Find a function $f$ over $\mathbb{R}$ that is $n$ times differentiable but $f^{(n)}$ is not continuous

Question

Let $n$ be a positive integer. Find a function $f$ over $\mathbb{R}$ that is $n$ times differentiable but $f^{(n)}$ is not continuous.

We know that all the derivatives $f',f'',...,f^{(n-1)}$ are all continuous, but how do we find a function where the last derivative is not continuous? The function we make up can't have absolute values obviously, so I was thinking it is going to have to be piecewise.

Plutoro · Answer 1 · 2016-04-01T20:47:51.600

4

Hint:

$$f(x)=\begin{cases}x^2\sin(1/x) & x\neq 0\\ 0 & x=0\end{cases}$$ is once differentiable. The derivative exists for all $x$, but the derivative is not continuous. Try higher powers of $x$ to get a function that is more times differentiable.

edited Apr 01 '16 at 20:47

answered Apr 01 '16 at 20:40

Plutoro

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How do you know the derivative exists for all $x$? What about $x = 0$? – user19405892 Apr 01 '16 at 20:45
Use the definition of the derivative together with the squeeze theorem. – Plutoro Apr 01 '16 at 20:48
How do you know that the derivative is not continuous? – user19405892 Apr 01 '16 at 20:53
The limit as $x\to 0$ of $f'(x)$ is not defined. – Plutoro Apr 01 '16 at 20:53
Also, does $n$-times differentiable mean that the function is differentiable everywhere for each $n$? – user19405892 Apr 01 '16 at 20:57
Yes, that's what it means. – Plutoro Apr 01 '16 at 20:58
Okay, so I have the function $x^n \sin \frac{1}{x}$ for $x \neq 0$ and $f(x) = 0$ for $x = 0$ and need to show it is $n$-times differentiable. Is there an easy way to show that? – user19405892 Apr 01 '16 at 21:00
Let us continue this discussion in chat. – Plutoro Apr 01 '16 at 21:01

score 2 · Answer 2 · edited Apr 13 '17 at 12:20

Consider $$ g(x) = \begin{cases} x^2 \sin(1/x) &\mbox{if } x \neq 0 \\ 0 & \mbox{if } x=0. \end{cases} $$ This is a differentiable function with discontinuous derivative (see Discontinuous derivative.). Then you can integrate $g$. (Check Prove that integral of continuous function is continuously differentiable)

In other words you find $f$ such that $f^{(n-1)}=g$. So $f$ is $n$ times differentiable and $f^{(n)}=g'$ is discontinuous.

Find a function $f$ over $\mathbb{R}$ that is $n$ times differentiable but $f^{(n)}$ is not continuous

2 Answers2