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

  1. $\{f_k\} \subset L^p$ (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.

Davide Giraudo
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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](http://en.wikipedia.org/wiki/Hanner%27s_inequalities) (and [Clarkson's inequalities](http://en.wikipedia.org/wiki/Clarkson%27s_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

81

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](http://dx.doi.org/10.1007%2FBF01457637) 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.](http://www.univie.ac.at/NuHAG/FEICOURS/ws0506/Riesz-Fisher.pdf) – 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
51

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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