Let $f: \mathbb R \to \mathbb R$ be an integrable function, for each $h>0$ let $$f_h(t)=\dfrac{1}{h}\int_{t-\frac{h}{2}}^{t+\frac{h}{2}}f(x)dx$$
Suppose $f \in L^P$, prove the following
(1) $f_h \in L^p$ and $||f_h||_p \leq ||f_p||$.
(2) For each $r \geq p \geq 1$, $||f_h||_r \leq h^{\frac{1}{r}-\frac{1}{p}}||f||_p$.
(3) $||f_h-f||_p \to 0$ when $h \to 0$.
I got stuck trying to show these results, I'll write what I could work on:
For (1), we have $$||f_h||_p=\left(\int_{\mathbb R}\left|\dfrac{1}{h}\int_{t-\frac{h}{2}}^{t+\frac{h}{2}}f(x)dx\right|^p\,dt\right)^{\frac{1}{p}}$$$$=\left(\int_{\mathbb R}\left|\dfrac{1}{h}\int_{\mathbb R}f(x)\chi_{[t-\frac{h}{2},t+\frac{h}{2}]}(x)dx\right|^p\,dt\right)^{\frac{1}{p}}$$
By Minkowski's integral inequality, this last expression is less than or equal to $$\int_{\mathbb R}\dfrac{1}{h}\left(\int_{\mathbb R}|f(x)|^p\chi_{[t-\frac{h}{2},t+\frac{h}{2}]}(x)dt\right)^{\frac{1}{p}}dx$$
At this point I got stuck with the integral.
For (2) I've arrived to a similar expression and had the same problem.
As for (3) I thought of the following, take $f \in L^p$ and take $g$ an arbitrary continuous function with compact support, we have $$||f_h-f||_p \leq ||f_h-g_h||_p+||g_h-g||_p+||f-g||_p$$
Using the inequality in (1), we get $$||f_h-g_h||_p+||g_h-g||_p+||f-g||_p \leq 2||f-g||_p+||g_h-g||_p$$
Since the space of continuous functions of compact support is dense in $L^p$, it follows it is sufficient to show $||f_h-f||_p \to 0$ when $h \to 0$ for continuous functions of compact support. So take a function $f$ of the sort, then $$|f_h-f||_p \leq \left(\int_{\mathbb R}\left(\dfrac{1}{h}\int_{t-\frac{h}{2}}^{t+\frac{h}{2}}\left|f(x)-f(t)\right|\,dx\right)^p\,dt\right)^{\frac{1}{p}}$$
I've tried to used the fact that $f$ is uniformly continuous and that outside a compact set $K$ is $0$ in order to show that this integral tends to $0$ but I couldn't.
I would really appreciate some help with these problems. Thanks in advance.