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Let $1\leq p < \infty$. Suppose that

  1. $\{f_k, f\} \subset L^p$ (the domain here does not necessarily have to be finite),
  2. $f_k \to f$ almost everywhere, and
  3. $\|f_k\|_{L^p} \to \|f\|_{L^p}$.

Why is it the case that $$\|f_k - f\|_{L^p} \to 0?$$

A statement in the other direction (i.e. $\|f_k - f\|_{L^p} \to 0 \Rightarrow \|f_k\|_{L^p} \to \|f\|_{L^p}$ ) follows pretty easily and is the one that I've seen most of the time. I'm not how to show the result above though.

Mittens
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user1736
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  • Miscellaneous notes. The norm of $L^p$ is uniformly convex $(1 < p < \infty)$. And $f_k$ converges weakly in $L^p$ to $f$. – GEdgar May 15 '11 at 12:37
  • Does $f_k$ converges weakly in $L^p$ to $f$ implies $f_k$ converges $L^p$ to $f$? – Topologieeeee May 15 '11 at 13:00
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    Nope, not at all. e.g. $f_{k}(x) = e^{2\pi i k x}$ converges weakly to zero in $L^{p}([0,1])$. However, what GEdgar is getting at: if $f_{k} \to f$ weakly and $|f_{k}|{p} \to |f|{p}$ then $f_{p} \to f$ due to uniform convexity of $L^{p}$ for $1 \lt p \lt \infty$. Can you do the case $p = 2$ (which is a lot easier)? Then look up Hanner's inequalities (and Clarkson's inequalities) for uniform convexity of $L^p$. – t.b. May 15 '11 at 13:39
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    An almost identical question has been merged into this one. I've cleaned up the comments a bit. – Willie Wong Jul 16 '11 at 01:52
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    Is my counterexample wrong ? here it is : $$$$ Let $f_n(x)=\chi_{[0,n]}(x)$ which converges, pointwisely, to $f(x)=1$. Then $\lim_{n\rightarrow\infty}||f_n||^p=\infty=||f||_p$, but $f_n\not\xrightarrow{L^p} f$. – Fardad Pouran Nov 24 '15 at 22:20
  • Someone would say going to intinity is NOT of the concept of convergence. But I say how if in this question $||f||_p=\infty$? In this way how can you redefine the convergence of the hypothesis of the OP's problem ? – Fardad Pouran Nov 24 '15 at 22:25
  • This is Theorem 7 of Section 7.3 in Real Analysis, fourth edition by Royden and Fitzpatrick. – user0 Apr 11 '22 at 14:29

2 Answers2

87

This is a theorem by Riesz.

Observe that $$|f_k - f|^p \leq 2^p (|f_k|^p + |f|^p),$$

Now we can apply Fatou's lemma to $$2^p (|f_k|^p + |f|^p) - |f_k - f|^p \geq 0.$$

If you look well enough you will notice that this implies that

$$\limsup_{k \to \infty} \int |f_k - f|^p \, d\mu = 0.$$

Hence you can conclude the same for the normal limit.

JT_NL
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    You're very fast :) – t.b. Jul 14 '11 at 21:02
  • Wow, thanks for the lightning fast response! I actually can't even mark the answer correct yet. Also, just out of curiosity, in what text is the result attributed to Riesz? – user1736 Jul 14 '11 at 21:09
  • @user1736: I'm not sure but the other theorem about the completeness of $L^p$ is called Riesz-Fischer, and I have seen this one named after Riesz. Thanks Theo, I hope I don't make too big errors in the process ;-). – JT_NL Jul 14 '11 at 21:11
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    @user1736, Jonas: Here's Riesz's foundational paper on $L^p$-spaces, where he proves (among many other things) completeness of $L^p$ for $1\leq p \lt \infty$ and some weak sequential compactness results which he then applies to solve some integral equations. The Riesz-Fischer theorem is called this way as it was proved simultaneously and independently by both of them. Both papers appeared in the Comptes rendus de l'Académie des sciences 144: 615–619 (Riesz) and 1022–1024 (Fischer). The result here is not proved but can easily be extracted... – t.b. Jul 14 '11 at 21:28
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    ... I don't know its history and where it appeared first, but it is very likely that it is due to Riesz. If you really want to know you should check Dieudonné's history of functional analysis. – t.b. Jul 14 '11 at 21:29
  • I should have said that the original Riesz-Fischer theorem only concerned completeness of $\ell^2$, strictly speaking. – t.b. Jul 14 '11 at 21:43
  • Ah, thanks for the backstory! I wish I could understand the paper, but alas, my foreign linguistic skills are not up to the task.. – user1736 Jul 14 '11 at 23:04
  • @user1736: That's a pity, but you can have a look at this. – t.b. Jul 14 '11 at 23:57
  • @JonasTeuwen why is the first inequality true? I've seen this result before, but can't remember why. The triangle inequality doesn't seem to get me anywhere. – anegligibleperson Jan 05 '13 at 21:30
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    @anegligibleperson: Convexity. Or, $$|f+g|^p \leq (2\max(|f|,|g|))^p = 2^p \max(|f|^p,|g|^p) \leq 2^p (|f|^p + |g|^p)>.$$ Note that the latter works for any $p \geq 0$. – cardinal Jan 05 '13 at 21:50
  • How does this use the a.s convergence condition? Would you mind giving a counter-example to show why the a.e. convergence is necessary? – user428487 May 31 '18 at 06:35
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    @anegligibleperson It is also a "weaker" version of parallelogram law on Banach space. There are many variants in this paper: https://pdfs.semanticscholar.org/324b/afba4fb564de640bc00ec644dcdc29d5dd02.pdf – Daniel Li Aug 16 '18 at 00:01
  • Worth adding that this solution is much more elementary than the one through uniform convexity. My measure theory class in grad school didn't even talk about uniform convexity. – Joshua Siktar Jan 06 '22 at 17:56
  • @JT_NL How do you get from the $\lim \inf$ in Fatou to $\lim \sup$? – Anon Jan 09 '22 at 11:11
54

Consider $g_k = 2^p(|f_k|^p + |f|^p) - |f_k - f|^p$.

Since $g_k \geq 0$ (why?), and $g_k \to 2^{p+1}|f|^p$ a.e., we can apply Fatou's Lemma: $$\int \liminf g_k \leq \liminf \int g_k$$ so that $$\int 2^{p+1}|f|^p \leq \liminf \left(\int 2^p |f_k|^p + \int 2^p |f|^p - \int |f_k - f|^p \right),$$ and I'll let you take it from here.

Jesse Madnick
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