For all $\epsilon >0$ ,exists $\delta>0$ such that if $||t||<\delta$ then: $ \int_{\mathbb{R^d}} |f(x-t)-f(x)| \,dx< \epsilon $

Question

Let $f:\mathbb{R^d}\to \mathbb{R}$ be an integrable function. I need to prove that fo all $\epsilon >0$ there exists $\delta>0$ sucha that if $||t||<\delta$ then, $$ \int_{\mathbb{R^d}} |f(x-t)-f(x)| \,dx < \epsilon $$ Any suggestions?

So far, I know that if $\delta$ is sufficiently small then $ \int_{||t||<\delta} |f(x-t)-f(x)| \,dx $ can be made small because $f$ is integrable ,but I don't know how to bound the complement. So I've trying to find a set in which the integral can be small and its complement as well

Answer 1

Let $\phi:\mathbb{R}^{d}\to \mathbb{R}$ continuous and compactly supported such that $||f-\phi||_{1}<\frac{\varepsilon}{3}$. Then, $$ \begin{align*} \int_{\mathbb{R}^{d}}|f(x-t)-f(x)|dx &\leq \int_{\mathbb{R}^{d}}|f(x-t)-\phi(x-t)|dx\, + \\ &+ \int_{\mathbb{R}^{d}}|\phi(x-t)-\phi(x)|dx + \int_{\mathbb{R}^{d}}|\phi(x)-f(x)|dx \\ &<\frac{2\varepsilon}{3}+\int_{\mathbb{R}^{d}}|\phi(x-t)-\phi(x)|dx. \end{align*} $$

Since $\phi$ is continuous and compactly supported, it is uniformly continuous, and you can find $\delta>0$ such that for all $x\in \mathbb{R}^{d}$, $$ |\phi(x-t)-\phi(x)| <\frac{\varepsilon}{3\cdot m(\text{supp }\phi)}, $$

where $m$ indicates the Lebesgue measure. Then $\int_{\mathbb{R}^{d}}|\phi(x-t)-\phi(x)|dx < \varepsilon/3$, finishing the problem.

Answer 2

Approximate the $L^{1}$ function $f$ by $\varphi\in C_{0}$: $\|\varphi-f\|_{L^{1}}<\epsilon$. Then We are to show that $\|\tau^{t}(\varphi)-f\|_{L^{1}}\rightarrow \|\varphi-f\|_{L^{1}}$, where $\tau^{t}$ is the translation operator. But that is simply by Lebesgue Dominated Convergent Theorem since $|\varphi(x-t)-f(x)|\rightarrow|\varphi(x)-f(x)|$ pointwise and $|\varphi(x-t)-f(x)|\leq\chi_{|x|\leq M}\varphi+|f|$, where $\{|x|\leq M\}$ covers the support of $\varphi$.