Isometric embeddings of a class of separable metric spaces into Banach spaces

TL;DR

This paper proves that certain bounded, countable metric spaces with a specific triangle inequality can be isometrically embedded into any Banach space containing an isomorphic copy of __.

Contribution

It establishes a new class of metric spaces that can be isometrically embedded into Banach spaces with __, expanding understanding of metric space embeddings.

Findings

Such metric spaces can be embedded into Banach spaces with __.

The embedding preserves distances exactly.

The result applies to all Banach spaces containing an isomorphic __.

Abstract

Let $(M,d)$ be a bounded countable metric space and $c>0$ a constant, such that $d(x,y)+d(y,z)-d(x,z) \ge c$, for any pairwise distinct points $x,y,z$ of $M$. For such metric spaces we prove that they can be isometrically embedded into any Banach space containing an isomorphic copy of $\ell_\infty$.

Equations

$$\phi :x\in M\mapsto (d(x,x_n)-d(x_1,x_n){)}_{n\ge 1}\in {\ell }_{\infty }$$

$$\sigma :M\to {\mathbb{R}}^{\mathbb{N}}\text{with}\sigma (x)=(d(x,x_n){)}_{n\ge 1}$$

$$\phi (x)=\sigma (x)-\sigma (x_1),x\in X$$

$$d\left(({\ell }_{\infty },\Vert \cdot {\Vert }_{\infty }),({\ell }_{\infty },\Vert \cdot \Vert )\right)\le 1+\delta$$

$$p_1(\epsilon )=\left(d(x_1,x_1)-d(x_1,x_1),d(x_1,x_2)-d(x_1,x_2),\dots ,d(x_1,x_n)-d(x_1,x_n),\dots \right)=(0,\dots ,0,\dots )$$

$$p_2(\epsilon )=\left(d(x_2,x_1)-d(x_1,x_1)+{\epsilon }_{(2,1)},d(x_2,x_2)-d(x_1,x_2),\dots ,d(x_2,x_n)-d(x_1,x_n),\dots \right)$$

$$p_n(\epsilon )=\left(d(x_n,x_1)-d(x_1,x_1)+{\epsilon }_{(n,1)},\dots ,d(x_n,x_{n-1})-d(x_1,x_{n-1})+{\epsilon }_{(n,n-1)},d(x_n,x_n)-d(x_1,x_n),\dots \right)$$

$$\Vert p_n(\epsilon )-p_m(\epsilon ){\Vert }_{\infty }=\underset{k}{\sup }|d(x_n,x_k)+{\epsilon }_{(n,k)}-(d(x_m,x_k)+{\epsilon }_{(m,k)})|$$

$$d(x_n,x_k)-d(x_m,x_k)+{\epsilon }_{(n,k)}-{\epsilon }_{(m,k)}\le d(x_n,x_m)-\eta +{\epsilon }_{(n,k)}-{\epsilon }_{(m,k)}\le d(x_n,x_m).$$

$$\epsilon =\left({\epsilon }_{(m,n)}\right)\in K\stackrel{\phi }{\mapsto }\phi (\epsilon )=\left({\phi }_{(m,n)}(\epsilon )\right)\in K,$$

$$d(x_n,x_m)+{\epsilon }_{(m,n)}=\Vert p_n(\epsilon )-p_m(\epsilon ){\Vert }_{\infty }\le (1+\delta )\Vert p_n(\epsilon )-p_m(\epsilon )\Vert \Rightarrow$$

$$\frac{1}{1+\delta }(d(x_n,x_m)+{\epsilon }_{(m,n)})\le \Vert p_n(\epsilon )-p_m(\epsilon )\Vert .$$

$${\phi }_{(m,n)}(\epsilon )=d(x_n,x_m)+{\epsilon }_{(m,n)}-\Vert p_n(\epsilon )-p_m(\epsilon )\Vert$$

$$\le d(x_n,x_m)+{\epsilon }_{(m,n)}-\frac{1}{1+\delta }\left(d(x_n,x_m)+{\epsilon }_{(m,n)}\right)$$

$$=\frac{\delta }{1+\delta }\left(d(x_n,x_m)+{\epsilon }_{(m,n)}\right).$$

$$d\left((Z,\Vert \cdot \Vert ),({\ell }_{\infty },\Vert \cdot {\Vert }_{\infty })\right)\le 1+\delta$$

Full text

††2010 Mathematics Subject Classification: Primary 46B20, 46E15;Secondary 46B26,54D30.

