assume $X$ is a normed space with separable dual $X'$. Show that $X$ is a separable space.

Question

What have i proven already:

1) If $C$ is a countable subset of $X$, such that $\overline{Sp}(C) = X$, then $X$ is separable.

2) If $A$ is a subset of $X, \text{ }^\circ(A^\circ)= \overline{Sp}(C)$.

So what i think we must do is assume we have a countable dense subset $D$ of $X'$ and relate it somehow to a countable set $C$ such that $\text{ }^\circ(C^\circ) = X$.

If $C^\circ = \{0\}$, we get $\text{ }^\circ(C^\circ) = \overline{Sp}(C) = X$.

So my conclusion is: Find a countable set $C$ such that $C^{\circ} = 0$?

If anyone has a better tip please help me :D

Kees

score 3 · Accepted Answer · answered Aug 15 '16 at 15:14

It is spelled separable. This is a well-known theorem, see e.g. Theorem 3.26 in Brezis' functional analysis book. Here is a quick outline of the proof:

Let $\{f_n\}$ be a countable dense subset of $X^*$. For each $n$ there exists $x_n \in X$ with $\|x_n\| = 1$ satisfying $f_n(x_n) \ge \frac 12 \|f_n\|$.

Let $D$ denote the set of all finite linear combinations of the $x_n$'s, and let $E$ denote the set of all finite linear combinations of the $x_n$'s with rational coefficients. Then: $D$ is a subspace of $X$, $E$ is countable, and $E$ is dense in $D$.

Suppose that $f \in X^*$ and $f(x) = 0$ for all $x \in D$. For each $n$ you have $$\|f|| \le \|f - f_n\| + \|f_n\|$$ and $$\|f_n\| \le 2 f_n(x_n) = 2(f_n - f)(x_n) \le 2\|f - f_n\|.$$ Thus $\|f\| \le 3 \|f - f_n\|$ for all $n$, and since $\{f_n\}$ is dense in $X^*$ it follows that $\|f\| = 0$. This means that $D$ (and consequently the countable set $E$) is dense in $X$.

Can u explain the statement in the second paragraph? – Kees Til Aug 15 '16 at 15:23 — Kees Til, Aug 15 '16 at 15:23
Look at the definition of the norm of a functional. – Umberto P. Aug 15 '16 at 15:24 — Umberto P., Aug 15 '16 at 15:24

assume $X$ is a normed space with separable dual $X'$. Show that $X$ is a separable space.

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