Why is every real harmonic function the real part of an holomorphic function?

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Why is every real harmonic function the real part of an holomorphic function? Can anyone give me a proof please? Is this always the case or are there certain conditions? I think we need a simply connected domain but I don't know why and how to prove it?

Anyone an idea?

thanks for the help

Kind regards.

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Every real harmonic function is locally the real part of a holomorphic function.

In fact, if $g$ is a real harmonic function on an open domain $D$, then for every open ball $B \subset D$, there exists a holomorphic function $F$ on $B$ such that $g = \text{Re}(F)$.

To prove this, we'll use two facts:

Fact 1: Let $f$ be a real continuous function, defined on the circle $\partial B$ that forms the boundary of the ball $B$. There exists a real function $F$ such that:

  • $F$ is continuous on the closed ball $\bar B$.
  • There exists a holomorphic function $g$ defined on $B$ such $F = \text{Re}(g)$ on $B$. (Hence $F$ is harmonic on $B$.)
  • $F$ is equal to $f$ on $\partial B$.

This is $F$ can be constructed using the Poisson integral formula, as you know.

Fact 2: Let $f$ be a real continuous function defined on the boundary circle $\partial B$. Suppose there exist two real functions $F_1$ and $F_2$ defined on $\bar B$ such that

  • $F_1$ and $F_2$ are continuous on $\bar B$
  • $F_1$ and $F_2$ are harmonic inside $B$.
  • $F_1$ and $F_2$ are equal to $f$ on $\partial B$.

Then $F_1$ and $F_2$ must be equal, i.e. $F_1 = F_2$ on $\bar B$.

This is the uniqueness theorem for harmonic functions.


Okay, so suppose that $F$ is a real harmonic function on the open domain $D$, and suppose that $B$ is an open ball such that $\bar B \subset D$.

Let $f = F|_{\partial B}$ be the restriction of $F$ to the circle $\partial B$. By fact 1, there exists a function $F'$ that is continuous on $\bar B$, equal to the real part of some holomorphic function on $B$, and equal to $f$ on $\partial B$. This $F'$ is harmonic in $B$.

But $F$ and $F'$ are two functions defined on $\bar B$ that are continuous in $\bar B$, harmonic in $B$ and equal to $f$ on $\partial B$. By fact 2, it must be the case that $F = F'$ on $\bar B$.

Since $F'$ is the real part of some holomorphic function on $B$, and $F = F'$, we find that $F$ is the real part of some holomorphic function on $B$.

[If we now weaken the hypotheses so that $B \subset D$ but $\bar B$ is not necessarily in $D$, then we can apply the same logic to disks of smaller radius. The holomorphic functions we get will patch together by the identity theorem. You can work through this.]


Finally, I'll give you an example of a real harmonic function is locally but not globally the real part of a holomorphic function.

Let $D$ be the domain $\mathbb C \setminus \{ 0 \}$. Let $f(z) = \log |z|$. This is a real harmonic function.

We want to say that $f(z)$ is the real part of the holomorphic function $\log z$. But the problem is that $\log z$ is not continuously defined on the whole of $D$, because it has a branch cut! $\log z$ is only continuously defined on local neighbourhoods in $D$. On any open ball, we can pick a branch for $\log z$ such that the branch cut doesn't intersect that open ball; then we can say that $f$ is the real part of that branch of $\log z$. But this doesn't work globally.