}’s are mutually disjoint.

Proof.

If r<1r<1, then

(1) xi(t)r+xj(t)rdt=xirr+xjrr=(xi(t)+xj(t))r𝑑t\displaystyle\int x_{i}(t)^{r}+x_{j}(t)^{r}\ dt=\|x_{i}\|_{r}^{r}+\|x_{j}\|_{r}^{r}=\int(x_{i}(t)+x_{j}(t))^{r}\ dt

Now observe that for all tt, (xi(t)+xj(t))rxi(t)r+xj(t)r(x_{i}(t)+x_{j}(t))^{r}\leq x_{i}(t)^{r}+x_{j}(t)^{r}, with equality iff either xi(t)=0x_{i}(t)=0 or xj(t)=0x_{j}(t)=0, so (xi+xj)rxirxjrL1+(x_{i}+x_{j})^{r}-x_{i}^{r}-x_{j}^{r}\in L_{1}+. Combined with the above equality in line (1), since (xi+xj)rxirxjr1=0,\|(x_{i}+x_{j})^{r}-x_{i}^{r}-x_{j}^{r}\|_{1}=0, it follows that xi(t)r+xj(t)r=(xi(t)+xj(t))rx_{i}(t)^{r}+x_{j}(t)^{r}=(x_{i}(t)+x_{j}(t))^{r} a.e., so xix_{i} must be disjoint from xjx_{j} when iji\neq j.

If r>1r>1, proceed as in the proof for r<1r<1, but with the inequalities reversed, given that in this instance xi(t)r+xj(t)r(xi(t)+xj(t))rx_{i}(t)^{r}+x_{j}(t)^{r}\leq(x_{i}(t)+x_{j}(t))^{r} for all tt.

Remark 2.4.

The above implies that a BLpLqBL_{p}L_{q} lattice XX is base atomless if it contains no bands lattice isometric to some LpL_{p} or LqL_{q} space. Indeed, if there were a base atom ee, then any two 0xye0\leq x\perp y\leq e would have to have NN-norms multiple to each other, so <x,y><x,y> is lattice isometric to q2\ell_{q}^{2}. Resultantly, the band generated by ee is an LqL_{q} space. Similarly, if ee is a fiber atom, then any 0xye0\leq x\perp y\leq e is also base disjoint, which implies that the band generated by ee is an LpL_{p} space.

We now conclude with the proof of Proposition 2.2:

Proof of Proposition 2.2.

Observe that for each appropriate pair (k,j)(k,j),

(01N[e(k,j)]p(s)𝑑s)q/p=Nq[e(k,j)]p/q=1\bigg{(}\int_{0}^{1}N[e(k,j)]^{p}(s)\ ds\bigg{)}^{q/p}=\|N^{q}[e(k,j)]\|_{p/q}=1

For notational ease, let E(k,j)=Nq[e(k,j)]E(k,j)=N^{q}[e(k,j)]. Pick j1,,jnj_{1},...,j_{n} with each jkmkj_{k}\leq m_{k}. Then, by disjointness of the e(k,j)e(k,j)’s, for all (ak)k0(a_{k})_{k}\geq 0 and all x=kake(k,jk)x=\sum_{k}a_{k}e(k,j_{k}),

ake(k,jk)q\displaystyle\|\sum a_{k}e(k,j_{k})\|^{q} =(01(kakqE(k,jk)(s))p/q𝑑s)q/p\displaystyle=\bigg{(}\int_{0}^{1}\bigg{(}\sum_{k}a_{k}^{q}E(k,j_{k})(s)\bigg{)}^{p/q}\ ds\bigg{)}^{q/p}
=akqE(k,jk)p/q.\displaystyle=\bigg{|}\bigg{|}\sum a_{k}^{q}E(k,j_{k})\bigg{|}\bigg{|}_{p/q}.

Now since the e(k,jk)e(k,j_{k})’s are isometric to p\ell_{p},

akqE(k,jk)p/qp/q=iakp=k(akq)p/q=kakqE(k,jk)p/qp/q.\displaystyle\bigg{|}\bigg{|}\sum a_{k}^{q}E(k,j_{k})\bigg{|}\bigg{|}_{p/q}^{p/q}=\sum_{i}a_{k}^{p}=\sum_{k}(a_{k}^{q})^{p/q}=\sum_{k}\|a_{k}^{q}E(k,j_{k})\|_{p/q}^{p/q}.

Since the E(k,jk)E(k,j_{k})’s are all positive and pqp\neq q, by Lemma 2.3, the E(k,jk)E(k,j_{k})’s are disjoint, that is, the e(k,jk)se(k,j_{k})^{\prime}s are base disjoint.

For 1kN1\leq k\leq N, let A(1),,A(n)A(1),...,A(n) be mutually disjoint measurable sets each supporting each E(k,j)E(k,j) for 1jnk1\leq j\leq n_{k}. Then each e(k,j)e(k,j) is supported by A(k)×[0,1]A(k)\times[0,1]. Now we prove (2). Fix kk, Then using similar computations as above, and since the e(k,j)e(k,j)’s for fixed kk generate qmk\ell_{q}^{m_{k}}:

jaje(k,j)q\displaystyle\|\sum_{j}a_{j}e(k,j)\|^{q} =jajqE(k,j)p/q=jajq=jajqE(k,j)p/q\displaystyle=\bigg{|}\bigg{|}\sum_{j}a^{q}_{j}E(k,j)\bigg{|}\bigg{|}_{p/q}=\sum_{j}a_{j}^{q}=\sum_{j}a_{j}^{q}\|E(k,j)\|_{p/q}

By Minkowski’s inequality, as pqp\neq q, equality occurs only when E(k,j)(s)=E(k,j)(s)E(k,j)(s)=E(k,j^{\prime})(s) a.e. for all 1j,jni1\leq j,j^{\prime}\leq n_{i}.

To show the converse, it is enough to give the computation:

k,ja(k,j)e(k,j)\displaystyle\|\sum_{k,j}a(k,j)e(k,j)\| =(01[(k,ja(k,j)e(k,j)(s,t))q𝑑t]p/q𝑑s)1/p\displaystyle=\bigg{(}\int_{0}^{1}\bigg{[}\int\bigg{(}\sum_{k,j}a(k,j)e(k,j)(s,t)\bigg{)}^{q}\ dt\bigg{]}^{p/q}\ ds\bigg{)}^{1/p}
=(k01[j=1ni|a(k,j)|qE(k,j)(s)]p/q𝑑s)1/p\displaystyle=\bigg{(}\sum_{k}\int_{0}^{1}\bigg{[}\sum_{j=1}^{n_{i}}|a(k,j)|^{q}E(k,j)(s)\bigg{]}^{p/q}\ ds\bigg{)}^{1/p}
=(k[j=1nk|a(k,j)|q]p/q01E(k,1)p/q(s)𝑑s)1/p\displaystyle=\bigg{(}\sum_{k}\bigg{[}\sum_{j=1}^{n_{k}}|a(k,j)|^{q}\bigg{]}^{p/q}\int_{0}^{1}E(k,1)^{p/q}(s)\ ds\bigg{)}^{1/p}
=(k[j=1nk|a(k,j)|q]p/q)1/p\displaystyle=\bigg{(}\sum_{k}\bigg{[}\sum_{j=1}^{n_{k}}|a(k,j)|^{q}\bigg{]}^{p/q}\bigg{)}^{1/p}

The following results will allow us to reduce homogeneity diagrams to those in which the atoms e(k,j)e(k,j) of some EB𝒦p,qE\in B\mathcal{K}_{{p},{q}} are mapped by both embeddings to characteristic functions of measurable A(k,j)[0,1]2A(k,j)\subseteq[0,1]^{2}. In fact, we can further simplify such diagrams to cases where EE is generated by such e(k,j)e(k,j)’s which additionally are base-simple, i.e., N[e(k,j)]N[e(k,j)] is a simple function.

Proposition 2.5.

Let 1pq<1\leq p\neq q<\infty and let eS(Lp(Lq))+e\in S(L_{p}(L_{q}))_{+} be an element with full support over [0,1]2[0,1]^{2}. Then there exists a lattice automorphism ϕ\phi from Lp(Lq)L_{p}(L_{q}) to itself such that ϕ(𝟏)=e\phi(\mathbf{1})=e. Furthermore, ϕ\phi can be constructed to bijectively map both simple functions to simple functions and base-simple functions to base-simple functions.

Proof.

The proof is an expansion of the technique used in Lemma 3.3 from [5]. Given a function g(y)Lq+g(y)\in{L_{q}}_{+}, define g~(y)q\tilde{g}(y)_{q} by g~(y)q=0yg(t)q𝑑t\tilde{g}(y)_{q}=\int_{0}^{y}g(t)^{q}\ dt, and for notation, use ex(y)=e(x,y)e_{x}(y)=e(x,y). Since ee has full support, we may assume that for all 0x10\leq x\leq 1, N[e](x)>0N[e](x)>0. From there, Define ϕ\phi by

ϕ(f)(x,y)=f(N[e]~(x)p,e~x(y)qNq[e](x))e(x,y)\phi(f)(x,y)=f\bigg{(}\widetilde{N[e]}(x)_{p},\frac{\tilde{e}_{x}(y)_{q}}{N^{q}[e](x)}\bigg{)}e(x,y)

e0e\geq 0 and the rest of the function definition is a composition, so ϕ\phi is a lattice homomorphism. To show it is also an isometry, simply compute the norm, using substitution in the appropriate places:

ϕ(f)p=\displaystyle\|\phi(f)\|^{p}= 01|01f(N[e]~(x)p,e~x(y)qNq[e](x))qe(x,y)q𝑑y|p/q𝑑x\displaystyle\int_{0}^{1}\bigg{|}\int_{0}^{1}f\bigg{(}\widetilde{N[e]}(x)_{p},\frac{\tilde{e}_{x}(y)_{q}}{N^{q}[e](x)}\bigg{)}^{q}e(x,y)^{q}\ dy\bigg{|}^{p/q}\ dx
=\displaystyle= 01|01f(N[e]~(x)p,y)q𝑑y|p/qNp[e](x)𝑑x\displaystyle\int_{0}^{1}\bigg{|}\int_{0}^{1}f(\widetilde{N[e]}(x)_{p},y)^{q}\ dy\bigg{|}^{p/q}N^{p}[e](x)\ dx
=\displaystyle= 01N[f](N[e]~(x)p)pNp[e](x)𝑑x\displaystyle\int_{0}^{1}N[f](\widetilde{N[e]}(x)_{p})^{p}N^{p}[e](x)\ dx
=\displaystyle= 01Np[f](x)𝑑x=fp.\displaystyle\int_{0}^{1}N^{p}[f](x)\ dx=\|f\|^{p}.

To show surjectivity, let B[0,1]2B\subseteq[0,1]^{2} be a measurable set. Note that any (x,y)[0,1]2(x^{\prime},y^{\prime})\in[0,1]^{2} can be expressed as (N[e]~(x)p,e~x(y)qNq[e](x))(\widetilde{N[e]}(x)_{p},\frac{\tilde{e}_{x}(y)_{q}}{N^{q}[e](x)}) for some x,yx,y, since N[e]~(x)p\widetilde{N[e]}(x)_{p} is an increasing continuous function from 0 to 1, while e~x(y)q\tilde{e}_{x}(y)_{q} is continuously increasing from 0 to Nq[e](x)N^{q}[e](x). Thus there exists BB^{\prime} such that ϕ(𝟏B)=𝟏Be\phi(\mathbf{1}_{B^{\prime}})=\mathbf{1}_{B}\cdot e, implying that ϕ\phi’s image is dense in the band generated by 𝜷(e)=Lp(Lq)\boldsymbol{\beta}(e)=L_{p}(L_{q}) since ee has full support. Therefore, ϕ\phi is also surjective.

Finally, ϕ\phi consists of function composition into ff multiplied by ee, so if ee and ff are simple, then it has a finite image, so if ff is simple, then the product is also simple, ϕ\phi maps simple functions to simple functions, Conversely, if ϕ(f)\phi(f) is simple, then ϕ(f)/e\phi(f)/e is also simple. Thus f(N[e]~(x)p,e~x(y)qN[e](x))f\bigg{(}\widetilde{N[e]}(x)_{p},\frac{\tilde{e}_{x}(y)_{q}}{N[e](x)}\bigg{)} has a finite image. It follows that ff itself has a finite image.

Using similar reasoning, if N[e]N[e] is simple, then whenever N[f]N[f] is simple, N[ϕ(f)]N[\phi(f)] must also be simple, and likewise the converse is true, since by the computation above, N[ϕ(f)](x)=N[f](N[e]~(x)p)N[e](x)N[\phi(f)](x)=N[f](\widetilde{N[e]}(x)_{p})\cdot N[e](x). ∎

3. Approximate Ultrahomogeneity of Lp(Lq)L_{p}(L_{q}) over BLpLqBL_{p}L_{q} spaces

In this section, we show that for any 1pq<1\leq p\neq q<\infty, Lp(Lq)L_{p}(L_{q}) is AUH over B𝒦p,qB\mathcal{K}_{{p},{q}}.

Let 𝐟:=(f1,,fn)\mathbf{f}:=(f_{1},...,f_{n}) and 𝐠:=(g1,,gn)\mathbf{g}:=(g_{1},...,g_{n}) be sequences of measurable functions and let λ\lambda be a measure in \mathbb{R}. Then we say that 𝐟\mathbf{f} and 𝐠\mathbf{g} are equimeasurable if for all λ\lambda-measurable BnB\subseteq\mathbb{R}^{n},

λ(t:𝐟(t)B)=λ(t:𝐠(t)B)\lambda(t:\mathbf{f}(t)\in B)=\lambda(t:\mathbf{g}(t)\in B)

We also say that functions 𝐟\mathbf{f} and 𝐠\mathbf{g} in Lp(Lq)L_{p}(L_{q}) are base-equimeasurable if N(𝐟)N(\mathbf{f}) and N(𝐠)N(\mathbf{g}) are equimeasurable.

