Answer to Question #151528 in Functional Analysis for anjali

Question

Theorem: Let X be normed liner space over a filed K and Let X' be its dual space of X. If X' is separable then X is separable

Answer 1

Suppose that $X'$ is separable. By definition, $\exists (f_n)_{n\in\mathbb{N}} \in (X')^{\mathbb{N}}, \overline{(f_n)} = X'$ . For all $f_n\neq 0$ we can find an $x_n \in X, f(x_n)\geq \frac{1}{2} ||f_n||$ . We claim that $F = Vect_{\mathbb{Q}} ((x_n)_{n\in\mathbb{N}})$ ($Vect_{\mathbb{Q}+i\mathbb{Q}}$ if $K$ is complex) is dense in $X$. Suppose that it is not true, then $\exists x\in X, x\neq0, \epsilon>0, \forall y\in Vect_{\mathbb{Q}}((x_n)), ||x-y|| \geq \epsilon$ . By Hahn-Banach theorem, $\exists f\in X', f(x) = 1 = ||f||, f(\overline{F})=f(F) = \{ 0\}$ , as $f|_{\overline{Vect(x, (x_n))}}$ is a continuous linear operator. Now let's consider $||f-f_n||$ for $f_n\neq 0$ :

$||f-f_n|| \geq \frac{|f(x_n)-f(x_n)|}{||x_n||} \geq \frac{1}{2}||f_n||$

But as $(f_n)$ is dense, $\forall \epsilon'>0, \exists n$ tel que $||f-f_n||<\epsilon' \Rightarrow ||f_n||<2\epsilon', ||f||<3\epsilon'$. It is a contradiction, as $||f||=1.$

For $f_n=0, ||f-f_n||=||f|| = 1 >0$

It is a contradiction, because $f \notin \overline{f_n}, f\in X'$ . Therefore $F$ is dense in $X$ and thus $X$ is separable.