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Let f be an integrable function on $\mathbb{R}$. Show that $\lim_{t\rightarrow 0} \int_{\mathbb{R}}|f(x + t) -f(x)|dx = 0$.

I can make it work once it is shown to be true for $f\in C_c(\mathbb{R})$ but I am having trouble proving this case.

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duplicate? –  David Mitra Jun 13 '12 at 21:32

2 Answers 2

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If $f\in C_c(\Bbb R)$, then the support of $f$ is contained in $[-R,R]$ for some $R>0$. Fix $\varepsilon>0$, and $\delta<1$ such that if $|t|\leq \delta$ and $x\in \Bbb R$ then $|f(x+t)-f(x)|\leq \varepsilon$ (it can be quite easily shown that $f$ is uniformly continuous on $\Bbb R$, since it is on $[-R-1,R+1]$. We have for $|t|\leq \delta$ that $$\int_{\Bbb R}\left|f(x+t)-f(x)\right|dx=\int_{[-R-1,R+1]}\left|f(x+t)-f(x)\right|dx\leq 2\cdot \varepsilon (R+1).$$

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If $f\in C_c(\mathbb{R})$, there exists $k>0$ with $f=0$ outside $[-k,k]$. Note that $$ \int_{\mathbb{R}} f(x+t)-f(x)\,dx=\int_{\mathbb{R}} f(x+t)\,dx - \int_{\mathbb{R}} f(x)\,dx =\int_{[-k,k]} f(x+t)\,dx - \int_{[-k,k]} f(x)\,dx =\int_{[-k+t,k+t]} f(x)\,dx - \int_{[-k,k]} f(x)\,dx =\int_{[-k,-k+t]\cup[k,k+t]}f(x)\,dx. $$ Since $[-k,k]$ is compact, $f$ is bounded: i.e. there exists $K>0$ with $|f|\leq K$. Then $$ \left|\int_{\mathbb{R}} f(x+t)-f(x)\,dx\right|=\left|\int_{[-k,-k+t]\cup[k,k+t]}f(x)\,dx\right|\leq\int_{[-k,-k+t]\cup[k,k+t]}|f(x)|\,dx\leq 2tK. $$ Choosing $t$ small enough, the left hand side goes to zero.

(note that I wrote things assuming $t>0$; for $t<0$ one does minor tweaks with the intervals but gets the same estimate but with $|t|$)

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