1
$\begingroup$

I can't seem to find any counterexamples to the statement "all functions that are continuous and differentiable at every point of the reals are smooth," nor can I find anyone asserting or proving this statement. Are there known functions that are continuous and differentiable at every point (with no holes / discontinuities / bounded domain) but are not smooth, that is, after some number of derivatives the derivative function is no longer fully differentiable?

$\endgroup$
5
  • $\begingroup$ Forgot to specify that I am talking about real functions only, if that wasn't clear - complex functions that are discontinuous off the real axis are not interesting counterexamples! $\endgroup$
    – jmarvin_
    Mar 24 '19 at 15:04
  • $\begingroup$ What do you mean by "smooth" --- continuous derivative or infinitely differentiable (or something else)? FYI, both uses of "smooth" occur quite often here, and nearly always without the questioner saying what "smooth" means until asked. $\endgroup$ Mar 24 '19 at 15:46
  • $\begingroup$ Dave: infinitely differentiable was the one I was using, since that's the limit of "smoothness" and thus what I thought was the unequivocal meaning of "smooth" (I specified this in the question at the end - it is "not smooth" if it is no longer totally differentiable after some derivative) $\endgroup$
    – jmarvin_
    Mar 24 '19 at 18:07
  • $\begingroup$ Actually, when I saw the part about totally differentiable, I thought you were talking about this, although I did wonder why you used the term in a non-multivariable setting (but not enough to think carefully about what you might have intended, since your question already had several answers). Incidentally, this answer might be of interest. $\endgroup$ Mar 24 '19 at 18:37
  • $\begingroup$ Ah, yes, I was using the word "totally" as a non-technical synonym for "completely." My bad - it's been years since I took multivariable calc. $\endgroup$
    – jmarvin_
    Mar 25 '19 at 19:03
3
$\begingroup$

Sure. Take, for instance$$f(x)=\begin{cases}x^2&\text{ if }x\geqslant0\\-x^2&\text{ otherwise.}\end{cases}$$

$\endgroup$
2
  • $\begingroup$ Thanks - I wasn't thinking about just composing a piecewise function like this. $\endgroup$
    – jmarvin_
    Mar 24 '19 at 18:18
  • $\begingroup$ I'm glad I could help. $\endgroup$ Mar 24 '19 at 18:21
3
$\begingroup$

An example is

$$f(x) = \begin{cases}0 & \text{for } x<0\\x^2 & \text{for } x\geq 0 \end{cases}.$$

It is clear that the function is continuous and differentiable for all $x\in \mathbb{R}$. But $f'(x)$ is not differentiable at $x=0$.

$\endgroup$
2
  • $\begingroup$ @TomislavOstojich I think you didn't notice the prime. I said that $f(x)$ differentiable. And $f'(x)$ (the derivative of f(x)$ is not differentiable. $\endgroup$ Mar 24 '19 at 15:27
  • $\begingroup$ @TomislavOstojich No contradiction.. $f$ differentiable always, but $f'$ (not $f$) isn't always. $\endgroup$
    – coffeemath
    Mar 24 '19 at 15:27
1
$\begingroup$

Let $W$ be the continuous, nowhere differentiable Weierstrass function. Then $f(x)=\int_0^x W(t)\,dt$ is continuously differentiable on $\mathbb R,$ but $f''(x)$ fails to exist for every $x.$

$\endgroup$
1
  • $\begingroup$ This is an interesting answer, probably the most extreme one easily available. Thanks! $\endgroup$
    – jmarvin_
    Mar 24 '19 at 18:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.