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Let $X$ be a normed linear space, $\psi \in X^{*}$ and $\displaystyle \{\psi_n\}_{n \in \Bbb N}$ a sequence in $X^{*}$. Show that if $\displaystyle \{\psi_n\}_{n \in \Bbb N}$ converges weak-${*}$ to $\psi$ then:

$$\|\psi\| \le \lim \sup \|\psi_n\|$$

Any suggestions or help will be greatly appreciated. I am not well acquainted with weak-$*$ convergence. Thank you in advance!

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2 Answers

up vote 2 down vote accepted

Weak-$*$-convergence gives you

$$\|\psi(x)\| \le \lim \sup \|\psi_n(x)\|$$

for every $x\in X$. Can you proceed from here?

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I don't think so... –  user44069 Mar 1 '13 at 16:50
    
@Stefan Strange. You've already accepted an answer, and asked me to take a look at it... –  Coiacy Mar 2 '13 at 6:13
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By definition of $weak^*$ convergence, $$ \forall x\in X,\quad \psi_n(x)\to \psi(x) $$ Usually we assume $\psi\in X^*:\ X \to \mathbb{C}$ $$ \psi_n(x)\to \psi(x)\quad \Rightarrow \quad |\psi(x)|=\lim_{n\to \infty}|\psi_n(x)| $$ Since $|\psi_n(x)|\le \|\psi_n\|\cdot \|x\|,$ $$\ \lim_{n\to \infty}|\psi_n(x)|=\limsup_{n\to \infty}|\psi_n(x)|\le \limsup_{n\to \infty}|\|\psi_n\|\cdot \|x\|=\|x\|\cdot \limsup_{n\to \infty}|\|\psi_n\| $$ $$ \Rightarrow \quad|\psi(x)|=\lim_{n\to \infty}|\psi_n(x)|\le \|x\|\cdot\limsup_{n\to \infty}|\|\psi_n\| $$

By definition, we obtain $$ \|\psi\|\le \limsup_{n\to \infty}\|\psi_n\| $$

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I am immensely grateful! Thank you for your time and for taking the trouble to help me! –  user44069 Mar 2 '13 at 6:13
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