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I am trying to solve the following exercise of Brezis' book on Functional Analysis.

Let $(f_n)_{n \in \mathbb{N}}$ be a sequence in $L^p(\Omega)$ with $1 < p < \infty$ and let $f \in L^p(\Omega)$. Assume that $f_n \rightharpoonup f$ weakly $\sigma(L^p, L^{p'})$ and $\lim_{n \to \infty} ||f_n||_p = ||f||_p$, then $f_n \rightarrow f$ strongly in $L^p(\Omega)$.

I tried to show that $(f_n)_{n \in \mathbb{N}}$ is a Cauchy sequence in $L^p(\Omega)$ using that $(||f_n||_p)_{n \in \mathbb{N}}$ is Cauchy since, by hypothesis, it is convergent. However, I was not able to do it. Could anyone give me a hint to continue?

Moreover, in Brezis' exercise 4.19, it is given a counterexample for this claim when $p=1$. Is there a counterexample for $p = \infty$?

User
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1 Answers1

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For $p=\infty$ consider the space $([0,\infty),\mathscr{B}([0,\infty)),\lambda)$ where $\lambda$ is Lebesgue's measure. Let $f_n=\mathbb{1}_{[0,n)}$. $\|f_n\|_\infty=1=\|f\|_\infty$ where $f\equiv1$. in the topology $\sigma(L_\infty,L_1)$ we have that $$\langle f_n,g\rangle=\int^n_0g\xrightarrow{n\rightarrow\infty}\int^\infty_0g=\langle 1,g\rangle$$ However $$\|f_n-f\|_\infty=1$$

A solution to the problem with $1<p<\infty$ can be found in this posting or here

Mittens
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  • Is there a name for that theorem ($1\lt p\lt\infty$)? I'm certain it's in my real analysis notes but I can't remember what it's called, how to find it – FShrike May 22 '22 at 19:30
  • @FShrike: All this kinds of theorems are due to F. Riesz – Mittens May 22 '22 at 19:36
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    For $p \in (1, \infty)$, the space $L^p (\Omega)$ is uniformly convex. The result then follows from Proposition 3.32 in the same book, i.e., assume that $E$ is a uniformly convex Banach space. Let $\left(x_n\right)$ be a sequence in $E$ such that $x_n \rightarrow x$ weakly $\sigma\left(E, E^{\star}\right)$ and $$ \limsup_n \left|x_n\right| \leq|x| . $$ Then $x_n \rightarrow x$ strongly. – Akira Apr 04 '23 at 20:52