Proving convergence in $L^p$

Question

Suppose $f \in L^p(\mathbb{R}) , 1 \leq p < \infty$. Prove that $$\lim_{h\rightarrow 0} \int_{\Bbb R} |f(x+h)-f(x)|^p = 0$$

I was thinking about showing that the sequence $\{f(x+\frac{1}{n}) - f(x)\}_n$ converges to zero in $L^p(\mathbb{R})$ so that I could pass the limit under the integral and conclude that the limit is zero, but I am having problems showing this convergence.

I'm also not sure what other approach to take/what other convergence theorems I could use.

Answer 1

Here's a more detailed version of the hint in my comment: the big theorem we shall use is $C_c(\Bbb R)$ is dense in $L^p(\Bbb R)$, by whose virtue we are entitled to choose some compactly supported continuous $g$ such that $\|f(x)-g(x)\|_{L^p}=\|f(x+h)-g(x+h)\|_{L^p}$ are both very small. Now we only care about small $h$, so we may assume both $g(x)$ and $g(x+h)$ are supported within $[-K,K]$ for some $K$. Thus, we see that $g(x)$ and $g(x+h)$ are continuous on the compact interval $[-K,K]$, so they must be uniformly continuous. By uniform continuity we immediately know that $\|g(x)-g(x+h)\|_{L^p}\to 0$ as $h\to 0$. Finally, combine the above results with the triangle inequality in $L^p(\Bbb R)$: $$\|f(x+h)-f(x)\|_{L^p}\le \|f(x+h)-g(x+h)\|_{L^p}+ \|g(x+h)-g(x)\|_{L^p}+ \|g(x)-f(x)\|_{L^p}.$$

Answer 2

Let $f=1_[a,b]$ where $A \subseteq \mathbb{R}$.

$\forall h \in (0,1)$ we have that $(\int_{\mathbb{R}}|1_{[a,b]}(x+h)-1_{[a,b]}(x)|^p)^{1/p}=(\int_{\mathbb{R}}|1_{[a,b]-h}(x)-1_{[a,b]}(x)|^p)^{1/p}=(\int_{([a,b]-h) \triangle [a,b]})^{1/p}=m(([a,b]-h) \triangle [a.b])^{1/p} \leqslant m([a-h,a) \cup(b-h,b])^{1/p} \leqslant (2h)^{1/p} \rightarrow 0 $

In the same way you can prove the above $\forall h \in (-1,0)$

Now if $f$ is a step function with compact support then its easy to see that $\int_{\mathbb{R}}|f(x+h)-f(x)|^p \rightarrow 0$ from the fist step.

Now we know that the set of step functions with compact support is dence in $L^p(\mathbb{R})$.

Let $\epsilon >0$ and $f \in L^p(\mathbb{R})$

Exist a step function $h$ with compact support such that $||f(x)-s(x)||_p \leqslant \epsilon$

$\forall h \in \mathbb{R}$ we have that $$\||f(x+h)-f(x)||_p \leqslant$$ $$\||f(x+h)-s(x+h)||_p+||s(x+h)-s(x)||_p+||f(x)-s(x)||_p$$ $$=2||f(x)-s(x)||_p+||s(x+h)-s(x)||_p$$

Thus $$\limsup_{h \rightarrow 0}||f(x+h)-f(x)||_p \leqslant 2 \epsilon +0=2 \epsilon$$

Therefore $\int_{\mathbb{R}}|f(x+h)-f(x)|^p \rightarrow 0$ as $h \rightarrow 0$