# Uniform Continuity and Differentiation

Is the following true or false?:

Let $f\colon [0,1) \to \mathbb{R}$ be a function differentiable in $[0,1)$ (where the derivative at zero means "right derivative") such that both $f$ and $f'$ are uniformly continuous in $(0,1)$. Then $f'$ is continuous.

Note that the mistery lies at $x=0$. So the question is: can we say with these hypotheses that $f'(0)=\lim_{x\to 0^{+}}f'(x)$ (which exists thanks to the uniform continuity of $f'_{\mid (0,1)}$). Note also that the uniform continuity of $f'_{\mid (0,1)}$ makes redundant the analogous requirement for $f$ (which will even more become a Lipschitz function).

It's true. Since you know how to prove existence of the limit of $f'$ I'll focus on the fact that $f'(0)$ exists and is equal to the limit - it's not a big thing in fact.
Note that Lagrange mean value theorem (following from Roll theorem) applies here. So we have $\frac{f(x)-f(0)}{x} = f'(\theta_x \cdot x)$ for some $\theta_x \in (0,1)$. Consequently: $$f'(0)=\lim_{x\to 0^+}\frac{f(x)-f(0)}{x} = \lim_{x\to 0^+} f'(\theta_x \cdot x)=\lim_{y\to 0+}f'(y).$$
• Note that the assumption of the existence of $f'(0)$ is not needed in the original post. Nice! Jan 6, 2012 at 23:06
For $x>0$, consider $${f(x)-f(0)\over x}={1\over x}\int^x_0 f^\prime(y)\, dy.$$ As $x\downarrow 0$, the left hand side converges to $f^\prime(0)$, while the right hand side converges to $\lim_{x\to 0^+} f^\prime(x).$ This limit exists because $f^\prime$ has a continuous extension to $[0,1)$, by uniform continuity on $(0,1)$.
As you stated, thanks to the uniform continuity of $f'$, $L=\lim\limits_{x\rightarrow0^+} f'(x)$ exists (uniformly continuous functions map Cauchy sequences to Cauchy sequences). If $L\ne f'(0)$, then a contradiction to Darboux's Theorem can be obtained.
Darboux's Theorem: If $f$ is differentiable on $I=[a,b]$ and if $k$ is a number between $f'(a)$ and $f'(b)$, then there is at least one point $c\in(a,b)$ such that $f'(c)=k$.