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You can use the proof of the Schwarz reflection principle to show that if $D$ is a domain in the complex plane, where $D^+$ denotes the subset of D above the real line, and $D^-$ the subset below, if $f:D\to C$ which is analytic on $D^+$ and $D^-$, and continuous on $D$, then $f$ is analytic on $D$.

I want to generalize this as follows: Let $D$ be a domain in the complex plane, and $D'$ an open dense subset of $D$. If $f$ is a function analytic on $D'$, and continuous on $D$, then $f$ is analytic on $D$.

The proof of the Schwarz reflection principle uses integration along arbitrary triangles, and since the intersection of a triangle with the real line is really simple, you can "cut the real line out" so to speak when you are taking the integral. But if all you know is that $f$ is analytic on an open dense set what can you do?

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@Trevor: Perhaps you could look at circles instead of lines. –  PEV Jan 17 '11 at 21:30
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@Trevor, @Trevor: This is quite amusing, but would be less confusing if one of you would change your username. –  Jonas Meyer Jan 17 '11 at 21:33
    
@Jonas Meyer: How do you change your user name? –  PEV Jan 17 '11 at 21:49
    
@Trevor: If you go to your user page when logged in and click "edit" to the right of where it says "Registered User", this is one of the things you can change. –  Jonas Meyer Jan 17 '11 at 21:52
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1 Answer

up vote 2 down vote accepted

It is not true in general. Counterexamples and references to counterexamples of this and related questions can be found in the MathOverflow questions Continuous + holomorphic on a dense open => holomorphic? and The Cauchy-Riemann equations and analyticity.

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Will change when stackexchange lets me (have to wait one month after I open my account) and I will look up the counter-example, thanks. –  Trevor Jan 18 '11 at 15:16
    
@Trevor: Don't worry, the other Trevor is now PEV, so we no longer have Trevor writing a message to Trevor. –  Jonas Meyer Jan 18 '11 at 21:25
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