Lusky’s main proof in [10] of linear approximate ultrahomogeneity in Lp(0,1)L_{p}(0,1) for p4,6,8,p\neq 4,6,8,... hinges on the equimeasurability of generating elements for two copies of some E=<e1,,en>E=<e_{1},...,e_{n}> in LpL_{p} containing 𝟏\mathbf{1}. But when p=4,6,8,p=4,6,8,..., there exist finite dimensional EE such that two linearly isometric copies of EE in LpL_{p} do not have equimeasurable corresponding basis elements. However, if homogeneity properties are limited to EE with mutually disjoint basis elements, then EE is linearly isometric to pn\ell_{p}^{n}, and for all 1p<1\leq p<\infty, LpL_{p} is AUH over all pn\ell_{p}^{n} spaces. Note that here, an equimeasurability principle (albeit a trivial one) also applies: Any two copies of pn=<e1,,en>\ell_{p}^{n}=<e_{1},...,e_{n}> into Lp(0,1)L_{p}(0,1) with kek=n1/p𝟏\sum_{k}e_{k}=n^{1/p}\cdot\mathbf{1} have (trivially) equimeasurable corresponding basis elements to each other as well.

In the Lp(Lq)L_{p}(L_{q}) setting, similar results arise, except rather than comparing corresponding basis elements fi(e1),,fi(en)f_{i}(e_{1}),...,f_{i}(e_{n}) of isometric copies fi(E)f_{i}(E) of EE, equimeasurability results hold in the LqL_{q}-norms N[fi(ej)]N[f_{i}(e_{j})] under similar conditions, with finite dimensional BLpLqBL_{p}L_{q} lattices taking on a role like pn\ell_{p}^{n} does in LpL_{p} spaces.

The following shows that equimeasurability plays a strong role in the approximate ultrahomogeneity of Lp(Lq)L_{p}(L_{q}) by showing that any automorphism fixing 𝟏\mathbf{1} preserves base-equimeasurability for characteristic functions:

Proposition 3.1.

Suppose pqp\neq q, and let T:Lp(Lq)T:L_{p}(L_{q}) be a lattice automorphism with T(𝟏)=𝟏T(\mathbf{1})=\mathbf{1}. Then there exists a function ϕLp(Lq)\phi\in L_{p}(L_{q}) and a measure preserving transformation ψ\psi over LpL_{p} such that for a.e. x[0,1]x\in[0,1], ϕ(x,)\phi(x,\cdot) is also a measure preserving transformation inducing an isometry over LqL_{q}, and for all ff,

Tf(x,y)=f(ψ(x),ϕ(x,y)).Tf(x,y)=f(\psi(x),\phi(x,y)).

Furthermore, for all measurable B1,,Bn[0,1]2B_{1},...,B_{n}\subseteq[0,1]^{2} with 𝟏Bi\mathbf{1}_{B_{i}}’s mutually disjoint, (𝟏B1,,𝟏Bn)(\mathbf{1}_{B_{1}},...,\mathbf{1}_{B_{n}}) and (T𝟏B1,,T𝟏Bn)(T\mathbf{1}_{B_{1}},...,T\mathbf{1}_{B_{n}}) are base-equimeasurable.

Proof.

By the main result in [13], there exists a strongly measurable function Φ:[0,1]B(Lq)\Phi:[0,1]\rightarrow B(L_{q}), a set isomorphism Ψ\Psi over LpL_{p} (see [13] for a definition on set isomorphisms), and some e(x)Lpe(x)\in L_{p} related to the radon-Nikodym derivative of Ψ\Psi such that

Tf(x)(y)=Φ(x)(e(x)Ψf(x))(y),Tf(x)(y)=\Phi(x)(e(x)\Psi f(x))(y),

and for a.e. xx, Φ(x)\Phi(x) is a linear isometry over LqL_{q}. Observe first that TT sends any characteristic function 1A×[0,1]Lp(Lq)1_{A\times[0,1]}\in L_{p}(L_{q}) constant over yy to characteristic function 𝟏ψ(A)×[0,1]\mathbf{1}_{\psi(A)\times[0,1]} for some ψ(A)[0,1]\psi(A)\subseteq[0,1], so since 1A×[0,1]Lp(Lq)1_{A\times[0,1]}\in L_{p}(L_{q}) is constant over yy, we can just refer to it as 𝟏A\mathbf{1}_{A}. Also, since TT is a lattice isometry, μ(A)=μ(ψ(A))\mu(A)=\mu(\psi(A)), so ψ\psi is measure preserving. Finally, observe that N[𝟏A]=𝟏AN[\mathbf{1}_{A}]=\mathbf{1}_{A}. Thus, for any simple function g:=ci𝟏AiLp(Lq)+g:=\sum c_{i}\mathbf{1}_{A_{i}}\in L_{p}(L_{q})_{+} constant over yy with the AiA_{i}’s mutually disjoint, we have N[g]=gN[g]=g, and Tg=gTg=g^{\prime}. Then for all xx,

N[g](x)=N[Tg](x)=N[Φ(x)(eg)](x)=e(x)N[Φ(x)(g)][x]=|e(x)|N[g](x)N[g^{\prime}](x)=N[Tg](x)=N[\Phi(x)(eg^{\prime})](x)=e(x)N[\Phi(x)(g^{\prime})][x]=|e(x)|N[g^{\prime}](x)

It follows that |e(x)|=1|e(x)|=1. We can thus adjust Φ\Phi by multiplying by 1-1 where e(x)=1e(x)=-1. Note also that Φ\Phi acts as a lattice isometry over LpL_{p} when restricted to elements constant over yy, so by Banach’s theorem in [1], the map Φf(x)\Phi f(x) can be interpreted as Φ(x)(f(ψ(x)))\Phi(x)(\ f(\psi(x))\ ), where ψ\psi is a measure preserving transformation over [0,1][0,1] inducing Ψ\Psi. By Banach’s theorem again for Φ(x)\Phi(x), this Φ\Phi can be interpreted by Φf(x,y)=e(x,y)f(ψ(x),ϕ(x,y))\Phi f(x,y)=e^{\prime}(x,y)f(\psi(x),\phi(x,y)), with ϕ(x,)\phi(x,\cdot) a measure preserving transformation for a.e. xx. But since T𝟏=𝟏T\mathbf{1}=\mathbf{1}, this e(x,y)=1e^{\prime}(x,y)=1 as well.

It remains to prove equimeasurability. Let 𝟏𝐁=(𝟏B1,,𝟏Bn)\mathbf{1}_{\mathbf{B}}=(\mathbf{1}_{B_{1}},...,\mathbf{1}_{B_{n}}), and observe that since for a.e. xx, ϕ(x,)\phi(x,\cdot) is a measure preserving transformation inducing a lattice isometry over LqL_{q}, it follows that

Nq[𝟏Bi](x)=μ(y:(x,y)Bi)=μ(y:(x,ϕ(x,y))Bi),N^{q}[\mathbf{1}_{B_{i}}](x)=\mu(y:(x,y)\in B_{i})=\mu(y:(x,\phi(x,y))\in B_{i}),

While

Nq[T𝟏Bi](x)=μ(y:(ψ(x),ϕ(x,y))Bi)\displaystyle N^{q}[T\mathbf{1}_{B_{i}}](x)=\mu(y:(\psi(x),\phi(x,y))\in B_{i})
=μ(y:(ψ(x),y)Bi)=Nq[𝟏Bi](ψ(x)).\displaystyle=\mu(y:(\psi(x),y)\in B_{i})=N^{q}[\mathbf{1}_{B_{i}}](\psi(x)).

Thus for each A=iAiA=\prod_{i}A_{i} with Ai[0,1]A_{i}\subseteq[0,1] measurable, since ψ\psi is also a measure preserving transformation,

μ(x:(Nq[𝟏𝐁](x)A)=μ(x:(Nq[𝟏𝐁](ψ(x))A)=μ(x:(Nq[T𝟏𝐁](x)A),\displaystyle\mu(x:(N^{q}[\mathbf{1}_{\mathbf{B}}](x)\in A)=\mu(x:(N^{q}[\mathbf{1}_{\mathbf{B}}](\psi(x))\in A)=\mu(x:(N^{q}[T\mathbf{1}_{\mathbf{B}}](x)\in A),

and we are done.

The following theorem describes a comparable equimeasurability property of certain copies of LpLqL_{p}L_{q} in Lp(Lq)L_{p}(L_{q}) for any 1pq<1\leq p\neq q<\infty:

Theorem 3.2.

Let 1pq<1\leq p\neq q<\infty, and suppose that fi:ELp(Lq)f_{i}:E\rightarrow L_{p}(L_{q}) are lattice embeddings with EB𝒦p,qE\in B\mathcal{K}_{{p},{q}} generated by a (k,j)(k,j)-indexed collection of atoms 𝐞:=(e(k,j))k,j\mathbf{e}:=(e(k,j))_{k,j} with 1kn1\leq k\leq n and 1jmk1\leq j\leq m_{k} as described in Proposition 2.2. Suppose also that f(k,je(k,j))=𝟏e(k,j).f(\sum_{k,j}e(k,j))=\mathbf{1}\cdot\|\sum e(k,j)\|. Then (f1(𝐞))(f_{1}(\mathbf{e})) and (f2(𝐞))(f_{2}(\mathbf{e})) are base-equimeasurable.

Proof.

Let η=k,je(k,j)\eta=\|\sum_{k,j}e(k,j)\|, and note first that each 1ηfi(e(k,j))\frac{1}{\eta}f_{i}(e(k,j)) is of the form 𝟏Ai(k,j)\mathbf{1}_{A_{i}(k,j)} for some measurable Ai(k,j)[0,1]2A_{i}(k,j)\subseteq[0,1]^{2}. Second, Nq[𝟏Ai(k,j)](s)=μ(Ai(k,j)(s))N^{q}[\mathbf{1}_{A_{i}(k,j)}](s)=\mu(A_{i}(k,j)(s)) with Ai(k,j)(s)[0,1]A_{i}(k,j)(s)\subseteq[0,1] measurable for a.e. ss, so by Proposition 2.2, for each fixed kk and each j,jj,j^{\prime}, μ(Ai(k,j)(s))=μ(Ai(k,j)(s))=1mk𝟏Ai(k)(s)\mu(A_{i}(k,j)(s))=\mu(A_{i}(k,j^{\prime})(s))=\frac{1}{m_{k}}\mathbf{1}_{A_{i}(k)}(s) with Ai(1),,Ai(n)[0,1]A_{i}(1),...,A_{i}(n)\subseteq[0,1] almost disjoint. It follows that for each appropriate k,jk,j, 1η=1mk1/qμ(Ai(k))1/p\frac{1}{\eta}=\frac{1}{m_{k}^{1/q}}\mu(A_{i}(k))^{1/p}, so μ(Ai(k))=(mk1/qη)p\mu(A_{i}(k))=\bigg{(}\frac{m_{k}^{1/q}}{\eta}\bigg{)}^{p}.

To show equimeasurability, observe that for a.e. tt, we have Nq[𝟏Ai(k,j)](s)=1mkN^{q}[\mathbf{1}_{A_{i}(k,j)}](s)=\frac{1}{m_{k}} iff sAi(k)s\in A_{i}(k), and 0 otherwise. Let 𝐁kmk\mathbf{B}\subseteq\prod_{k}\mathbb{R}^{m_{k}} be a measurable set. Note then that any (k,j)(k,j)-indexed sequence (N[fi(𝐞)](s))(N[f_{i}(\mathbf{e})](s)) is of the form 𝐜𝐬𝐢kmk\mathbf{c^{i}_{s}}\in\prod_{k}\mathbb{R}^{m_{k}} with csi(k,j)=(1mk)1/qc^{i}_{s}(k,j)=\bigg{(}\frac{1}{m_{k}}\bigg{)}^{1/q} for some unique kk, and csi(k,j)=0c^{i}_{s}(k,j)=0 otherwise. It follows then that for some I1,,nI\subseteq{1,...,n},

μ(s:𝐜𝐬𝐢𝐁)=kIμ(Ai(k))=kI(mk1/qη).\mu(s:\mathbf{c^{i}_{s}}\in\mathbf{B})=\sum_{k\in I}\mu(A_{i}(k))=\sum_{k\in I}\bigg{(}\frac{m_{k}^{1/q}}{\eta}\bigg{)}.

Since the above holds independent of our choice of ii, we are done.

Remark 3.3.

The above proof shows much more than base-equimeasurability for copies of B𝒦p,qB\mathcal{K}_{{p},{q}} lattices in Lp(Lq)L_{p}(L_{q}). Indeed, if 𝟏E=<(e(k,j))k,j>\mathbf{1}\in E=<(e(k,j))_{k,j}> with EB𝒦p,qE\in B\mathcal{K}_{{p},{q}}, then each atom is in fact base-simple, and e(k,j)=η𝟏\sum e(k,j)=\eta\cdot\mathbf{1} where η=(kmkp/q)1/p\eta=(\sum_{k}m_{k}^{p/q})^{1/p}. Furthermore, there exist measurable sets A(1),,A(n)A(1),...,A(n) partitioning [0,1][0,1] with μ(A(k))=mkp/qηp\mu(A(k))=\frac{m_{k}^{p/q}}{\eta^{p}} such that N[e(k,j)]=ηmk1/q𝟏A(k)N[e(k,j)]=\frac{\eta}{m_{k}^{1/q}}\mathbf{1}_{A(k)}. Based on this, we can come up with a ”canonical” representation of EE, with e(k,j)η𝟏Wk×Vk,je(k,j)\mapsto\eta\cdot\mathbf{1}_{W_{k}\times V_{k,j}}, where

Wk=[l=1k1μ(A(l)),l=1kμ(A(l))], and Vk,j=[j1mk,jmk].W_{k}=\big{[}\sum_{l=1}^{k-1}\mu(A(l)),\sum_{l=1}^{k}\mu(A(l))\big{]}\text{, and }V_{k,j}=\bigg{[}\frac{j-1}{m_{k}},\frac{j}{m_{k}}\bigg{]}.

This canonical representation will become relevant in later results.

Having characterized representations of lattice in B𝒦p,qB\mathcal{K}_{{p},{q}}, we now move towards proving the AUH result. Before the final proof, we use the following perturbation lemma.

Lemma 3.4.

