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Suppose that $f_n,f \in L^p(\mathbb{R}^d)$ and $f_n \to f$ pointwise everywhere. In addition, assume there is some $C_p > 0$ independent of $n$ so that $||f_n||_p \leq C_p ||f||_p$ for all $n$. Is it true that $f_n \to f$ in $L^p$?

I am not sure whether we can use the dominated convergence theorem since we do not have $|f_n| \leq g$ with $g \in L^p$ here.

Context: I am studying Fourier Integrals in Classical Analysis, 2nd Edition by Sogge. In the proof of Corollary 2.3.2, it seems to me that we need the proposition above.

Thank you for your help!

David C. Ullrich
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b04902072
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  • Are you sure they are using that result, because it is not true in general. – copper.hat Dec 13 '19 at 06:43
  • There is a generalised DCT but that is not in play here. – copper.hat Dec 13 '19 at 06:45
  • May be requiring some monotonicity is enough ? (But the inequality hypothesis seems too much in this case) – nicomezi Dec 13 '19 at 06:46
  • For $1 < p < \infty$, the hypotheses guarantee that $f_n$ converges weakly to $f$. You cannot conclude norm-convergence. Let $g \neq 0$ be a continuous function with compact support, and $f_n(x) = f(x) + g(x - ny)$ for a fixed $y \neq 0$. Then $f_n \to f$ pointwise, and $\lVert f_n - f\rVert_p = \lVert g\rVert_p$ for all $n$. – Daniel Fischer Dec 13 '19 at 13:21
  • @DanielFischer Thank you for your comment. This is a nice counterexample. – b04902072 Dec 14 '19 at 03:05
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    @copper.hat Now I think the proposition in the question is not true. I will rethink how they prove the corollary there. – b04902072 Dec 14 '19 at 03:06
  • I think the result should be $f_k$ converges weakly to $f$. see https://math.stackexchange.com/questions/4347232/weak-convergence-in-lp-space?noredirect=1&lq=1 – gaoqiang Dec 10 '24 at 13:05

1 Answers1

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I suspect there is something wrong with the statement of the proposition Let $p=1$. On $(0,1)$ define $f_n(x)= n +x$ if $0<x<\frac 1 n$ and $f_n(x)=x$ otherwise. Let $f(x)=x$ for all $n$. Then the hypothesis is satisfied but $\int |f_n-f| =1$ for all $n$.

Kavi Rama Murthy
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