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?
Best Answer
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 $$\lim_{h\rightarrow 0}\frac{F_\delta(x+h)-F_\delta(x)}{h} = \frac{f(\delta+x)-f(-\delta+x)}{2\delta}$$