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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.

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  • $\begingroup$ Miscellaneous notes. The norm of $L^p$ is uniformly convex $(1 < p < \infty)$. And $f_k$ converges weakly in $L^p$ to $f$. $\endgroup$
    – GEdgar
    Commented May 15, 2011 at 12:37
  • $\begingroup$ Does $f_k$ converges weakly in $L^p$ to $f$ implies $f_k$ converges $L^p$ to $f$? $\endgroup$
    – Topologieeeee
    Commented May 15, 2011 at 13:00
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    $\begingroup$ 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 $L^p$. $\endgroup$
    – t.b.
    Commented May 15, 2011 at 13:39
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    $\begingroup$ An almost identical question has been merged into this one. I've cleaned up the comments a bit. $\endgroup$
    – Willie Wong
    Commented Jul 16, 2011 at 1:52
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    $\begingroup$ 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 $f_n\not\xrightarrow{L^p} f$. $\endgroup$
    – Fardad Pouran
    Commented Nov 24, 2015 at 22:20

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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    $\begingroup$ You're very fast :) $\endgroup$
    – t.b.
    Commented Jul 14, 2011 at 21:02
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    $\begingroup$ @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... $\endgroup$
    – t.b.
    Commented Jul 14, 2011 at 21:28
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    $\begingroup$ ... 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. $\endgroup$
    – t.b.
    Commented Jul 14, 2011 at 21:29
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    $\begingroup$ @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 $p \geq 0$. $\endgroup$
    – cardinal
    Commented Jan 5, 2013 at 21:50
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    $\begingroup$ @JT_NL How do you get from the $\lim \inf$ in Fatou to $\lim \sup$? $\endgroup$
    – Anon
    Commented Jan 9, 2022 at 11:11
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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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  • $\begingroup$ A point that wasn't addressed so far: $L^1$-convergence (in the title) is not the same thing as pointwise convergence a.e. $\endgroup$
    – t.b.
    Commented May 15, 2011 at 20:55
  • $\begingroup$ True. Even in finite measure spaces, neither implies the other. $\endgroup$
    – Jesse Madnick
    Commented May 15, 2011 at 21:03
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    $\begingroup$ I love how the combining of two questions makes it seem like I answered this question a whole two months before it was asked :-) $\endgroup$
    – Jesse Madnick
    Commented Apr 19, 2012 at 6:24
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    $\begingroup$ I guess in light of the first comment to the other answer, that makes you very, very fast. (+1) $\endgroup$
    – cardinal
    Commented Jan 5, 2013 at 22:16
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    $\begingroup$ @FardadPouran: I think that in writing $\Vert f_k \Vert_p \to \Vert f \Vert_p$, we are implicitly assuming that $\Vert f \Vert_p < \infty$. $\endgroup$
    – Jesse Madnick
    Commented Nov 25, 2015 at 1:22

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