Let f:ELp(Lq)f:E\rightarrow L_{p}(L_{q}) be a lattice embedding of a lattice E=<e1,,en>E=<e_{1},...,e_{n}>. Then for all ε>0\varepsilon>0, there exists an embedding g:ELp(Lq)g:E\rightarrow L_{p}(L_{q}) such that g(E)g(E) fully supports Lp(Lq)L_{p}(L_{q}) and fg<ε\|f-g\|<\varepsilon.

Proof.

Let Mk=supp(N[f(ek)])\supp(N[f(1n1ek)])M_{k}=supp\big{(}N[f(e_{k})]\big{)}\backslash supp\big{(}N[f(\sum_{1}^{n-1}e_{k})]\big{)}. For each eke_{k}, we will construct eke^{\prime}_{k} disjoint from f(E)f(E) with support in Mk×[0,1]M_{k}\times[0,1]. Let MM^{\prime} be the elements in [0,1]2[0,1]^{2} disjoint from f(E)f(E). Starting with n=1n=1, Observe that MM^{\prime} can be partitioned by MMk×[0,1]:=MkM^{\prime}\cap M_{k}\times[0,1]:=M^{\prime}_{k}. Let

ηk(x,y)=ε1/qN[f(ek)](x)μ(Mk(x))1/q𝟏Mk(x,y).\eta_{k}(x,y)=\varepsilon^{1/q}\frac{N[f(e_{k})](x)}{\mu(M^{\prime}_{k}(x))^{1/q}}\mathbf{1}_{M^{\prime}_{k}}(x,y).

When μ(Mk(x))=0\mu(M^{\prime}_{k}(x))=0, let ηk(x,y)=0\eta_{k}(x,y)=0 as well. Now, let g:ELp(Lq)g^{\prime}:E\rightarrow L_{p}(L_{q}) be the lattice homomorphism induced by

g(ek)=(1ε)1/qf(ek)𝟏Mk+ηn+f(ek)𝟏Mkc.g^{\prime}(e_{k})=(1-\varepsilon)^{1/q}f(e_{k})\cdot\mathbf{1}_{M_{k}}+\eta_{n}+f(e_{k})\cdot\mathbf{1}_{M_{k}^{c}}.

First, we show that gg^{\prime} is an embedding. Observe that for each kk,

Nq[g(ek)](x)=\displaystyle N^{q}[g^{\prime}(e_{k})](x)= ηkq(x,y)+(1ε)f(ek)q(x,y)dy\displaystyle\int\eta^{q}_{k}(x,y)+(1-\varepsilon)f(e_{k})^{q}(x,y)\ dy
=\displaystyle= εNq[f(ek)](x)μ(Mk(x))𝟏Mk(x,y)+(1ε)f(ek)q(x,y)dy\displaystyle\int\varepsilon\frac{N^{q}[f(e_{k})](x)}{\mu(M^{\prime}_{k}(x))}\cdot\mathbf{1}_{M^{\prime}_{k}}(x,y)+(1-\varepsilon)f(e_{k})^{q}(x,y)\ dy
=\displaystyle= εNq[f(ek)](x)+(1ε)f(ek)q(x,y)𝑑y\displaystyle\varepsilon N^{q}[f(e_{k})](x)+(1-\varepsilon)\int f(e_{k})^{q}(x,y)\ dy
=\displaystyle= εNq[f(ek)](x)+(1ε)Nq[f(ek)](x)=Nq[f(ek)](x).\displaystyle\varepsilon N^{q}[f(e_{k})](x)+(1-\varepsilon)N^{q}[f(e_{k})](x)=N^{q}[f(e_{k})](x).

It easily follows that g(E)g^{\prime}(E) is in fact isometric to f(E)f(E), and thus to EE. Furthermore, for every kk,

f(ek)g(ek)=\displaystyle\|f(e_{k})-g^{\prime}(e_{k})\|= 𝟏Mk[(1(1ε)1/q)f(ek)+ηk]\displaystyle\|\mathbf{1}_{M_{k}}[(1-(1-\varepsilon)^{1/q})f(e_{k})+\eta_{k}]\|
\displaystyle\leq (1(1ε)1/q)+ε.\displaystyle(1-(1-\varepsilon)^{1/q})+\varepsilon.

The above can get arbitrarily small.

Now, if supp(N(ek))=[0,1]supp(N(\sum e_{k}))=[0,1], let g=gg=g^{\prime}, and we are done. Otherwise, let M~=kMk\tilde{M}=\cup_{k}M_{k}, and observe that g(ek)\sum g^{\prime}(e_{k}) fully supports Lp(M~;Lq)L_{p}(\tilde{M};L_{q}). Observe also that Lp(Lq)=Lp(M~;Lq)pLp(M~c;Lq)L_{p}(L_{q})=L_{p}(\tilde{M};L_{q})\oplus_{p}L_{p}(\tilde{M}^{c};L_{q}). However, both Lp(M~;Lq)L_{p}(\tilde{M};L_{q}) and Lp(M~c;Lq)L_{p}(\tilde{M}^{c};L_{q}) are lattice isometric to Lp(Lq)L_{p}(L_{q}) itself. So there exists an isometric copy of EE fully supporting Lp(M~c;Lq)L_{p}(\tilde{M}^{c};L_{q}). Let e1,,enLp(M~c;Lq)e^{\prime}_{1},...,e^{\prime}_{n}\in L_{p}(\tilde{M}^{c};L_{q}) be the corresponding basic atoms of this copy, and let g(ei)=(1εp)1/pg(ei)+εeng(e_{i})=(1-\varepsilon^{p})^{1/p}g^{\prime}(e_{i})+\varepsilon\cdot e^{\prime}_{n}. Then for xEx\in E,

g(x)p=(1ε)g(x)p+εxp=xp.\|g(x)\|^{p}=(1-\varepsilon)\|g^{\prime}(x)\|^{p}+\varepsilon\|x\|^{p}=\|x\|^{p}.

Using similar reasoning as in the definition of gg^{\prime}, one also gets gg<(1(1ε)1/p)+ε\|g-g^{\prime}\|<(1-(1-\varepsilon)^{1/p})+\varepsilon, so gg can also arbitrarily approximate ff.

Observe that the lemma above allows us to reduce the approximate homogeneity question down to cases where the copies of a B𝒦p,qB\mathcal{K}_{{p},{q}} lattice fully support Lp(Lq)L_{p}(L_{q}). Combined with Proposition 2.5, we can further reduce the possible scenarios to cases where for each ii, fi(x)=𝟏f_{i}(x)=\mathbf{1} for some xEx\in E. It turns out these reductions are sufficient for constructing a lattice automorphism that makes the homogeneity diagram commute as desired:

Theorem 3.5.

Suppose 1pq<1\leq p\neq q<\infty, and for i=1,2i=1,2, let fi:ELp(Lq)f_{i}:E\rightarrow L_{p}(L_{q}) be a lattice embedding with E:=<(e(k,j))k,j>B𝒦p,qE:=<(e(k,j))_{k,j}>\in B\mathcal{K}_{{p},{q}} and 1kn1\leq k\leq n and 1jmk1\leq j\leq m_{k}. Suppose also that each fi(E)f_{i}(E) fully supports Lp(Lq)L_{p}(L_{q}). Then there exists a lattice automorphism ϕ\phi over Lp(Lq)L_{p}(L_{q}) such that ϕf1=f2\phi\circ f_{1}=f_{2}.

Proof.

Let η=k,je(k,j)\eta=\|\sum_{k,j}e(k,j)\|; by Proposition 2.5, we can assume that for both ii’s, we have fi(k,je(k,j))=η𝟏f_{i}(\sum_{k,j}e(k,j))=\eta\cdot\mathbf{1}. For notation’s sake, let ei(k,j):=fi(e(k,j))e_{i}(k,j):=f_{i}(e(k,j)). By Proposition 2.2, for each ii there exist mutually disjoint sets Ai(1),,Ai(n)A_{i}(1),...,A_{i}(n) partitioning [0,1][0,1] such that for each 1jmk1\leq j\leq m_{k}, supp(N[ei(k,j)])=Ai(k)supp(N[e_{i}(k,j)])=A_{i}(k). In addition, for the sets Ai(k,1),,Ai(k,mk)A_{i}(k,1),...,A_{i}(k,m_{k}), where Ai(k,j):=supp(ei(k,j))A_{i}(k,j):=supp(e_{i}(k,j)), partition Ai(k)×[0,1]A_{i}(k)\times[0,1]. It follows also from the statements in Remark 3.3 that μ(A1(k))=μ(A2(k))\mu(A_{1}(k))=\mu(A_{2}(k)) for each kk and Nq[ei(k,j)](x)=ηqmk𝟏Ai(k)(x)N^{q}[e_{i}(k,j)](x)=\frac{\eta^{q}}{m_{k}}\mathbf{1}_{A_{i}(k)}(x).

To prove the theorem, it is enough to generate lattice automorphisms ϕi\phi^{i} mapping each band 𝜷(ei(k,j))\boldsymbol{\beta}(e_{i}(k,j)) to a corresponding band 𝜷(𝟏Wk×Vk,j)\boldsymbol{\beta}(\mathbf{1}_{W_{k}\times V_{k,j}}) where WkW_{k} and Vk,jV_{k,j} are defined as in Remark 3.3, with 𝟏Ai(k,j)𝟏Wk×Vk,j\mathbf{1}_{A_{i}(k,j)}\mapsto\mathbf{1}_{W_{k}\times V_{k,j}}.

To this end, we make a modified version of the argument in [7, Proposition 2.6] and adopt the notation in Proposition 2.5: construct lattice isometries ψk,ji\psi^{i}_{k,j} from Lp(Ai(k));Lq(Vk,j))L_{p}(A_{i}(k));L_{q}(V_{k,j})) to 𝜷(ek,ji)\boldsymbol{\beta}(e^{i}_{k,j}) with

ψk,ji(f)(x,y)=f(x,(𝟏~Ai(k,j))x(y)q+j1mk)𝟏Ai(k,j)(x,y)\psi^{i}_{k,j}(f)(x,y)=f\bigg{(}x,\big{(}\widetilde{\mathbf{1}}_{A_{i}(k,j)}\big{)}_{x}(y)_{q}+\frac{j-1}{m_{k}}\bigg{)}\mathbf{1}_{A_{i}(k,j)}(x,y)

By similar reasoning as in the proof of Proposition 2.5, ψk,ji\psi^{i}_{k,j} is a lattice embedding. Surjectivity follows as well. Indeed, since Nq[𝟏Ai(k,j)](x)=1mkN^{q}[\mathbf{1}_{A_{i}(k,j)}](x)=\frac{1}{m_{k}}, for a.e. xAi(k)x\in A_{i}(k) the function (𝟏~Ai(k,j))x(y)q+j1mk\big{(}\widetilde{\mathbf{1}}_{A_{i}(k,j)}\big{)}_{x}(y)_{q}+\frac{j-1}{m_{k}} matches [0,1][0,1] continuously to Vk,jV_{k,j} with supp(ei(k,j)(x,))supp(e_{i}(k,j)(x,\cdot)) mapped a.e. surjectively to Vk,jV_{k,j}. So ψk,ji\psi^{i}_{k,j}’s image is dense in 𝜷(ei(k,j))\boldsymbol{\beta}(e_{i}(k,j)).

Observe that ψk,ji\psi^{i}_{k,j} also preserves the random norm NN along the base (that is: N[f]=N[ψk,ji(f)]N[f]=N[\psi^{i}_{k,j}(f)]. Resultantly, the function ψki:=jψj,ki\psi^{i}_{k}:=\oplus_{j}\psi^{i}_{j,k} mapping Lp(Ai(k),Lq(0,1))L_{p}(A_{i}(k),L_{q}(0,1)) to j𝜷(ei(k,j))\oplus_{j}\boldsymbol{\beta}(e_{i}(k,j)) is also a lattice automorphism. Indeed, for f=1mkfjf=\sum_{1}^{m_{k}}f_{j} with fj𝜷(ei(k,j))f_{j}\in\boldsymbol{\beta}(e_{i}(k,j)), one gets

ψki(f)\displaystyle\|\psi^{i}_{k}(f)\| =N[jψk,ji(fj)]p=(jNq[ψk,ji(fj)])1/qp\displaystyle=\bigg{|}\bigg{|}N[\sum_{j}\psi_{k,j}^{i}(f_{j})]\bigg{|}\bigg{|}_{p}=\bigg{|}\bigg{|}\big{(}\sum_{j}N^{q}[\psi^{i}_{k,j}(f_{j})]\big{)}^{1/q}\bigg{|}\bigg{|}_{p}
=(jNq[fj])1/qp=N[jfj]p=f\displaystyle=\bigg{|}\bigg{|}\big{(}\sum_{j}N^{q}[f_{j}]\big{)}^{1/q}\bigg{|}\bigg{|}_{p}=\bigg{|}\bigg{|}N[\sum_{j}f_{j}]\bigg{|}\bigg{|}_{p}=\|f\|

Now let ψi=kψki\psi^{i}=\oplus_{k}\psi^{i}_{k}, and observe that given f=1nfkf=\sum_{1}^{n}f_{k} with fkLp(Ai(k),Lq(0,1))f_{k}\in L_{p}(A_{i}(k),L_{q}(0,1)), since the fkf_{k}’s are base disjoint, we have

ψifp=1nψkifkp=1nfkp=fp.\|\psi^{i}f\|^{p}=\sum_{1}^{n}\|\psi^{i}_{k}f_{k}\|^{p}=\sum_{1}^{n}\|f_{k}\|^{p}=\|f\|^{p}.

Thus ψi\psi^{i} is a lattice automorphism over Lp(Lq)L_{p}(L_{q}) mapping each 1Ai(k)×Vk,j1_{A_{i}(k)\times V_{k,j}} to 𝟏Ai(k,j)\mathbf{1}_{A_{i}(k,j)}.

