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Suppose I have $f \in L^1\cap L^\infty$, and I want to take the following limit $$\lim_{h\to0} \int|f(x+h)-f(x)|dx$$

Does this follow from the dominated convergence theorem? Since $|f(x+h)-f(x)| \leq |f(x+h)|+|f(x)|$, and $\|f(x+h)\|_{L^1(\mathbb{R})}<\infty$ and $\|f(x)\|_{L^1(\mathbb{R})}<\infty$, we then have $$\lim_{h\to0} \int|f(x+h)-f(x)|dx = \int\lim_{h\to0}|f(x+h)-f(x)|dx = 0$$?

I'm not convinced, and I think a better argument would be to involve the density of continuous function in $L^1$. If the approach in this post does not work, can anyone explain why?

reiman
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  • To use the DCT you would need $f$ to be continuous ae. The density of continuous functions would be a better approach. (This has been answered here before, but I am unable to find the answer.) – copper.hat Aug 02 '14 at 17:15
  • @reiman: $\left|f(x+h)\right|+\left|f(x)\right|$ still depends on $h\in\mathbb{R}$. In the usual dominated convergence theorem, the bounding $L^{1}(\mathbb{R})$ function $g$ is independent of $h$. – Matt Rosenzweig Aug 02 '14 at 17:16
  • If you take $f(x)=1_{\mathbb{Q}}(x) e^{-|x|}$, then the limit $|f(x+h)-f(x)|$ doesn't exist anywhere. – copper.hat Aug 02 '14 at 17:30
  • @copper.hat: In your example, $f=0$ a.e. So the integral is zero. – Matt Rosenzweig Aug 02 '14 at 17:39
  • @Matt: I know, but the OP asked about their approach to a solution, which involved passing the limit inside the integral. I was just pointing out that you can't in general just take limits. And I meant to put $\mathbb{Q}^c$ in there :-(. – copper.hat Aug 02 '14 at 17:44
  • @reiman You don't need $f$ to be essentially bounded (in $L^{\infty}$) for the limit to be zero; $f \in L^1$ is enough. – GaussTheBauss Dec 24 '15 at 19:17

2 Answers2

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Let $g\in L^{1}$ be continuous such that $\left\|g-f\right\|_{L^{1}(\mathbb{R})}<\epsilon$. Let $N>0$ be sufficiently large so that $$\max\left\{\int_{[-N,N]^{c}}\left|f\right|,\int_{[-N,N]^{c}}\left|g\right|\right\}<\epsilon$$ for $\epsilon>0$ given. For $0\leq x, h<N$, $\left|g(x+h)\right|\leq \sup_{x\in [-2N,2N]}\left|g(x)\right|=:M$. So $\left|g(x+h)-g(x)\right|\leq 2M$. By the DCT, $$\lim_{h\rightarrow 0}\int_{-N}^{N}\left|g(x+h)-g(x)\right|dx=0$$ Observe that $$\int_{[-N,N]^{c}}\left|g(x+h)-g(x)\right|dx\leq 2M\epsilon$$

This should get you started.

Matt Rosenzweig
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(I marked the question as a duplicate, but then realised that you were asking about the approach to a proof.)

You can use the DCT as part of the proof, but it doesn't follow directly.

To use the DCT, you need an integrable bound and a limit function.

In the above, you need to provide an integral upper bound that is 'independent' of $h$. And the limit doesn't necessarily exist unless the function is continuous ae. For example, $f(x)=1_{\mathbb{Q}^c}(x) e^{-|x|}$ is nowhere continuous.

An approach using compactly supported continuous approximations is fairly standard.

copper.hat
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