# Confusing step in a proof of Weyl's Lemma

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.

• Just trivial chain rule by writing $w=\zeta +z$ – Conrad Apr 12 at 12:10
• @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? – Alex Ortiz Apr 12 at 18:33
• I will make it a solution since the commentary is too long – 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.