Use [5, Lemma 3.3] to construct a lattice isometry ρi:LpLp\rho_{i}:L_{p}\rightarrow L_{p} such that for each kk, ρi(𝟏Wk)=𝟏Ai(k)\rho_{i}(\mathbf{1}_{W_{k}})=\mathbf{1}_{A_{i}(k)}. By [1, Ch. 11 Theorem 5.1] this isometry is induced by a measure preserving transformation ρi¯\bar{\rho_{i}} from [0,1] to itself such that ρi(f)(x)=f(ρi¯(x))\rho^{i}(f)(x)=f(\bar{\rho_{i}}(x)). It is easy to show that ρi\rho_{i} induces a lattice isometry with f(x,y)f(ρi¯(x),y)f(x,y)\mapsto f(\bar{\rho_{i}}(x),y). In particular, we have N[ρif](x)=N[f](ρi¯(x))N[\rho_{i}f](x)=N[f](\bar{\rho_{i}}(x)) , and ρi(𝟏Wk×Vk,j)=𝟏Ai(k)×Vk,j\rho_{i}(\mathbf{1}_{W_{k}\times V_{k,j}})=\mathbf{1}_{A_{i}(k)\times V_{k,j}}, now let ϕi(f)=(ψiρi)(f)\phi^{i}(f)=(\psi^{i}\circ\rho^{i})(f), and we are done.

Using the above, we can now show:

Theorem 3.6.

For 1pq<1\leq p\neq q<\infty, the lattice Lp(Lq)L_{p}(L_{q}) is AUH for the class B𝒦p,qB\mathcal{K}_{{p},{q}}.

Proof.

Let fi:ELp(Lq)f_{i}:E\rightarrow L_{p}(L_{q}) as required, and suppose ε>0\varepsilon>0. use Lemma 3.4 to get copies EiE^{\prime}_{i} of fi(E)f_{i}(E) fully supporting Lp(Lq)L_{p}(L_{q}) such that for each atom ekEe_{k}\in E and corresponding atoms ekiEie^{i}_{k}\in E^{\prime}_{i}, we have fi(ek)eki<ε/2\|f_{i}(e_{k})-e^{i}_{k}\|<\varepsilon/2. now use Theorem 3.5 to generate a lattice automorphism ϕ\phi from Lp(Lq)L_{p}(L_{q}) to itself such that ϕ(ek1)=ek2\phi(e^{1}_{k})=e^{2}_{k}. Then

ϕ(f1(ek))f2(ek))ϕ(f1(ek)ek1)+e2kf2(ek)<ε\|\phi(f_{1}(e_{k}))-f_{2}(e_{k}))\|\leq\|\phi(f_{1}(e_{k})-e^{1}_{k})\|+\|e^{2}_{k}-f_{2}(e_{k})\|<\varepsilon

Remark 3.7.

Observe that the doubly atomless Lp(Lq)L_{p}(L_{q}) space is unique among separable BLpLqBL_{p}L_{q} spaces that are AUH over B𝒦p,qB\mathcal{K}_{{p},{q}}. Indeed, this follows from the fact that such a space must be doubly atomless to begin with: let EE be a one dimensional space generated by atom ee and suppose XX is not doubly atomless. Suppose also that EE is embedded by some f1f_{1} into a part of XX supported by some LpL_{p} or LqL_{q} band, and on the other hand is embedded by some f2f_{2} into F:=p2(q2)F:=\ell_{p}^{2}(\ell_{q}^{2}) with f2(e)f_{2}(e) a unit in FF. Then one cannot almost extend f1f_{1} to some lattice embedding g:FXg:F\rightarrow X with almost commutativity.

One can also expand this approximate ultrahomogeneity to separable sublattices with a weaker condition of almost commutativity in the diagram for generating elements: for any BLpLqBL_{p}L_{q} sublattice EE generated by elements <e1,,en>L<e_{1},...,e_{n}>_{L}, for any ε>0\varepsilon>0, and for all lattice embedding pairs fi:ELp(Lq)f_{i}:E\rightarrow L_{p}(L_{q}), there exists a lattice automorphism g:Lp(Lq)Lp(Lq)g:L_{p}(L_{q})\rightarrow L_{p}(L_{q}) such that for all j=1,,nj=1,...,n, g(f2(ej))f1(ej)<ε\|g(f_{2}(e_{j}))-f_{1}(e_{j})\|<\varepsilon.

Theorem 3.8.

For all 1pq<1\leq p\neq q<\infty, The lattice Lp(Lq)L_{p}(L_{q}) is AUH for the class of finitely generated BLpLqBL_{p}L_{q} lattices.

Proof.

Let E=<e1,en>LE=<e_{1},...e_{n}>_{L}, and let fi:ELp(Lq)f_{i}:E\rightarrow L_{p}(L_{q}) be lattice embeddings. We can assume that ek1\|e_{k}\|\leq 1 for each 1in1\leq i\leq n. By Proposition 2.1, EE is the inductive limit of lattices in B𝒦p,qB\mathcal{K}_{{p},{q}}. Given ε>0\varepsilon>0, pick a B𝒦p,qB\mathcal{K}_{{p},{q}} lattice E=<e1,,em>EE^{\prime}=<e^{\prime}_{1},...,e^{\prime}_{m}>\subseteq E such that for each eke_{k}, there is some xkB(E)x_{k}\in B(E^{\prime}) such that xkek<ε3\|x_{k}-e_{k}\|<\frac{\varepsilon}{3}. Each fi|Ef_{i}|_{E^{\prime}} is an embedding into Lp(Lq)L_{p}(L_{q}), so pick an automorphism ϕ\phi over Lp(Lq)L_{p}(L_{q}) such that ϕf1|Ef2|E<ε3\|\phi\circ f_{1}|_{E^{\prime}}-f_{2}|_{E^{\prime}}\|<\frac{\varepsilon}{3}. Then

ϕf1(ek)f2(ek)ϕf1(ekxk)+ϕf1(xk)f2(xk)+f2(xkek)<ε.\|\phi f_{1}(e_{k})-f_{2}(e_{k})\|\leq\|\phi f_{1}(e_{k}-x_{k})\|+\|\phi f_{1}(x_{k})-f_{2}(x_{k})\|+\|f_{2}(x_{k}-e_{k})\|<\varepsilon.

We can also expand homogeneity to include not just lattice embeddings but also disjointness preserving linear isometries, that is, if embeddings fi:ELp(Lq)f_{i}:E\rightarrow L_{p}(L_{q}) are not necessarily lattice homomorphisms but preserve disjointness, then there exists a disjointness preserving linear automorphism ϕ\phi over Lp(Lq)L_{p}(L_{q}) satisfying almost commutativity:

Corollary 3.9.

Lp(Lq)L_{p}(L_{q}) is AUH over finitely generated sublattices in BLp(Lq)BL_{p}(L_{q}) with disjointness preserving embeddings.

Proof.

Use the argument in [5, Proposition 3.2] to show that Lp(Lq)L_{p}(L_{q}) is disjointness preserving AUH over B𝒦p,qB\mathcal{K}_{{p},{q}}. From there, proceed as in the argument in Theorem 3.8 to extend homogeneity over B𝒦p,qB\mathcal{K}_{{p},{q}} to that over BLpLqBL_{p}L_{q}. ∎

4. Approximate Ultrahomogeneity of Lp(Lq)L_{p}(L_{q}) when p/qp/q\notin\mathbb{N}

The above results largely focused approximate ultrahomogeneity over BLpLqBL_{p}L_{q} lattices. What can be said, however, of sublattices of LpLqL_{p}L_{q} spaces? The answer to this question is split into two cases: first, the cases where p/qp/q\notin\mathbb{N}, and the second is when p/qp/q\in\mathbb{N}. We address the first case in this section. It turns out that if p/qp/q\notin\mathbb{N}, then Lp(Lq)L_{p}(L_{q}) is AUH for the class of its finitely generated sublattices. The argument involves certain equimeasurability properties of copies of fixed finite dimensional lattices in Lp(Lq)L_{p}(L_{q}). Throughout, we will refer to the class of sublattices of spaces in B𝒦p,qB\mathcal{K}_{{p},{q}} as simply 𝒦p,q\mathcal{K}_{{p},{q}}, and let 𝒦p,q¯\overline{\mathcal{K}_{{p},{q}}} be the class of finitely generated sublattices of Lp(Lq)L_{p}(L_{q}).

The following result appeared as [7, Proposition 3.2], which is a multi-dimensional version based on Raynaud’s proof for the case of n=1n=1 (see [11, lemma 18]). The approach taken here is a multi-dimensional version of the proof of Lemma 2 in [8].

Theorem 4.1.

Let r=p/qr=p/q\notin\mathbb{N}, and suppose fi:ELp(Lq)f_{i}:E\rightarrow L_{p}(L_{q}) are lattice isometric embeddings with E=<e1,,en>E=<e_{1},...,e_{n}>. Suppose also that f1(x)=f2(x)=𝟏f_{1}(x)=f_{2}(x)=\mathbf{1} for some xE+x\in E_{+}. Then f1(𝐞)f_{1}(\mathbf{e}) and f2(𝐞)f_{2}(\mathbf{e}) are base-equimeasurable.

Throughout the proof, let μ\mu be a measure in some interval InC:=+nI^{n}\subseteq C:=\mathbb{R}^{n}_{+}. To this end, we first show the following:

Lemma 4.2.

Suppose 0<r0<r\notin\mathbb{N}, and α,β\alpha,\beta are positive finite Borel measures on CC such that for all 𝐯C\mathbf{v}\in C with v0>0v_{0}>0,

C|v0+𝐯𝐳|r𝑑α(𝐳)=C|v0+𝐯𝐳|r𝑑β(𝐳)<.\int_{C}|v_{0}+\mathbf{v}\cdot\mathbf{z}|^{r}\ d\alpha(\mathbf{z})=\int_{C}|v_{0}+\mathbf{v}\cdot\mathbf{z}|^{r}\ d\beta(\mathbf{z})<\infty.

Then α=β\alpha=\beta.

Proof.

It is equivalent to prove that the signed measure ν:=αβ=0\nu:=\alpha-\beta=0. First, observe that since |ν|α+β|\nu|\leq\alpha+\beta, and for any 𝐯0\mathbf{v}\geq 0, |v0+𝐯𝐳|rd|ν|(𝐳)<\int|v_{0}+\mathbf{v}\cdot\mathbf{z}|^{r}\ d|\nu|(\mathbf{z})<\infty.

Now, we show by induction on polynomial degree that for all kk\in\mathbb{N}, 𝐯0\mathbf{v}\geq 0, and for all multivariate polynomials P(𝐳)P(\mathbf{z}) of degree kkk^{\prime}\leq k,

C|v0+𝐯𝐳|rkP(𝐳)dν(𝐳)=0.*\int_{C}|v_{0}+\mathbf{v}\cdot\mathbf{z}|^{r-k}P(\mathbf{z})\ d\nu(\mathbf{z})=0.

This is true for the base case k=0k=0 by assumption. Now assume it is true for kk\in\mathbb{N} and let k:=likk^{\prime}:=\sum l_{i}\leq k with 𝐥n\mathbf{l}\in\mathbb{N}^{n}. For notational ease, let 𝐳𝐥=z1l1znln\mathbf{z}^{\mathbf{l}}=z_{1}^{l_{1}}...z_{n}^{l_{n}}. Then for each viv_{i} and 0<t<10<t<1,

R+n𝐳𝐥(v0+𝐯𝐳+zit)rk(v0+𝐯𝐳)rkt𝑑ν(𝐳)=0.\int_{R^{n}_{+}}\mathbf{z}^{\mathbf{l}}\frac{(v_{0}+\mathbf{v}\cdot\mathbf{z}+z_{i}t)^{r-k}-(v_{0}+\mathbf{v}\cdot\mathbf{z})^{r-k}}{t}\ d\nu(\mathbf{z})=0.

Now, if k+1<rk+1<r and t(0,1)t\in(0,1), then

|𝐳𝐥(v0+𝐯𝐳+zit)rk(v0+𝐯𝐳)rkt|𝐳𝐥zi(rk)(v0+𝐯𝐳+vi)rk1\displaystyle\bigg{|}\mathbf{z}^{\mathbf{l}}\frac{(v_{0}+\mathbf{v}\cdot\mathbf{z}+z_{i}t)^{r-k}-(v_{0}+\mathbf{v}\cdot\mathbf{z})^{r-k}}{t}\bigg{|}\leq\mathbf{z}^{\mathbf{l}}z_{i}(r-k)(v_{0}+\mathbf{v}\cdot\mathbf{z}+v_{i})^{r-k-1}
\displaystyle\leq rk𝐯𝐥vi(v0+𝐯𝐳+vi)r\displaystyle\frac{r-k}{\mathbf{v}^{\mathbf{l}}v_{i}}(v_{0}+\mathbf{v}\cdot\mathbf{z}+v_{i})^{r}

Since in this case, 0<rk1<r0<r-k-1<r and |ν|<|\nu|<\infty, the right hand side must also be |ν||\nu|-integrable. On the other hand, If k+1>rk+1>r, then we have

|𝐳𝐥(v0+𝐯𝐳+vit)rk(v0+𝐯𝐳)rkt|<|rk|v0r𝐯𝐥vi\bigg{|}\mathbf{z}^{\mathbf{l}}\frac{(v_{0}+\mathbf{v}\cdot\mathbf{z}+v_{i}t)^{r-k}-(v_{0}+\mathbf{v}\cdot\mathbf{z})^{r-k}}{t}\bigg{|}<|r-k|\frac{v_{0}^{r}}{\mathbf{v}^{\mathbf{l}}v_{i}}

which is also |ν||\nu|-integrable. So now we apply Lebesgue’s differentiation theorem over viv_{i} to get, for any kk\in\mathbb{N} and for each 1in1\leq i\leq n:

C𝐳𝐥zi|v0+𝐯𝐳|rk1𝑑ν(𝐳)=0,\int_{C}\mathbf{z}^{\mathbf{l}}z_{i}|v_{0}+\mathbf{v}\cdot\mathbf{z}|^{r-k-1}\ d\nu(\mathbf{z})=0,

since rr\notin\mathbb{N}. A similar argument, deriving over v0v_{0}, can be made to show that

C|v0+𝐯𝐳|rk1𝑑ν(𝐳)=0\int_{C}|v_{0}+\mathbf{v}\cdot\mathbf{z}|^{r-k-1}\ d\nu(\mathbf{z})=0

One can make linear combinations of the above, which implies line *.

