# exercise, Hahn-banach theorem

I have this exerciose:

Let $\Omega$ be a normed space. Prove that $\Omega$ is seperable if $\Omega^*$ is.

It is in the chapter with the Hahn-Banach theorem, so I think I should use that theorem.

My attempt:

We know that there is a countable set $K \subset \Omega^*$, such that for any $x^* \in \Omega^*$ there is a sequence of bounded lienar functionals in K, such that they converge to $x^*$ in the operator-norm.

So I need to associate the set $K\in \Omega^*$ with some set in $\Omega$, and show that this set is dense in $\Omega$.

The only thing I can think of is to start the other way around. For any $x \in \Omega$, we have the space $\{ax| a \in \mathbb{C}\}$, this is a subspace of $\Omega$, and it has a natural bounded linear functional $P_x(ax)=a$. by the Hahn-Banach theorem, this functional can be extended to a functional on the entire set $\Omega$, so we can assume that $P_x \in \Omega^*$.

Now we have a sequence in $\Omega^*$: $k_n\rightarrow P_x$, where $k_n \in K$, and the convergence is in the operator norm.

The problem now is to associate a proper element in $\Omega$ to each $k_n$, and show that this sequence converges to $x$.

Could this approach work, if so, how do I finish it? If not, could you please help me some other way?

• In this way, I think it will not work. The point is that you want a countable set in $\Omega$ that can approximate every vector. The $k_n$ you take depend on $x$; so even if you associate a point in each to each $k_n$ in a suitable way, you will have uncountably many sequences in $\Omega$ (one for each $x$). – Etienne Oct 3 '15 at 8:14

Start with a countable set $D$ which is dense in the unit sphere of $\Omega^*$, say $D=\{ x_i^*;\; i\in\mathbb N\}$.
For each $i\in\mathbb N$, one find a point $x_i\in\Omega$ such that $\Vert x_i\Vert=1$ and $\vert \langle x_i^*,x_i\rangle\vert>1/2$.
To show that $\Omega$ is separable, it is enough to prove that the linear span of the vectors $x_i$ is dense in $\Omega$ (because then, the linear combinations with rational coefficients, a countable set, will be dense in $\Omega$).
To do this, it is enough (by Hahn-Banach) to show that if $x^*\in\Omega^*$ is such that $\langle x^*,x_i\rangle=0$ for every $i\in\mathbb N$, then $x^*=0$.
Assume that $x^*\neq 0$, so that (by scaling) we may in fact assume that $\Vert x^*\Vert=1$. Since $D$ is dense in the unit sphere of $\Omega^*$, one can choose $i\in\mathbb N$ such that $\Vert x_i^*-x^*\Vert\leq1/2$. Since $\langle x^*,x_i\rangle=0$ and $\Vert x_i\Vert=1$, we then have $$\vert \langle x_i^*,x_i\rangle\vert=\vert\langle x_i^*-x,x_i\rangle\vert \leq \Vert x_i^*-x^*\Vert\leq1/2\, ,$$ which is a contradiction.