Non-expansive bijections between unit balls of Banach spaces (A technical version with some boring proofs included)

TL;DR

This paper extends known results about non-expansive bijections being isometries from specific Banach spaces to more general cases involving mappings between different Banach spaces, including finite-dimensional, strictly convex, or  spaces.

Contribution

It generalizes the class of Banach spaces for which non-expansive bijections are necessarily isometries, covering mappings between different spaces.

Findings

Non-expansive bijections between certain Banach spaces are isometries.

Extension of isometry results to mappings between different Banach spaces.

Applicable to finite-dimensional, strictly convex, and  spaces.

Abstract

It is known that if $M$ is a finite-dimensional Banach space, or a strictly convex space, or the space $\ell_1$, then every non-expansive bijection $F: B_M \to B_M$ is an isometry. We extend these results to non-expansive bijections $F: B_E \to B_M$ between unit balls of two different Banach spaces. Namely, if $E$ is an arbitrary Banach space and $M$ is finite-dimensional or strictly convex, or the space $\ell_1$ then every non-expansive bijection $F: B_E \to B_M$ is an isometry.

Equations

$$F(0)=0,\qquad F(x)=\|x\|\,f\left(x/\|x\|\right)\quad\bigl{(}x\in X\setminus\{0\}\bigr{)}.$$

$$D(x,a)=a\{x+y\in B_X:x-\frac{a}{1-a}y\in B_X\}.$$

$$u^{\ast }(v)=\underset{a\to 0^{+}}{\lim }\frac{1}{a}\left(\Vert u+av{\Vert }_X-\Vert u{\Vert }_X\right).$$

$$u^{\ast }(y_1)+u^{\ast }(y_2)\ge u^{\ast }(y_1+y_2),u^{\ast }(ty)=t{u}^{\ast }(y),\text{for}t\ge 0.$$

$$u^{\ast }(y_1+y_2)$$

$$=\underset{a\to 0^{+}}{\lim }\frac{1}{2a}\left(2\Vert u+2a(y_1+y_2){\Vert }_X-2\Vert u{\Vert }_X\right)=$$

$$=\underset{b\to 0^{+}}{\lim }\frac{1}{b}\left(2\Vert u+b(y_1+y_2){\Vert }_X-2\Vert u{\Vert }_X\right)\le$$

$$\le \underset{b\to 0^{+}}{\lim }\frac{1}{b}\left(\Vert u+b{y}_1{\Vert }_X-\Vert u{\Vert }_X\right)+$$

$$+\underset{b\to 0^{+}}{\lim }\frac{1}{b}\left(\Vert u+b{y}_2{\Vert }_X-\Vert u{\Vert }_X\right)=u^{\ast }(y_1)+u^{\ast }(y_2).$$

$$t{u}^{\ast }(y)$$

$$=\underset{b\to 0^{+}}{\lim }\frac{1}{b}\left(\Vert u+bty{\Vert }_X-\Vert u{\Vert }_X\right)=u^{\ast }(ty).$$

$$u^{\ast }(v)\ge -u^{\ast }(-v).$$

$$u^{\ast }(y_1)-u^{\ast }(y_2)\le \Vert y_1-y_2{\Vert }_X.$$

$$u^{\ast }(y_1)-u^{\ast }(y_2)$$

$$\le \underset{a\to 0^{+}}{\lim }\frac{1}{a}\left(\Vert u+a{y}_1-u-a{y}_2{\Vert }_X\right)=\Vert y_1-y_2{\Vert }_X.$$

$$(F(u){)}^{\ast }(-F(v))$$

$$=\underset{a\to 0^{+}}{\lim }\frac{1}{a}\left(\Vert F(u)-F(av){\Vert }_Y-\Vert u{\Vert }_X\right)$$

$$\le \underset{a\to 0^{+}}{\lim }\frac{1}{a}\left(\Vert u-av{\Vert }_X-\Vert u{\Vert }_X\right)=u^{\ast }(-v),$$

$$(F(u){)}^{\ast }(-F(-v))\le u^{\ast }(v).$$

$$(F(u){)}^{\ast }(F(v))$$

$$\Vert F(y_1)-F(y_2){\Vert }_Y$$

$$F\left(\sum _{n\le N}{a}_n{g}_n\right)=\sum _{n\le N}{a}_n{e}_n.$$

$$\Vert x\Vert =\sum _{i=1}^N|{\alpha }_i|.$$

$$\Vert \sum _{i=1}^{N-1}{\alpha }_i{g}_i\Vert \le \sum _{i=1}^{N-1}\Vert {\alpha }_i{g}_i\Vert =\sum _{i=1}^{N-1}|{\alpha }_i|\le \sum _{i=1}^N|{\alpha }_i|\le 1,$$

