Comments on complex analysis and de Branges' theorem

The undergraduate course in complex analysis that I took had the regrettable feature of being contaminated with the applied mathematician's addiction to computing contour integrals so much so that it put off quite a lot of good students from ever touching the subject again. I think computing the occasional integral can be amusing, but it is a shame when the beauty of the underlying subject is obscured or interesting results are omitted because so much time is devoted to making students compete with Wolfram Alpha. As with the previous post on representation theory, I would like to record here a number of results that I would have liked to have been lectured. Arguably these are less applicable than the techniques I was taught, but said techniques have left no lasting impression on me and I feel that I could easily have learned those as needed  (which I have so far not).

Let's start with the most puzzling omission: the Gauss-Lucas theorem, which states that if $P$ is a non-constant polynomial with complex coefficients, all zeros of $P'$ belong to the convex hull of the set of zeros of $P$. This is very doable as an exercise for an undergraduate and is in my opinion good practice in manipulating complex numbers/polynomials.

The Gauss-Lucas theorem, pretty as it is, isn't just a curiosity. It has been used time and again to prove interesting results. Here are two consequences that can be found on Tao's blog:

Grace-Heawood Theorem: Let $f: {\mathbb{C}} \rightarrow {\mathbb{C}}$ be a polynomial of degree $n \geq 1$ such that ${f(1)=f(-1)}$. Then ${f'}$ contains a zero in the closure of ${D( 0, \cot \frac{\pi}{n} )}$.

Perpendicular bisector theorem:  Let $f: {\mathbb{C}} \rightarrow {\mathbb{C}}$ be a polynomial such that ${f(z_1)=f(z_2)}$ for some distinct ${z_1,z_2}$. Then the zeroes of ${f'}$ cannot all lie on one side of the perpendicular bisector of ${z_1,z_2}$. 

(Tao's blog post discusses other theorems which really ought to be more widely known. Also while browsing Tao's blog I found here the following problem which, though very appealing, I could at least understand why it wouldn't be appropriate for an undergraduate course: 

Let ${a_0, a_1, \dots}$ be a bounded sequence of real numbers, and suppose that the power series \[ f(x) := \sum_{n=0}^\infty a_n\frac{x^n}{n!} \] (which has an infinite radius of convergence) decays like ${O(e^{-x})}$ as ${x \rightarrow +\infty}$, in the sense that the function ${e^x f(x)}$ remains bounded as ${x \rightarrow +\infty}$. Must the sequence ${a_n}$ be of the form ${a_n = C (-1)^n}$ for some constant ${C}$?)

A variant of the Gauss-Lucas theorem is also used in this MO post where an ingenious answer to a generalisation of a Putnam problem is given.

de Branges' theorem

A bit more ambitiously, a course could have in theory covered the proof of the Bieberbach conjecture, now de Branges' theorem. 

Recall that a holomorphic function on an open subset of the complex plane is called univalent if it is injective, and a function defined on the open unit disk which is holomorphic and univalent with Taylor series of the form ${\displaystyle f(z)=z+\sum _{n\geq 2}a_{n}z^{n}}$ is called schlicht (German for simple or plain). The theorem states that for a schlicht function as above, $|a_n|\leq n$ for all $n \geq 2$. 

Since $a_n=n$ defines the schlicht function $k(z):= \frac{z}{(1-z)^2}$ known as the Koebe function, this theorem is best possible. For a long time I thought this was a very difficult theorem, and in some sense it still is since it was open for so many decades and the original proof is nearly 400 pages long. It is therefore all the more surprising and satisfying that relatively short and elementary proofs of this have been found, one of which I will now attempt to give an account of following this pdf, which also contains a lot of the history of the problem and outlines of alternative approaches.

The attempted proofs of the Bieberbach conjecture led to a series of reductions to what was known as the Milin conjecture. I chose to follow the chronological pathway, mainly because it doesn't seem very clear to me what the pedagogically most instructive order is. In what follows we denote by $S$ the set of all univalent (i.e. one-to-one) analytic functions $f$ defined on the disk $|z| < 1$, with $f(0) = 0$ and $f'(0) = 1$

The Robertson Conjecture

The Robertson conjecture: Let $p(z) = z + c_3 z^3 + c_5 z^5 + \cdots \in S$ be odd. Then (letting $c_1 = 1$), we have $|c_1|^2 + |c_3|^2 + \cdots + |c_{2n-1}|^2 \leq n$, for $n = 2,3,4,\ldots$. Furthermore, equality occurs for any one $n$ only when $p$ satisfies $p(z)^2 = r(z^2)$, where $r$ is a rotation of the Koebe function, i.e. for some $\alpha$ such that $|\alpha|=1$ we have that $r(z)=\alpha k(\alpha^{-1}z)$.

