I am trying to show these two statements:
Let $f \in L^p(\mathbb R^d)$, $1 \leq p < \infty$, then
\begin{align}(a) & \quad \left(\int_{\mathbb R^d}|f(x-h)-f(x)|^p\,dx\right)^{\frac{1}{p}} \to 0 \text{ when } \|h\| \to 0 \\ (b) & \quad \left(\int_{\mathbb R^d}|f(x-h)+f(x)|^p\,dx\right)^{\frac{1}{p}} \to 2^{\frac{1}{p}}\|f\|_p \text{ when } \|h\| \to \infty \end{align}
I am having some difficulty with this problem.
For (a), it is sufficient to show that $\|f(x+h)-f(x)\|_p^p \to 0 $ when $\|h\| \to 0 $. First I've tried to prove this for continuous functions of compact support. So take $f$ such that $f(x)=0$ for all $x \in K^c$ with $K$ a compact set.
Now, $\int_{\mathbb R^d}|f(x-h)-f(x)|^p\,dx=\int_{K^c}|f(x-h)-f(x)|^p\,dx+\int_K|f(x-h)-f(x)|^p\,dx$.
If I could show that for $\|h\|$ sufficiently small, $x-h \in K^c$ for all $x \in K^c$ and $x-h \in K$ for all $x \in K$, then I can use the fact that $f$ is uniformly continuous on $K$ and that $f=0$ on $K^c$ to prove that $\int_{\mathbb R^d}|f(x-h)-f(x)|^p\,dx$ tends to $0$. I don't know what to do to show this.
As for part (b) I am completely lost.
Any help would be greatly appreciated. Thanks in advance.
Hint: Suppose first $f$ is continuous with compact support. For (a) note the integral in question is over a fixed bounded set if $|h|<1.$ So the result will hold by the uniform continuity of $f.$ For (b) observe that the supports of $f(x-h), f(x)$ are disjoint for $|h|$ large. Thus the integral equals, exactly, $2^{1/p}\|f\|_p$ for $|h|$ large. For the general $f\in L^p$ use the fact that continuous functions with compact support are dense in $L^p.$