# Uniform convergence of $(f(x+h)-f(x))/h$ to $f^{\prime}(x)$

Let $$f:\mathbb{R}\to\mathbb{R}$$ be continuously differentiable and let $$f^{\prime}$$ be Lipschitz continuous on $$[0,1]$$. Why is it that $$$$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=f^{\prime}(x)\ \ \text{uniformly on [0,1]}\ ?$$$$

I don't see how the Lipschitz continuity of $$f^{\prime}$$ helps to conclude the uniform convergence. Help is very much appreciated!

$$f(x+h)-f(x)=h f'(z_h)$$ 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 $$|f'(z_h)-f'(x)|\le L |z_h-x|\le L|h|$$.
$$\left|\frac{f(x+h ) -f(x)}{h} -f'(x) \right|=|f'(\theta ) -f'(x)|\leq M|\theta -x| \leq M|h|$$ Since $$\theta \in [x, x+h]$$ by Lagrange theorem.
So $$\lim_{h\to 0} \sup_{x} \left|\frac{f(x+h ) -f(x)}{h} -f'(x) \right| \leq \lim_{h\to 0} M|h| =0$$ which means that the convergence is uniform.
• I've just come across the posting again... What if the Lipschitz continuity is only available on $[0,1]$? I am having trouble to estimate $\lvert f^{\prime}(\theta)-f^{\prime}(x)\rvert$ as the point $\theta\in[x,x+h]$ need not necessarily belong to $[0,1]$, no? Commented Oct 14, 2021 at 14:19