Lemma 1: For each $n = 2,3,4,\ldots$, the Robertson conjecture for $n$ implies the Bieberbach conjecture for $n$.

Proof: Assume that the Robertson conjecture holds for $n$, and let

\[f(z) = z + a_2 z^2 + a_3 z^3 + \cdots \in S.\]

Let $p(z)$ be the unique odd function analytic in $|z| < 1$ such that

\[p(z) = z + c_3 z^3 + c_5 z^5 + \cdots,\]

and such that

\[p(z)^2 = f(z^2).\]

We may write

\[p(z) = \sqrt{f(z^2)}\]

provided we choose the appropriate branch of the square root for each $z$. Now, $p$ is univalent, for if $p(z_1) = p(z_2)$, then $f(z_1^2) = f(z_2^2)$, so since $f$ is univalent, $z_1 = \pm z_2$. But then the oddness of $p$ implies $z_1 = z_2$ (if $p(z_1) = p(z_2) = 0$, then $z_1 = z_2 = 0$ by the univalence of $f$). Hence, $p$ is an odd function in $S$, and the Robertson conjecture applies.

Since

\[f(z^2) = p(z)^2,\]

comparing coefficients shows

\[a_n = c_1 c_{2n-1} + c_3 c_{2n-3} + \cdots + c_{2n-1} c_1,\].

 Letting

 \[\mathbf{v} := (|c_1|, |c_3|, \ldots, |c_{2n-1}|),\] and

\[\mathbf{w} := (|c_{2n-1}|, |c_{2n-3}|, \ldots, |c_1|)\], we get

\[|a_n| \leq |c_1| |c_{2n-1}| + |c_3| |c_{2n-3}| + \cdots + |c_{2n-1}| |c_1| = \mathbf{v} \cdot \mathbf{w}.\]

and applying the Cauchy-Schwarz inequality to the final expression shows that

\[|a_n| \leq \|\mathbf{v}\| \|\mathbf{w}\| = \|\mathbf{v}\|^2 = |c_1|^2 + |c_3|^2 + \cdots + |c_{2n-1}|^2.\]

The Robertson conjecture thus implies that $|a_n| \leq n$. Furthermore, if $|a_n| = n$, then $|c_1|^2 + |c_3|^2 + \cdots + |c_{2n-1}|^2 = n$, so the Robertson conjecture implies that $p(z)^2 = r(z^2)$, where $r$ is a rotation of the Koebe function. But then $f(z^2) = r(z^2)$, so comparing power series shows $f(z) = r(z)$, and $f$ is a rotation of the Koebe function. $\blacksquare$

The Milin conjecture

Definition: Given a function $f \in S$, we define its logarithmic coefficients $\{\gamma_n\}$ by

\[\log \frac{f(z)}{z} = 2 \sum_{n=1}^\infty \gamma_n z^n, \quad |z| < 1.\]

(Note that, letting $g(z) = \frac{f(z)}{z}$, we have

\[g(z) = 1 + a_2 z + a_3 z^2 + \cdots,\]

so $g(0) = 1 \neq 0$. Furthermore, $g(z)$ is not zero elsewhere in $|z| < 1$ by the univalence of $f$. Hence, we may formally define $\log \frac{f(z)}{z}$ as the integral from $0$ to $z$ of $g'(z)/g(z)$, which is single-valued. The integral is independent of path since the disk $|z| < 1$ is simply connected. Thus $\log \frac{f(z)}{z}$ is analytic in $|z| < 1$.)

The Milin conjecture: For each function $f \in S$, its logarithmic coefficients satisfy

\[\sum_{m=1}^n \sum_{k=1}^m \left(k|\gamma_k|^2 - \frac{1}{k}\right) \leq 0,\]

or equivalently

\[\sum_{k=1}^n (n - k + 1) \left(k|\gamma_k|^2 - \frac{1}{k}\right) \leq 0,\]

for $n=1,2,3,\ldots$. Furthermore, we have equality for any one $n$ only when $f$ is a rotation of the Koebe function.

