Convergence in $L^1$ problem.

Question

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.

Answer 1

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 $g$.

Answer 2

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.