Proving that $F_{\delta}(x)=\frac{1}{2\delta}\int_{-\delta}^{\delta} f(x+t) dt$ is continuously differentiable for continuous $f(x)$ Let $f(x)$ be continuous. Show that for every $\delta>0$ the following function is continuously differentiable:
$$F_{\delta}(x)=\frac{1}{2\delta}\int_{-\delta}^{\delta} f(x+t) dt$$
(The purpose of this exercise is to show that every continuous function on a closed interval can be uniformly approximated to arbitrary precision by a continuously differentiable function.)
I tried using the definition of the derivative and the basic properties of definite integrals:
$$\frac{d F_\delta(x)}{dx}=\lim_{h\to 0}\frac{F_{\delta}(x+h)-F_{\delta}(x)}{h}=\frac{1}{2\delta}\lim_{h\to 0}\int_{-\delta}^{\delta}\frac{f(x+t+h)-f(x+t)}{h} dt$$
and if I can take the limit inside the integral then assuming $f(x)$ is differentiable I get $f'(x+t)$ inside the integral, but since we do not know that $f(x)$ is differentiable then the limit inside the integral does not necessarily exist.
I also thought of using the mean value theorem for definite integrals but this seems to lead to the very same problem where to conclude anything I have to rely on differentiability of $f(x)$.
This is supposed to be solved using only the elementary properties of the Riemann integral. I'd appreciate any hint.
EDIT: I was offered this great solution in the comments: by change of variables $F_\delta(x)=\frac{1}{2\delta}\int_{x-\delta}^{x+\delta} f(u)du$ and by the fundamental theorem of calculus we have that $F'(x)=\frac{f(x+\delta)-f(x-\delta)}{2\delta}$ which is clearly continuous.
But this question is from lecture notes where the fundamental theorem only appears a few pages later. So I still wonder: is there a simple way to prove the same thing without relying on the fundamental theorem?
 A: OP explicitly stated in the comments that they're not allowed to use the fundamental theorem of calculus.
So going back to the definition of the derivative:
$$\begin{split}
2\delta \cdot\frac{F_\delta(x+h)-F_\delta(x)}{h} &= \frac 1 h\int_{-\delta}^{\delta}f(x+t+h)dt-\frac 1 h\int_{-\delta}^{\delta}f(x+t)dt\\
&= \frac 1 h\int_{-\delta+x+h}^{\delta+x+h}f(u)du-\frac 1 h\int_{-\delta+x}^{\delta+x}f(v)dv&\,\,&\text{ (1)}\\
&= \frac 1 h\int_{\delta+x}^{\delta+x+h}f(t)dt-\frac 1 h\int_{-\delta+x}^{-\delta+x+h}f(t)dt&\,\,&\text{ (2)}\\
&= f(\delta+x)+\frac 1 h\int_{\delta+x}^{\delta+x+h}(f(t)-f(\delta+x))dt\\
&-f(-\delta+x)-\frac 1 h\int_{-\delta+x}^{-\delta+x+h}(f(t)-f(-\delta+x))dt&\,\,&\text{ (3)}\\
\end{split}$$
where $(1)$ is obtained by changing the variable in the integrals and $(2)$ by rearranging the integral domains. Also, ​$(3)$ stems from the fact that
$$f(\delta+x)=\frac 1 h \int_{\delta+x}^{\delta+x+h}f(\delta+x)dt\text{ and }f(-\delta+x)=\frac 1 h \int_{-\delta+x}^{-\delta+x+h}f(-\delta+x)dt$$
Now, because $f$ is continuous, for any arbitrarily small $\varepsilon>0$, there exists $H_1>0$ such that for all $|h|<H_1$ and all $t \in[\delta+x, \delta+x+h]$, we have $$|f(\delta+x)-f(t)|<\varepsilon$$
Likewise, there exists $H_2>0$ such that for all $|h|<H_2$ and all $t \in[-\delta+x, -\delta+x+h]$, we have $$|f(-\delta+x)-f(x)|<\varepsilon$$
Thus, for all $|h|<\min(H_1, H_2)$, the two integrals in equation $(3)$ are both less than $\varepsilon$, and you can conclude that
$$\lim_{h\rightarrow 0}\frac{F_\delta(x+h)-F_\delta(x)}{h} = \frac{f(\delta+x)-f(-\delta+x)}{2\delta}$$