We will show that the Milin conjecture implies the Robertson conjecture using the following:

Lebedev-Milin Inequality: Let

\[\varphi(z) = \alpha_1 z + \alpha_2 z^2 + \alpha_3 z^3 + \cdots\]

be any complex power series with radius of convergence $R > 0$. Write

\[e^{\varphi(z)} = 1 + \beta_1 z + \beta_2 z^2 + \cdots, \quad |z| < R.\]

Then for $n = 1,2,3,\ldots$, we have (letting $\beta_0 = 1$)

\[\frac{1}{n+1} \sum_{k=0}^n |\beta_k|^2 \leq \exp\left(\frac{1}{n+1} \sum_{m=1}^n \sum_{k=1}^m \left(k|\alpha_k|^2 - \frac{1}{k}\right)\right).\]

This can be proved by completely elementary means and has been useful in many other areas since it doesn't require a univalent hypothesis. For a short and elementary proof of a generalisation the reader is referred to this paper.

Lemma 2: For each $n=1,2,3,\ldots$, the Milin conjecture for $n$ implies the Robertson conjecture for $n+1$.

Proof: Assume the Milin conjecture holds for $n$, and let $h \in S$ be odd. Then $h(z)^2$ is even, so

\[h(z)^2 = f(z^2) \]

for some function $f$ analytic in $|z| < 1$. Furthermore, $f$ is univalent, for if $f(z_1) = f(z_2)$, then choosing $\zeta_1, \zeta_2$ with $\zeta_1^2 = z_1$ and $\zeta_2^2 = z_2$, we have $h(\zeta_1)^2 = h(\zeta_2)^2$. Then the univalence and oddness of $h$ implies $\zeta_1 = \pm \zeta_2$, so $z_1 = z_2$. Hence, $f \in S$. Let $\{\gamma_n\}$ be its logarithmic coefficients, so that

\[\log \frac{f(z)}{z} = 2 \sum_{n=1}^\infty \gamma_n z^n.\]

Write

\[h(z) = z + c_3 z^3 + c_5 z^5 + \cdots\]

Then

\[\frac{h(\sqrt{z})}{\sqrt{z}} = 1 + c_3 z + c_5 z^2 + \cdots = \sum_{n=0}^\infty c_{2n+1} z^n \quad (\text{where } c_1 = 1).\]

Now,

\[\log \frac{h(\sqrt{z})}{\sqrt{z}} = \log \sqrt{\frac{f(z)}{z}} = \frac{1}{2} \log \frac{f(z)}{z} = \sum_{n=1}^\infty \gamma_n z^n.\]

Hence,

\[\sum_{n=0}^\infty c_{2n+1} z^n = \frac{h(\sqrt{z})}{\sqrt{z}} = \exp\left(\sum_{n=1}^\infty \gamma_n z^n\right). \]

The Lebedev-Milin Inequality then says that

\[\frac{1}{n+1} \sum_{k=0}^n |c_{2k+1}|^2 \leq \exp\left(\frac{1}{n+1} \sum_{m=1}^n \sum_{k=1}^m \left(k|\gamma_k|^2 - \frac{1}{k}\right)\right),\]

and Milin's conjecture then implies that

\[\frac{1}{n+1} \sum_{k=0}^n |c_{2k+1}|^2 \leq 1,\]

which is the inequality of Robertson's conjecture for $n+1$. Furthermore, if

\[\frac{1}{n+1} \sum_{k=0}^n |c_{2k+1}|^2 = 1,\]

then we must have

\[\sum_{m=1}^n \sum_{k=1}^m \left(k|\gamma_k|^2 - \frac{1}{k}\right) = 0.\]

The Milin conjecture then implies that $f$ is a rotation of the Koebe function, so that

\[h(z)^2 = f(z^2),\]

with $f$ a rotation of the Koebe function as required. $\blacksquare$

The proof of the Milin conjecture

Definition: Given a function $F$ in one complex variable $z$, a Loewner chain for $F$ is a family of functions $\{f(z,t) \mid t \geq 0\}$ with

\[f(z,t) = e^t z + \sum_{n=2}^{\infty} a_n(t) z^n, \quad (z \in \mathbb{D}, t \geq 0, a_n(t) \in \mathbb{C} \ (n \geq 2)),\]

which start with $f(z,0) = F(z)$, and for which the relation

\[\operatorname{Re}\left(\frac{\dot{f}(z,t)}{z f'(z,t)}\right) > 0 \quad (z \in \mathbb{D}) \tag{1}\]

is satisfied. Here $f'$ and $\dot{f}$ denote the partial derivatives w.r.t $z$ and $t$, respectively.

