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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.

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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 $f$.2011-05-15
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    Does $f_k$ converges weakly in $L^p$ to $f$ implies $f_k$ converges $L^p$ to $f$?2011-05-15
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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 $L^p$.2011-05-15
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    An almost identical question has been merged into this one. I've cleaned up the comments a bit.2011-07-16
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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 $f_n\not\xrightarrow{L^p} f$.2015-11-24
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    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 ?2015-11-24

2 Answers 2

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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.

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    You're *very* fast :)2011-07-14
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    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?2011-07-14
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    @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 ;-).2011-07-14
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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...2011-07-14
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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.2011-07-14
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    I should have said that the original Riesz-Fischer theorem only concerned completeness of $\ell^2$, strictly speaking.2011-07-14
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    Ah, thanks for the backstory! I wish I could understand the paper, but alas, my foreign linguistic skills are not up to the task..2011-07-14
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    @user1736: That's a pity, but you can [have a look at this.](http://www.univie.ac.at/NuHAG/FEICOURS/ws0506/Riesz-Fisher.pdf)2011-07-14
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    @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.2013-01-05
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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* $p \geq 0$.2013-01-05
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    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?2018-05-31
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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.pdf2018-08-16
41

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.

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    A point that wasn't addressed so far: $L^1$-convergence (in the title) is *not* the same thing as pointwise convergence a.e.2011-05-15
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    True. Even in finite measure spaces, neither implies the other.2011-05-15
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    I love how the combining of two questions makes it seem like I answered this question a whole two months before it was asked :-)2012-04-19
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    I guess in light of the first comment to the other answer, that makes you very, *very* fast. (+1)2013-01-05
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    @JesseMadnick, What if $||f||_p=\infty$? Then your $RHS$ would be $\infty$2015-11-24
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    @FardadPouran: I think that in writing $\Vert f_k \Vert_p \to \Vert f \Vert_p$, we are implicitly assuming that $\Vert f \Vert_p < \infty$.2015-11-25