Letting $x_n\to x$ weakly in a normed vector space $X$, prove the following statement.

Question

"Let $x_n\to x$ weakly in a normed vector space $X$. Show,that there exists a sequence of finite convex combinations of the $x_n$'s converging in norm to $x$."

Here is what I have so far:

For a normed vector space $X$, and it's dual $X^*$, we characterise weak convergence in the following manner. For each $f\in X^*$, we define a seminorm on $X$ by $p_f(x)=|f(x)|$, where $x\in X$.

Then, $p_f(x_n-x)=|f(x_n-x)|\le\|f\|\|x_n-x\|\to0$ as $n\to\infty$ since $x_n$ converges weakly to $x$.

Now I'm wondering how exactly this gives rise to a "finite convex combination of the $x_n$'s". Such a combination is of the form $\alpha_1x_1+...+\alpha_nx_n$ where $\alpha_i\ge0$ and $\sum_{i=1}^n\alpha_i=1$. Is the idea that I should somehow insert this convex combination into the $\|x_n-x\|$ part? Or have I gone about it the completely wrong way so far?

Answer 1

Theorem 3.12 of Rudin's FA says the the weak closure of a convex set is equal to the norm closure. Let $E$ be the convex hull of $\{x_1,x_2,...\}$. Then x is in the weak closure of this set. Hence it is in the norm closure and this is what is being asserted. Rudin's Theorem is based on Hahn - Banach type theorems and I don't think you can answer this question without that theorem.

Answer 2

Something looks pretty wrong. We cannot say that $\|x_n-x\|\to0$ if $x_n\to x$ weakly. Besides, if you could show that $\|x_n-x\|\to0$ then you would be done, as the sequence $(x_n)$ is a sequence of finite convex combinations of $\{x_n\}$.

Perhaps the easiest way to solve this is using the fact that for convex sets, the norm and weak closures coincide.