The definition of $C^\infty$ at one point $x$ of a function $f$ is that the derivative $D^{(k)}$ of arbitrary order $k$ exists in a neighbourhood of $x$. But we may have a smaller and smaller neighbourhood as $k$ increases. So whether there is a function that is $C^{\infty}$ at one point $x$ but not in any small deleted neighbourhood $U$? I want to construct it by convergence but don't know where to start.

  • $\begingroup$ $f(x) = \begin{cases} 0 & x \text{ is rational} \\ e^{-1/x^2} & x \text{ is irrational}\end{cases}$ is one example. Another would be to multiply $e^{-1/x^2}$ by a bounded nowhere-differentiable function. $\endgroup$ Oct 12, 2015 at 3:36
  • $\begingroup$ @PaulSinclair that doesn't appear to be $C^\infty$ at a point in the sense described by the question, since neither of those examples have derivatives of any order in any neighborhood of $0$. It works in the sense that there is a sequence of values for which $f$ satisfies Taylor-like approximations but that is not what appears to be meant $\endgroup$
    – jxnh
    Oct 12, 2015 at 14:29
  • $\begingroup$ @JHance - $f$ has a derivative of the first order at $0$ (as does the other example), but you are correct that the higher derivatives fail to exist, as the 2nd derivative requires the first derivative to exist at more than a single point in order to exist itself. I overlooked that consequence. $\endgroup$ Oct 12, 2015 at 16:34

1 Answer 1


Let $f_0$ be a bounded continuous nowhere-differentiable function. Define inductively $f_n(x) = \int_0^x f_{n-1}(t) dt$. So $f_n \in C^n(\Bbb R)$ but is not $n+1$ times differentiable at any point. Define $$f(x) = \begin{cases} 0 & x = 0\\ e^{-1/x^2}f_n(x) & {1\over n+1} \le |x| < {1\over n},\; n \in \Bbb N_+\end{cases}$$

This example is not continuous (unless perhaps $f_0$ has just the right properties), but that would not take much to fix.

  • $\begingroup$ Even if fixed to continuous,it appears that f is still not $C^k$ in this question at 0 since for each n,at x=1/n,f has different left and right derivative of each order. $\endgroup$
    – T.Jassen
    Oct 13, 2015 at 0:38
  • $\begingroup$ Yes it is. Because $f_0$ is bounded, so are $f_n$ on $[-1, 1]$ for all $n$, by the same bound. $e^{-1/x^2}$ times any bounded function is infinitely differentiable at $0$. $\endgroup$ Oct 13, 2015 at 2:38
  • $\begingroup$ Not true, even if we demand the bounded function to be analytic. See this for an example of such a function which isn't $C^1$ at 0: math.stackexchange.com/a/107695/210610 $\endgroup$ Oct 23, 2016 at 1:52
  • $\begingroup$ @ElliotGlazer - True. I really should have thought about this a little harder. $e^-1/x^2$ is powerful enough to overcome rational poles, but you don't get nowhere-differentiable with only rational growth. $\endgroup$ Oct 27, 2016 at 22:23

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