$$\Vert x\Vert =\Vert \sum _{i=1}^N{\alpha }_i{g}_i\Vert \le \sum _{i=1}^N|{\alpha }_i|.$$

$$\Vert x\Vert$$

$$U_N=\left\{\sum _{n\le N}{a}_n{g}_n:\sum _{n\le N}|a_n|\le 1\right\},\partial U_N=\left\{\sum _{n\le N}{a}_n{g}_n:\sum _{n\le N}|a_n|=1\right\}.$$

$$F(U_N)\subset V_N.$$

$$x_1=\frac{{\sum }_{i=1}^{N-1}{\alpha }_i{g}_i}{{\sum }_{i=1}^{N-1}|{\alpha }_i|}$$

$$F(x_1)=\frac{{\sum }_{i=1}^{N-1}{\alpha }_i{e}_i}{{\sum }_{i=1}^{N-1}|{\alpha }_i|}.$$

$$2$$

$$=\Vert F(x_1)-F(x)\Vert +\Vert F(x)-\frac{{\alpha }_N}{|{\alpha }_N|}{e}_N\Vert -2\sum _{i=N+1}^{\infty }|y_i|$$

$$\le \Vert F(x_1)-F(x)\Vert +\Vert F(x)-F\left(\frac{{\alpha }_N}{|{\alpha }_N|}{g}_N\right)\Vert \le \Vert x_1-x\Vert +\Vert x-\frac{{\alpha }_N}{|{\alpha }_N|}{g}_N\Vert$$

$$=\sum _{j=1}^{N-1}\left|{\alpha }_j-\frac{{\alpha }_j}{{\sum }_{i=1}^{N-1}|{\alpha }_i|}\right|+|{\alpha }_N|+\sum _{j=1}^{N-1}|{\alpha }_j|+\left|{\alpha }_N-\frac{{\alpha }_N}{|{\alpha }_N|}\right|$$

$$=\sum _{j=1}^{N-1}|{\alpha }_j|\left(1+\left|1-\frac{1}{{\sum }_{i=1}^{N-1}|{\alpha }_i|}\right|\right)+|{\alpha }_N|\left(1+\left|1-\frac{1}{|{\alpha }_N|}\right|\right)=2.$$

$$F(U_N)\supset \partial V_N.$$

Full text

1. Introduction

Let $M$ be a metric space. A map $F:M\to M$ is called non-expansive, if $\rho(F(x),F(y))\leq\rho(x,y)$ for all $x,y\in M$. The space $M$ is called expand-contract plastic (or simply, an EC-space) if every non-expansive bijection from $M$ onto itself is an isometry.

It is known [7, Theorem 1.1] that every compact (or even totally bounded) metric space is expand-contract plastic, so in particular every bounded subset of ${\mathbb{R}}^{n}$ is an EC-space.

The situation with bounded subsets of infinite dimensional spaces is different. On the one hand, there is a non-expand-contract plastic bounded closed convex subset of a Hilbert space [2, Example 2.7] (in fact, that set is an ellipsoid), but on the other hand, the unit ball of a Hilbert space, and in general the unit ball of every strictly convex Banach space is an EC-space [2, Theorem 2.6]. It is unknown whether the strict convexity condition in [2, Theorem 2.6] can be omitted, that is, in other words, the following problem arises.

Problem 1.1**.**

For what Banach spaces $Y$ every bijective non-expansive map $F:B_{Y}\to B_{Y}$ is an isometry? Is this true for every Banach space?

Outside of strictly convex spaces, Problem 1.1 is solved positively for all finite-dimensional spaces (because of the compactness of the unit ball), and for the space $\ell_{1}$ [6, Theorem 1].

To the best of our knowledge, the following natural extension of Problem 1.1 is also open.

Problem 1.2**.**

Is it true that for every pair $(X,Y)$ of Banach spaces, every bijective non-expansive map $F:B_{X}\to B_{Y}$ is an isometry?

An evident bridge between these two problems is the third one, which we also are not able to solve.

