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Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be non-constant and differentiable with a sequence of $(x_n)_{n \in \mathbb{N}}$ such that $f'(x_n)=0 \, \forall \, n$ and $x_n \rightarrow c$ for some $c \in \mathbb{R}$. Can $f$ be analytic?

I'm looking for either a proof that $f$ cannot be analytic or an example of an analytic $f$.

(I'm most of the way through an answer to this question, trying to prove that an analytic function of a continuous random variable is itself a continuous random variable. But I'm stuck and need to prove that an $f$ as described cannot be analytic.)

José Carlos Santos
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Malkin
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No, such a function cannot be analytic, unless it is constant. If it was analytic, $f'$ would be analytic too. But $f'(c)=\lim_{n\to\infty}f'(x_n)=0$. Therefore, by the identity theorem, $f'$ is the null function and so $f$ is constant.

José Carlos Santos
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  • Thanks. I had thought of the identity theorem, but thought that it only applied to holomorphic functions. Is there a version for real analytic functions? – Malkin Aug 10 '18 at 22:51
  • @Malkin Yes. The identity theorem is a theorem about analytic functions. It doesn't matter whether we're talking about real-valued or about complex-valued functions. – José Carlos Santos Aug 10 '18 at 22:55
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    @Malkin: Any real-analyic function in a neighborhood of a point of $\mathbb{R}$ extends to a complex-analytic function in a neighborhood of that point in $\mathbb{C}$: just use the same power series! – Eric Wofsey Aug 10 '18 at 22:57