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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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  • $\begingroup$ 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 May 15 '11 at 12:37
  • $\begingroup$ Does $f_k$ converges weakly in $L^p$ to $f$ implies $f_k$ converges $L^p$ to $f$? $\endgroup$ – Topologieeeee May 15 '11 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. May 15 '11 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 Jul 16 '11 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 Nov 24 '15 at 22:20
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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. Jul 14 '11 at 21:02
  • $\begingroup$ 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? $\endgroup$ – user1736 Jul 14 '11 at 21:09
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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. Jul 14 '11 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. Jul 14 '11 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 Jan 5 '13 at 21:50
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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. May 15 '11 at 20:55
  • $\begingroup$ True. Even in finite measure spaces, neither implies the other. $\endgroup$ – Jesse Madnick May 15 '11 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 Apr 19 '12 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 Jan 5 '13 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 Nov 25 '15 at 1:22

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