Problem 1.3**.**

Let $X,Y$ be Banach spaces that admit a bijective non-expansive map $F:B_{X}\to B_{Y}$. Is it true that spaces $X$ and $Y$ are isometric?

In fact, if one solves Problem 1.2 in positive, one evidently solves also Problem 1.3. On the other hand, for a fixed pair $(X,Y)$ the positive answers to Problems 1.1 and 1.3 would imply the same for Problem 1.2.

The aim of this article is to demonstrate that for all spaces $Y$ where Problem 1.1 is known to have a positive solution (i.e. strictly convex spaces, $\ell_{1}$, and finite-dimensional spaces), Problem 1.2 can be solved in the positive for all pairs of the form $(X,Y)$. In fact, our result for pairs $(X,Y)$ with $Y$ being strictly convex repeats the corresponding proof of the case $X=Y$ from [2, Theorem 2.6] almost word-to-word. The proof for pairs $(X,\ell_{1})$ on some stage needs additional work comparing to its particular case $X=\ell_{1}$ from [6, Theorem 1]. The most difficult one is the finite-dimensional case, where the approach from [7, Theorem 1.1] is not applicable for maps between two different spaces, because it uses iterations of the map. So, for finite-dimensional spaces we had to search for a completely different proof. Our proof in this case uses some ideas from [2] and [6] but elaborates them a lot.

There is another similar circle of problems that motivates our study.

In 1987, D. Tingley [11] proposed the following question: let $f$ be a bijective isometry between the unit spheres $S_{X}$ and $S_{E}$ of real Banach spaces $X$, $E$ respectively. Is it true that $f$ extends to a linear (bijective) isometry $F:X\to E$ of the corresponding spaces?

Let us mention that this is equivalent to the fact that the following natural positive-homogeneous extension $F:X\to E$ of $f$ is linear:

[TABLE]

Since according to P. Mankiewicz’s theorem [8] every bijective isometry between convex bodies can be uniquely extended to an affine isometry of the whole spaces, Tingley’s problem can be reformulated as follows:

Problem 1.4**.**

Let $F:B_{X}\to B_{E}$ be a positive-homogeneous map, whose restriction to $S_{X}$ is a bijective isometry between $S_{X}$ and $S_{E}$. Is it true that $F$ is an isometry itself?

There is a number of publications devoted to Tingley’s problem (see [3] for a survey of corresponding results) and, in particular, the problem is solved in the positive for many concrete classical Banach spaces. Surprisingly, for general spaces this innocently-looking question remains open even in dimension two. For finite-dimensional polyhedral spaces the problem is solved in the positive by V. Kadets and M. Martín in 2012 [5], and the positive solution for the class of generalized lush spaces was given by Dongni Tan, Xujian Huang, and Rui Liu in 2013 [10]. A step in the proof of the latter result was a lemma (Proposition 3.4 of [10]) which in our terminology says that if the map $F$ in Problem 1.4 is non-expansive, then the problem has a positive solution. So, the problem which we address in our paper (Problem 1.2) can be considered as a much stronger variant of that lemma.

2. Preliminaries

In the sequel, the letters $X$ and $Y$ always stand for real Banach spaces. We denote by $S_{X}$ and $B_{X}$ the unit sphere and the closed unit ball of $X$ respectively. For a convex set $A\subset X$ denote by $\mathop{\rm ext}\nolimits(A)$ the set of extreme points of $A$; that is, $x\in\mathop{\rm ext}\nolimits(A)$ if $x\in A$ and for every $y\in X\setminus\{0\}$ either $x+y\not\in A$ or $x-y\not\in A$. Recall that $X$ is called strictly convex if all elements of $S_{X}$ are extreme points of $B_{X}$, or in other words, $S_{X}$ does not contain non-trivial line segments. Strict convexity of $X$ is equivalent to the strict triangle inequality $\|x+y\|<\|x\|+\|y\|$ holding for all pairs of vectors $x,y\in X$ that do not have the same direction. For subsets $A,B\subset X$ we use the standard notation $A+B=\{x+y{:}\allowbreak\ x\in A,y\in B\}$ and $aA=\{ax{:}\allowbreak\ x\in A\}$.

Now let us reformulate the results of [2] on the case of two different spaces. In order to do it we give the following lemmas from the same source. For $x\in S_{X}$ and $a\in(0,1)$ let $D(x,a):=aB_{X}\cap(x+(1-a)B_{X})$.

