Does convergence in $L^p$ imply convergence almost everywhere?

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 $$x$$?

• 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 Apr 28 '12 at 14:10
• 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! Apr 28 '12 at 14:25
• Look at $\chi_{[0,1]}$, $\chi_{[0,1/2]}$, $\chi_{[1/2,1]}$, $\chi_{[0,1/4]}$, $\chi_{[1/4,1/2]}$, $\ldots\,$. Apr 28 '12 at 14:36
• 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? May 26 '14 at 13:36
• If we repalce $\mathbb R$ by a compact region, and assume that $f_n$ are smooth, does the statement hold?
– DLIN
Mar 21 '18 at 13:57

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 $f$.

• This is also sometimes called the typewriter sequence. Dec 29 '16 at 12:09

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.

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.

• Does $p=\infty$ also counts here? Oct 6 '18 at 16:28
• @Error404 Of course, $p=\infty$ is not included here. Since converging in $L^\infty$-norm means converging uniformly a.e.; hence pointwise a.e. Oct 10 '18 at 14:22