# Every separable Banach space is isometrically isomorphic to a subspace of $\ell_\infty$

I want to prove exactly what is written in the title, using the following idea:

Let $$X$$ be a separable Banach space. We have that $$X$$ is isometrically isomorphic to a subspace $$M$$ of $$C(B_{X^\ast})$$, where $$B_{X^\ast} = \{x^\ast\in X^\ast:\ \|x^\ast\| = 1 \}$$ and $$C(B_{X^\ast})$$ is the space of continuous functions from $$B_{X^\ast}$$ to $$\mathbb{K}$$ ( we have $$\mathbb{K} = \mathbb{R}$$ or $$\mathbb{C}$$).

I want to prove that $$M$$ is isometrically isomorphic to some subspace of $$\ell_\infty$$. Doing this it's sufficient to finish the proof, because then $$X\cong M\cong$$ some subspace of $$\ell_\infty$$. The first isometric isomorphism is already proven, the second one is the problem. I came up with a solution, but I'm not sure if it's right because the Banach hypothesis was not used (or I just missed where it's used). I appreciate your help, thanks.

My solution: $$B_{X^\ast}$$ is $$w^\ast$$-separable so let $$(x^\ast_n)_{n\in\mathbb{N}}$$ be a $$w^\ast$$-dense sequence in $$B_{X^\ast}$$ and define $$\Phi:M\to\ell_\infty$$ such that $$\Phi(f) = (f(x^\ast_1),\ldots,f(x^\ast_n),\ldots)$$. It's not hard to see $$\Phi$$ is well defined, for $$|f(x^\ast_n)| \leq \|f\| \cdot \|x^\ast_n\| \leq \|f\|$$ for all $$n\in\mathbb{N}$$, so $$\Phi(f)\in \ell_\infty$$ for all $$f\in C(B_{X^\ast})$$.

It's not hard also to prove that $$\Phi$$ is a linear injection, so the main problem is to prove that $$\Phi$$ is an isometry. Take any $$f\in C(B_{X^\ast})$$, we want to prove $$\|\Phi(f)\| = \|f\|$$, and for this we just need to prove that $$\sup_{n\in\mathbb{N}}|f(x^\ast_n)| = \sup_{x^\ast\in B_{X^\ast}} |f(x^\ast)|$$ ( for now, all we know is $$\sup_{n\in\mathbb{N}}|f(x^\ast_n)| \leq \sup_{x^\ast\in B_{X^\ast}} |f(x^\ast)|$$).

First of all, note that $$B_{X^\ast}$$ is $$w^\ast$$-compact (by Banach-Alaoglu theorem), then $$\sup_{x^\ast\in B_{X^\ast}} |f(x^\ast)| = \max_{x^\ast\in B_{X^\ast}} |f(x^\ast)| = |f(y^\ast)|$$ for some $$y^\ast \in B_{X^\ast}$$. Also, we know $$f$$ is continuous in $$y^\ast$$, so for any $$\varepsilon > 0$$ there is $$\delta > 0$$ such that $$\|x^\ast - y^\ast\| < \delta \implies |f(x^\ast) - f(y^\ast)| < \varepsilon$$.

We don't know if there is some $$x^\ast_n$$ close enough to $$y^\ast$$, for the separability is about the $$w^\ast$$ topology, not the norm, so let's work with that. Given any $$w^\ast$$-neighborhood $$V$$ of $$0$$ in $$B_{X^\ast}$$, we know $$V$$ is limited, so consider the collection of $$w^\ast$$-neighborhoods $$(\frac{1}{m}V)_{m\in\mathbb{N}}$$. Take $$m\in\mathbb{N}$$ such that $$\frac{1}{m} < \delta$$, since $$(x^\ast_n)_{n\in\mathbb{N}}$$ is $$w^\ast$$-dense in $$B_{X^\ast}$$, there is some $$x^\ast_n$$ such that $$x^\ast_n \in y^\ast+\frac{1}{m}V \implies x^\ast_n - y^\ast \in\frac{1}{m}V \subset \delta V \subset \delta B_{X^\ast} \implies \|x^\ast_n - y^\ast\| <\delta .$$

From this we conclude that $$|f(x^\ast_n) - f(y^\ast)| < \varepsilon$$, but $$\varepsilon > 0$$ is arbitrary, so there is functionals $$x^\ast_n$$ such that $$f(x^\ast_n)$$ is arbitrarily close to $$f(y^\ast)$$. With this we can conclude that

$$\sup_{n\in\mathbb{N}} |f(x^\ast_n)| = |f(y^\ast)|,$$

and this proves $$\Phi$$ is a isometry.

PS: There is this thread. But it looks like a different approach.

• It doesn't matter for your main argument, but $X$ is of course again isomorphic to a subspace of $C(B)$, not to the whole space. – user138530 Nov 14 '15 at 0:58
• Also, your question is answered in the question you linked too. – user138530 Nov 14 '15 at 1:02
• You are right about $X$ being isomorphic to a subspace, not the whole space, I'm going to fix this. About the the other answer, it goes in another direction. They don't prove the way I want to prove. – Integral Nov 14 '15 at 17:58
• Can you have the same argument by replacing $C(B_{X^*})$ with $X^{**}$? – Hashimoto Feb 17 at 4:46

## 1 Answer

I have two comments.

1. I'm not sure that $\Phi$ is surjective. However, since you don't want to prove that $X$ is isometrically isomorphic to $\ell_\infty$ (but to a subspace of $\ell_\infty$), you can replace "linear bijection" by "linear injection".

2. Every separable set has a countable norming set (Lemma 6.7 here). And any normed space with a countable norming set is isometric to a subspace of $\ell_\infty$ (Corollary 6.8 here). Thus, the completeness (that seems to be not used in your argument) is not really needed.

So, your solution looks fine to me.

• Thank you for your observations! – Integral Nov 19 '15 at 1:27