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If I know $\| f_n - f \|_{L^p(\mathbb{R})} \to 0$ as $n \to \infty$, do I know that $\lim_{n \to \infty}f_n(x) = f(x)$ for almost every $x$?

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    $\begingroup$ For sure along a subsequence. This is the first step when you prove that $L^p$ is a complete normed space. See johndcook.com/modes_of_convergence.html $\endgroup$
    – Siminore
    Commented Apr 28, 2012 at 14:10
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    $\begingroup$ Choosing suitable intervals with length going to zero you can exhibit a sequence of L^p, but pointwise the sequence doesn't converges at any point! $\endgroup$
    – checkmath
    Commented Apr 28, 2012 at 14:25
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    $\begingroup$ Look at $\chi_{[0,1]}$, $\chi_{[0,1/2]}$, $\chi_{[1/2,1]}$, $\chi_{[0,1/4]}$, $\chi_{[1/4,1/2]}$, $\ldots\,$. $\endgroup$ Commented Apr 28, 2012 at 14:36
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    $\begingroup$ If we add a condition: $(f_n)_{n=1}^\infty$ is a sequence of nonnegative functions and $f_{n+1}(x)\geq f(x)$, then will $L^p$-norm convergence imply pointwise convergence? $\endgroup$
    – Shiquan
    Commented May 26, 2014 at 13:36
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    $\begingroup$ If we repalce $\mathbb R$ by a compact region, and assume that $f_n$ are smooth, does the statement hold? $\endgroup$
    – DLIN
    Commented Mar 21, 2018 at 13:57

3 Answers 3

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Let $\rho(x)=\chi_{[0,1]}$ be the characteristic function of the interval $[0,1]$. Then take the "dancing" sequence

$$ f_n(x) = \rho(2^mx-k) $$

where $n=2^m+k$ with $0\leq k<2^m$. This sequence converges to $0$ in $L^p$ but for any $x\in(0,1)$ we have $f_n(x)$ is not convergent.

However, it is a general fact that one can always extract a subsequence converging almost everywhere to $f$.

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    $\begingroup$ This is also sometimes called the typewriter sequence. $\endgroup$
    – Watson
    Commented Dec 29, 2016 at 12:09
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Consider the "typewriter sequence" defined by the formula: $$f_n:=1_{\left[\frac{n-2^k}{2^k},\frac{n-2^k+1}{2^k}\right]}$$ where $k$ is an integer such that $2^k\leq n<2^{k+1}$, the sequence converges to zero in $L^p$ norm, but not pointwise.

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  • $\begingroup$ Does $p=\infty$ also counts here? $\endgroup$
    – Error 404
    Commented Oct 6, 2018 at 16:28
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    $\begingroup$ @Error404 Of course, $p=\infty$ is not included here. Since converging in $L^\infty$-norm means converging uniformly a.e.; hence pointwise a.e. $\endgroup$
    – Xiang Yu
    Commented Oct 10, 2018 at 14:22
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    $\begingroup$ Thanks for your reply and nice answer! $\endgroup$
    – Error 404
    Commented Oct 10, 2018 at 16:55
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For $n$ integer and $0\leq k\leq 2^n-1$, denote $f_{n,k}:=n\chi_{[k2^{-n},(k+1)2^{-n}]}$. Consider $f_{1,0},f_{1,1},f_{2,0},f_{2,1},f_{2,2},f_{2,3},\ldots$ (put $g_n:=f_{\alpha_n,n-\alpha_n}$ where $\alpha$ is an integer such that $1+\ldots+2^{\alpha_n}\leq n<1+\ldots+2^{\alpha_n+1}$). Then $g_n$ doesn't converge to $0$ for any $x$, but converge to $0$ in $L^p$ for $1\leq p<\infty$. However, we can find an almost everywhere converging subsequence (here we can pick $g_{1+\ldots+2^n}$).

More generally, if $f_n\to 0$ in $L^p$, with $1\leq p<\infty$, we get can find $n_k$ increasing to $+\infty$ such that $\lVert f_{n_k}\rVert_{L^p}\leq 2^{-k}$. Let $g_k:=|f_{n_k}|$. Then $\mu\{x:g_k(x)\geq n^{-1}\}\leq \frac {2^{-kp}}{n^{-p}}$, so by Borel-Cantelli $\mu\{\limsup_k g_k\geq \frac 1n\}=0$ for each $n$ and $g_k\to 0$ almost everywhere.

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