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Imagine we have a function $f : \mathbb{R} \to \mathbb{R}$ which is differentiable over an interval $[a,b]$. Does differentiability over $[a,b]$ imply that the derivative $f'$ is continuous over $(a,b)$?

Intuitively, this seems right since differentiability implies that $f$ is "locally linear". And thus $f$ should have roughly the same slope at all points in a small region. But I haven't been able to prove this statement. A direct proof is a little technical, involving the absolute difference of limits of ratios.

More generally, does the same hold true for functions differentiable over a region in $\mathbb{R}^n$? Intuition again tells me yes, but there could be counter examples.

And finally, assuming the statements above are true, how might I prove them, or alternatively, where might I find a proof?

ted
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    It can be even worse than the answers thus far posted suggest. All the answers have the property that there is a subinterval on which the function is continuously differentiable. However, $f: {\mathbb R} \rightarrow {\mathbb R}$ can be (finitely) differentiable at each point (in fact, we can even have the derivative bounded between $-1$ and $1,$ so we're not having the derivative be unbounded in some places), and yet there is no nonempty open interval on which the derivative is continuous. – Dave L. Renfro Dec 30 '17 at 18:58

4 Answers4

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No, take $f(x)=x^{2}\sin(1/x)$ for $x\ne 0$ and $f(0)=0$. One has $f'(0)=0$ and $f'(x)=2x\sin(1/x)-\cos(1/x)$ for $x\ne 0$ and $\lim_{x\rightarrow 0}f'(x)$ does not exist. You may consider the interval $[-1,1]$ in this example.

user284331
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No, this result is false.

The standard example of a differentiable-but-not-continuously-differentiable function will do, $f:\mathbb{R}\to\mathbb{R}$ defined by

$$f(x)=\begin{cases}x^2\sin(1/x) & x\neq 0\\ 0 & x=0\end{cases}$$

You can verify that $f$ is differentiable on $\mathbb{R}$ but $f'$ is not continuous at $0$.

symplectomorphic
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No. A textbook counterexample is $f\colon\mathbb{R}\to\mathbb{R}$, $f(x)=x^2\sin(1/x)$ for $x\neq 0$, $f(0)=0$.

Taneli Huuskonen
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No, as the $x \to x^2 \sin {1 \over x} $ example show.

However, there is a continuity like property given by Darboux's theorem which is a sort of intermediate value theorem that shows that $[f'(a),f'(b)] \subset f'([a,b])$.

copper.hat
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