Analytic continuation via square of absolute value - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T06:29:34Z http://mathoverflow.net/feeds/question/44115 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/44115/analytic-continuation-via-square-of-absolute-value Analytic continuation via square of absolute value Alexander Braverman 2010-10-29T11:58:23Z 2010-10-29T14:49:53Z <p>Is the following fact true (I think that I can prove it but I don't trust myself on these matters): let $f(z)$ be an analytic function defined on some open subset $U$ of ${\mathbb C}$. Assume that the function $|f(z)|^2$ extends as a real-analytic function to some bigger simply connected open subset $V$ of ${\mathbb C}$. Then $f$ extends analytically to $V$.</p> <p>Is there a reference for this fact?</p> http://mathoverflow.net/questions/44115/analytic-continuation-via-square-of-absolute-value/44126#44126 Answer by Robin Chapman for Analytic continuation via square of absolute value Robin Chapman 2010-10-29T13:00:22Z 2010-10-29T13:00:22Z <p>Rather obviously not: if $f(z)=\sqrt{z}$ on $U$, the plane slit along the negative real axis, then $|f(z)|^2=|z|$ is real analytic on $V$ the plane with the origin removed but $f$ does not analytically continue from $U$ to $V$.</p> http://mathoverflow.net/questions/44115/analytic-continuation-via-square-of-absolute-value/44139#44139 Answer by Willie Wong for Analytic continuation via square of absolute value Willie Wong 2010-10-29T14:43:45Z 2010-10-29T14:43:45Z <p>Trivially not. Let $\psi$ be a smooth function on $\mathbb{R}$ such that $\psi(x) = 0$ if $x &lt; 1$ and $\psi(x) = 1$ if $x > 2$ and $\psi(x)$ monotonic. </p> <p>Let $f(z) = e^{i\psi(|z|)}$. $f(z)$ is complex analytic in the unit disk, $|f(z)|^2 = 1$ is real analytic on the entire plane, but $f(z)$ is not analytic on any open sets strictly containing the unit disk. </p> <p>(I defined $\psi$ just so that you see even if you upgrade some a priori assumption on the regularity of $f$, it is not enough. Otherwise you can just take $f$ to equal 1 in the unit disk and -1 outside and get a discontinuous counterexample.)</p> http://mathoverflow.net/questions/44115/analytic-continuation-via-square-of-absolute-value/44140#44140 Answer by Gerald Edgar for Analytic continuation via square of absolute value Gerald Edgar 2010-10-29T14:49:53Z 2010-10-29T14:49:53Z <p>This needs some details. For example, what about points with $f(z) = 0$? </p> <p>Assume $|f(z)|^2$ is real analytic in a larger domain. Then $u(z) = \log |f(z)|$ is real analytic. It is harmonic in the original domain, so (?) deduce it is harmonic in the larger domain. Construct a harmonic conjugate $v(z)$ so that $g(z) = u(z)+iv(z)$ is analytic, using simple connectivity. And then your extension is $exp(g(z))$. </p> <p>On the other hand, if (?) doesn't work, it shows how to do a counterexample...</p>