Help in trying to prove that $f’_n(x) \to f’(x)$ uniformly Let $f:\Bbb{R} \to \Bbb{R}$ be a function of class $C^1$ and $I= [a,b] $ a closed and bounded interval. We define  $ f_n(x) = \frac{n}{2} \int_{x-\frac{1}{n}}^{x+\frac{1}{n}}f(t)dt$. Prove that $f’_n(x) \to f’(x) $ uniformly convergent on $I$.
I dont know how to solve this . I guess I need to apply the Fundamental Theorem, but I dont know how to apply it to series of functions and with these integral limits . All help will be appreciated. Thanks!
Edit : Is $f’(x)$ uniformly continous? Why? I know that a fuction class $C^1$ in a closed intervals uniformly continous (with $f’(x)$ bounded). But why $f’(x)$ uniformly continous?
 A: Note that the $f_n$ are differentiable on all of $\mathbb{R}$, and you can take their derivatives by splitting them as such: $f_n(x) = \frac{n}{2}\left[\int_0^{x+1/n} f(t)dt - \int_0^{x-1/n} f(t)dt\right]$. Then use the Fundamental Theorem (applicable since $f$ is continuous) to get $f'_n(x) = \frac{n}{2}\left[f(x+\frac{1}{n}) - f(x-\frac{1}{n})\right]$. Now since $f$ is differentiable, we can rewrite a bit and see
$f'_n(x) = \dfrac{n}{2}\left[f(x+\frac{1}{n}) -f(x) + f(x)- f(x-\frac{1}{n})\right] = \dfrac{1}{2}\left[\dfrac{f(x+1/n) -f(x)}{1/n} + \dfrac{f(x)- f(x-1/n)}{1/n}\right]$.
Thus we have $f'_n(x)$ in terms of difference quotients of $f$. Hopefully it is clear from here how to show the uniform convergence.
A: Need to show that $\forall \epsilon>0,\exists N$ such that $$n>N\Rightarrow |f_n’(x)-f’(x)|<\epsilon,\forall x\in I.$$
To prove the above statement, let $\epsilon$ be given and $I’=[a-1,b+1].$ Since $f$ is $C^1$, $f’$ is uniformly continuous on $I’,$ namely, there exists $\delta>0$ such that $$x,y\in I’~{\rm and~}|x-y|<\delta \Rightarrow |f’(x)-f'(y)|<\epsilon.$$
By the FTC and MVT, $\forall x\in I$, $$f_n’(x)=\frac n2\left(f\left(x+\frac 1n\right)-f\left(x-\frac 1n\right)\right)=f’(\xi_{x,n}),$$ for some $$\xi_{x,n}\in \left(x-\frac 1n,x+\frac 1n\right)\subseteq I’.$$
Now let $N=\lfloor{\frac 2\delta}\rfloor.$ Then for $n>N,$ one has $$|\xi_{x,n}-x|<\frac 2 n<\delta,$$ hence $$|f_n’(x)-f’(x)|=|f’(\xi_{x,n})-f’(x)|<\epsilon,
\forall x\in I,$$ as required. QED
A: All we need to show is that $a_{n}=\sup_{x\in I}|f_{n}'(x)-f'(x)|\rightarrow 0$ as $n\rightarrow\infty$.
This is equivalent to show that, for any subsequence $(n_{k})$, there is a further subsequence $(n_{k_{l}})$ such that $a_{n_{k_{l}}}\rightarrow 0$.
Now we fix a subsequence $(n_{k})$. For the sake of simplicity, we will denote $(n_{k})$ just by the usual sequence $(n)$ and the so called further subsequence that we are looking for just by $(n_{k})$.
We see that
\begin{align*}
f_{n}'(x)=\dfrac{n}{2}(f(x+1/n)-f(x-1/n))
\end{align*}
and hence by the usual mean value theorem,
\begin{align*}
|f_{n}'(x)|\leq\dfrac{n}{2}\dfrac{2}{n}|f'(\xi_{x,n})|\leq\sup_{x\in 2I}|f'(x)|,
\end{align*}
where $2I$ denotes the doubly expansion of the interval $I$.
Given $\varepsilon>0$, we have by the uniform continuity of $f'$ on $2I$ some $\delta>0$ such that
\begin{align*}
|f'(u)-f'(v)|<\varepsilon,\quad|u-v|<\delta,\quad u,v\in I.
\end{align*}
On the other hand, for any $n$, $x,y\in I$ with $|x-y|<\delta$,
\begin{align*}
|f_{n}'(x)-f_{n}'(y)|=|f'(\xi_{x,n})-f'(\eta_{y,n})|<\varepsilon.
\end{align*}
The above estimate shows that $(f_{n}')$ is uniformly bounded, uniformly equicontinuous, and we conclude by Arzela-Ascoli theorem some subsequence $(n_{k})$ such that $a_{n_{k}}\rightarrow 0$, we are done.
