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.



Does a bounded real function have an analytic continuation
$f(x) = \sqrt{1-x^2}$ 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 $0$.