Now for fixed 𝐯>0\mathbf{v}>0, v0>0v_{0}>0 we define a measure Λ\Lambda on CC, where for measurable B+n,B\subseteq\mathbb{R}^{n}_{+},

Λ(B)=ϕ1(B)|v0+𝐯𝐳|r𝑑ν(𝐳).\Lambda(B)=\int_{\phi^{-1}(B)}|v_{0}+\mathbf{v}\cdot\mathbf{z}|^{r}\ d\nu(\mathbf{z}).

where ϕ(𝐳)=1v0+𝐯𝐳𝐳\phi(\mathbf{z})=\frac{1}{v_{0}+\mathbf{v}\cdot\mathbf{z}}\mathbf{z}. It is sufficient to show that Λ=0\Lambda=0. Observe first that ϕ\phi is continuous and injective; indeed, if ϕ(𝐳)=ϕ(𝐰)\phi(\mathbf{z})=\phi(\mathbf{w}), then it can be shown that 𝐯𝐰=𝐯𝐳\mathbf{v}\cdot\mathbf{w}=\mathbf{v}\cdot\mathbf{z}. Thus 𝐰v0+𝐯𝐳=𝐳v0+𝐯𝐳\frac{\mathbf{w}}{v_{0}+\mathbf{v}\cdot\mathbf{z}}=\frac{\mathbf{z}}{v_{0}+\mathbf{v}\cdot\mathbf{z}}, implying that 𝐰=𝐳\mathbf{w}=\mathbf{z}. Resultantly, ϕ(B)\phi(B) for any Borel BB is also Borel, hence we will have shown that for any such BB, ν(B)=0\nu(B)=0 as well, so ν=0\nu=0.

Observe that by choice of 𝐯>0\mathbf{v}>0 and and since (v0+𝐯𝐳)>0(v_{0}+\mathbf{v}\cdot\mathbf{z})>0 for all 𝐳+n\mathbf{z}\in\mathbb{R}^{n}_{+}, have

|Λ|(B)=ϕ1(B)|v0+𝐯𝐳|rd|ν|(𝐳).|\Lambda|(B)=\int_{\phi^{-1}(B)}|v_{0}+\mathbf{v}\cdot\mathbf{z}|^{r}\ d|\nu|(\mathbf{z}).

Using simple functions and the definition of Λ\Lambda, one can show both that for each ii, we have

mi(k):=Cwikd|Λ|(𝐰)=C(v0+𝐯𝐳)rkzikdν|(𝐳)<\displaystyle**\ \ m_{i}(k):=\int_{C}w_{i}^{k}\ d|\Lambda|(\mathbf{w})=\int_{C}(v_{0}+\mathbf{v}\cdot\mathbf{z})^{r-k}z_{i}^{k}\ d\|\nu|(\mathbf{z})<\infty

and also that

Cwik𝑑Λ(𝐰)=C|v0+𝐯𝐳|rkzik𝑑ν(𝐳)=0,\int_{C}w_{i}^{k}\ d\Lambda(\mathbf{w})=\int_{C}|v_{0}+\mathbf{v}\cdot\mathbf{z}|^{r-k}z_{i}^{k}\ d\nu(\mathbf{z})=0,

More generally, if k=ilik=\sum_{i}l_{i}, then

C𝐰𝐥𝑑Λ(𝐰)=C𝐳𝐥|v0+𝐯𝐳|rk𝑑ν(𝐳)=0,\displaystyle\int_{C}\mathbf{w}^{\mathbf{l}}\ d\Lambda(\mathbf{w})=\int_{C}\mathbf{z}^{\mathbf{l}}|v_{0}+\mathbf{v}\cdot\mathbf{z}|^{r-k}\ d\nu(\mathbf{z})=0,

So it follows that CP(𝐰)𝑑Λ(𝐰)=0\int_{C}P(\mathbf{w})\ d\Lambda(\mathbf{w})=0 for all polynomials P(𝐰)P(\mathbf{w}).

Now if k>rk>r and ν0\nu\neq 0, since vi>0v_{i}>0, we then we have

mi(k)\displaystyle m_{i}(k) =C|v0+𝐯𝐳|rkzikdν|(𝐳)\displaystyle=\int_{C}|v_{0}+\mathbf{v}\cdot\mathbf{z}|^{r-k}z_{i}^{k}\ d\|\nu|(\mathbf{z})
C|v0+𝐯𝐳|rvikd|ν|(𝐳)vik|Λ|(C)<\displaystyle\leq\int_{C}|v_{0}+\mathbf{v}\cdot\mathbf{z}|^{r}v_{i}^{-k}\ d|\nu|(\mathbf{z})\leq v_{i}^{-k}|\Lambda|(C)<\infty

so

mi(k)1/2kvi1/2|Λ|(C)1/2k\displaystyle m_{i}(k)^{-1/2k}\geq v_{i}^{1/2}|\Lambda|(C)^{-1/2k}

Thus for each 1in1\leq i\leq n, kmi(k)1/2k=0\sum_{k}m_{i}(k)^{-1/2k}=0. So by [4, Theorem 5.2], |Λ||\Lambda| is the unique positive measure over CC with moment values mi(k)m_{i}(k). Since |Λ|+Λ|\Lambda|+\Lambda yields the same values, and by **, CP(𝐰)d(|Λ|+Λ)(𝐰)=CP(𝐰)d|Λ|(𝐰)\int_{C}P(\mathbf{w})\ d(|\Lambda|+\Lambda)(\mathbf{w})=\int_{C}P(\mathbf{w})\ d|\Lambda|(\mathbf{w}), it follows that Λ=0\Lambda=0, so ν=0\nu=0.

Now we are ready to prove Theorem 4.1.

Proof.

For simplicity of notation, let Fji=Nq[fi(ej)]F^{i}_{j}=N^{q}[f_{i}(e_{j})] and I=[0,1]I=[0,1]. By definition of NN, the support of FjiF^{i}_{j} as well as of μ\mu is the unit interval. Define positive measures αj\alpha_{j} by

αi(B)=μ({tI:𝐅i(t)B})=μ((𝐅i)1(B)).\alpha_{i}(B)=\mu(\{t\in I:\mathbf{F}^{i}(t)\in B\})=\mu((\mathbf{F}^{i})^{-1}(B)).

Now, for any measurable BCB\subseteq C, we have

C𝟏B(𝐳)𝑑ai(𝐳)=αi(B)=μ((𝐅i)1(B))=I(𝟏B𝐅i)(t)𝑑t\displaystyle\int_{C}\mathbf{1}_{B}(\mathbf{z})\ da_{i}(\mathbf{z})=\alpha_{i}(B)=\mu((\mathbf{F}^{i})^{-1}(B))=\int_{I}(\mathbf{1}_{B}\circ\mathbf{F}^{i})(t)\ dt

so for any simple function σ\sigma over CC,

Cσ(𝐳)𝑑αi=01σ𝐅i(t)𝑑t\int_{C}\sigma(\mathbf{z})\ d\alpha_{i}=\int_{0}^{1}\sigma\circ\mathbf{F}^{i}(t)\ dt

Using simple functions to approximate |v0+𝐯𝐳|r|v_{0}+\mathbf{v}\cdot\mathbf{z}|^{r}, and given that |v0+𝐯𝐳|r|v_{0}+\mathbf{v}\cdot\mathbf{z}|^{r} is in L1(C,μ)L_{1}(C,\mu) and the support of μ\mu is the unit interval, it follows that

C|1+𝐯𝐳|r𝑑αi(𝐳)=01|1+𝐯𝐅i(t)|r𝑑t.\int_{C}|1+\mathbf{v}\cdot\mathbf{z}|^{r}\ d\alpha_{i}(\mathbf{z})=\int_{0}^{1}|1+\mathbf{v}\cdot\mathbf{F}^{i}(t)|^{r}\ dt.

It is sufficient now to show that for all 𝐯+n\mathbf{v}\in\mathbb{R}^{n}_{+},

01|1+𝐯𝐅1(t)|r𝑑t=01|1+𝐯𝐅2(t)|r𝑑t.\int_{0}^{1}|1+\mathbf{v}\cdot\mathbf{F}^{1}(t)|^{r}\ dt=\int_{0}^{1}|1+\mathbf{v}\cdot\mathbf{F}^{2}(t)|^{r}\ dt.

.

For i,ji,j and s[0,1]s\in[0,1], let Mij={(s,t):(s,t)supp(fi(ej))}M_{i}^{j}=\{(s,t):(s,t)\in supp(f_{i}(e_{j}))\}, and let Mij(s)={t:(s,t)Mij}M_{i}^{j}(s)=\{t:(s,t)\in M_{i}^{j}\}. By assumption, x=jxjejx=\sum_{j}x_{j}e_{j} with xj>0x_{j}>0, so 𝟏=Nq[fi(x)]=jxjqFji\mathbf{1}=N^{q}[f_{i}(x)]=\sum_{j}x_{j}^{q}F^{i}_{j}. Therefore, since each fif_{i} is an embedding, for all 𝐜+n\mathbf{c}\in\mathbb{R}^{n}_{+},

jcjejp=\displaystyle\|\sum_{j}c_{j}e_{j}\|^{p}= (jcjqFji(s))1/qp\displaystyle\bigg{|}\bigg{|}\big{(}\sum_{j}c_{j}^{q}F^{i}_{j}(s)\big{)}^{1/q}\bigg{|}\bigg{|}_{p}
=\displaystyle= (𝟏+j(cjqxjq)Fji(s))1/qp\displaystyle\bigg{|}\bigg{|}\big{(}\mathbf{1}+\sum_{j}(c_{j}^{q}-x_{j}^{q})F^{i}_{j}(s)\big{)}^{1/q}\bigg{|}\bigg{|}_{p}

Let vj:=cjqxjqv_{j}:=c^{q}_{j}-x_{j}^{q}: then in particular it follows that for all 𝐯0\mathbf{v}\geq 0, we have

01(1+𝐯𝐅1(s))p/q𝑑s=01(1+𝐯𝐅2(s))p/q𝑑s.\int_{0}^{1}\bigg{(}1+\mathbf{v}\cdot\mathbf{F}^{1}(s)\bigg{)}^{p/q}\ ds=\int_{0}^{1}\bigg{(}1+\mathbf{v}\cdot\mathbf{F}^{2}(s)\bigg{)}^{p/q}\ ds.

By Lemma 4.2, we can conclude that α1=α2\alpha_{1}=\alpha_{2}, so 𝐅1\mathbf{F}^{1} and 𝐅2\mathbf{F}^{2} are equimeasurable. ∎

Using Theorem 4.1, we can uniquely characterize lattices in 𝒦p,q\mathcal{K}_{{p},{q}} in a way that parallels Proposition 2.2.

Theorem 4.3.

Suppose that p/qp/q\notin\mathbb{N}, and let ELp(Lq)E\subseteq L_{p}(L_{q}) with E=<e1,,em>E=<e_{1},...,e_{m}>. Then the following hold:

  • E𝒦p,qE\in\mathcal{K}_{{p},{q}} iff there exist mutually disjoint measurable functions ϕ(k,j)S(Lp(Lq))+\phi(k,j)\in S(L_{p}(L_{q}))_{+}, with 1jn1\leq j\leq n and 1kL1\leq k\leq L such that for each jj, ej<(ϕ(k,j))k>=pne_{j}\in<(\phi(k,j))_{k}>=\ell_{p}^{n}, and <(ϕ(k,j))k,j>B𝒦p,q<(\phi(k,j))_{k,j}>\in B\mathcal{K}_{{p},{q}}.

  • Suppose fi:ELp(Lq)f_{i}:E\rightarrow L_{p}(L_{q}) is a lattice embedding with i=1,2i=1,2 and E𝒦p,q.E\in\mathcal{K}_{{p},{q}}. Then there exist embeddings fi:ELp(Lq)f^{\prime}_{i}:E^{\prime}\rightarrow L_{p}(L_{q}) extending fif_{i} such that EB𝒦p,qE^{\prime}\in B\mathcal{K}_{{p},{q}}.

    Lp(Lq){L_{p}(L_{q})}   Lp(Lq){L_{p}(L_{q})}   E{E^{\prime}}      E{E}   f1\scriptstyle{f^{\prime}_{1}}f2\scriptstyle{f^{\prime}_{2}}f1\scriptstyle{f_{1}}f2\scriptstyle{f_{2}}ι\scriptstyle{\iota}
Proof.

For part 1, clearly the reverse direction is true. To prove the main direction, we can suppose that EE fully supports Lp(Lq)L_{p}(L_{q}). If not, recall that the band generated by EE is itself doubly atomless, and hence is lattice isometric to Lp(Lq)L_{p}(L_{q}) itself. Thus, if under these conditions, there is a BLpLqBL_{p}L_{q} sublattices extending EE as in the statement of the theorem, it will also be the case in general.

By Proposition 2.5, we can also suppose that jej=η𝟏\sum_{j}e_{j}=\eta\cdot\mathbf{1}. Now by assumption, since E𝒦p,qE\in\mathcal{K}_{{p},{q}}, then there is an embedding ψ:EE~B𝒦p,q\psi:E\rightarrow\widetilde{E}\in B\mathcal{K}_{{p},{q}} such that each ψ(ej)=kx(k,j)e~(k,j)\psi(e_{j})=\sum_{k}x(k,j)\tilde{e}(k,j), with 1kmk1\leq k\leq m^{\prime}_{k}. Without loss of generality we may also drop any e~(k,j)\tilde{e}(k,j)’s disjoint from ψ(E)\psi(E) and assume that ψ(E)\psi(E) fully supports E~\widetilde{E}. Now E~\widetilde{E} is a B𝒦p,qB\mathcal{K}_{{p},{q}} lattice admitting a canonical representation in Lp(Lq)L_{p}(L_{q}) as described in Theorem 3.2 and Remark 3.3.

