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Consider a differentiable (and hence continuous) function of order $n-1$. Is the $n$th derivative of such a function always continuous?

As an example, is the $n-th$ derivative of the function $f(s) = s \exp (n-1) (s-1)$ evaluated at $s=1$ continuous ?

3 Answers3

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A derivative can be discontinuous but it always has the intermediate value property. Thus for instance $f(x)=\operatorname{sign}(x)$ is not the derivative of any function which is differentiable on $[-1,1]$; the "antiderivative" of $f$ is $F(x)=|x|$ which is simply not differentiable at all at $0$. On the other hand, the function

$$g(x)=\begin{cases} 0 & x=0 \\ \int_0^x \sin(1/y) dy & x \neq 0 \end{cases}$$

is differentiable everywhere on $[-1,1]$, with a derivative which is not continuous at zero. (It is an interesting exercise to show that $g'(0)=0$.)

Reference: Wikipedia article on Darboux's theorem

Ian
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  • Could you provide a hint about how to compute $g'(0)$? – hunter Aug 30 '15 at 21:28
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    @hunter Intuitively, by splitting appropriately, the integral on a small interval $[0,x]$ is something like $\sum_{n=N}^\infty \frac{(-1)^n}{n^2 \pi}$, where $N=O(1/x)$. This is $O(1/N^2)$ by the error estimate for the alternating series. Dividing by $x$ gets you something which is $O(1/N)$, so the difference quotient is $O(x)$ so it goes to zero. It might be easier to compute all of this by writing $\int_0^x \sin(1/y) dy = \int_{1/x}^\infty \frac{\sin(u)}{u^2} du$. – Ian Aug 30 '15 at 21:48
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You may verify yourself that the function $f$ given by $$f(x) = \begin{cases} x^2 \sin\tfrac1x &\text{if } x \neq 0 \\ 0 & \text{if } x=0.\end{cases}$$ is continuous, is differentiable everywhere, but the derivative is not continuous at $0$. If you integrate this function $n-1$ times, you obtain (using the main theorem of integration) a function that is $n$ times differentiable and the $n$th derivative is not continuous at $0$.

Hagen von Eitzen
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  • Thanks guys for your responses, i have edited the question by adding a specific example. May you now please consider the question in its recent form. Thank you. –  Aug 30 '15 at 21:36
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To go with the counterexamples given by Hagen von Eitzen and Ian, it is good to remember that there are explicitly differentiability classes for functions $f:\mathcal{I}\to\mathbb{R}$ on an interval $\mathcal{I}$ (with the case $\mathcal{I}=\mathbb{R}$ included). The class $C^n(\mathcal{I})$ comprises all functions that are $n$-fold differentiable and have continuous $n^{th}$ derivatives. Naturally $C^k(\mathcal{I})\subset C^j(\mathcal{I})$ whenever $k>j$ (a function $k$ times differentiable is also $j$ times differentiable when $j< k$) and the inclusion is strict. That is, there are always examples of functions whose $n^{th}$ derivatives are continuous, but whose $(n+1)^{th}$ derivatives are not. Differentiating a function always lowers its differentiability class by one.

You also see the notations $C^\infty(\mathcal{I})$ for infinitely differentiable functions ("Smooth functions") and $C^\omega(\mathcal{I})$ for analytic functions (those whose values are given by convergent Taylor series in the interval). Note that $C^\infty\subset C^\omega$: there are smooth functions without convergent Taylor series: $$f:\mathbb{R}\to[0,\,1);\,f(x) = \left\{\begin{array}{ll}\exp\left(-\frac{1}{x}\right);&x>0\\0;&x\leq0\end{array}\right.$$

being a standard example. See here

Community
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Selene Routley
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