# Does a bounded real function have an analytic continuation [closed]

Consider the function $f:[0,1]\rightarrow\mathbb{R}$, where

• $f$ is real-analytic on the open interval $(0,1)$
• $f$ is bounded on the closed interval $[0,1]$ (ie. there is some constant $C$ such that $-C\leq f(x)\leq C$ for $x\in[0,1]$).

Is it true that there is a real-analytic continuation of $f$ to the interval $[-\epsilon, 1+\epsilon]$ for some small positive $\epsilon$? If not, what conditions can be added to make it true?

Suggestions for books (or other references) where I could have learned to answer this myself would also be appreciated.

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## closed as off topic by quid, Emil Jeřábek, Qiaochu Yuan, Andreas Blass, GH from MOSep 21 '11 at 17:29

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I suggest to reask this question on math.stackexchange.com a similar site but with a broader scope. Here I vote to close. – quid Sep 21 '11 at 15:03
While others have provided quite smooth counterexamples, I note that your conditions do not even force $f$ to be continuous at the end-points. – Emil Jeřábek Sep 21 '11 at 15:33
@quid I wasn't aware of the difference between math.stackexchange and mathoverflow. Now that I've looked that up, I agree, this question would have been better on math.stackexchange. – Essex Sep 21 '11 at 16:35
Essex, thanks for the response. No problem, this is a frequent phenomenon. – quid Sep 21 '11 at 16:39

$f(x) = \sqrt{1-x^2}$ is real analytic on $(-1,1)$, bounded and continuous on $[-1,1]$, but of course not even one-sided differentiable at the endpoints. But then Igor's example is one-sided differentiable of all orders at the endpoint $0$, but still not real analytic in any neighborhood of $0$.
What about $e^{-1/x^2}?$