Lemma 2.1** (Lemma 2.1 of [2]).**

For every $x\in S_{X}$ and $a\in(0,1)$

[TABLE]

When $x$ is an extreme point of $B_{X}$, then $D(x,a)=\{ax\}$. When $x$ is not an extreme point of $B_{X}$, then $D(x,\frac{1}{2})$ consists of more than one point.

Lemma 2.2** (Lemma 2.2 of [2]).**

$D_{1}(x)=-x+\{x+y\in B_{X}:x-y\in B_{X}\}$. When $x$ is an extreme point of $B_{X}$, then $D_{1}(x)=\{0\}$.

The following theorem generalizes [2, Theorem 2.3], where the case $X=Y$ was considered. It can be demonstrated repeating the proof of [2, Theorem 2.3] almost word to word.

Theorem 2.3**.**

Let $F:B_{X}\to B_{Y}$ be a non-expansive bijection. In the above notations the following hold.

1. (1)

$F(0)=0$. 2. (2)

$F^{-1}(S_{Y})\subset S_{X}$. 3. (3)

$F(D(x,a))\subset D(F(x),a)$* for all $x\in F^{-1}(S_{Y})$ and $a\in(0,1)$.* 4. (4)

If $F(x)$ is an extreme point of $B_{Y}$, then $F(ax)=aF(x)$ for all $a\in(0,1)$. 5. (5)

If $F(x)$ is an extreme point of $B_{Y}$, then $x$ is also an extreme point of $B_{X}$. 6. (6)

If $F(x)$ is an extreme point of $B_{Y}$, then $F(-x)=-F(x)$.

Moreover, if $Y$ is strictly convex, then

1. (i)

$F$* maps $S_{X}$ bijectively onto $S_{Y}$;* 2. (ii)

$F(ax)=aF(x)$* for all $x\in S_{X}$ and $a\in(0,1)$;* 3. (iii)

$F(-x)=-F(x)$* for all $x\in S_{X}$.*

Proof.

(1) The point 0 has a special property: it is the unique point of the ball whose distance to all other points of the ball is less than or equal to 1. Evidently, a non-expansive bijection preserves this property.

(2) Notice that (1) implies that if $\|x\|_{X}<1$, then $\|F(x)\|_{Y}<1$.Hence $F$ maps interior points into interior points. So $F^{-1}(S_{Y})\subset S_{X}$.

(3) $D(x,a)$ is the set of those points $z$ such that $\|z\|_{X}\leq a$ and $\|x-z\|_{X}\leq 1-a$. So for every $z\in D(x,a)$, $F(z)$ must satisfy the conditions $\|F(z)\|_{Y}\leq a$ and $\|F(x)-F(z)\|_{Y}\leq 1-a$, that is, $F(z)\in D(F(x),a)$.

(4) Let $F(x)\in\mathop{\rm ext}\nolimits(B_{Y})$. Then $D(F(x),a)=\{aF(x)\}$ by Lemma 2.1. Since for every $a\in(0,1)$ we have $ax\in D(x,a)$, it follows that $F(ax)\in D(F(x),a)=\{aF(x)\}$.

(5) Suppose $F(x)$ is an extreme point of $B_{Y}$. Then $D(F(x),1/2)$ consists of one point, so (3) and the injectivity of $F$ imply that $D(x,1/2)$ consists of one point, that is, $x\in\mathop{\rm ext}\nolimits(B_{X})$.

(6) Suppose that $F(x)$ is an extreme point of $B_{Y}$. By the surjectivity of $F$ there is a $y\in S_{X}$ such that $F(y)=-F(x)$. Then $\|x-y\|_{X}\geq\|F(x)-(-F(x))\|_{Y}=2$. Let $z=\frac{1}{2}(x+y)\in B_{X}$. Since $\|x-z\|_{X}=\|y-z\|_{X}\leq 1$, we have that $\|F(x)-F(z)\|_{Y}\leq 1$ and $\|F(y)-F(z)\|_{Y}=\|-F(x)-F(z)\|_{Y}\leq 1$. This means that $F(z)\in D_{1}(F(x))=\{0\}$, so $z=0$ and $y=-x$. Hence $F(-x)=-F(x)$.

Now, let us proceed with the case when Y is strictly convex.

(i) Assume to the contrary that $F(S_{X})\neq S_{Y}$. By (2) this means that $F(y)\not\in S_{Y}$ for some $y\in S_{X}$. By the surjectivity of $F$ there is an $x\in S_{X}$ such that $F(x)=F(y)/\|F(y)\|_{Y}$. Then from (4) it follows that $F(\|F(y)\|_{Y}x)=\|F(y)\|_{Y}F(x)=F(y)$, which contradicts the injectivity of $F$.

