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Problem: Let $f \in L^1(\mathbb{R},~\mu)$, where $\mu$ is the Lebesgue measure. For any $h \in \mathbb{R}$, define $f_h : \mathbb{R} \rightarrow \mathbb{R}$ by $f_h(x) = f(x - h)$. Prove that: $$\lim_{h \rightarrow 0} \|f - f_h\|_{L^1} = 0.$$

My attempt: So, I know that given $\epsilon > 0$, we can find a continuous function $g : \mathbb{R} \rightarrow \mathbb{R}$ with compact support such that $$\int_{\mathbb{R}} |f - g|d\mu < \epsilon.$$ We can then use the inequality $|f - f_h| \leq |f - g| + |g - g_h| + |g_h - f_h|$ to reduce the problem to the continuous case, so to speak, since the integral of the first and last terms will be $< \epsilon$. But now I'm stuck trying to show that $$\lim_{h \rightarrow 0} \|g - g_h\|_{L^1} = 0.$$ I tried taking a sequence $(h_n)_{n \in \mathbb{N}}$ converging to $0$ and considering $g_n := g_{h_n}$, but I don't have monotonicity and the convergence doesn't seem to be dominated either, so I don't know what to do.

Any help appreciated. Thanks.

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It follows from the uniform continuity of $g$ on its support. For a detailed proof, see Rudin, Real and complex analysis, Theorem 9.4. It also appears on Brezis' book (at least in the italian edition), but the main step of the proof is left to the reader! – Siminore Aug 16 '12 at 15:41
up vote 4 down vote accepted

By your construction, $g$ is continuous and compactly supported. Let $K$ be the support of $g$, and let $K_h=K\cup (h+K)$. Then we have $$ \int|g(x)-g(x-h)|\,\mathrm{d}x\leq |K_h|\|g-g_h\|_{L^\infty(K_h)}. $$ For all $h>0$ sufficiently small we have $K_h\subset K_1$, and you can invoke uniform continuity of $g$.

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I think you can use the fact that $f_h$ is in $L^1$ by translation invariance and the fact that $|f-f_h|\leq|f|+|f_h|$. So now you have a function $g_n=|f-f_h|$ which converges to 0 and is bounded by an integrable function.

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What is the integrable function that bounds $g_h$? – timur Aug 16 '12 at 15:47
What is a dominant function for $|f_h|$, independent of $h$? Of course it is a multiple of $|f|$ when $f$ is continuous, but then your suggestion reduces to the continuous case. – Siminore Aug 16 '12 at 15:48
I believe $|f|+|f_h|$ is integrable. – Dave Aug 16 '12 at 16:01
But $|f_h|$ is not fixed.. – timur Aug 16 '12 at 20:35

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