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If we have a normed vector space $X$, and $x_n \to x$ weakly, then it's clear how to show there's a sequence $y_n \to x$ strongly (since the weak closure and strong closure of a convex set coincide, just consider the closure of the convex hull of the sequence).

I'd like to know how to prove the second half of the lemma i.e. you can choose the $y_n$ to lie inside the convex hull of $\lbrace x_1,x_2,..,x_n \rbrace$. Why is this?

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The way I would do it is to notice that $$ \text{conv}\{x_n\}_{n\in\mathbb{N}}=\bigcup_{k=1}^\infty\text{conv}\{x_1,\ldots,x_k\}, $$ and then of course the closures are equal. Since $$ x\in\overline{\bigcup_{k=1}^\infty\text{conv}\{x_1,\ldots,x_k\}}, $$ we have $\lim_{n\to\infty}\text{dist}(x,\bigcup_{k=1}^n\text{conv}\{x_1,\ldots,x_k\})=0$ (the numbers inside the limit form a decreasing sequence of nonnegative numbers with inferior limit equal to zero). So we can choose a sequence $\{y_n\}$ with $y_n\in\text{conv}\{x_1,\ldots,x_n\}$ and $\lim_{n\to\infty}\|x-y_n\|=0$.

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You really meant $\lim_{n\to\infty}\text{dist}(x,\bigcup_{k=1}^n\{x_1,\ldots,x_k\})=0$ or it was supposed to be $\lim_{n\to\infty}\text{dist}(x,\bigcup_{k=1}^n\text{conv}\{x_1,\ldots,x_k\})=0$ ? – Integral Oct 31 '15 at 16:58
    
It was with "conv", of course. Thank you. – Martin Argerami Oct 31 '15 at 17:27
    
It is a very nice proof. However, I would like to ask you why "the numbers inside the limit form a decreasing sequence of nonnegative numbers with inferior limit equal to zero"? – hjinghao May 12 at 11:38
    
Decreasing because as $n $ grows the set grows, so there are more points available that can be closer to $x $. To zero because $x $ is in the closure of the union. – Martin Argerami May 12 at 11:43

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