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Hallo,

Let $f: U \rightarrow \mathbb{R}$ be a analytic function, where $U \subset \mathbb{C}^{n}$ is a open set (paracompact, starshaped or convex i.e. sufficiently nice). Does there exist a function $\varphi : U \rightarrow \mathbb{R}$ such that $f + i \varphi$ is holomorphic? Is this possible?

hapchiu

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  • $\begingroup$ Perhaps explain "analytic" vs. "holomorphic". $\endgroup$ Dec 4 '12 at 19:01
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Suppose $n=1$. Then, if $f=u+iv$ is holomorphic, $u,v$ are harmonic, i.e. $u_{xx}+u_{yy}=0$. Hence a necessary condition for $u$ to be the real part of a holomorphic function is that $u$ is harmonic. The converse is true if $U$ is simply connected: See Theorem 6.3 in http://www.math.binghamton.edu/sabalka/teaching/09Spring375/Chapter6.pdf.

In the general case ($n \ge 1$) one has to consider pluriharmonic functions: A function $u(x_1,y_1,...,x_n,y_n): U \subseteq \mathbb{R}^{2n} \to \mathbb{R}$ is called pluriharmonic, if it is $C^2$ and satisfies $u_{x_jx_k}+u_{y_jy_k}=0$ for all $1 \le j,k\le n$.

The real part of a holomorphic function $f: U \to \mathbb{C}$ is pluriharmonic and conversely, each pluriharmonic function is locally (in a disc) the real part of a holomorphic function. See Theorem 26 of http://www.dms.umontreal.ca/~gauthier/6140.pdf.

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  • $\begingroup$ in dimension $\geq 2$? does the same still hold? $\endgroup$
    – hapchiu
    Dec 4 '12 at 17:28
  • $\begingroup$ Yes, if one replaces "harmonic" by "pluriharmonic". $\endgroup$
    – Ralph
    Dec 4 '12 at 18:00

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