We define $F(x)$ = $\frac{1}{2a}$$\int_{-a}^{a} f(x+t) dt$,
for some $f(x)$ which is continuous on R. I need to show that $F(x)$ is differentiable and has a continuous derivative. I am having trouble showing it's differentiable. I'm just trying to calculate $\lim_{h \rightarrow 0}$ $\frac{F(x+h) - F(x)}{h}$ = $\frac{1}{2a}$ $\frac{\int_{-a}^{a} f(x+h+t) - f(x+t) dt}{h}$, but don't know where to go from here. Is there any way I can use the continuity of $f(x)$?
Make the substitution $y=x-t$ and note that this is the same thing as
$$\frac{1}{2a}\int_{t-a}^{a+t} f(y) dt=\frac{1}{2a}\int_{0}^{t+a}f(y)dt-\frac{1}{2a}\int_{0}^{t-a}f(y)dt$$
and note that it will suffice to show that $\frac{1}{2a} \int_0^{t+a}f(y)dt$ is continuous, since the the other integral will follow by a similar argument.
Here, you can apply the fundamental theorem of calculus as is done in this answer
Use change of variables, we see that \begin{align} \frac{1}{2a}\int^a_{-a} f(x+t)\ dt = \frac{1}{2a} \int^{x+a}_{x-a} f(u)\ du. \end{align} Then it follows \begin{align} \frac{F(x+h)-F(x)}{h} =&\ \frac{1}{2ah} \left(\int^{x+h+a}_{x+h-a} f(u)\ du - \int^{x+a}_{x-a} f(u)\ du \right)\\ =&\ \frac{1}{2ah}\left(\int^{x+h+a}_{x+a} + \int^{x+a}_{x+h-a} f(u)\ du - \int^{x+a}_{x+h-a}-\int^{x+h-a}_{x-a} f(u)\ du \right)\\ =&\ \frac{1}{2ah}\left(\int^{x+a+h}_{x+a} f(u)\ du -\int^{x-a+h}_{x-a} f(u)\ du \right). \end{align} Hence in the limit as $h\rightarrow 0$ you get \begin{align} F'(x) =\frac{1}{2a}\left(f(x+a)-f(x-a) \right) \end{align}
