# An identity involving the chain rule

I have a question concerning a demonstration in Pommerenke's Univalent Functions :

Let $\Phi(t)$ be twice continuously differentiable and

$\Psi(t) = t\frac{d}{dt}[t \Phi'(t)]$, $(0\leq t < \infty)$.

Let $f(z)$ be analytic in $D$ (open unit disk). Since $r\frac{d}{dr}|f(z)| = |f(z)|\text{Re}z \frac{f'(z)}{f(z)}$ and $\frac{d}{d\theta}|f(z)| = -|f(z)|\text{Im}z \frac{f'(z)}{f(z)}$, a short calculation shows that

$\left(r \frac{d}{dr}\right)^2 \Phi(|f(z)|) + \left(\frac{d}{d\theta}\right)^2 \Phi(|f(z)|) = \Psi(|f(z)|)\left|z \frac{f'(z)}{f(z)}\right|^2$.

This short calculation turns out to be rather confusing to me.

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I think it should be $\frac{d}{d\theta} |f(z)| = -|f(z)| \operatorname{Im} \left[z \frac{f'(z)}{f(z)} \right]$, without the factor of $r$. And similarly the factor of $r$ in front of the $\frac{d}{d\theta}$ in the line you are trying to prove should not be there, I guess. Does it work in this modified form? –  Lukas Geyer Nov 27 '12 at 1:00
It would make more sense without the factor $r$ in front of the $\theta$-derivative, too, since then the left side would be $r^2 \Delta (\Phi(|f|))$, a multiple of the Laplacian, whereas the line as written is not related to the Laplacian at all. –  Lukas Geyer Nov 27 '12 at 1:06
Copy/paste mistakes, sorry. –  M.G Nov 27 '12 at 1:44

## 2 Answers

Expanding the operators \begin{align} \left(r \frac{d}{d r}\right)^2 = \left(r \frac{d}{d r}\right) \left(r \frac{d}{d r }\right) = r \frac{d}{d r} \left(r \frac{d}{d r}\right)\\ \left(\frac{d}{d \theta}\right)^2 = \left(\frac{d}{d \theta}\right) \left(\frac{d}{d \theta}\right) = \frac{d}{d \theta} \left(\frac{d}{d \theta}\right) \end{align} we have \begin{align} \left(r \frac{d}{d r}\right)^2 \Phi\big(|f(z)|\big) &= r \frac{d}{d r} \left\{r \frac{d \Phi\big(|f(z)|\big)}{d r}\right\} = r \frac{d}{d r}\left\{r\,\Phi'\big(|f(z)|\big) \frac{d}{d r}|f(z)|\right\} \\ \\ &= r \frac{d}{d r}\left\{|f(z)|\,\Phi'\big(|f(z)|\big) \Re\left[\frac{zf'(z)}{f(z)}\right]\right\}\\ \\ &= |f(z)| \left\{|f(z)|\,\Phi'\big(|f(z)|\big)\right\}' \Re^2\left[\frac{zf'(z)}{f(z)}\right] \\ & \hskip2in + r |f(z)|\,\Phi'\big(|f(z)|\big) \Re'\left[\frac{zf'(z)}{f(z)}\right] \\ \\ \\ \left(\frac{d}{d \theta}\right)^2 \Phi\big(|f(z)|\big) &= \frac{d}{d \theta} \left\{\frac{d \Phi\big(|f(z)|\big)}{d \theta}\right\} = \frac{d}{d \theta}\left\{\Phi'\big(|f(z)|\big) \frac{d}{d \theta}|f(z)|\right\} \\ \\ &= -\frac{d}{d \theta}\left\{|f(z)|\,\Phi'\big(|f(z)|\big) \Im\left[\frac{zf'(z)}{f(z)}\right]\right\}\\ \\ &= |f(z)| \left\{|f(z)|\,\Phi'\big(|f(z)|\big)\right\}' \Im^2\left[\frac{zf'(z)}{f(z)}\right] \\ & \hskip2in - |f(z)|\,\Phi'\big(|f(z)|\big) \Im'\left[\frac{zf'(z)}{f(z)}\right] \end{align} Adding the two \begin{multline} \left(r \frac{d}{d r}\right)^2 \Phi\big(|f(z)|\big) + \left(\frac{d}{d \theta}\right)^2 \Phi\big(|f(z)|\big) = \\ \Psi\big(|f(z)|\big) \left|\frac{zf'(z)}{f(z)}\right|^2 + |f(z)|\,\Phi'\big(|f(z)|\big)\left\{\Re'\left[\frac{zf'(z)}{f(z)}\right] - \Im'\left[\frac{zf'(z)}{f(z)}\right]\right\} \end{multline}

Finally, using that $f(z)$ is analytic in $D$, you need to prove that the second term is zero. I'm very rusty on my complex analysis theorems, but I believe is pretty straigth forward since $$\Re'\left[\frac{zf'(z)}{f(z)}\right] - \Im'\left[\frac{zf'(z)}{f(z)}\right] = \frac{d}{dz} \left(\frac{\bar{z}\bar{f}'(z)}{\bar{f}(z)}\right) = \frac{d}{dz} \left(\frac{\bar{z} f'(\bar{z})}{f(\bar{z})}\right) = 0$$ Please doublecheck that the last argument is correct.

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You are a hard worker ! Thank you. I've doublecheck and it seems correct. –  M.G Nov 27 '12 at 2:07
No problem. The thing about chain rule is to be very patient and go step by step, no matter how silly it can be :) –  Pragabhava Nov 27 '12 at 2:08

First of all we get by a straightforward application of the chain rule $$r \frac{d}{dr} \Phi(|f(z)|) = \Phi'(|f(z)|) |f(z)| \operatorname{Re} \left[z \frac{f'(z)}{f(z)}\right] = \eta(|f(z)|) u(z)$$ and $$\frac{d}{d\theta} \Phi(|f(z)|) = -\Phi'(|f(z)|) |f(z)| \operatorname{Im} \left[z \frac{f'(z)}{f(z)}\right] = -\eta(|f(z)|) v(z)$$ where we write for ease of notation $$\eta(t) = \Phi'(t) t \quad \text{ and } \quad g(z) = z \frac{f'(z)}{f(z)} = u(z) + i v(z).$$ Then we get $$\left(r \frac{d}{dr}\right)^2 \Phi(|f(z)|) = \eta'(|f(z)|) |f(z)| u(z)^2 + \eta(|f(z)|) r \frac{du}{dr}(z)$$ and $$\left(\frac{d}{d\theta}\right)^2 \Phi(|f(z)|) = \eta'(|f(z)|) |f(z)| v(z)^2 - \eta(|f(z)|) \frac{dv}{d\theta}(z)$$ Now the polar form of the Cauchy-Riemann equations for $g=u+iv$ gives $r \frac{du}{dr} = \frac{dv}{d\theta}$, so $$\begin{split} \left(r \frac{d}{dr}\right)^2 \Phi(|f(z)|) + \left(\frac{d}{d\theta}\right)^2 \Phi(|f(z)|) &= \eta'(|f(z)|) |f(z)| (u(z)^2+v(z)^2) \\ &= \Psi(|f(z)|) |g(z)|^2 \end{split}$$ which is exactly the claim from Pommerenke's book.

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