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I have some problem that I can't figure out myself. Hope that someone can help me out. The problem is: Let $f : U \to \mathbb{R}$ be some real function on a simply-connected domain $U \subset \mathbb{C}^{n}$ with $\partial\bar{\partial}f = 0$ on $U$. Then $f$ is the real part of a holomorphic function on $U$.

The question is: Is this holomorphic function unique? If not what choices are there to choose such a function? Are there any choices? How does the proof go?

I hope that someone has some answers for me and I will be very happy for a lot of answers. I have to apologize if this question is too trivial for some of you :).

Best regards philippe

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1 Answer 1

Suppose $g_1, g_2 : U \to \mathbb{C}$ are holomorphic functions such that $\Re(g_1) = \Re(g_2) = f$. Then $\Re(g_1 - g_2) = 0$, that is $g_1 - g_2 = 0 + iv(x, y)$; that is, $g_1 - g_2$ is a purely imaginary function. Furthermore, as both $g_1$ and $g_2$ are holomorphic, so is $g_1 - g_2$. Hence we can apply the Cauchy-Riemann equations which tell us that $v(x, y)$ is locally constant.

In summary, if $U$ is connected, any two holomorphic functions with real part $f$ differ by an imaginary constant.

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Connected is not enough -- think of an annulus in the plane –  user16299 Sep 30 '12 at 9:35
    
@YemonChoi: Could you be a bit more specific? I can't see what is wrong with the argument above. –  Michael Albanese Sep 30 '12 at 13:02
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There is nothing wrong with your uniqueness argument. @Yemon remarked that one needs additional assumption to obtain existence, but apparently the OP did not ask about existence. –  user31373 Sep 30 '12 at 17:24
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@LVK is right - sorry for the unclear and somewhat muddled comment. As you point out: the question's assumption gives us a $g$ and asks about uniqueness; I was thinking of the variant of the problem where one is given a function and asked to find a harmonic conjugate. –  user16299 Sep 30 '12 at 23:03
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Existence argument can be found here. –  user31373 Sep 30 '12 at 23:38

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