So we can assume that ψ\psi embeds EE into Lp(Lq)L_{p}(L_{q}) in such a way that ψ(E)\psi(E) fully supports it and each ψ(ej)\psi(e_{j}) is both simple and base-simple. Now, use Proposition 2.5 to adjust ψ\psi into an automorphism over Lp(Lq)L_{p}(L_{q}) such that ψ(ej)=η𝟏\psi(\sum e_{j})=\eta\cdot\mathbf{1} in a way that preserves both simplicity and base-simplicity. By Theorem 4.1, ψ(𝐞)\psi(\mathbf{e}) and 𝐞\mathbf{e} are base-equimeasurable. Since the ψ(k,j)s\psi(k,j)^{\prime}s are base-simple, there exist tuples 𝐬𝟏,,𝐬𝐋m\mathbf{s^{1}},...,\mathbf{s^{L}}\in\mathbb{R}^{m} such that for a.e. t[0,1]t\in[0,1], there is some kLk\leq L such that N[𝐞](t)=𝐬𝐤N[\mathbf{e}](t)=\mathbf{s^{k}}. By equimeasurability, the same is true for N[ψ(𝐞)](t)N[{\psi(\mathbf{e})}](t).

Let 𝐒𝐤={t:N[𝐞](t)=𝐬𝐤}\mathbf{S^{k}}=\{t:N[\mathbf{e}](t)=\mathbf{s^{k}}\}, and let Sjk=𝐒𝐤×[0,1]supp(ej)S^{k}_{j}=\mathbf{S^{k}}\times[0,1]\cap supp(e_{j}). Let 𝐒¯𝐤={t:N[ψ(𝐞)](t)=𝐬𝐤}\mathbf{\overline{S}^{k}}=\{t:N[\psi(\mathbf{e})](t)=\mathbf{s^{k}}\} with S¯jk\overline{S}^{k}_{j} defined similarly. Note that each 𝟏Sjk\mathbf{1}_{S^{k}_{j}} is also base-characteristic, as N[𝟏Sjk]=cjk𝟏𝐒𝐤N[\mathbf{1}_{S^{k}_{j}}]=c^{k}_{j}\mathbf{1}_{\mathbf{S^{k}}} for some cjk>0c^{k}_{j}>0, so for fixed kk and for any j,jmkj,j^{\prime}\leq m_{k}, we must have that N[𝟏Sjk]N[\mathbf{1}_{S^{k}_{j}}] and N[𝟏Sjk]N[\mathbf{1}_{S^{k}_{j^{\prime}}}] are scalar multiples of each other. Thus for each appropriate pair (k,j)(k,j) with sjk>0s^{k}_{j}>0, define ϕ(k,j)\phi(k,j) by 𝟏Sjk𝟏Sjk\frac{\mathbf{1}_{S^{k}_{j}}}{\|\mathbf{1}_{S^{k}_{j}}\|}. By definition of 𝐒𝐤\mathbf{S^{k}}, for any kkk\neq k^{\prime} and any appropriate j,jj,j^{\prime}, ϕ(k,j)\phi(k,j) and ϕ(k,j)\phi(k^{\prime},j^{\prime}) are fiber-disjoint, and N[ϕ(k,j)]=N[ϕ(k,j)]N[\phi(k,j)]=N[\phi(k,j^{\prime})]. Thus by Proposition 2.2, <(ϕ(k,j))k,j>B𝒦p,q<(\phi(k,j))_{k,j}>\in B\mathcal{K}_{{p},{q}}.

To prove part 2, Observe first that we have already essentially proven part 2 in the case that f1=Idf_{1}=Id and f2=ψf_{2}=\psi. To show the general case, we first assume that for each ii, fi(ej)\sum f_{i}(e_{j}) maps to 𝟏\mathbf{1}. Now, by Theorem 4.1, f1(𝐞)f_{1}(\mathbf{e}) and f2(𝐞)f_{2}(\mathbf{e}) are also base-equimeasurable, but by the procedure for part 1, we also know that each fi(ej)f_{i}(e_{j}) is also base-simple. Define 𝐬𝟏,,𝐬𝐋\mathbf{s^{1}},...,\mathbf{s^{L}} as above, and Let 𝐒𝐤(i)={t:N[fi(𝐞)](t)=𝐬𝐤}\mathbf{S^{k}}(i)=\{t:N[f_{i}(\mathbf{e})](t)=\mathbf{s^{k}}\}. Define similarly Sjk(i)S^{k}_{j}(i) and the associated characteristic functions ϕi(k,j)\phi_{i}(k,j) for appropriate pairs k,jk,j such that 1kl1\leq k\leq l and sjk:=ϕi(k,j)fi(ej)>0s^{k}_{j}:=\|\phi_{i}(k,j)\wedge f_{i}(e_{j})\|>0. Note first that

fi(ej)=k:sk(j)>0sjkϕi(k,j).f_{i}(e_{j})=\sum_{k:s_{k}(j)>0}s^{k}_{j}\phi_{i}(k,j).

Second, observe that by equimeasurability, the eligible pairs (k,j)(k,j) are the same for i=1,2i=1,2. Let Ei=<(ϕi(k,j))k,j>E^{\prime}_{i}=<(\phi_{i}(k,j))_{k,j}>. Clearly EiB𝒦p,qE^{\prime}_{i}\in B\mathcal{K}_{{p},{q}}, and since the eligible pairs (k,j)(k,j) are the same, E1E^{\prime}_{1} and E2E^{\prime}_{2} are isometric to each other. Let EE^{\prime} be one of the EiE^{\prime}_{i}’s and let fi:ELp(Lq)f^{\prime}_{i}:E^{\prime}\rightarrow L_{p}(L_{q}) be the expected embedding mapping EE^{\prime} to EiE^{\prime}_{i}, and we are done. ∎

From here, we can now easily extend Theorem 3.5 to lattices in 𝒦p,q\mathcal{K}_{{p},{q}}:

Corollary 4.4.

Suppose p/qp/q\notin\mathbb{N} and suppose fi:ELp(Lq)f_{i}:E\rightarrow L_{p}(L_{q}) are lattice embeddings from EKp,qE\in K_{p,q} with fi(E)f_{i}(E) fully supporting Lp(Lq)L_{p}(L_{q}). Then there exists a lattice automorphism ϕ\phi over Lp(Lq)L_{p}(L_{q}) such that f2=ϕf1f_{2}=\phi\circ f_{1}.

Proof.

Use Theorem 4.3 to generate a B𝒦p,qB\mathcal{K}_{{p},{q}} lattice EE^{\prime} containing EE and lattice embeddings fi:ELp(Lq)f^{\prime}_{i}:E^{\prime}\rightarrow L_{p}(L_{q}) such that fi|E=fif^{\prime}_{i}|_{E}=f_{i}. Clearly each fi(E)f^{\prime}_{i}(E^{\prime}) fully supports Lp(Lq)L_{p}(L_{q}). Now apply Theorem 3.5 to generate an automorphism ϕ\phi over Lp(Lq)L_{p}(L_{q}) with ϕf1=f2\phi\circ f^{\prime}_{1}=f^{\prime}_{2}. Clearly ϕf1=f2\phi\circ f_{1}=f_{2} as well.

When p/qp/q\notin\mathbb{N}, using Theorem 4.3, we can show that the same holds with the more general class 𝒦p,q\mathcal{K}_{{p},{q}}. However, we can make an even stronger claim by showing that homogeneity holds for any finite dimensional sublattice of Lp(Lq)L_{p}(L_{q}). This is done using the following result, which gives a standard way of approximating finite dimensional sublattices of Lp(Lq)L_{p}(L_{q}) with lattices in 𝒦p,q\mathcal{K}_{{p},{q}}.

Lemma 4.5.

Suppose p/qp/q\notin\mathbb{N}, and let fi:ELp(Lq)f_{i}:E\rightarrow L_{p}(L_{q}) be embeddings with E=<e1,,en>E=<e_{1},...,e_{n}>. Then for all ε>0\varepsilon>0, there exists a 𝒦p,q\mathcal{K}_{{p},{q}} lattice E=<e1,,en>E^{\prime}=<e^{\prime}_{1},...,e^{\prime}_{n}> and embeddings gi:ELp(Lq)g_{i}:E^{\prime}\rightarrow L_{p}(L_{q}) such gi(E)g_{i}(E^{\prime}) fully supports Lp(Lq)L_{p}(L_{q}) and for each nn, fi(en)gi(en)<ε\|f_{i}(e_{n})-g_{i}(e^{\prime}_{n})\|<\varepsilon.

Proof.

We can assume each fi(E)f_{i}(E) fully supports Lp(Lq)L_{p}(L_{q}): given ε>0\varepsilon>0, use Lemma 3.4 to get copies of EE sufficiently close to each fi(E)f_{i}(E) with full support. We then also assume that fi(1nek)=𝟏f_{i}(\sum_{1}^{n}e_{k})=\mathbf{1} using Proposition 2.5.

By Theorem 4.1, f1(𝐞)f_{1}(\mathbf{e}) and f2(𝐞)f_{2}(\mathbf{e}) are base-equimeasurable. In particular, given any measurable CnC\in\mathbb{R}^{n}, one has μ(t:N[f1(𝐞)](t)C)=μ(t:N[f2(𝐞)](t)C)\mu(t:N[f_{1}(\mathbf{e})](t)\in C)=\mu(t:N[f_{2}(\mathbf{e})](t)\in C). Now pick an almost disjoint partition C1,,CmC_{1},...,C_{m} of S(1n)S(\ell_{1}^{n}), where each CiC_{i} is closed, has relatively non-empty interior, and is of diameter less than ε2n\frac{\varepsilon}{2n}. Let Dki={t:N[fi(𝐞)](t)Ci\ji1Cj}D_{k}^{i}=\{t:N[f_{i}(\mathbf{e})](t)\in C_{i}\backslash\cup_{j}^{i-1}C_{j}\}. Then by equimeasurability, μ(Dk1)=μ(Dk2)\mu(D_{k}^{1})=\mu(D_{k}^{2}). For each kk, pick some 𝐬𝐤=(s1k,,snk)Ck\mathbf{s^{k}}=(s^{k}_{1},...,s^{k}_{n})\in C_{k}, and for each xDkix\in D_{k}^{i}, let

e¯ji(x,y)=sjkN[fi(ej)](x)fi(ej)(x,y).\overline{e}^{i}_{j}(x,y)=\frac{s^{k}_{j}}{N[f_{i}(e_{j})](x)}f_{i}(e_{j})(x,y).

Observe that je¯jijfi(ej)<ε\|\sum_{j}\overline{e}^{i}_{j}-\sum_{j}f_{i}(e_{j})\|<\varepsilon, and N[e¯ji](x)=sjkN[\overline{e}^{i}_{j}](x)=s_{j}^{k} for xDkix\in D^{i}_{k}.

Consider now the lattice E=<e¯j1,,e¯n1>E^{\prime}=<\overline{e}^{1}_{j},...,\overline{e}^{1}_{n}>. Now, for any linear combination aje¯ji\sum a_{j}\overline{e}^{i}_{j}, we have, as in the argument in Proposition 2.5, that

aje¯jip=kM(j(ajsjk)q)p/q\displaystyle\|\sum a_{j}\overline{e}^{i}_{j}\|^{p}=\sum_{k}^{M}(\sum_{j}(a_{j}s^{k}_{j})^{q})^{p/q}

implying that aje¯j1=aje¯j2\|\sum a_{j}\overline{e}^{1}_{j}\|=\|\sum a_{j}\overline{e}^{2}_{j}\|. It follows both that EE^{\prime} embeds into pM(qn)\ell_{p}^{M}(\ell_{q}^{n}), implying that it is a 𝒦p,q\mathcal{K}_{{p},{q}} lattice, and it is isometric to the lattice generated by the e¯j2\overline{e}^{2}_{j}’s. Let ej=e¯j1e^{\prime}_{j}=\overline{e}^{1}_{j}, and define gi:ELp(Lq)g_{i}:E^{\prime}\rightarrow L_{p}(L_{q}) as the maps generated by gi(ej)=e¯jig_{i}(e^{\prime}_{j})=\overline{e}^{i}_{j}. Clearly these are lattice embeddings and fi(ej)gi(ej)<ε\|f_{i}(e_{j})-g_{i}(e^{\prime}_{j})\|<\varepsilon.

Theorem 4.6.

For all 1p,q<1\leq p,q<\infty with p/qp/q\notin\mathbb{N}, the lattice Lp(Lq)L_{p}(L_{q}) is AUH for the class of finite dimensional sublattices of LpLqL_{p}L_{q} lattices.

Proof.

It is sufficient to show that the result is true over generation by basic atoms. Let fi:ELp(Lq)f_{i}:E\rightarrow L_{p}(L_{q}) be two embeddings with E=<e1,,en>E=<e_{1},...,e_{n}>. Use Lemma 4.5 to find gi:ELp(Lq)g_{i}:E^{\prime}\rightarrow L_{p}(L_{q}), with E:=<e1,,en>𝒦p,qE^{\prime}:=<e^{\prime}_{1},...,e^{\prime}_{n}>\in\mathcal{K}_{{p},{q}}, gi(ek)fi(ek)<ε/2\|g_{i}(e^{\prime}_{k})-f_{i}(e_{k})\|<\varepsilon/2, and each gi(E)g_{i}(E^{\prime}) fully supporting Lp(Lq)L_{p}(L_{q}). Then by Lemma 4.4, there exists an automorphism ϕ:Lp(Lq)Lp(Lq)\phi:L_{p}(L_{q})\rightarrow L_{p}(L_{q}) such that ϕg1=g2\phi\circ g_{1}=g_{2}. Note then that ϕ(f1(ek))f2(ek)ϕ(f1(ek)g1(ek))+f2(ek)g2(ek)<ε\|\phi(f_{1}(e_{k}))-f_{2}(e_{k})\|\leq\|\phi(f_{1}(e_{k})-g_{1}(e^{\prime}_{k}))\|+\|f_{2}(e_{k})-g_{2}(e^{\prime}_{k})\|<\varepsilon. ∎

In a manner similar to that of Theorem 3.8, we can also extend the AUH property to finitely generated sublattices of Lp(Lq)L_{p}(L_{q}) as well:

Theorem 4.7.

For all 1p,q<1\leq p,q<\infty with p/qp/q\notin\mathbb{N}, The lattice Lp(Lq)L_{p}(L_{q}) is AUH for the class 𝒦p,q¯\overline{\mathcal{K}_{{p},{q}}} of its finitely generated lattices.

Proof.

