# Modification of Mazur's lemma

Working through Brézis and solving an exercise I have a question about my solution.

It's well known that, if $x_n$ converges weakly to $x$ in a Banach space $X$, then there exists a sequence $\{y_l\}$ of convex combinations of the $x_n$ converging strong to $x$. (Banach space is not needed)

Now in Brézis, as an exercise, I should show:

1). if $x_n \rightharpoondown x$ weakly then it exists a sequence $y_l$ such that $y_l \in \operatorname{conv}\left(\bigcup_{i=l}^\infty \{x_i\}\right) \forall l$ s.t. $y_l \to x$

2.) if $x_n \rightharpoondown x$ weakly then it exists a sequence $y_l$ s.t $y_l \in \operatorname{conv}(\bigcup_{i=1}^l \{x_i\}) \forall l$ s.t. $y_l \to x$ where $\operatorname{conv}$ is the convex hull.

My proof of 1) is like this:

I will use that every convex set is closed if and only if it's weakly closed. so proof of 1):

$x \in cl(\operatorname{conv}\{x_n\}_{n\ge 1}))$ therefore I can pick a $y_1$ s.t. $y_1 = \sum_{i=1}^{m_1} \lambda_i x_i$ which is in $\operatorname{conv}(\{x_n\}_{n \ge 1})$ and $\| y_1 - x\| < \frac{1}{2}$.

Then we know, that $x \in cl(\operatorname{conv}\{x_n\}_{n\ge m_1}))$, clearly choose $y_2$ s.t. $$y_2= \sum_{i=m_1+1}^{m_2} \lambda_i x_i \mbox{ such that } \|y_2-x\| < \frac{1}{4}.$$

Just as a note, $y_2 \in \operatorname{conv}(\{x_n\}_{n\ge m_1})$ Suppose we have chosen our sequence up to an index $k$. We know:

$x \in \operatorname{cl}(\operatorname{conv}\{x_n\}_{n\ge m_k}))$ again choose $y_{k+1}$ s.t :

$$y_{k+1}= \sum_{i=m_k+1}^{m_{k+1}} \lambda_i x_i\mbox{ such that } \|y_{k+1}-x\| < \frac{1}{2^{k+1}}.$$

and this should prove 1), if I'm not wrong. As a question at the point where I have chosen the $y_{k+1}$: the $x_i$ in the sum are not the same as for the sequence. To be precise, the $x_i$ in the sum is not the $i$-th element of the sequence, right? If not, why can I choose it in such a way?

Looking at 2) it's clear, that $y_1 = x_1$.

If it exists an index $N_0$ s.t. $x \in \operatorname{cl}(\operatorname{conv}\{x_n\}_{n\le N_0}))$, then I have no troubles since I can choose a $y_k$ as "near" to $x$ as I want. Therefore I get strong convergence. So should I just choose the $y_1, \dots y_{{N_0}-1}$ as an arbitrary convex combination, and start choosing $y_{N_0}$ in a way, such that the sequence converges (Since the finitely many elements at the beginning are not important for convergence). However I'm not sure any more about the existence of such an index $N_0$.

Edit: I have a new idea about 2).

define $\alpha_k:= inf \|x-y_k\|$ over all $y_k \in \operatorname{conv}(\{x_n\}_{n\le k})$. Obviously $\alpha_k \ge \alpha_{k+1}$.

Claim: $\alpha_k \to 0$.

proof: let $\epsilon > 0$, it exists a $z_\epsilon \in \operatorname{conv}(\{x_n\}_{n\ge_1})$,hence

$z_\epsilon = \sum_{i=1}^{m(\epsilon)} \lambda_ix_i$, s.t $\|x-z_\epsilon \| \le \epsilon$

Therefore we could find a $N \in \mathbb{N}$ s.t.

$x_1, \dots, x_{m(\epsilon)} \in \operatorname{conv}(\{x_n\}_{n\le N})$

so we conclude, $\alpha_{N+1} \le \epsilon$.

Is there another way to solve this? Are my proofs / thoughts correct?

Thanks.

math

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@ Davide Giraudo: thx for editing my question. Maybe a stupid question, but how do you get this blue line, what are the LaTeX commands? – math Dec 2 '11 at 21:03
Just type ">" at the beginning of the line. I will look at your work now. – Davide Giraudo Dec 3 '11 at 9:25