Because of strict convexity of Y the items (ii) and (iii) are now evident consequences of (4) and (6) respectively. ∎

Following notations from [2] for every $u\in S_{X}$ and $v\in X$ denote by $u^{*}(v)$ the directional derivative of the function $x\mapsto\|x\|_{X}$ at the point $u$ in the direction $v$:

[TABLE]

By the convexity of the function $x\mapsto\|x\|_{X}$, the directional derivative exists. If $E\subset X$ is a subspace and $u$ is a smooth point of $S_{E}$ then $u^{*}|_{E}$ (the restriction of $u^{*}$ to $E$) is the unique norm-one linear functional on $E$ that satisfies $u^{*}|_{E}(u)=1$ (the supporting functional at point $u$). In general $u^{*}:X\to{\mathbb{R}}$ is not linear, but it is sub-additive and positively homogeneous:

[TABLE]

For proving these facts we will use the triangle inequality:

[TABLE]

[TABLE]

If one substitutes in the subadditivity condition $y_{1}=v$ and $y_{2}=-v$, one gets

[TABLE]

Also $u^{*}$ possesses the following property: for arbitrary $y_{1},y_{2}\in X$

[TABLE]

Let us demonstrate the latter property:

[TABLE]

The next lemma generalizes in a straightforward way [2, Lemma 2.4].

Lemma 2.4**.**

Let $F:B_{X}\to B_{Y}$ be a bijective non-expansive map, and suppose that for some $u\in S_{X}$ and $v\in B_{X}$ we have $u^{*}(-v)=-u^{*}(v)$, $\|F(u)\|=\|u\|$ and $F(av)=aF(v)$ for all $a\in[-1,1]$. Then $(F(u))^{*}(F(v))=u^{*}(v)$.

Proof.

[TABLE]

so $(F(u))^{*}(-F(v))\leq u^{*}(-v)$. Substituting $-v$ in the place of $v$, we get also

[TABLE]

Using these two inequalities together with (2.3), we have

[TABLE]

So all the inequalities in this chain are equalities. ∎

The following result and Corollary 2.6 are extracted from the proof of [2, Lemma 2.5].

Lemma 2.5**.**

Let $F:B_{X}\to B_{Y}$ be a bijective non-expansive map such that $F(S_{X})=S_{Y}$. Let $V\subset S_{X}$ be such a subset that $F(av)=aF(v)$ for all $a\in[-1,1]$, $v\in V$. Denote $A=\{tx:x\in V,t\in[-1,1]\}$, then $F|_{A}$ is a bijective isometry between $A$ and $F(A)$.

Proof.

Fix arbitrary $y_{1},y_{2}\in A$. Let $E=\mathrm{span}\{y_{1},y_{2}\}$, and let $W\subset S_{E}$ be the set of smooth points of $S_{E}$ (which is dense in $S_{E}$). All the functionals $x^{*}$, where $x\in W$, are linear on $E$, so $x^{*}(-y_{i})=-x^{*}(y_{i})$, for $i=1,2$. Also, according to our assumption, $F(ay_{i})=aF(y_{i})$ for all $a\in[-1,1]$. Now we can apply Lemma 2.4.

[TABLE]

where on the last step we used the inequality (2.4). So $\|F(y_{1})-F(y_{2})\|=\|y_{1}-y_{2}\|$. ∎

Corollary 2.6**.**

If $F:B_{X}\to B_{Y}$ is a bijective non-expansive function that satisfies (i), (ii), and (iii) of Theorem 2.3, then $F$ is an isometry.

Proof.

We can apply Lemma 2.5 with $V=S_{X}$ and $A=B_{X}$. ∎

3. Main results

Theorem 3.1**.**

Let $F:B_{X}\to B_{Y}$ be a bijective non-expansive map. If $Y$ is strictly convex, then $F$ is an isometry.

Proof.

If $Y$ is strictly convex, then $F$ satisfies (i), (ii), and (iii) of Theorem 2.3, so Corollary 2.6 is applicable. ∎

Our next goal is to show that each non-expansive bijection from the unit ball of arbitrary Banach space to the unit ball of $\ell_{1}$ is an isometry. In the proof we will use the following three known results.

