2
$\begingroup$

Could you please help me find the mistake in the following reasoning? I may have become dumb since I've spent a lot of time thinking about it but can't see what's wrong.

Let $U\subset \mathbb{C}$ be open and let $f=(u,v):U\to \mathbb{C}$ be a function. It is easy to check that $f$ is harmonic in $U$ iff $$ \frac{\partial}{\partial \overline{z}}\frac{\partial}{\partial z} f=0 \qquad\text{in U.} $$ Now, from this it seems to me that one can deduce that $f$ is holomorphic (and even antiholomorphic, and thus constant!), which is clearly wrong since there are no reasons for two arbitrary harmonic functions $u$ and $v$ to satisfy the Cauchy-Riemann equations. The reasoning is the following: from $\frac{\partial}{\partial \overline{z}}\frac{\partial}{\partial z} f=0$ we see that $\frac{\partial}{\partial z} f$ is holomorphic in $U$, then locally it has a holomorphic primitive, which must coincide with $f$, thus $f$ is holomorphic.

$\endgroup$

1 Answer 1

2
$\begingroup$

When writing down the question, I realized that the problem is that it is not true that the holomorphic primitive of $\frac{\partial}{\partial z}f$ must coincide with $f$. In fact, a function $g$ is such that $\frac{\partial}{\partial z}g=\frac{\partial}{\partial z}f$ if and only if $g=f+h$ for some antiholomorphic function $h$.

$\endgroup$
1
  • 1
    $\begingroup$ correct and you deduced the local representation of harmonic functions $f=g+\bar h$ where $g,h$ are holomorphic $\endgroup$
    – Conrad
    Commented Feb 7, 2022 at 21:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .