Limit of an integral in $L^p$

Question

I am dealing with the following limit:

Let $\lambda$ be the Lebesgue measure on $\mathbb{R}^n$, let $f\in L^p(\lambda)$ for $p\in[0,\infty)$. Compute $$\lim_{h\to 0}\int_{\mathbb{R}^n} |f(x+h) - f(x)|^p\ d\lambda(x)$$ and use this fact to show that for any $A$ measurable set with $\lambda(A)>0$, the set $A-A:=\{x-y\ | x,y\in A\}$ contains open ball centered at 0.

As for the first part, I was trying to apply dominated convergence to put the limit inside, but I do not know how to bound the function $|f(x+h) - f(x)|^p$ properly. The second part is also confusing. I thought that the indicator function of $A$ might be of help but I am not sure.

Thanks!

Answer 1

Regarding the first part, a common technique in these kinds of questions is to first prove this question on indicator functions (and their linear combinations), then for positive integrible functions, and after that for general functions in $L_p$.

Here for example, lets try and prove that for indicator functions $$ \lim_{h \to 0} \int |1_{A}(x) - 1_{A}(x+h)|d \lambda = 0.$$

To do so we will use the following two properties of the Lebesgue measure:

1. Translation invariance.
2. Regularity.

Fix $h$:

$$ \int |1_A (x+h) - 1_A (x)|^p d\lambda = \int |1_A (x+h) - 1_A (x)| d\lambda $$ $$ = \int |1_{A+h}(x) - 1_A(x)|d \lambda.$$

Here $A+h$ is all the $x$ so that $x+h\in A$.

By the regularity of the Lebesgue measure, for every $\varepsilon > 0 $ we can find an open set $O \supseteq A$ so that $\lambda(O) < \lambda(A) + \varepsilon$ and a compact set $K \subseteq A$ so that $\lambda(K) > \lambda(A) - \varepsilon$. From translation invariance, the same relation applies to $K+h, A+h, O+h$.

By Lebesgue's covering lemma, if $K \subseteq O$ and $h$ is small enough then $K+h \subseteq O$.

Finally using the triangle inequality:

$$\int |1_A (x+h) - 1_A (x)| d\lambda \leq \int |1_A (x+h) - 1_{K}(x+h)|d\lambda + \int |1_K(x+h) - 1_O(x)|d\lambda + \int |1_O(x)-1_A(x)|d\lambda = (\lambda(A)-\lambda(K)) + (\lambda(O) - \lambda(K)) + (\lambda(O)-\lambda(A)).$$ Which goes to $0$ when we take $\varepsilon,h$ to $0$ (the function $|1_K(x+h) - 1_O(x)|$ is the indicator of $O \setminus (K+h)$, this set is of size $\lambda(O) - \lambda(K)$).

To generalize this to all functions, just use the fact that you can approximate any function $f \in L_p$ by linear combinations of indicator functions (please comment if you need further assistance).

For the second part, note that whenever $\lambda(A \cap (A+h))>0$ then $h \in A-A$ (since when there is some $x \in A\cap A+h$, this indicates that $x,x-h \in A$ and thus $h \in A-A$).

From the first part we have that $$ \lim_{h \to 0}\int|1_A(x)-1_{A+h}(x)|d\lambda = \lim_{h\to0}\lambda(A \cup (A+h)) - \lambda(A \cap (A+h)) = 0.$$ In particular, there exists some interval around $0$ where $\lambda(A \cap (A+h))>0$. The claim follows.