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Fix a real number $1\leq p<\infty$.

Is it true that if functions $f\in L^p(\mathbb{R})$ and $f_1,f_2,\ldots\in L^p(\mathbb{R})$ are such that $\|f_n-f\|_p\rightarrow 0$ as $n\rightarrow \infty$, then $f_n(y)\rightarrow f(y)$ as $n\rightarrow\infty$ for almost every $y\in\mathbb{R}$?

It really shouldn't be true, but what would be a counterexample?

Kunal
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1 Answers1

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This answer is wrong, as @asbjørn-holk has pointed out below.

Let $f_n(x) = \sin (2\pi nx) \cdot 1_{[0,2 \pi]}(x)$. $f_n \in L^p$, and $\int |f_n| \le \frac{2}{\pi n}$, hence $f_n \to 0$.

For $x \in [0, 2 \pi] \setminus \mathbb{Q}$, we have that $\{2\pi ( nx - \lfloor nx \rfloor ) \}_n$ is dense in $[0, 2 \pi]$, and hence $\{ f_n(x) \}_n$ is dense in $[-1,1]$. In particular, $f_n(x)$ does not have a limit, and $m ([0, 2 \pi] \setminus \mathbb{Q}) = 2 \pi >0$.

copper.hat
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  • I know this is a very old answer, but in case anyone else comes across this, looking for another example other than the typewriter sequence, there seems to be a mistake in the above. Namely, $f_n\not\to0$ in $L^1$. Instead, if $f_n$ is as written, $\int|f_n|\to4$. – muldyr Jun 20 '24 at 13:13
  • I am unable to delete an accepted answer. I will ping chat and see if some mod can assist. – copper.hat Jun 20 '24 at 17:05