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It is very possible that this question was asked before, but I cannot find it from the list.

Question. Let $f : [a, b] \rightarrow \mathbb{R}$ be infinitely differentiable on $(a, b)$. Can we extend this to a holomorphic function $F : D \subseteq \mathbb{C} \rightarrow \mathbb{C}$ where $D$ is an open disc such that $D \cap \mathbb{R} = (a, b)$?

Motivation to the question is simply that I want to organize the concept of Taylor expansion of real variable in terms of complex variable, and this question seems very plausible to me but not immediate.

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I implicitly meant $F|_{(a, b)} = f$ although it is not included. – user123454321 Oct 31 '12 at 16:06
Later realization: I was confused between when we can use Taylor's theorem and have analytic function. The hypothesis given in the question is enough for us to use Taylor series but we cannot expand it in an infinite sum because our remainder is too large. Old John's answer is a good example to see this. I might ask a new independent question as I was interested in expanding analytic function. – user123454321 Oct 31 '12 at 18:22
up vote 4 down vote accepted

You might want to consider the function $f(x)=e^{-1/x^2}$, which is infinitely differentiable as a real function. All the derivatives are zero at the origin, so it is not possible to get a Taylor series for the function which represents the function away from zero, and as a complex function, it has an essential singularity at the origin, so is not holomorphic in any region which includes the origin.

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"... it is not possible to get a Taylor series for the function" - You mean possible? – user123454321 Oct 31 '12 at 16:12
Thanks and I will accept this answer in several minutes as the system does not allow me. But are there extra assumptions that we can see where the answer to this question is "yes"? – user123454321 Oct 31 '12 at 16:13
I mean that since all the derivatives are zero, you get a Taylor series in which all the terms are zero, so the series cannot represent the function, since the function is clearly not identically zero. – Old John Oct 31 '12 at 16:13
you might want to wait a few hours before accepting - someone might have a better answer! – Old John Oct 31 '12 at 16:14
True. I will wait for extra answers. Thanks for the clarification! I will edit as I think this question is not exactly what I had in mind. – user123454321 Oct 31 '12 at 16:15

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