Suppose ELp(Lq)E\subseteq L_{p}(L_{q}) is finitely generated. Then since EE is order continuous and separable, it is the inductive limit of finite dimensional lattices as well, so pick a finite dimensional EE^{\prime} with elements sufficiently approximating the generating elements of EE, and proceed with the same proof as in Theorem 3.8. ∎

The argument used in Corollary 3.9 can also be used to show:

Corollary 4.8.

For p/qp/q\notin\mathbb{N}, Lp(Lq)L_{p}(L_{q}) is disjointness preserving AUH over 𝒦p,q¯\overline{\mathcal{K}_{{p},{q}}}.

Remark 4.9.

Lp(Lq)L_{p}(L_{q}) for p/qp/q\notin\mathbb{N} is AUH over the entire class of its finitely generated sublattices, a property which is equivalent to such a class being a metric Fraïssé class with Lp(Lq)L_{p}(L_{q}) as its Fraïssé limit. Recall that a class 𝒦\mathcal{K} of finitely generated lattices is Fraïssé if it satisfies the following properties:

  1. (1)

    Hereditary Property (HP): 𝒦\mathcal{K} is closed under finitely generated sublattices.

  2. (2)

    Joint Embedding Property (JEP): any two lattices in 𝒦\mathcal{K} lattice embed into a third in 𝒦\mathcal{K}.

  3. (3)

    Continuity Property (CP): any lattice operation symbol are continuous with respect to the Fraïssé pseudo-metric d𝒦d^{\mathcal{K}} in [2, Definition 2.11].

  4. (4)

    Near Amalgamation Property (NAP): for any lattices E=<e1,en>LE=<e_{1},...e_{n}>_{L}, F1F_{1} and F2F_{2} in 𝒦\mathcal{K} with lattice embeddings fi:EFif_{i}:E\rightarrow F_{i}, and for all ε>0\varepsilon>0, there exists a G𝒦G\in\mathcal{K} and embeddings gi:FiGg_{i}:F_{i}\rightarrow G such that g1f1(ek)g2f2(ek)<ε.\|g_{1}\circ f_{1}(e_{k})-g_{2}\circ f_{2}(e_{k})\|<\varepsilon.

  5. (5)

    Polish Property (PP): The Fraïssé pseudo-metric d𝒦d^{\mathcal{K}} is separable and complete in 𝒦n\mathcal{K}_{n} (the 𝒦\mathcal{K}-structures generated by nn many elements).

Now clearly the finitely generated sublattices of Lp(Lq)L_{p}(L_{q}) fulfill the first two properties, and the third follows from the lattice and linear operations having moduli of continuity independent of lattice geometry. In addition, if one can show that the class 𝒦\mathcal{K} has the NAPNAP, has some separable XX which is universal for 𝒦\mathcal{K}, and its NAP amalgamate lattices can be chosen so that they are closed under inductive limits, then one can prove that 𝒦\mathcal{K} also has the Polish Property (a technique demonstrated in [14, Theorem 4.1] and more generally described in Section 2.5 of [9]). The main difficulty in proving that a class of lattices 𝒦\mathcal{K} is a Fraïssé class is in showing that it has the NAP. However, thanks to Theorem 4.7, we have

Corollary 4.10.

𝒦p,q¯\overline{\mathcal{K}_{{p},{q}}} has the NAP.

Theorem 4.7 implies an additional collection of AUH Banach lattices to the currently known AUH Banach lattices: namely LpL_{p} for 1p<1\leq p<\infty,the Gurarij M-space \mathcal{M} discovered in [5], and the Gurarij lattice discovered in [14].

However, if one considers classes of finite dimensional Banach spaces with Fraïssé limits using linear instead of lattice embeddings, the only known separable AUH Banach spaces are the Gurarij space and LpL_{p} for p4,6,8,p\neq 4,6,8,..., and it is currently unknown if there are other Banach spaces that are AUH over its finite dimensional subspaces with linear embeddings. Certain combinations of pp and qq are also ruled out for Lp(Lq)L_{p}(L_{q}) as a potential AUH candidate as discussed in Problem 2.9 of [5]: in particular, when 1p,q<21\leq p,q<2, Lp(Lq)L_{p}(L_{q}) cannot be linearly AUH.

5. Failure of homogeneity for p/qp/q\in\mathbb{N}

Recall that when E=<e1,,en>B𝒦p,qE=<e_{1},...,e_{n}>\in B\mathcal{K}_{{p},{q}} is embedded into Lp(Lq)L_{p}(L_{q}) through f1,f2f_{1},f_{2}, then we can achieve almost commutativity for any pqp\neq q. However, the automorphism in Theorem 3.6 clearly preserves the equimeasurability of the generating basic atoms of fi(E)f_{i}(E) as it fixes 𝟏\mathbf{1}.

In this section, we show that the results of Section 4 do not hold when p/qp/q\in\mathbb{N}. The first results in this section show that when some eLp(Lq)+e\in L_{p}(L_{q})_{+} is sufficiently close to 𝟏\mathbf{1}, the automorphism originally used in the argument of Proposition 2.5 sending 𝟏\mathbf{1} to ee also perturbs selected functions piecewise continuous on their support in a controlled way. Second, Theorem 4.1 does not hold, and thus we cannot infer equimeasurability for arbitrary finite dimensional sublattices of Lp(Lq)L_{p}(L_{q}). Finally, we use these results to strengthen the homogeneity property for any Lp(Lq)L_{p}(L_{q}) lattice assumed to be AUH, and then show that when p/qp/q\in\mathbb{N}, Lp(Lq)L_{p}(L_{q}) does not fulfill this stronger homogeneity property, and thus cannot be AUH.

Lemma 5.1.

Let 1pq<1\leq p\neq q<\infty, and let <f1,,fn>Lp(Lq)<f_{1},...,f_{n}>\subseteq L_{p}(L_{q}) be such that fi=𝟏\sum f_{i}=\mathbf{1}. Suppose also that for a.e. xx, fk(x,)=𝟏[gk(x),gk+1(x)]f_{k}(x,\cdot)=\mathbf{1}_{[g_{k}(x),g_{k+1}(x)]} where each gkg_{k} has finitely many discontinuities. Let ε>0\varepsilon>0, and let eS(Lp(Lq))+e\in S(L_{p}(L_{q}))_{+} fully support Lp(Lq)L_{p}(L_{q}). Consider

ϕ(f)(x,y)=f(N[e]~(x)p,e~x(y)qNq[e](x))e(x,y)\phi(f)(x,y)=f\bigg{(}\widetilde{N[e]}(x)_{p},\frac{\widetilde{e}_{x}(y)_{q}}{N^{q}[e](x)}\bigg{)}e(x,y)

which is the lattice isometry defined in Proposition 2.5 mapping 𝟏\mathbf{1} to ee.

Then there exists δ\delta such that if 𝟏e<δ\|\mathbf{1}-e\|<\delta, then for each kk, we have that ϕ(fk)fk<ε\|\phi(f_{k})-f_{k}\|<\varepsilon.

Proof.

We can assume ε<1\varepsilon<1. Let K[0,1]K\subseteq[0,1] be a closed set such that for 1kn+11\leq k\leq n+1, gk|Kg_{k}|_{K} is continuous and μ(K)>1ε\mu(K)>1-\varepsilon. Pick δ<ε\delta^{\prime}<\varepsilon such that for any x,xKx,x^{\prime}\in K, if |xx|<δ|x-x^{\prime}|<\delta^{\prime}, then |gk(x)gk(x)|<ε/4|g_{k}(x)-g_{k}(x^{\prime})|<\varepsilon/4. Now, let δ<δ2p\delta<{\delta^{\prime}}^{2p} be such that 1δ4(1δ)p<(1+δ)p<1+δ41-\frac{\delta^{\prime}}{4}\leq(1-\delta)^{p}<(1+\delta)^{p}<1+\frac{\delta^{\prime}}{4}, and suppose 𝟏e<δ\|\mathbf{1}-e\|<\delta. Observe that for each xx, we have N[𝟏e]~(x)p<δ\widetilde{N[\mathbf{1}-e]}(x)_{p}<\delta. For each 1kn1\leq k\leq n, let

f~n(x,y)=f(N[e]~(x)p,e~x(y)qNq[e](x)).\widetilde{f}_{n}(x,y)=f\bigg{(}\widetilde{N[e]}(x)_{p},\frac{\widetilde{e}_{x}(y)_{q}}{N^{q}[e](x)}\bigg{)}.

Observe that f~kϕ(fk)<δ<ε/4\|\widetilde{f}_{k}-\phi(f_{k})\|<\delta<\varepsilon/4, so it is enough to show that f~kfk\|\widetilde{f}_{k}-f_{k}\| is sufficiently small as well.

To this end, first note that since ff is being composed with increasing continuous functions in both arguments, each f~n(x,)\widetilde{f}_{n}(x,\cdot) is also the characteristic function of an interval: indeed, we have piecewise continuous g~1,,g~n+1\widetilde{g}_{1},...,\widetilde{g}_{n+1} with g~k(x):=g(N[e]~(x)p)\widetilde{g}_{k}(x):=g(\widetilde{N[e]}(x)_{p}) and g~n+1(x)=1\widetilde{g}_{n+1}(x)=1 such that for each kk, f~k(x,y)=𝟏[g~k(x),g~k+1(x)](y)\widetilde{f}_{k}(x,y)=\mathbf{1}_{[\widetilde{g}_{k}(x),\widetilde{g}_{k+1}(x)]}(y). Also observe that for M:={xK:N[e1](x)<δ}M:=\{x\in K:N[e-1](x)<\delta\}, we have μ(M))>1δε\mu(M))>1-\delta^{\prime}-\varepsilon. In addition, as

fkf~kp=N[fkf~k]pp=μ(D(x))p𝑑x,\|f_{k}-\widetilde{f}_{k}\|^{p}=\|N[f_{k}-\widetilde{f}_{k}]\|^{p}_{p}=\int\mu(D(x))^{p}\ dx,

Where Dk(x)={y:fk(x,y)f~k(x,y)}D_{k}(x)=\{y:f_{k}(x,y)\neq\widetilde{f}_{k}(x,y)\}. The above set up, in combination with the triangle inequality properties of NN, leads us to the following inequalities:

  • For all 0x10\leq x\leq 1, |N[e]~(x)px|<δ|\widetilde{N[e]}(x)_{p}-x|<{\delta}.

  • For all xMx\in M , |N[e](x)1|<δ|N[e](x)-1|<\delta.

  • For all xMx\in M and 0y10\leq y\leq 1, |e~x(y)qy|<δ2|\widetilde{e}_{x}(y)_{q}-y|<\frac{\delta^{\prime}}{2}.

  • For all xMx\in M and 0y10\leq y\leq 1, if y:=e~x(y)qNq[e](x)y^{\prime}:=\frac{\widetilde{e}_{x}(y)_{q}}{N^{q}[e](x)}, then |yex(y)q|<δ2|y^{\prime}-e_{x}(y)_{q}|<\frac{\delta^{\prime}}{2} (which implies with the above that |yy|<δ|y-y^{\prime}|<\delta^{\prime}).

We now show that the above implies that Dk(x)<2εD_{k}(x)<2\varepsilon. Observe first that for all xMx\in M, if fk(x,y)f~k(x,y)f_{k}(x,y)\neq\widetilde{f}_{k}(x,y) it must be because, but y[g~k(x),g~k+1(x)]y^{\prime}\notin[\widetilde{g}_{k}(x),\widetilde{g}_{k+1}(x)], or vice versa. In either case, it can be shown that either |ygk(x)|<δ+ε4|y-g_{k}(x)|<\delta+\frac{\varepsilon}{4} or |ygk+1(x)|<δ+ε4|y-g_{k+1}(x)|<\delta+\frac{\varepsilon}{4}. Suppose y[gk(x),gk+1(x)]y\in[g_{k}(x),g_{k+1}(x)] and y<g~k(x)y^{\prime}<\widetilde{g}_{k}(x) (a similar proof will work in the case that y>g~k+1(x)y^{\prime}>\widetilde{g}_{k+1}(x). Then since y>gk(x)y>g_{k}(x), |yy|δ|y-y^{\prime}|\leq\delta^{\prime}, and |gk(x)g~k(x)|<ε4|g_{k}(x)-\widetilde{g}_{k}(x)|<\frac{\varepsilon}{4},

0ygk(x)=(yy)+(yg~k(x))+(g~k(x)gk(x))<δ+ε4.0\leq y-g_{k}(x)=(y-y^{\prime})+(y^{\prime}-\widetilde{g}_{k}(x))+(\widetilde{g}_{k}(x)-g_{k}(x))<\delta+\frac{\varepsilon}{4}.

It follows then that accounting for both ends of the interval [gk(x),gk+1(x)][g_{k}(x),g_{k+1}(x)] and for xMx\in M, we have Dk(x)<2εD_{k}(x)<2\varepsilon. Resultantly,

fkf~kp=Mμ(D(x))p𝑑x+Mcμ(D(x))p𝑑x<(2ε)p+δp<3εp,\displaystyle\|f_{k}-\widetilde{f}_{k}\|^{p}=\int_{M}\mu(D(x))^{p}\ dx+\int_{M^{c}}\mu(D(x))^{p}\ dx<(2\varepsilon)^{p}+\delta^{p}<3\varepsilon^{p},

which can be made arbitrarily small.

Theorem 5.2.

Let 1pq<1\leq p\neq q<\infty and suppose Lp(Lq)L_{p}(L_{q}) is AUH over its finite dimensional sublattices. Let fi:ELp(Lq)f_{i}:E\rightarrow L_{p}(L_{q}) be lattice embeddings with E=<e1,,en>E=<e_{1},...,e_{n}> such that fi(x)=𝟏f_{i}(x)=\mathbf{1} for some xEx\in E. Then for all ε>0\varepsilon>0, there exists an automorphism ϕ\phi fixing 𝟏\mathbf{1} such that ϕf1f2<ε\|\phi f_{1}-f_{2}\|<\varepsilon.

Proof.

