If $\|x\|$ converges weakly to $x$, then $\|x\| \leq \liminf_{n\in\mathbb{N}} \|x_n\|$.
Can someone explain this? Thanks.
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What exactly are you looking for? A proof? Or the intuition why it is plausible that it is true? What do you already know? – Severin Schraven Mar 25 '20 at 22:07
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Related – Pedro Mar 25 '20 at 22:31
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I don't understand the proof in the related post. I can see $|x| = \lim \inf |x_n|$. But where is the "$<$" from? – baNv Mar 26 '20 at 00:07
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@baNv Are you talking about this? If so, note that $|\phi(x_n)|\leq|\phi|_{\text{op}}|x_n|=|x_n|$ – Pedro Mar 27 '20 at 01:53
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I'm lost at this step: $\lim _{n \rightarrow \infty} |\phi (x_n) | \leq \liminf _{n \rightarrow \infty} |\phi (x_n) |$, why $\leq$? I thought it should be $\geq$. – baNv Mar 27 '20 at 16:09