2

I have the following question.

Let $E$ be a Banach space and separable, then exists an isometry form $E$ in $\ell^{\infty}$.

The hint is: Consider Since $K = B_{E^{\star}}$ is compac and metrizable in $\sigma(E^{\star},E)$ there is a countable subset $(t_n)$ in $K$. Consider $S: E \to \ell^{\infty}$ difinded by $$ Sx = ( t_1(x), t_2(x),\dots,t_n(x),\dots ). $$

My attempt. How $\{t_n\}_{n=1}^{\infty} \subset K$, then $\|t_{n}\| \leq 1$ for all $n \in \mathbb{N}$, so $|t_{n}x|\leq \|x\|$ it´s $Sx \in \ell ^{\infty}$ and $\|Sx\|_{\ell^{\infty}}\leq \|x\|$. How can i prove $\|x\| \leq \|Sx\|_{\ell^{\infty}}$?

Could this work? Isometric Embedding of a separable Banach Space into $\ell^{\infty}$.

Thanks.

Skinner.
  • 342

1 Answers1

0

I would try it like this: Let $x_n$ be the dense sequence of $B_E$. By Hanh-Banach, for every $n$, there is a linear functional $t_n$ such that $||t_n||=1$ and $|t_n(x_n)| = ||x_n||$. Then define $S(x) = (t_i(x))_{i \in \mathbb{N}}$. Then $$ ||Sx||_{\infty} = \sup_{n \in \mathbb{N}}| t_i(x)| \leq ||x||$$ and if $x$ is any element of $B_E$, $x = \lim x_{n_k}$ ($x_n$ be the dense sequence of $B_E$) $$ ||Sx_{n_k}||_{\infty} = \sup_{n \in \mathbb{N}}| t_i(x_{n_k})| \geq |t_{n_k}(x_{n_k})| = ||x_{n_k}||$$ then $||Sx||_{\infty} \geq ||x||$. For any $x\in E$, consider $\dfrac{x}{||x||} \in B_E$ and argue as before.

Ilovemath
  • 2,921