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Does a function exist so that

f(x) = 0    x <= 0
f(1) = 1

under the additional constraint that all derivatives of f shall be continuously differentiable in [0;1]?

I strongly suspect that this is impossible, but I can't come up with a proof.

Constructing an f(x) which is continuous up to its nth derivative (for arbitrary finite n) is obvious, but not what I'm looking for.

Matthias Urlichs
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    the standard example is $f(x)=\exp (-\frac 1{x^2})$ for $x>0$ scaled as you like to get the value at $x=1$. – lulu Sep 06 '17 at 18:49
  • See https://en.wikipedia.org/wiki/Non-analytic_smooth_function#Smooth_transition_functions – Chris Culter Sep 06 '17 at 18:49
  • Further details here on math.SE: https://math.stackexchange.com/questions/328868/how-to-build-a-smooth-transition-function-explicitly – Chris Culter Sep 06 '17 at 18:51
  • The example with $\exp(-\frac{1}{x^2}$ was invented( discovered) by Cauchy, it's rather amazing. – orangeskid Sep 06 '17 at 19:12

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