1

I know that analytic implies infinitely differentiable, but is the converse always true as well?

Roberto
  • 199
  • 4
    Nope. Look up smooth bump functions. – Qiaochu Yuan May 15 '16 at 04:17
  • 1
    https://en.wikipedia.org/wiki/Smoothness#Differentiability_classes This question should be closed because OP never bothered even to check Wikipedia – MathematicsStudent1122 May 15 '16 at 04:18
  • 1
    The answer depends on whether you're working over $\Bbb R$ (in which case no) or $\Bbb C$ (in which case yes). – Rolf Hoyer May 15 '16 at 04:18
  • https://en.wikipedia.org/wiki/Non-analytic_smooth_function – Roberto May 15 '16 at 04:23
  • 1
    @MathematicsStudent1122: While "checking Wikipedia" is a natural step for you, knowing a certain body of useful search terms to put material in a useful context, this may not be true of all Community members, or there may be some who have less confidence in Wikipedia than in Math.SE on a given subject. A better response would be to check for possible duplicates here. – hardmath May 15 '16 at 04:39

1 Answers1

3

Our old friend $$f(x) = e^{-\frac{1}{x^2}}$$ is infinitely differentiable and in particular at $x=0$ all its derivatives are zero. But it is not analytic at $x=0$.

Mark Fischler
  • 42,297
  • Isn't this function undefined at 0? Doesn't this make it not continuous and therefore not differentiable at 0? Am I missing something? – B flat Oct 31 '22 at 19:59
  • Yes, to turn it into a function on the whole real line (actually, the whole complex plain) one must add to its definition $f(0)=0$. Then it is infinitely differentiable everywhere, but not analytic at $0$. – Mark Fischler Nov 05 '22 at 02:03