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Let $X$ be a normed space and $(x'_n) \subseteq X'$ a sequence of functionals where $x_n'$ has $x'$ has its limit in the *-weak topology in $X'$. Show that $$ ||x'|| \le \operatorname{lim inf}_{n\to \infty} ||x'_n||. $$ I have no glue how to show this, do you have any hints?

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3 Answers 3

Fix $N$ an integer, and let $v_N$ of norm $1$ such that $\lVert x'(v_N)\rVert\geqslant \lVert x'\rVert -N^{—1}$.

Then write $$\lVert x'\rVert\leqslant N^{-1}+\liminf_{n\to +\infty}\lVert x'_n(v_N)\rVert$$

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Let $L=\liminf_{n\to\infty}\|x_n\|$. Consider a subsequence $x_{n_k}'$ so $\|x_{n_k}\|<L+\epsilon$. This converges to *-weakly $x'$. By Banach-Alaoglu, $\|x'\|\leq L+\epsilon$. Of course, $\epsilon$ can be chosen arbitrarily small. [This is less elementary than the first solution proposed, but can easily be adapted to nets.]

[It is false that $\|x_n'\|\to \|x'\|$, as suggested below. Consider any orthonormal sequence in an infinite-dimensional Hilbert space, which necessarily *-weakly converges to $0$.]

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Here's an sketch of what you have to do:

First try and prove that $x'$ is indeed linear, that's easy since you can see it as $\lim x_n'$.

From $x_n' \rightarrow x$ follows that $||x_n'|| \rightarrow ||x'||$ so in particular $\{x_n'(y): n \in \mathbb{N}\}$ is bounded for every $y \in X$. Now apply the $\textit{Banach-Steinhaus Principle}$ to conclude that $\{x_n'\}$ is (uniformly) bounded by a constant $C$.

The proof is straightforward from here...

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$x_n'$ converges to $x'$ only in the weak-star sense. How do you know $\|x_n'\|\to\|x'\|$? –  timur Apr 3 '13 at 3:10
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