Surjective linear isometries on $\ell_\infty(\mathbb{N})$

Question

In Volume 1 of "Classical Banach Spaces" Lindenstrauss and Tzafriri note that all surjective linear isometries on $\ell_\infty$ are of the from $(a_i) \mapsto (\varepsilon_i a_{\pi(i)})$ where $\pi$ is a permutation of $\mathbb{N}$ and $\varepsilon_i \in\{\pm 1\}$. They mention that the proof is a consequence of the more general result on isometries on $C(K)$ spaces that it will appear in Volume 3 (which does not exist). Does anyone know a reference for the proof of this fact?

Answer 1

I don't know a reference (other users will probably provide one). Here's a (quite immediate) proof anyway. I'll replace $\mathbf{N}$ with an arbitrary set $X$ since the integers play no role.

Fact The extremal points of the closed 1-ball of $\ell^\infty(X)$ precisely consists of the maps $X\to\{-1,1\}$.

Proof: Given $f,g,u:X\to [-1,1]$ if $u\in [f,g]$ and for some $x\in X$, $u(x)\in\{-1,1\}$, then we see that $f(x)=g(x)=u(x)$. Hence, all $u:X\to\{-1,1\}$ are extremal. Conversely, let $u$ not be of this form. So $-1<u(y)<1$ for some $y$. For a scalar $\varepsilon$, define $u_\pm(y)=u(y)\pm\varepsilon$ and $u_\pm(x)=u(x)$ for $x\neq y$. Then $u=\frac12(u_++u_-)$ and for $\varepsilon>0$ small enough, both $u_\pm$ belong to the closed 1-ball. So $u$ is not extremal.

Corollary: every bijective self-isometry of $\ell^\infty$ has the required form (call this "standard", and "diagonal-standard" when the permutation is trivial).

Proof: Write $W_A=1_A-1_{A^c}$. Then, by the fact, the isometry group acts on the power set $2^X$ by $g\cdot W_A=W_{gA}$.

Given an isometry, after composing with a diagonal-standard isometry, we can suppose that $f(1_X)=1_X$, i.e., $f\in G$, where $G$ is the stabilizer of $1_X$. We have to show that $G$ is reduced to the permutation group.

Now for $A,B,C\subseteq X$, write $F(A,B,C)$ if $W_A+W_B-W_C=1_X$; since $G$ acts linearly and fixes $1_X$, this ternary relation is preserved by $G$. Also it is easy to check that $F(A,B,C)$ holds if and only if $C=A\cap B$. Hence, we see that $G$ (acting on $2^X$ by $f(W_A)=W_{gA})$ preserves the intersection operation. In particular, it preserves the ordering of $2^X$, and hence is an automorphism of the Boolean algebra $2^X$. Such an automorphism is necessarily induced by a permutation.

Now for $f\in G$, composing with a permutation (which is standard), we can suppose that $f$ pointwise fixes all maps $X\to\{-1,1\}$. By a $\mathbf{Q}$-linear combination, it therefore preserves all characteristic functions (=maps $X\to\{0,1\}$), and in turn, it therefore preserves all simple functions (= maps $X\to\mathbf{R}$ with finite image). Since these form a dense subspace of $\ell^\infty(X)$, the resulting isometry is the identity, which means that the original isometry is standard.

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Note: an analogous result also holds for all $p\in [1,\infty]\smallsetminus\{2\}$. The case $1$ is similarly easy, but in the other cases this easy argument doesn't work since then the all sphere consists of extremal points.

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As pointed out by @GiorgioMetafune the initial argument was incorrect and some combinatorial argument was needed.

Answer 2

This is a simple consequence of the Banach-Stone theorem. Identify $l^\infty$ with $C(\beta \mathbb{N})$ and note that any self-homeomorphism of $\beta \mathbb{N}$ must take $\mathbb{N}$ to itself, since this is the set of isolated points of $\beta\mathbb{N}$.