A Third Derivative in Ahlfors' `Some Remarks on Teichmuller Space'

Question

I'm slugging my way through Ahlfors' "Some Remarks on Teichmuller Space" and am stuck on the calculation of equation (1.18). The problem here is to compute the third derivative, on the unit disk $\mathbb{D}$, of the function $$ \Phi(\zeta) = \frac{-2}{\pi} \iint_{\mathbb{D}} \bar{\nu}(z) \frac{\zeta^3}{1-\bar{z}\zeta} dx dy. $$ Here, $\nu$ is some bounded, measurable function. Ahlfors claims that the third derivative is $$ (1) \;\;\;\; \Phi'''(\zeta) = \frac{-12}{\pi} \iint_{\mathbb{D}} \frac{\bar{\nu}(z)}{(1-\bar{z}\zeta)^4} dx dy $$ I believe the idea here is to use Dominated Convergence, since $\nu$ is bounded, to justify the commuting of $\frac{\partial}{\partial \zeta}$ with the integral. Upon doing this, we see $$ \Phi'''(\zeta) = \frac{-2}{\pi} \iint_{\mathbb{D}} \bar{\nu}(z) \frac{\partial^3}{\partial \zeta^3} \left( \frac{\zeta^3}{1-\bar{z}\zeta} \right) dx dy $$ I've checked the calculation of this derivative on Wolfram, and it comes out to $$ \Phi'''(\zeta) = \frac{-2}{\pi} \iint_{\mathbb{D}} \bar{\nu}(z) \left( \frac{6}{1-\bar{z}\zeta} + \frac{18\zeta \bar{z}}{(1- \bar{z}\zeta)^2} + \frac{18 \zeta^2 \bar{z}^2}{(1-\bar{z}\zeta)^3} + \frac{6 \zeta^3 \bar{z}^3}{(1-\bar{z}\zeta)^4} \right) dx dy $$ Doing some simplifying algebra, we then see that $$(2) \;\;\;\; \Phi'''(\zeta) = \frac{-12}{\pi} \iint_{\mathbb{D}} \frac{\bar{\nu}(z)}{(1-\bar{z}\zeta)^4}dxdy \, +\\ \frac{-12}{\pi} \iint_{\mathbb{D}} \bar{\nu}(z) \left( \frac{1}{\zeta^3 \bar{z}^3(1-\bar{z}\zeta)} + \frac{3}{\zeta^2 \bar{z}^2(1-\bar{z}\zeta)^2} + \frac{3}{\zeta \bar{z} (1-\bar{z}\zeta)^3} dx dy\right)$$ Thus, verification of Ahlfors' formula (1) now boils down to showing the second integral in equation (2) is zero. On the other hand, we know nothing besides boundedness of $\nu$. Hence, I see no reason that this integral should vanish. Any help would be greatly appreciated.

Answer 1

It's just a rearrangement. Get a common denominator, and take out the factor $6$ and the denominator: \begin{align} \frac{6}{1 - \overline{z}\zeta} + \frac{18\zeta\overline{z}}{(1- \overline{z}\zeta)^2} &{}+ \frac{18\zeta^2\overline{z}^2}{(1 - \overline{z}\zeta)^3} + \frac{6\zeta^3\overline{z}^3}{(1 - \overline{z}\zeta)^4} \\ &= \frac{6(1 - \overline{z}\zeta)^3}{(1 - \overline{z}\zeta)^4} + \frac{18\zeta\overline{z}(1 - \overline{z}\zeta)^2}{(1 - \overline{z}\zeta)^4} + \frac{18\zeta^2\overline{z}^2(1 - \overline{z}\zeta)}{(1 - \overline{z}\zeta)^4} + \frac{6\zeta^3\overline{z}^3}{(1 - \overline{z}\zeta)^4} \\ &= \frac{6}{(1 - \overline{z}\zeta)^4}\Bigl((1 - \overline{z}\zeta)^3 + 3(1 - \overline{z}\zeta)^2(\zeta\overline{z}) + 3(1 - \overline{z}\zeta)(\zeta\overline{z})^2 + (\zeta\overline{z})^3\Bigr) \\ &= \frac{6}{(1 - \overline{z}\zeta)^4}\bigl((1 - \overline{z}\zeta) + (\zeta\overline{z})\bigr)^3 \\ &= \frac{6}{(1 - \overline{z}\zeta)^4}\,. \end{align}

Computing the derivative is easier when one expands the denominator in a geometric series: $$\frac{\zeta^3}{1 - \overline{z}\zeta} = \sum_{k = 0}^{\infty} \overline{z}^k \zeta^{k+3}$$ leads to the third derivative $$\sum_{k = 0}^{\infty} (k+3)(k+2)(k+1)(\overline{z}\zeta)^k = \frac{6}{(1 - \overline{z}\zeta)^4}$$ from the familiar derivatives of the geometric series $\frac{1}{1-w} = \sum_{k = 0}^{\infty}w^k$.

Answer 2

Starting with $$g(\zeta):={\zeta^3\over1-\bar z\zeta}$$ Mathematica gives me $$g'''(\zeta)={6\over (1-\bar z\zeta)^4}\ .$$ There is no problem.