Let $f:\mathbb{R}\to\mathbb{R}$ be continuously differentiable and let $f^{\prime}$ be Lipschitz continuous on $[0,1]$. Why is it that
\begin{equation}
\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=f^{\prime}(x)\ \
\text{uniformly on $[0,1]$}\ ?
\end{equation}
I don't see how the Lipschitz continuity of $f^{\prime}$ helps to conclude the uniform convergence. Help is very much appreciated!
Best Answer
$f(x+h)-f(x)=h f'(z_h)$ where $z_h\in[x,x+h]$ (or $[x+h,x]$ if $h<0$). So $$\frac{f(x+h)-f(x)}{h}=f'(z_h).$$
Now $|f'(z_h)-f'(x)|\le L |z_h-x|\le L|h|$.