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I'm working on the following problem from a real analysis qualifying exam from Texas A&M from Spring 2001, and I'm not sure how to proceed:

Suppose $f\in L^p(\mathbb{R})$ and $1\leq p<\infty$. Prove that $\lim_{h\to 0}\int_{\mathbb{R}}|f(x+h)-f(x)|^p\,dx=0$.

José Carlos Santos
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Alex
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1 Answers1

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This is a very standard result in real analysis done by so called "density argument". One should be able to check it in any advanced real analysis textbook. See for instance Folland's Real Analysis:

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Or Tao's excellent note (Proposition 19) regarding the density argument.

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  • The proof is making sense; you wouldn't happen to have Lemma 8.4 on hand as well, would you? That seems to be my main hangup and the missing piece in my "$\epsilon/3$" argument. – Alex Jun 17 '17 at 21:36
  • Lemma 8.4 says that if $f\in C_c({\bf R}^n)$ then $f$ is uniformly continuous, which is a very standard result in undergraduate real analysis. –  Jun 17 '17 at 21:39
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    @AlexBates: Lemma 8.4 added. –  Jun 17 '17 at 23:00
  • @Jack I know this is from long ago, but if we restrict ourselves to $L^1$ is unif. conv. of $f$ necessary? Applying the same argument, $\forall \vert y\vert \le 1$, the functions $\tau_y f$ are all supported some compact $K$. But $$\vert \tau_y f(x)\vert = \mathbf{1}K \cdot \vert f(x-y)\vert \le M\cdot 1_K\in L^1,$$ where $M=\sup{x\in K}\vert f(x)\vert<\infty$. Since $f$ is cts, $f(x-y)\rightarrow f(x)$ as $y\rightarrow 0$. Then take a seq. $y_n\rightarrow 0$ s.t. $\vert y_n\vert\le 1$ & DCT would give $|\tau_{y_n}f(x)-f(x)|_1\rightarrow 0$. So is unif. conv. only important for $p>1$? – Satana Feb 04 '18 at 22:55
  • Do you mean "uniform continuity" by "unif. cont."? @Satana –  Feb 05 '18 at 00:39
  • You can check the density argument in the proof of Proposition 19 in this set of notes, where the uniform continuity is not used (explicitly, at least). –  Feb 05 '18 at 00:46
  • @Jack Yes, I meant uniform continuity, I just exhausted my characters. Is my proof valid for $L^1$? I don't explicitly use it, I merely use DCT. I checked out the notes, and while he doesn't explicitly use it, he does mention it, which makes me believe that it is necessary. But I'm not sure what is wrong with the argument I gave above. Thanks! – Satana Feb 05 '18 at 04:40
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    Thanks. Before I check the details and (may be able to) answer your question in a comment, you can certainly post a question about this to see if some one would give you a quick response. :-) @Satana –  Feb 05 '18 at 14:43
  • @Jack Thanks a lot. I'll likely post a question later. I certainly understand why $f$ is uniformly continuous, & where it is used in the proof of Proposition 8.5 above. However, it is not clear to me if the comment I gave is incorrect (at least in the $L^1$ case), and would be useful to know! I simply apply DCT by finding an $L^1$ function bounding $\tau_y f$ for small $y$ and then use pointwise convergence of $\tau_y f$ to $f$ as $y\rightarrow x$ (from continuity). I don't see an obvious error in this, so I want to know if it is necessary to use a similar type of argument as in Folland gives. – Satana Feb 05 '18 at 17:03