Key words and phrases: concave metric space, isometric embedding, separated set.

Introduction

Let $(M,d)$ be a metric space; following [4] we will call it concave, when the triangle inequality is strict, i.e. when $d(x,y)+d(y,z)>d(x,z)$ for any pairwise distinct points $x,y,z$ of $M$.

In this note we are interested in (concave) metric spaces satisfying the stronger property: there is a constant $c>0$, such that $d(x,y)+d(y,z)-d(x,z)\geq c$, for any pairwise distinct points $x,y,z$. Let us call these spaces strongly concave metric spaces.

The main result we prove is an infinite dimensional version of Theorem 4.3 of [4], that is, if a Banach space $X$ contains an isomorphic copy of $\ell_{\infty}$, then $X$ contains isometrically any bounded countable strongly concave metric space (Th.2). An immediate consequence of this result is that any Banach space containing an isomorphic copy of $c_{0}$, admits an infinite equilateral set (Th.3). This result was first proved (by similar methods) in [5] (Th.2).

A subset $S$ of a metric space $(M,d)$ is said to be equilateral, if there is a $\lambda>0$ such that for $x\neq y\in S$ we have $d(x,y)=\lambda$; we also call $S$ a $\lambda$-equilateral set (see [8]).

If $X$ is any (real) Banach space, then $B_{X}$ and $S_{X}$ denote its closed unit ball and unit sphere respectively. $X$ is said to be strictly convex, if for any $x\neq y\in S_{X}$ we have $\|x+y\|<2$. The Banach-Mazur distance between two isomorphic Banach spaces $X$ and $Y$ is $d(X,Y)=\inf\{\|T\|\cdot\|T^{-1}\|:T\>\text{is an isomorphism}\}$.

Strongly concave metric spaces

We start by presenting some examples of concave metric spaces

Examples 1

(1) a) Let $(M,d)$ be a discrete metric space (i.e. $d(x,y)=1$ when $x\neq y$). Clearly $1=d(x,z)<d(x,y)+d(y,z)=2$ for any pairwise distinct triplet $x,y,z\in M$. Therefore $(M,d)$ is a concave metric space. In particular, every $\lambda$-equilateral subset of any metric space is a concave metric space.

b) More generally, every ultrametric space is concave. This holds since for any $x,y,z$ pairwise distinct points we have $d(x,z)\leq\max\{d(x,y),d(y,z)\}<d(x,y)+d(y,z)$.

(2) Let $(X,\|\cdot\|)$ be a strictly convex Banach space. As is well known, if $x,y,z$ are non collinear points of $X$ then $\|x-z\|<\|x-y\|+\|y-z\|$.

It then follows that the unit sphere $S_{X}$ and every affinely independent subset $A$ of $X$ with the norm metric are concave metric spaces (in any case no three pairwise distinct points are collinear).

(3) Let $(X,\|\cdot\|)$ be a Banach space and $A\subseteq B_{X}$ such that $x\neq y\in A\Rightarrow\|x-y\|>1$ (see [3]). Then for any $x,y,z$ pairwise distinct points of $A$ we have $\|x-y\|+\|y-z\|-\|x-z\|>1+1-\|x-z\|\geq 1+1-2=0$. Hence $A$ with the norm metric is concave.

(4) Let $(M,d)$ be any metric space and $p\in(0,1)$. Then it is rather easy to show that $d^{p}$ is a concave metric on $M$. This follows from the fact that, given $a,b,c>0$ with $a\leq b+c$ then $a^{p}<b^{p}+c^{p}$. The metric $d^{p}$ is then called the snowflaked version of $d$ (see [6]).

