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The following statement can be obtain in the proof of Proposition $7.1$

Since every Banach space is isometric to a subspace of an $L_{\infty}(\mu)$-space (e.g. an $l_{\infty}(T)$ for suitable $T$), the proposition is trivial if $p=\infty$.

Question: How to show that 'every Banach space is isometric to a subspace of an $L_{\infty}(\mu)$-space'?

Idonknow
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  • http://math.stackexchange.com/questions/112619/isometric-embedding-of-a-separable-banach-space-into-ell-infty – A Blumenthal Dec 11 '15 at 03:54
  • @ABlumenthal: The question in your link involves separable Banach space, but in my question separability is not assumed. – Idonknow Dec 11 '15 at 03:58

1 Answers1

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Let $X$ be a Banach space and let $B_{X^*}$ be the unit ball of $X^*$. Define a map $T\colon X\to \ell_\infty(B_{X^*})$ by $$Tx = (\langle f, x\rangle)_{f\in B_{X^*}}\quad (x\in X).$$ Then $T$ is linear. It follows from the Hahn–Banach theorem that $T$ is isometric.

In some sense the answer is optimal because spaces which do not admit a strictly convex renorming (such as $\ell_\infty/c_0$) do not embed into $L_\infty(\mu)$ for $\mu$ being $\sigma$-finite.

Tomasz Kania
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