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?


  • $\begingroup$ Perhaps explain "analytic" vs. "holomorphic". $\endgroup$ Dec 4 '12 at 19:01

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.