We are interested in concave metric spaces $(M,d)$ satisfying the stronger property: there is a constant $c>0$ such that for any pairwise distinct points $x,y,z$ of $M$ we have $d(x,y)+d(y,z)-d(x,z)\geq c$, equivalently $d(x,z)+c\leq d(x,y)+d(y,z)$. Let us call these spaces strongly concave spaces.

Lemma 1**.**

Every strongly concave metric space is separated (or uniformly discrete).

Proof.

Assume that $(M,d)$ is a $c$-strongly concave metric space. We claim that $x\neq y\in M\Rightarrow d(x,y)\geq\frac{c}{2}$. Assume for the purpose of contradiction that there is a pair $\{x,y\}\subseteq M$ with $d(x,y)<\frac{c}{2}$. Let also $z\in M\setminus\{x,y\}$. We then have $d(x,y)+d(y,z)\leq d(x,y)+(d(y,x)+d(x,z))=2d(x,y)+d(x,z)\Rightarrow$ $d(x,y)+d(y,z)-d(x,z)\leq 2d(x,y)<2\frac{c}{2}=c$. The last inequality clearly contradicts the fact that $M$ is $c$-strongly concave.

∎

The following are examples of strongly concave metric spaces.

Examples 2

(1) Every finite concave metric space is clearly strongly concave.

(2) Let $A$ be a a $\lambda$-equilateral  subset of any metric space $(M,d)$. For any pairwise distinct points $x,y,z$ of $A$ we have $d(x,y)+d(y,z)-d(x,z)=\lambda+\lambda-\lambda=\lambda$, so $A$ is a $\lambda$-strongly concave metric subspace of $(M,d)$.

(3) Let $(X,\|\cdot\|)$ be a Banach space. Also let $A\subseteq B_{X}$ with the property that $x\neq y\in A\Rightarrow\|x-y\|\geq 1+\varepsilon$, where $\varepsilon>0$ is a constant. Then we have $\|x-y\|+\|y-z\|-\|x-z\|>(1+\varepsilon)+(1+\varepsilon)-2=2\varepsilon$ (cf. Examples 1(3)). Therefore $A$ with the norm metric is a $2\varepsilon$-strongly concave metric space.

Note that if $\dim X=\infty$, then by a result of Elton and Odell ([2]) there is $A\subseteq S_{X}$ infinite and $\varepsilon>0$ such that $x\neq y\in A\Rightarrow\|x-y\|\geq 1+\varepsilon$.

Remarks 1 (1) Clearly every separable strongly concave metric space $M$ is at most countable (this is so because $M$ is separated, hence it has the discrete topology).

(2) Every subspace of a concave (resp. strongly concave) space has the same property.

The following result is classical (see [6]).

Theorem 1**.**

(Frechèt) Every separable metric space $(M,d)$ embeds isometrically into $\ell_{\infty}$.

Proof.

Let $(x_{n})\subseteq M$ be a dense sequence in $M$. Then the map

[TABLE]

satifies our claim.

∎

Remark 2 Let $(M,d)$ be a separable metric space. We define a map

[TABLE]

where $(x_{n})$ is any dense sequence in $M$. Then the Frechèt embedding of $M$ into $\ell_{\infty}$ is the map

[TABLE]

Note that if the space $(M,d)$ is bounded (that is, there is $k>0$ such that $d(x,y)\leq k$ for all $x,y\in M$), then the map $\sigma$ is already an isometric embedding of $M$ into $\ell_{\infty}$, which we will still call the Frechèt embedding of $M$ into $\ell_{\infty}$.

Proposition 1**.**

Let $(M,d)$ be a bounded countable infinite metric space. Then there is an infinite subset $N$ of $M$ such that the Frechèt embedding of $N$ into $\ell_{\infty}$ takes values into the space $\mathbf{c}$.

Proof.

Let $\{x_{1},x_{2},\dots,x_{n},\dots\}$ be a one-to-one enumeration of $M$. Then $\sigma(x_{k})=(d(x_{k},x_{n}))_{n\geq 1}\in\ell_{\infty}$, for $k\in\mathbb{N}$, since $d$ is a bounded metric. We construct by induction a subsequence $\{x^{\prime}_{1},x^{\prime}_{2},\dots,x^{\prime}_{n},\dots\}$ of $(x_{n})$ satisfying our claim.

