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I know this does hold in $L^2$, since it's a Hilbert space.

I suspect that this is not true, but I cannot think of a counterexample.

Specifically, I want to know if $f_n \xrightarrow{w} f$ and $\Vert f_n \Vert \rightarrow \Vert f \Vert$ implies $\Vert f_n -f \Vert \rightarrow 0$.

jackson5
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  • https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S0004972700003932 This seems to answer your question, without needing convergence of norm. – herb steinberg Feb 04 '19 at 17:06

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It $\textit{is}$ true in $L^2:\ \|f-f_n\|^2=\|f\|^2-2\langle f,f_n\rangle+\|f_n\|^2\to 0$ as $n\to \infty.$ In fact, the result is true for $f\in L^p:\ 1<p<\infty$ because these spaces are reflexive.

For a nice counterexample in $L^1([0,2])$, see the comment here.

Matematleta
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  • Ah sorry, I meant it is true in $L^2$ since I had already shown that it holds for Hilbert spaces – jackson5 Feb 04 '19 at 01:31
  • OK. The proof for general $1<p<\infty$ is a great exercise! – Matematleta Feb 04 '19 at 01:48
  • @Matematleta: For the $L^p$ case: It is really sufficient to know that these spaces are reflexive? I always thought that you need the uniform convexity of these spaces? – gerw Feb 04 '19 at 06:44
  • Heyy, really like your answer, if you could help with an exercise, would really appreciate it :) https://math.stackexchange.com/questions/3096022/normed-vector-space-schauder-basis-exercise – Homaniac Feb 04 '19 at 08:43
  • @Homaniac Sure. I will look at it. – Matematleta Feb 04 '19 at 16:23
  • @gerw For some reason, I thought the two notions were equivalent in $L^p;\ 1<p<\infty.$ Maybe because I know a proof that does not even use uniform convexity. But uniform convexity is stronger, Thanks! – Matematleta Feb 04 '19 at 16:34