most recent 30 from http://mathoverflow.net 2013-06-19T03:13:36Z http://mathoverflow.net/feeds/question/76063 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/76063/does-a-bounded-real-function-have-an-analytic-continuation Essex 2011-09-21T14:37:31Z 2013-04-21T19:34:07Z

<p>Consider the function $f:[0,1]\rightarrow\mathbb{R}$, where</p> <ul> <li>$f$ is real-analytic on the open interval $(0,1)$</li> <li>$f$ is bounded on the closed interval $[0,1]$ (ie. there is some constant $C$ such that $-C\leq f(x)\leq C$ for $x\in[0,1]$). </li> </ul> <p>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?</p> <p>Suggestions for books (or other references) where I could have learned to answer this myself would also be appreciated.</p>

http://mathoverflow.net/questions/76063/does-a-bounded-real-function-have-an-analytic-continuation/76064#76064 Igor Rivin 2011-09-21T14:44:00Z 2011-09-21T14:44:00Z

<p>What about $e^{-1/x^2}?$</p>

http://mathoverflow.net/questions/76063/does-a-bounded-real-function-have-an-analytic-continuation/76066#76066 Gerald Edgar 2011-09-21T15:16:30Z 2011-09-21T15:16:30Z

<p>$f(x) = \sqrt{1-x^2}$ 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 $0$.</p>