Since $(d(x_{1},x_{n}))_{n\geq 1}$ is a bounded sequence of real numbers, there is $A_{1}\subseteq\mathbb{N}$ infinite, such that $d(x_{1},x_{n})\stackrel{{\scriptstyle n\in A_{1}}}{{\longrightarrow}}\alpha_{1}$. Set $n_{1}=1$.

Let $n_{2}=\min A_{1}$, for which we may assume that $n_{2}>n_{1}$. Then for the sequence $(d(x_{n_{2}},x_{n}))_{n\in A_{1}}$, there is $A_{2}\subseteq A_{1}$ infinite with $n_{3}=\min A_{2}>n_{2}$ such that $d(x_{n_{2}},x_{n})\stackrel{{\scriptstyle n\in A_{2}}}{{\longrightarrow}}\alpha_{2}$.

Then for the sequence $(d(x_{n_{3}},x_{n}))_{n\in A_{2}}$, there is $A_{3}\subseteq A_{2}$ infinite with $n_{4}=\min A_{3}>n_{3}$ such that $d(x_{n_{3}},x_{n})\stackrel{{\scriptstyle n\in A_{3}}}{{\longrightarrow}}\alpha_{3}$.

The inductive process should be clear. Now set $A=\{n_{1}<n_{2}<\dots<n_{k}<\dots\}$. Clearly $\{n_{k},n_{k+1},\dots\}\subseteq A_{k}$ for $k\geq 1$ and hence $d(x_{n_{k}},x_{n})\stackrel{{\scriptstyle n\in A}}{{\longrightarrow}}\alpha_{k}$ for all $k\geq 1$. It is clear that the set $N=\{x^{\prime}_{k}=x_{n_{k}}:k\geq 1\}$ satisfies our requirements. ∎

The following theorem is the main result of this note; its proof resembles the proof of Theorem 4.3 of [4] and the proof of Theorem 2 of [5] (we use Schauder’s fixed point theorem the same way we did in [5]). The origins of these ideas can be traced in Brass (see [1] and [8]) and Swanepoel and Villa (see [9] and [10]).

Theorem 2**.**

Let $X$ be any Banach space containing an isomorphic copy of $\ell_{\infty}$. Then $X$ contains isometrically any bounded separable strongly concave metric space.

Proof.

We shall use a kind of non distortion property of $\ell_{\infty}$ proved independently by Talagrand ([11]) and Partington ([7]). Let us denote by $\|\cdot\|_{\infty}$ the usual norm of $\ell_{\infty}$.

Claim**.**

Let $(M,d)$ be any bounded separable strongly concave metric space. There is $\delta>0$, such that if $\|\cdot\|$ is any equivalent norm on $\ell_{\infty}$ with Banach Mazur distance

[TABLE]

then the space $(M,d)$ embeds isometrically into $(\ell_{\infty},\|\cdot\|)$.

Proof of the Claim: Since $(M,d)$ is strongly concave, there is $\eta>0$ such that $d(x,y)+d(y,z)-d(x,z)\geq\eta$, for each triplet $x,y,z$ of pairwise distinct points of $M$. We may assume that $\|x\|\leq\|x\|_{\infty}\leq(1+\delta)\|x\|$ for $x\in\ell_{\infty}$, where $\delta>0$ is to be determined.

Let $I=\{(m,n):n<m,\;n,m\in\mathbb{N}\}$; denote by $K$ the compact cube $[0,\eta]^{I}$. Since $M$ is (strongly concave and) separable, it is at most countable, so let $M=\{x_{1},x_{2},\dots,x_{n},\dots\}$. For $\varepsilon=(\varepsilon_{(m,n)})\in K$ set

[TABLE]

[TABLE]

⋮

[TABLE]

⋮

(Note that $x_{n}\mapsto p_{n}(0)$ is the Frechèt embedding of $M$ into $(\ell_{\infty},\|\cdot\|_{\infty})$).

