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

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    For sure along a subsequence. This is the first step when you prove that $L^p$ is a complete normed space. See http://www.johndcook.com/modes_of_convergence.html2012-04-28
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    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!2012-04-28
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    Look at $\chi_{[0,1]}$, $\chi_{[0,1/2]}$, $\chi_{[1/2,1]}$, $\chi_{[0,1/4]}$, $\chi_{[1/4,1/2]}$, $\ldots\,$.2012-04-28
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    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?2014-05-26
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    If we repalce $\mathbb R$ by a compact region, and assume that $f_n$ are smooth, does the statement hold?2018-03-21

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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    This is also sometimes called the typewriter sequence.2016-12-29
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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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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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    Does $p=\infty$ also counts here?2018-10-06
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    @Error404 Of course, $p=\infty$ is not included here. Since converging in $L^\infty$-norm means converging uniformly a.e.; hence pointwise a.e.2018-10-10
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    Thanks for your reply and nice answer!2018-10-10