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In Lax`s Functional Analysis he affirms that the following result due to Mazur:

Let $K$ be a closed convex subset of $X$ a normed linear space, and $x_n \rightharpoonup x$, then $x \in K$

implies that if $x_n \rightharpoonup x$ then $|x| \leq \liminf |x_n|$, but i cannot see why is it true.

I have attempted a proof, but I only managed to find that $|x_n| \leq C$ where C is some constant, which would imply $|x| \leq C$. My other attempts fall into a proof that does not use Mazur`s result.

I appreciate any help

Francisco Maion
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1 Answers1

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Let $a=\liminf|x_n|$ and $\varepsilon>0$. Then there exists $N\in\mathbb N$ such that, for all $n\geq N$, $|x_n|<a+\varepsilon$.

Set now $$K_N=\overline{{\rm conv}}\{x_N,x_{N+1},\dots\},$$ the closure of the convex hull of the $x_n$, for $n\geq N$. Since $x_n\rightharpoonup x$ and $K_N$ is convex, we obtain that $x\in K_N$. But, since all the $x_n$ for $n\geq N$ belong to the ball $B(a+\varepsilon)$, centered at $0$ and with radius $a+\varepsilon$, and this ball is convex, we obtain that $$K_N\subseteq\overline{B(a+\varepsilon)}.$$ This implies that $x\in \overline{B(a+\varepsilon)}$, so $|x|\leq a+\varepsilon$ for all $\varepsilon>0$. Hence $|x|\leq a$.

detnvvp
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  • Just one thing: In your first affirmation you would have to use a subsequence, correct? But I can see that the rest of the proof would continue to be valid anyway for a subsequence. – Francisco Maion Feb 15 '18 at 12:20
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    Yes, you are right, the subsequence here would be $(x_{N+n})_{n\in\mathbb N}$. – detnvvp Feb 15 '18 at 13:05