Assume the above, and pick E=<e1,,em>Lp(Lq)E^{\prime}=<e^{\prime}_{1},...,e^{\prime}_{m}>\subseteq L_{p}(L_{q}), where ek=ak𝟏Ak×Bke^{\prime}_{k}=a_{k}\cdot\mathbf{1}_{A_{k}\times B_{k}} with AkA_{k} and BkB_{k} intervals such that k𝟏Ak×Bk=𝟏\sum_{k}\mathbf{1}_{A_{k}\times B_{k}}=\mathbf{1} and for each eke_{k} there is xkS(E)+x_{k}\in S(E^{\prime})_{+} such that xkf2(ek)<ε4n\|x_{k}-f_{2}(e_{k})\|<\frac{\varepsilon}{4n}.

Since Lp(Lq)L_{p}(L_{q}) is AUH, there exists an automorphism ψ\psi such that ψf1f2<δ\|\psi f_{1}-f_{2}\|<\delta, where δ\delta satisfies the conditions for ε4mn\frac{\varepsilon}{4mn} and each of the eke^{\prime}_{k}’s in EE^{\prime} in Lemma 5.1. Now pick the automorphism ϕ\phi^{\prime} over Lp(Lq)L_{p}(L_{q}) mapping 𝟏\mathbf{1} to ψf1(x)\psi f_{1}(x) as defined in Lemma 5.1. It follows that for each eke^{\prime}_{k}, ϕ(ek)ek<ε4mn\|\phi^{\prime}(e^{\prime}_{k})-e^{\prime}_{k}\|<\frac{\varepsilon}{4mn}, so ϕ(xk)xk<ε4n\|\phi^{\prime}(x_{k})-x_{k}\|<\frac{\varepsilon}{4n}. Thus for each ekEe_{k}\in E,

ϕf2(ek)ψf1(ek)\displaystyle\|\phi^{\prime}f_{2}(e_{k})-\psi f_{1}(e_{k})\|\leq ϕ(f2(ek)xk)+ϕ(xk)xk\displaystyle\|\phi^{\prime}(f_{2}(e_{k})-x_{k})\|+\|\phi^{\prime}(x_{k})-x_{k}\|
+\displaystyle+ xkf2(ek)+f2(ek)ψf1(ek)<εn,\displaystyle\|x_{k}-f_{2}(e_{k})\|+\|f_{2}(e_{k})-\psi f_{1}(e_{k})\|<\frac{\varepsilon}{n},

Now let ϕ=ϕ1ψ\phi={\phi^{\prime}}^{-1}\circ\psi to obtain the desired automorphism; then ϕf1f2<ε.\|\phi f_{1}-f_{2}\|<\varepsilon.

The above can be used to show that if Lp(Lq)L_{p}(L_{q}) is AUH and fi(E)f_{i}(E) contains 𝟏\mathbf{1} for i=1,2i=1,2, then we can induce almost commutativity with automorphisms fixing 𝟏\mathbf{1} as well. This will allow us to reduce possible automorphisms over Lp(Lq)L_{p}(L_{q}) to those that in particular fix 𝟏\mathbf{1}. The importance of this result is that these particular homomorphisms fixing 𝟏\mathbf{1} must always preserve base-equimeasurability for characteristic functions, as shown in Proposition 3.1. Thus a natural approach in disproving that Lp(Lq)L_{p}(L_{q}) is AUH would involve finding sublattices containing 𝟏\mathbf{1} which are lattice isometric but whose generating elements are not base-equimeasurable. The following results do exactly that:

Lemma 5.3.

Lemma 4.2 fails when r:=p/qr:=p/q\in\mathbb{N}. In particular, there exists a non-zero measure ν:=αβ\nu:=\alpha-\beta, with α\alpha and β\beta positive measures such that for all polynomials PP of degree jrj\leq r,

01P(x)𝑑ν(x)=0.\int_{0}^{1}P(x)\ d\nu(x)=0.
Remark 5.4.

It is already known that a counter-example exists for Lr(0,)L_{r}(0,\infty) for all rr\in\mathbb{N}, with

dν(x)=eu14sin(u14)dud\nu(x)=e^{-u^{\frac{1}{4}}}\sin(u^{\frac{1}{4}})\ du

(see [12] and [8] for more details).

.

Here we provide another example over the unit interval:

Proof.

Fix such an rr, and define a polynomial g(x)g(x) of degree r+1r+1 with g(x)=0r+1aixig(x)=\sum_{0}^{r+1}a_{i}x^{i} such that for all 0jr0\leq j\leq r, 01xjg(x)𝑑x=0\int_{0}^{1}x^{j}g(x)\ dx=0. This can be done by finding a non-trivial a0,,ar+1a_{0},...,a_{r+1} in the null set of the (n+1)×(n+2)(n+1)\times(n+2) size matrix AA with A(i,j)=1i+j+1A(i,j)=\frac{1}{i+j+1}. Then let dν(x)=g(x)dxd\nu(x)=g(x)\ dx. Let α=ν+\alpha=\nu_{+} and β=ν\beta=\nu_{-}. Clearly α\alpha and β\beta are finite positive Borel measures, but since g0g\neq 0, αβ\alpha\neq\beta. ∎

Lemma 5.5.

Let p/qp/q\in\mathbb{N}. Then there exists a two dimensional lattice E=<e1,e2>E=<e_{1},e_{2}> and lattice embeddings fi:ELp(Lq)f_{i}:E\rightarrow L_{p}(L_{q}) with 𝟏E\mathbf{1}\in E such that g1(𝐞)g_{1}(\mathbf{e}) and g2(𝐞)g_{2}(\mathbf{e}) are not base-equimeasurable.

Proof.

Let f(x)f(x) be a polynomial of degree at least r+1r+1 as defined in Lemma 5.3 such that for all 0kr0\leq k\leq r, 01tkf(t)𝑑t=0\int_{0}^{1}t^{k}f(t)\ dt=0, and 01|f(x)|𝑑x=1\int_{0}^{1}|f(x)|\ dx=1. Let h1(x)=12+f(x)+h_{1}(x)=\frac{1}{2}+f(x)_{+}, and let h2(x)=12+f(x)h_{2}(x)=\frac{1}{2}+f(x)_{-}. Note that each hi(x)>0h_{i}(x)>0, and furthermore that 01hi(t)𝑑t=1\int_{0}^{1}h_{i}(t)\ dt=1. Additionally, each map Hi(x)=0xhi(t)𝑑tH_{i}(x)=\int_{0}^{x}h_{i}(t)\ dt is strictly increasing with Hi(0)=0H_{i}(0)=0 and Hi(1)=1H_{i}(1)=1. Now we will construct characteristic functions fjiLp(Lq)f^{i}_{j}\in L_{p}(L_{q}) such that the linear map fj1fj2f_{j}^{1}\mapsto f_{j}^{2} induces an isometry, but N𝐟1N\mathbf{f}^{1} and 𝐟2\mathbf{f}^{2} are not base-equimeasurable. From there, we let ej=fj1fjie_{j}=\frac{f^{1}_{j}}{\|f^{i}_{j}\|}, and let gig_{i} be the lattice isometry induced by gi(ej)=fjifjig_{i}(e_{j})=\frac{f^{i}_{j}}{\|f^{i}_{j}\|},

To this end, let

F1i(x):=Hi1(x), and F2i(x):=1F1i(x).F^{i}_{1}(x):=H_{i}^{-1}(x),\text{ and }F^{i}_{2}(x):=1-F^{i}_{1}(x).

Observe that F11(x)F12(x)F^{1}_{1}(x)\neq F^{2}_{1}(x). Indeed, one can show that the associated push forwards dF1i#μd{F^{i}_{1}}_{\#}\mu for each F1iF^{i}_{1} have the corresponding equivalence:

dF1i#μ(x)=hi(x)dxd{F^{i}_{1}}_{\#}\mu(x)=h_{i}(x)\ dx

So (F11,F21)(F_{1}^{1},F_{2}^{1}) and (F12,F22)(F_{1}^{2},F_{2}^{2}) are not equimeasurable. However, For 0jr0\leq j\leq r, ujhi(u)du=ujdF1i#(u)=F1i(x)jdxu^{j}h_{i}(u)\ du=u^{j}\ d{F_{1}^{i}}_{\#}(u)=F_{1}^{i}(x)^{j}\ dx, so it follows from the construction of the hih_{i}’s that

01F11(x)j𝑑x=01F12(x)j𝑑x.\int_{0}^{1}F_{1}^{1}(x)^{j}\ dx=\int_{0}^{1}F_{1}^{2}(x)^{j}\ dx.

Thus for any v1,v2>0v_{1},v_{2}>0, since F1iF^{i}_{1} and F2iF^{i}_{2} are both positive, we have

01|v1F11(x)+v2F21(x)|r𝑑x=01((v1v2)F11(x)+v2)r𝑑x\displaystyle\int_{0}^{1}|v_{1}F^{1}_{1}(x)+v_{2}F^{1}_{2}(x)|^{r}\ dx=\int_{0}^{1}((v_{1}-v_{2})F^{1}_{1}(x)+v_{2})^{r}\ dx
=\displaystyle= 0r(rj)(v1v2)jv2rj01F11(x)j𝑑x=01|v1F12(x)+v2F22(x)|r𝑑x\displaystyle\sum_{0}^{r}\binom{r}{j}(v_{1}-v_{2})^{j}v_{2}^{r-j}\int_{0}^{1}F^{1}_{1}(x)^{j}\ dx=\int_{0}^{1}|v_{1}F^{2}_{1}(x)+v_{2}F^{2}_{2}(x)|^{r}\ dx

To conclude the proof, let f1i(x,y)=𝟏[0,F1i(x)](y)f^{i}_{1}(x,y)=\mathbf{1}_{[0,F^{i}_{1}(x)]}(y), and let f2i=𝟏f1if^{i}_{2}=\mathbf{1}-f^{i}_{1}. Clearly N[fji]=FjiN[f^{i}_{j}]=F^{i}_{j}. ∎

Theorem 5.6.

If p/qp/q\in\mathbb{N} and pqp\neq q, then Lp(Lq)L_{p}(L_{q}) is not AUH for the class of its finite dimensional sublattices.

Proof.

Fix p/qp/q\in\mathbb{N}, and let EE be the 2-dimensional lattice generated in Lemma 5.5, with fi:ELp(Lq)f_{i}:E\rightarrow L_{p}(L_{q}) embeddings mapping to copies of E=<e1,e2>E=<e_{1},e_{2}> such that f1(𝐞)f_{1}(\mathbf{e}) and f2(𝐞)f_{2}(\mathbf{e}) are not base-equimeasurable. In addition, by assumption 𝟏E\mathbf{1}\in E. For notational ease, let Fji=N[fi(ej)]F^{i}_{j}=N[f_{i}(e_{j})].

Suppose for the sake of contradiction that Lp(Lq)L_{p}(L_{q}) is AUH. Pick some measurable C[0,1]2C\subseteq[0,1]^{2} and ε>0\varepsilon>0 such that

𝐅#2μ(C)>𝐅#1μ(C+ε)+ε,*\quad\mathbf{F}^{2}_{\#}\mu(C)>\mathbf{F}^{1}_{\#}\mu(C+\varepsilon)+\varepsilon,

where

C+ε={𝐭[0,1]2:𝐭𝐬<ε for some 𝐬C}.C+\varepsilon=\{\mathbf{t}\in[0,1]^{2}:\|\mathbf{t-s}\|_{\infty}<\varepsilon\text{ for some }\mathbf{s}\in C\}.

By Theorem 5.2, there is some lattice automorphism ϕ:Lp(Lq)Lp(Lq)\phi:L_{p}(L_{q})\rightarrow L_{p}(L_{q}) fixing 𝟏\mathbf{1} such that ϕf1f2<ε2\|\phi\circ f_{1}-f_{2}\|<\varepsilon^{2}. Let ϕFji=N[ϕfi(ej)]\phi F^{i}_{j}=N[\phi f_{i}(e_{j})]. By Proposition 3.1, ϕ\phi preserves base-equimeasurability, so for any measurable BB,

ϕ𝐅#1μ(B)=𝐅#1μ(B).\quad\phi\mathbf{F}^{1}_{\#}\mu(B)=\mathbf{F}^{1}_{\#}\mu(B).

By the properties of NN, we also have ϕFj1Fj2pϕf1(ej)f2(ej)\|\phi F^{1}_{j}-F^{2}_{j}\|_{p}\leq\|\phi f_{1}(e_{j})-f_{2}(e_{j})\|. It also follows that

μ(t:ϕ𝐅1(t)𝐅2(t)>ε)<ε,\mu(t:\|\phi\mathbf{F}^{1}(t)-\mathbf{F}^{2}(t)\|_{\infty}>\varepsilon)<\varepsilon,

so ϕ𝐅#1μ(C+ε)+ε>𝐅#2μ(C)\phi\mathbf{F}^{1}_{\#}\mu(C+\varepsilon)+\varepsilon>\mathbf{F}^{2}_{\#}\mu(C), but this contradicts the assumption (*). So Theorem 5.2 cannot apply, implying that Lp(Lq)L_{p}(L_{q}) is not AUH as desired. ∎

Remark 5.7.

For p/qp/q\in\mathbb{N}, Lp(Lq)L_{p}(L_{q}) is the unique lattice that is separably AUH over finitely generated BLpLqBL_{p}L_{q} spaces, since up to isometry it is the unique doubly atomless BLpLqBL_{p}L_{q} space. In light of Theorem 5.6, this implies that the class of finitely generated sublattices of Lp(Lq)L_{p}(L_{q}) is not a Fraïssé class as defined in [2], as Lp(Lq)L_{p}(L_{q}) is the only possible candidate as a Fraïssé limit.

In particular, LpLqL_{p}L_{q} lacks the NAP. Indeed, otherwise, one can use that NAP with BLpLqBL_{p}L_{q} amalgamate lattices and [7, Proposition 2.8] to situate a d𝒦d^{\mathcal{K}}-Cauchy sequence into a Cauchy-sequence of generating elements in an ambient separable BLpLqBL_{p}L_{q} lattice. Thus 𝒦p,q¯\overline{\mathcal{K}_{{p},{q}}} would also have the Polish Property, implying that 𝒦p,q¯\overline{\mathcal{K}_{{p},{q}}} is a Fraïssé class. Since the only possible candidate Fraïssé limit space is Lp(Lq)L_{p}(L_{q}) itself, this would contradict Theorem 5.6.

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