Proposition 3.2** (P. Mankiewicz’s [8]).**

If $A\subset X$ and $B\subset Y$ are convex with non-empty interior, then every bijective isometry $F:A\to B$ can be extended to a bijective affine isometry $\tilde{F}:X\to Y$.

Taking into account that in the case of $A$, $B$ being the unit balls every isometry maps 0 to 0, this result implies that every bijective isometry $F:B_{X}\to B_{Y}$ is the restriction of a linear isometry from $X$ onto $Y$.

Proposition 3.3** (Brower’s invariance of domain principle [1]).**

Let $U$ be an open subset of ${\mathbb{R}}^{n}$ and $f:U\to{\mathbb{R}}^{n}$ be an injective continuous map, then $f(U)$ is open in ${\mathbb{R}}^{n}$.

Proposition 3.4** (Proposition 4 of [6]).**

Let $X$ be a finite-dimensional normed space and $V$ be a subset of $B_{X}$ with the following two properties: $V$ is homeomorphic to $B_{X}$ and $V\supset S_{X}$. Then $V=B_{X}$.

Now we give the promised theorem.

Theorem 3.5**.**

Let $X$ be a Banach space, $F:B_{X}\to B_{\ell_{1}}$ be a bijective non-expansive map. Then $F$ is an isometry.

Proof.

Denote $e_{n}=(\delta_{i,n})_{i\in{\mathbb{N}}}$, $n=1,2,\ldots$ the elements of the canonical basis of $\ell_{1}$ (here, as usual, $\delta_{i,n}=0$ for $n\neq i$ and $\delta_{n,n}=1$). It is well-known and easy to check that $ext(B_{\ell_{1}})=\{\pm e_{n},i=1,2,...\}$.

Denote $g_{n}=F^{-1}e_{n}$. According to item (5) of Theorem 2.3 each of $g_{n}$ is an extreme point of $B_{X}$.

One more notation: for every $N\in{\mathbb{N}}$ and $X_{N}=\mathrm{span}\{g_{k}\}_{k\leq N}$ denote $U_{N}$ and $\partial U_{N}$ the unit ball and the unit sphere of $X_{N}$ respectively and analogously for $Y_{N}=\mathrm{span}\{e_{k}\}_{k\leq N}$ denote $V_{N}$ and $\partial V_{N}$ the unit ball and the unit sphere of $Y_{N}$ respectively.

Claim. For every $N\in{\mathbb{N}}$ and every collection $\{a_{k}\}_{k\leq N}$ of reals with $\|\sum_{n\leq N}a_{n}g_{n}\|\leq 1$

[TABLE]

Proof of the Claim. We will use induction on $N$. For $N=1$, the Claim follows from items (4) and (6) of Theorem 2.3. Now assume the validity of the Claim for $N-1$, and let us prove it for $N$. At first, for every $x=\sum_{i=1}^{N}{\alpha_{i}g_{i}}$ we will show that

[TABLE]

Note that, due to the positive homogeneity of the norm, it is sufficient to consider $x=\sum_{i=1}^{N}{\alpha_{i}g_{i}}$, with $\sum_{i=1}^{N}{|\alpha_{i}|}\leq 1$. In such a case $x\in U_{N}$. So

[TABLE]

and $\sum_{i=1}^{N-1}{\alpha_{i}g_{i}}\in U_{N}$. On one hand,

[TABLE]

On the other hand, by the induction hypothesis $F(\sum_{i=1}^{N-1}{\alpha_{i}g_{i}})=\sum_{i=1}^{N-1}{\alpha_{i}e_{i}}$ and by items (4) and (6) of Theorem 2.3 $F(-\alpha_{N}g_{N})=-\alpha_{N}e_{N}$. Consequently,

[TABLE]

and (3.1) is demonstrated. That means that

[TABLE]

The remaining part of the proof of the Claim, and of the whole theorem repeats almost literally the corresponding part of the proof of [6, Thorem 1], so we present it here only for the reader’s convenience.

