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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 $V$.

Is there a reference for this fact?

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3 Answers 3

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 $V$.

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But $|z|$ is not a real-analytic function on ${\mathbb C}$, so this is not a counterexample. –  Alexander Braverman Oct 29 '10 at 13:21
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It is a counterexample: $z\mapsto|z|$ is a real analytic function on the set $V$. –  Robin Chapman Oct 29 '10 at 13:27
    
Ok, you are right. I guess I was thinking about $U$ and $V$ being discs. Let me add "simply connected" to the question. Is it true then? –  Alexander Braverman Oct 29 '10 at 13:31

Trivially not. Let $\psi$ be a smooth function on $\mathbb{R}$ such that $\psi(x) = 0$ if $x < 1$ and $\psi(x) = 1$ if $x > 2$ and $\psi(x)$ monotonic.

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.

(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.)

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(Let me add the comment that the word "trivial" does not mean that your question is trivial, but that the counterexample is a "trivial" function.) –  Willie Wong Oct 29 '10 at 15:00
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At first sight, this looked nice, but now I'm not sure. This $f$ is constant in the unit disc, but while $f$ is not an analytic extension of $f$ from the unit disc to $\mathbb{C}$ the constant function $F(z)=1$ is. –  Robin Chapman Oct 29 '10 at 15:23
    
@Robin Chapman: I see your point. I was treating the statement as an "a priori" statement: if there exists an extension such that $|f(z)|^2$ is real analytic, must the extension be complex analytic? Another way to read the question is whether there exists any analytic extension at all, which is a question I didn't treat. For that I defer to Gerald's answer. –  Willie Wong Oct 29 '10 at 16:00
    
No, the question was about existence of analytic continuation -- I am not given any preferred extension of $f$ to the larger domain. –  Alexander Braverman Oct 30 '10 at 7:56

This needs some details. For example, what about points with $f(z) = 0$?

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 $exp(g(z))$.

On the other hand, if (?) doesn't work, it shows how to do a counterexample...

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... began writing before Willie Wong posted ... but I will leave it for what it is worth. –  Gerald Edgar Oct 29 '10 at 14:51

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