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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?

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  • $\begingroup$ 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 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. $\endgroup$ – Dave L. Renfro Mar 24 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 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. $\endgroup$ – Dave L. Renfro Mar 24 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 at 19:03
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Sure. Take, for instance$$f(x)=\begin{cases}x^2&\text{ if }x\geqslant0\\-x^2&\text{ otherwise.}\end{cases}$$

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  • $\begingroup$ Thanks - I wasn't thinking about just composing a piecewise function like this. $\endgroup$ – jmarvin_ Mar 24 at 18:18
  • $\begingroup$ I'm glad I could help. $\endgroup$ – José Carlos Santos Mar 24 at 18:21
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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 $x=0$.

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  • $\begingroup$ @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. $\endgroup$ – MachineLearner Mar 24 at 15:27
  • $\begingroup$ @TomislavOstojich No contradiction.. $f$ differentiable always, but $f'$ (not $f$) isn't always. $\endgroup$ – coffeemath Mar 24 at 15:27
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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 $x.$

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

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