Prove $f_n\to f'$ uniformly on $[0,1]$ Let $f:[0,2]\to \Bbb{R}$ be a continuously differentiable function. Let us define $f_n:[0,1]\to \Bbb{R}$ by $f_n(x)=n(f(x+{1
\over n})-f(x))$. Prove $f_n\to f'$ uniformly on $[0,1]$. I know that $f_n(x)\to f'(x)$, but I can't seem to understand what to do with it. Can't I, for instance, assume there is $x_0$ for which $\sup_{x\in [0,1]}|f_n-f|=|f_n(x)-f'(x)|$? It will converge to 0 at infinity, won't it? I could use some help here. 
 A: Let me expand on my comment.
Since $f'$ is a continuous function on a compact set, it is uniformly continuous.
Therefore there is a modulus of continuity $\omega$ so that $|f'(x+h)-f'(x)|\leq\omega(h)$ for any $x$ and $h$ (provided the $f'$s are defined).
The function $\omega$ is increasing and $\omega(t)\to0$ as $t\to0$.
Using the fundamental theorem of calculus, we have
$$
\begin{split}
|f_n(x)-f'(x)|
&=
|n[f(x+1/n)-f(x)]-f'(x)|
\\&=
\left|n\int_{x}^{x+1/n}f'(y)dy-f'(x)\right|
\\&=
\left|n\int_{x}^{x+1/n}[f'(y)-f'(x)]dy\right|
\\&\leq
n\int_{x}^{x+1/n}|f'(y)-f'(x)|dy
\\&\leq
n\int_{x}^{x+1/n}\omega(y-x)dy
\\&\leq
n\int_{x}^{x+1/n}\omega(1/n)dy
\\&=
\omega(1/n).
\end{split}
$$
Therefore
$$
\sup_{x\in[0,1]}|f_n(x)-f'(x)|
\leq
\omega(1/n).
$$
This bound decays to zero as $n\to\infty$.
A: For any $\;x\in [0,1]\;$ , and using the MVT to pass from the first to second line, we have that
$$\left|n\left(f\left(x+\frac1n\right)-f(x)\right)-f'(x)\right|=\left|\frac{f\left(x+\frac1n\right)-f(x)}{\left(x+\frac1n\right)-x}-f'(x)\right|=$$
$$=\left|f'(c_n)-f'(x)\right|\;\;,\;\;c_n\in\left(x,\,x+\frac1n\right)$$
By continuity of $\;f'(x)\;$ , we then get that 
$$\left|f'(c_n)-f'(x)\right|\xrightarrow[n\to\infty]{}f'(x)-f'(x)=0$$
and we're done
