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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).

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3 Answers 3

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: $$f'(0)=\lim_{x\to 0^+}\frac{f(x)-f(0)}{x} = \lim_{x\to 0^+} f'(\theta_x \cdot x)=\lim_{y\to 0+}f'(y).$$

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Note that the assumption of the existence of $f'(0)$ is not needed in the original post. Nice! –  David Mitra Jan 6 '12 at 23:06
    
Right: the MVT needs differentiability in the open set only, and the first equality is the definition of the derivative at zero (which is shown to exist via the second and third equalities). –  David Jan 6 '12 at 23:13

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 $(0,1)$.

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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 $f'(c)=k$.

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