I am reading Kra's and Farkas' book on Riemann surfaces, and Theorem II.2.1 is Weyl's Lemma:

Let $\varphi$ be a measurable square integrable function on the unit disk $D$. The function $\varphi$ is harmonic if and only if $$ \iint_D\varphi\,\Delta\eta = 0 $$ for every $C^\infty$ function $\eta$ on $D$ with compact support.

I am confused by a step in the proof of the sufficiency (that is, if the integral condition holds, then $\varphi$ is harmonic).

For now, let $\epsilon > 0$ and $\mu$ be a $C^\infty$ function with support in $D_{1-2\epsilon}$. For $r > 0$, define $\omega(r) = \frac{1}{2\pi}\rho(r)\log r$, where $\rho\colon[0,\infty)\to\mathbb R$ is a $C^\infty$ function such that $0\leqslant \rho \leqslant 1$, $\rho(r) = 0$ if $r > \epsilon$, and $\rho(r) = 1$ if $0 \leqslant r < \epsilon/2$.

I am stuck at the following equality:

\begin{align*} \iint_{\mathbb C} \omega(|\zeta|)\color{red}{\frac{\partial}{\partial \overline z}}\mu(\zeta + z)\frac{d\zeta\wedge d\overline \zeta}{-2i} = \iint_{\mathbb C} \omega(|\zeta|)\color{red}{\frac{\partial}{\partial \overline \zeta}}\mu(\zeta + z)\frac{d\zeta\wedge d\overline \zeta}{-2i}. \end{align*}

My question is simply why can we replace $\partial_{\overline z}$ with $\partial_{\overline \zeta}$? I assume this comes down to some sort of holomorphic change of coordinates, but I can't see what it is.

  • $\begingroup$ Just trivial chain rule by writing $w=\zeta +z$ $\endgroup$ – Conrad Apr 12 at 12:10
  • $\begingroup$ @Conrad: Thanks for your reply. I tried to apply the chain rule, but I am new to manipulations involving complex derivatives like this. Would you care to elaborate? $\endgroup$ – Alex Ortiz Apr 12 at 18:33
  • $\begingroup$ I will make it a solution since the commentary is too long $\endgroup$ – Conrad Apr 12 at 21:44

If you are comfortable with the complex chain rule, set $w = \zeta + z$, so $\overline w = \overline\zeta + \overline z$. Thus, $$ \frac{\partial \overline w}{\partial \overline z}=\frac{\partial \overline w}{\partial \overline \zeta}=1,\quad\text{and}\quad \frac{\partial w}{\partial \overline z}=\frac{\partial w}{\partial \overline \zeta}=0. $$ Hence, \begin{align*} \frac{\partial}{\partial \overline z}\big\{\mu(\zeta + z)\big\} &= \frac{\partial\mu}{\partial \overline w}\frac{\partial \overline w}{\partial \overline z} + \frac{\partial\mu}{\partial w}\frac{\partial w}{\partial \overline z} \\ &=\frac{\partial\mu}{\partial \overline w}\frac{\partial \overline w}{\partial \overline \zeta} + \frac{\partial\mu}{\partial w}\frac{\partial w}{\partial \overline \zeta} \\ &= \frac{\partial}{\partial \overline \zeta}\big\{\mu(\zeta + z)\big\}. \end{align*}

If not, use the usual partial derivatives in real coordinates and show that $\partial_{1}$ is the same for both $z, \zeta$ (pretty much by definition) and the same with $\partial_{2}$ and then use $\overline \partial = \frac{1}{2}(\partial_{1}+i\partial_{2})$ but the notation sometimes can get tricky, so I much prefer using the complex chain rule.

  • $\begingroup$ I hope my edits are within reason. I wanted to link to to the complex chain rule for future reference if anyone needed it, and to format it differently. Thank you for your comment and your answer; this is precisely what I needed. I had forgotten all about this version of the chain rule. $\endgroup$ – Alex Ortiz Apr 13 at 3:01
  • $\begingroup$ Happy to be of help. $\endgroup$ – Conrad Apr 13 at 11:47

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