For $n<m$ we have

[TABLE]

where we set $\varepsilon_{(k,l)}=0$, for $l\geq k$. This supremum is equal to $d(x_{n},x_{m})+\varepsilon_{(m,n)}$, as for $k\neq n,m$ we have

[TABLE]

We define a function

[TABLE]

by the rule $\varphi_{(m,n)}(\varepsilon)=d(x_{n},x_{m})+\varepsilon_{(m,n)}-\|p_{n}(\varepsilon)-p_{m}(\varepsilon)\|$. Note that $\varphi_{(m,n)}(\varepsilon)\geq d(x_{n},x_{m})+\varepsilon_{(m,n)}-\|p_{n}(\varepsilon)-p_{m}(\varepsilon)\|_{\infty}=0$ (using the computation above and the fact that the norm $\|\cdot\|_{\infty}$ dominates $\|\cdot\|$). We also have

[TABLE]

[TABLE]

Therefore

[TABLE]

[TABLE]

[TABLE]

It then follows from (this inequality and) the fact that $M$ is bounded that if $\delta$ is quite small, then $\varphi_{(m,n)}(\varepsilon)\leq\eta$, for $\varepsilon\in K$.

Since each coordinate function $\varphi_{(m,n)}$ is continuous (as dependent on finite coordinates, i.e. from the set $\{(k,l):1\leq l<k\leq m\}$) it follows that $\varphi$ is also continuous. By a classical result of Schauder, $\varphi$ has a fixed point $\varepsilon^{\prime}=\left(\varepsilon^{\prime}_{(m,n)}\right)\in K$, that is $\varphi(\varepsilon^{\prime})=\varepsilon^{\prime}$, which implies $\|p_{n}(\varepsilon^{\prime})-p_{m}(\varepsilon^{\prime})\|=d(x_{n},x_{m})$, for all $n,m\in\mathbb{N}$. The proof of the Claim is complete.

Denote by $\|\cdot\|$ the norm of $X$ and let $Y$ be a subspace of $X$ isomorphic to $\ell_{\infty}$. By the non distortion property of $(\ell_{\infty},\|\cdot\|_{\infty})$ there is a subspace $Z\subseteq Y$ (isomorphic to $\ell_{\infty}$) such that

[TABLE]

(this is the $\delta>0$ postulated in the Claim). It follows immediately from the Claim that the space $(Z,\|\cdot\|)$ contains an isometric copy of $(M,d)$.

∎

In the special case when $(M,d)$ is the countable infinite discrete metric space we get the following result first proved in [5] (Th.2), essentially with the same method

Theorem 3**.**

Every Banach space $X$ containing an isomorphic copy of $c_{0}$ admits an infinite equilateral set.

Proof.

Take in the proof of the previous theorem $(M,d)$ to be the countable infinite discrete space. Then $\eta=1$ and the resulting family $(p_{n}(\varepsilon))_{n\geq 1},\;\varepsilon\in K=\left[0,1\right]^{I}$ takes values in $c_{0}$ (remember that $x_{n}\mapsto p_{n}(0)$ is the Frechèt embedding of $(M,d)$ into $c_{0}$). Since $(c_{0},\|\cdot\|_{\infty})$ is non distortable, we get the conclusion. ∎

Theorem 2 can be improved in the following way

Theorem 4**.**

Let $(M,d)$ be an infinite bounded separable strongly concave metric space. Then there is $N\subseteq M$ infinite such that the metric space $(N,d)$ can be isometrically embedded into any Banach space containing an isomorphic copy of the space $c_{0}$.

Proof.

By Proposition 1, there is $N\subseteq M$ infinite such that the Frechèt embedding $\sigma:N\rightarrow\ell_{\infty}$ takes values into $\mathbf{c}$. Then the proof of Theorem 2 gives us a family of embeddings $(p_{n}(\varepsilon))_{n\geq 1},\;\varepsilon\in K=[0,\eta]^{I}$ taking values into $\mathbf{c}$. Since $\mathbf{c}$ is isomorphic to $c_{0}$, we are done. ∎