3
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ Perhaps explain "analytic" vs. "holomorphic". $\endgroup$ Dec 4, 2012 at 19:01

1 Answer 1

5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ in dimension $\geq 2$? does the same still hold? $\endgroup$
    – hapchiu
    Dec 4, 2012 at 17:28
  • $\begingroup$ Yes, if one replaces "harmonic" by "pluriharmonic". $\endgroup$
    – Ralph
    Dec 4, 2012 at 18:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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