Let $ \displaystyle{ f \in L^p (\mathbb R^n), 1\leq p <\infty }$ and let $ \upsilon \in \mathbb R^n$. For $h>0$ define $\displaystyle{ f_h(x) = \frac{1}{h} \int_0^h f(x+s \upsilon) ds }$.
Prove that: $\displaystyle{ f_h \to f}$, in $L^p (\mathbb R^n)$ as $h \to 0^+ .$
I tried to do it by approximating $f$ by functions of compact support, but I didn't succeed to end it. I also tried to use to estimate using the operator $\tau_\alpha f (x):= f(x+ \alpha \upsilon)$ and then using Jensen's inequality and Fubini's theorem, I got to that:
$\displaystyle{ \| f_h -f\|_{L^p}^p \leq \int_0^1 \|\tau_{sh} - f\|_{L^p}^p ds }$
I think that it isn't that hard, but I am missing something...
Indeed, we have to prove that $$\int_0^1\lVert\tau_{shv}(f)-f\rVert_{\mathbb L^p}^p\mathrm ds\to 0\mbox{ as }h\to 0.$$ Let $g$ be a continuous function with compact support such that $\lVert f-g\rVert_{\mathbb L^p}^p\lt \varepsilon$, where $\varepsilon$ is a positive arbitrary but fixed number. We have \begin{align}\lVert\tau_{shv}(f)-f\rVert_{\mathbb L^p}^p&\leqslant 2^p(\lVert\tau_{shv}(f)-\tau_{shv}(g)\rVert_{\mathbb L^p}^p+\lVert\tau_{shv}(g)-g\rVert_{\mathbb L^p}^p+\lVert g-f\rVert_{\mathbb L^p}^p)\\ &\leqslant 2^p(2\varepsilon+\lVert\tau_{shv}(g)-g\rVert_{\mathbb L^p}^p). \end{align} Using the fact that $g$ has a compact support and is uniformly continuous, we have $$\lim_{h\to 0}\int_0^1\lVert\tau_{shv}(g)-g\rVert_{\mathbb L^p}^p\mathrm ds=0.$$