Limit in $L^{1}(\mathbb{R})$

Question

I have the following problem concerning a limit in $L^{1}(\mathbb{R}$), the class of all Lebesgue integrable functions on $\mathbb{R}$ with the Lebesgue measure $m$.

Let $f\in L^{1}(\mathbb{R})$. Does $$\lim_{h\to 0}\int_{\mathbb{R}}|f(x-h)-f(x)|~dx$$ exist? If it does, what is its value?

My intuition is that the limit does exist. However, I have no idea how to prove it. I noticed that $\int_{\mathbb{R}}|f(x-h)-f(x)|~dx=||f(x-h)-f(x)||_{1}$, and I know that since $f\in L^{1}(\mathbb{R})$, we have $||f(x)||_{1}<\infty$. I think this needs to used somewhere, but I'm not sure where. Any help is appreciated.

Answer 1

Let $f \in L^1$ and let $\epsilon > 0$ then there exists a continuous function with compact support $g$ such that

$||f -g||_{L^1} \leq \epsilon/3$.

Now,

$||f(x-h)-f(x)||_{L^1} \leq ||f(x-h)-g(y-h)||_{L^1} + ||g(y-h) - g(y)||_{L^1}+ ||g(y)-f(x)||_{L^1}$

Since $g$ is continuous with compact support, $\int_K |g(y-h) - g(y)|dy \leq m(K)\epsilon/3$ for sufficiently small $h$.

Also, using the fact that Lebesgue measure is translation invariant, we have that

$||f(x-h) - f(y-h)||_{L^1} = ||f(x) - g(x)||_{L^1}$. Hence each term in the

inequality above is less than $\epsilon/3$ we are done.

Answer 2

Hint: use the fact that, since $f \in L^1(\mathbb{R})$, there exists a continuous function $g$ with compact support such that $\int |f-g| < \epsilon$.

You will get that $\lim_{h \rightarrow \infty} \int|f(x+h) - f(x)| dm(x) = 0$.