Let us show that

[TABLE]

To this end, consider $x\in U_{N}$. If $x$ is of the form $\alpha g_{N}$ the statement follows from Theorem 2.3. So we must consider $x=\sum_{i=1}^{N}{\alpha_{i}g_{i}}$, $\sum_{i=1}^{N}{|\alpha_{i}|}\leq 1$ with $\sum_{i=1}^{N-1}|\alpha_{i}|\neq 0$. Denote the expansion of $F(x)$ by $F(x)=\sum_{i=1}^{\infty}{y_{i}e_{i}}$. For the element

[TABLE]

by the induction hypothesis

[TABLE]

So we may write the following chain of inequalities:

[TABLE]

This means that all the inequalities in between are in fact equalities, so in particular $\sum_{i=N+1}^{\infty}|y_{i}|=0$, i.e. $F(x)=\sum_{i=1}^{N}{y_{i}e_{i}}\in V_{N}$ and (3.2) is proved.

Now, let us demonstrate that

[TABLE]

Assume on the contrary that there is a $y\in\partial V_{N}\setminus F(U_{N})$. Denote $x=F^{-1}(y)$. Then, $\|x\|=1$ (by (2) of Theorem 2.3) and $x\notin U_{N}$. For every $t\in[0,1]$ consider $F(tx)$. Let $F(tx)=\sum_{n\in{\mathbb{N}}}b_{n}e_{n}$ be the corresponding expansion. Then,

[TABLE]

so $\sum_{n>N}|b_{n}|=0$. This means that $F(tx)\in V_{N}$ for every $t\in[0,1]$. On the other hand, $F(U_{N})$ contains a relative neighborhood of 0 in $V_{N}$ (here we use that $F(0)=0$ and Proposition 3.3), so the continuous curve $\{F(tx):t\in[0,1]\}$ in $V_{N}$ which connects 0 and $y$ has a non-trivial intersection with $F(U_{N})$. This implies that there is a $t\in(0,1)$ such that $F(tx)\in F(U_{N})$. Since $tx\notin U_{N}$ this contradicts the injectivity of $F$. Inclusion (3.3) is proved.

Now, inclusions (3.2) and (3.3) together with Proposition 3.4 imply $F(U_{N})=V_{N}$. Observe that by (3.1) $U_{N}$ is isometric to $V_{N}$ and, by finite dimensionality, $U_{N}$ and $V_{N}$ are compacts. So, $U_{N}$ and $V_{N}$ can be considered as two copies of one the same compact metric space, and Theorem 1.1 of [7] implies that every bijective non-expansive map from $U_{N}$ onto $V_{N}$ is an isometry. In particular, $F$ maps $U_{N}$ onto $V_{N}$ isometrically. Finally, the application of Proposition 3.2 gives us that the restriction of $F$ to $U_{N}$ extends to a linear map from $X_{N}$ to $Y_{N}$, which completes the proof of the Claim.

Now let us complete the proof of the theorem. At first, passing in (3.1) to limit as $N\to\infty$ we get

[TABLE]

for every $z=\sum_{n=1}^{\infty}z_{n}g_{n}$ with $\sum_{n=1}^{\infty}|z_{n}|<\infty$. The continuity of $F$ and the claim imply that for every $x=\sum_{n=1}^{\infty}x_{n}e_{n}\in B_{\ell_{1}}$

[TABLE]

Consequently, for every $x,y\in B_{\ell_{1}}$, $x=\sum_{n=1}^{\infty}x_{n}e_{n}$ and $y=\sum_{n=1}^{\infty}y_{n}e_{n}$ the following equalities hold true:

[TABLE]

So, $F^{-1}$ is an isometry, consequently the same is true for $F$. ∎

Our next (and the last) goal is to demonstrate that each non-expansive bijection between unit balls of two different finite dimensional Banach spaces is an isometry. Below we recall the definitions and well-known properties of total and norming subsets of dual spaces that we will need further.

A subset $V\subset S_{X^{*}}$ is called total if for every $x\neq 0$ there exists $f\in V$ such that $f(x)\neq 0$. $V$ is called norming if $\sup|f(x)|_{f\in V}=\|x\|$ for all $x\in X$. We will use the following easy exercise.

Lemma 3.6** ([4], Exercise 9, p. 538).**

Let $A\subset S_{X}$ be dense in $S_{X}$, for every $a\in A$ let $f_{a}$ be a supporting functional at $a$. Then $V=\{f_{a}:a\in A\}$ is norming (and consequently total).

The following known fact is an easy consequence of the bipolar theorem.

Lemma 3.7**.**

Let $X$ be a reflexive space. Then $V\subset S_{X^{*}}$ is norming if and only if $\overline{{\mathrm{aconv}}}(V)={B_{X^{*}}}$.

Now we can demonstrate the promised result.

