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Let $U \subset \mathbb{R}$ be an open interval and let $x_0 \in U$. Let $f: U \to \mathbb{R}$ be defined by a power series around $x_0$ with radius of convergence $R > 0$,

$$f(x) = \sum_{n=0}^{\infty} a_n (x - x_0)^n$$

where $(x_0 - R, x_0 + R) \subseteq U$.

Clearly, $f$ is $C^\infty$ on $(x_0 - R, x_0 + R)$, but:

With no other assumptions, is it true that $f$ is analytic at $x_0$?

If yes, how to prove it? If not, what would be a counter-example?


My thoughts:

For a moment this seemed trivially "yes", but then I paused. Let's look at the definition on Wikipedia:

A function $f$ defined on some subset of the real line is said to be real analytic at a point $x_0$ if there is a neighborhood $D$ of $x_0$ on which $f$ is real analytic.

I know that $f$ is equal to its Taylor Series centered at $x_0$, but according to the definition above, the function $f$ must also agree to other Taylor Series, centered at the nearby points as well. But what do I know about the points near $x_0$? Do the taylor series centered at them also agree to the function?

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  • $\begingroup$ The answer is yes, and the proof is the same as in the complex case. Note, however, that being a real differentiable function tells nothing about its analyticity. There are smooth but not analytic real functions. $\endgroup$ – Qiyu Wen May 31 '16 at 2:20
  • $\begingroup$ @QiyuWen thanks for your comment, but I did not fully understand. I don't know how the proof could be the same as in the complex case, since what I would use for that one is "f is defined by a power series, therefore holomorphic, therefore analytic", and the last step is quite nontrivial and uses strong facts on the complex numbers. I don't see how that could be possibly similar to the real case; most likely you are referring to something else that I did not catch up. Could you (or someone else) possibly elaborate on this, and maybe write a complete answer when you have some time? Thank you! $\endgroup$ – Pedro A May 31 '16 at 11:27
  • $\begingroup$ To say a function is holomorphic/analytic is to say that it is complex differentiable in a region of the complex plane. That's why many smooth real variable functions are not analytic - complex differentiability is much stronger. To say a real variable function is analytic is essentially to say it's complex differentiable when extended to a complex variable function, e.g., in the obvious way. As for a power series, its radius of convergence is indifferent to whether the variable is taken real or complex. $\endgroup$ – Qiyu Wen May 31 '16 at 16:18

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