Theorem 3.8**.**

Let $X$, $Y$ be Banach spaces, $Y$ be finite-dimensional, $F:B_{X}\to B_{Y}$ be a bijective non-expansive map. Then $F$ is an isometry.

Proof.

Take an arbitrary finite-dimensional subspace $Z\subset X$. Then the restriction of $F$ to $B_{Z}$ is a bijective and continuous map between two compact sets $B_{Z}$ and $F(B_{Z})$, so $B_{Z}$ and $F(B_{Z})$ are homeomorphic. Thus, Brower’s invariance of domain principle (Proposition 3.3) implies that $\dim Z\leq\dim Y$. By arbitrariness of $Z\subset X$ this implies that $\dim X\leq\dim Y$. Consequently, $F$ being bijective and continuous map between compact sets $B_{X}$ and $B_{Y}$, is a homeomorphism. Another application of Proposition 3.3 says that $\dim X=\dim Y$, $F$ maps interior points in interior points, and $F(S_{X})=S_{Y}$.

Let $G$ be the set of all $x\in S_{X}$ such that norm is differentiable both at $x$ and $F(x)$. According to [9, Theorem 25.5], the complement to the set of differentiability points of the norm is meager. Consequently, $G$ being an intersection of two comeager sets, is dense in $S_{X}$. Recall that $F$ is a homeomorphism, so $F(G)$ is dense in $S_{Y}$. Thus, Lemma 3.6 ensures that $A:=\{x^{*}:x\in G\}$ and $B:=\{F(x)^{*}:x\in G\}=\{y^{*}:y\in F(G)\}$ are norming subsets of $X^{*}$ and $Y^{*}$ respectively, and consequently by Lemma 3.7

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Denote $K=F^{-1}(\mathop{\rm ext}\nolimits B_{Y})\subset\mathop{\rm ext}\nolimits B_{X}$. Note that for all $x\in G$ the corresponding $(F(x))^{*}$ and $x^{*}$ are linear, and Lemma 2.4 implies that for all $x\in G$ and $z\in K$ the following equality holds true:

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Let us define the map $H:A\to B$ such that $H(x^{*})=(F(x))^{*}$. For the correctness of this definition it is necessary to verify for all $x_{1},x_{2}\in G$ the implication

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Assume for given $x_{1},x_{2}\in G$ that ${x_{1}}^{*}={x_{2}}^{*}$. In order to check equality ${F(x_{1})}^{*}={F(x_{2})}^{*}$ it is sufficient to verify that ${F(x_{1})}^{*}y={F(x_{2})}^{*}y$ for $y\in\mathop{\rm ext}\nolimits B_{Y}$, i.e. for $y$ of the form $y=F(x)$ with $x\in K$. Indeed,

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Let us extend $H$ by linearity to $\tilde{H}:X^{*}=\mathrm{span}(x^{*},x\in G)\to Y^{*}$. For $x^{*}=\sum_{k=1}^{N}\lambda_{k}{x_{k}}^{*}$, $x_{k}\in G$ let $\tilde{H}(x^{*})=\sum_{k=1}^{N}\lambda_{k}H({x_{k}}^{*})$. To verify the correctness of this extension we will prove that

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Again we will prove equality $\sum_{k=1}^{N}\lambda_{k}H({x_{k}}^{*})=\sum_{k=1}^{M}\mu_{k}H({y_{k}}^{*})$ of functionals only on elements of the form $y=F(x)$ with $x\in K$.

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Observe that, according to (3.4), $\tilde{H}(X^{*})=\mathrm{span}H(A)=\mathrm{span}B=Y^{*}$, so $\tilde{H}$ is surjective, and consequently, by equality of corresponding dimensions, is bijective. Recall, that $\tilde{H}(A)=H(A)=B$, so $\tilde{H}$ maps $A$ to $B$ bijectively. Applying again (3.4) we deduce that $\tilde{H}(B_{X^{*}})=B_{Y^{*}}$ and $X^{*}$ is isometric to $Y^{*}$. Passing to the duals we deduce that $Y^{**}$ is isometric to $X^{**}$ (with $\tilde{H}^{*}$ being the corresponding isometry), that is $X$ and $Y$ are isometric. So, $B_{X}$ and $B_{Y}$ are two copies of the same compact metric space, and the application of EC-plasticity of compacts [7, Theorem 1.1] completes the proof. ∎

Acknowledgement. The author is grateful to her scientific advisor Vladimir Kadets for constant help with this project.