The above equation is referred to as the Loewner differential equation. It geometrically states that---as a result of Schwarz' lemma, again---the image domains of $f(\mathbb{D},t)$ expand as $t$ increases. Hence Loewner investigated the dynamics of univalent functions. The subset of $S$ that he considered consists of those functions whose image domains are all of $\mathbb{C}$ besides a slit which is a Jordan arc. In this case, the Loewner chain consists of those functions whose slit gets shorter as $t$ increases.

The Loewner differential equation characterizes univalent functions, since it turns out that for all $f \in S$ there is a Loewner chain. Furthermore, one can prove that under rather general conditions for the function

\[p(z, t) := \frac{\dot{f}(z,t)}{z f'(z,t)}, \tag{2}\]

the univalence of $f(z,0)$ implies the univalence of $f(z,t)$ ($t > 0$).

We will need the following representation of the $t$-derivatives of the logarithmic coefficients $d_{k}(t)$ of the Loewner chain ($\zeta:=re^{i\theta}$, $r\in(0,1)$):

\begin{align*}d_{k}(t) &\stackrel{\text{(Cauchy formula)}}{=} \frac{1}{2\pi} \int\limits_{0}^{2\pi}\frac{f(\zeta,t)\,d\theta}{f(\zeta,t)\,\zeta^{k}} \\&\stackrel{(2)}{=} \frac{1}{2\pi}\int\limits_{0}^{2\pi}p(\zeta,t)\frac{\zeta f'(\zeta,t)\,d\theta}{f(\zeta,t)\,\,\,\zeta^{k}} \\&= \frac{1}{2\pi}\int\limits_{0}^{2\pi}p(\zeta,t)\left(1+\sum\limits_{j=1}^{\infty}jd_{j}(t)\zeta^{j}\right)\frac{d\theta}{\zeta^{k}} \\&= \lim\limits_{r\to 1}\frac{1}{2\pi}\int\limits_{0}^{2\pi}p(\zeta,t) \left(1+\sum\limits_{j=1}^{\infty}jd_{j}(t)\zeta^{j}\right)\frac{d\theta}{\zeta^{k}}. \tag{3}\end{align*}

Let now $z\in\mathbb{D}$ and set $\varphi(z,t):=K^{-1}\Bigl{(}e^{-t}K(z)\Bigr{)}$, hence

\[\frac{z}{(1-z)^{2}}=e^{t}\frac{\varphi}{(1-\varphi)^{2}}\quad(t\geq 0). \tag{4}\]

The mapping $\varphi:\mathbb{D}\times\mathbb{R}_{\geq 0}\to\mathbb{D}$ is a special Loewner family of bounded univalent functions whose image domains are the unit disk with a radial slit on the negative real axis that increases with $t$. Taking derivative in the defining equation (4) leads to the relation

\[\dot{\varphi}=-\varphi\frac{1-\varphi}{1+\varphi}. \tag{5}\]

Consider a generating function of the Milin expression $\sum_{k=1}^n (n - k + 1) \left(k|\gamma_k|^2 - \frac{1}{k}\right) \leq 0$. Because of the compactness of $S$ this generating function converges absolutely and locally uniformly in $\mathbb{D}$, and changing the order of summation yields

\[\omega(z):=\sum\limits_{n=1}^{\infty}\left(\sum\limits_{k=1}^{n}(n+1-k)\left(\frac{4}{k}-k\,|d_{k}(0)|^{2}\right)\right)z^{n+1}=\frac{z}{(1-z)^{2}}\sum\limits_{k=1}^{\infty}\left(\frac{4}{k}-k\,|d_{k}(0)|^{2}\right)z^{k}.\]

Note that this procedure automatically generates the Koebe function as a factor! Using (4) and the fundamental theorem of calculus implies with $\varphi(z,\infty)=0$ and $\varphi(z,0)=z$

\[\omega(z)=\int\limits_{0}^{\infty}-\frac{e^{t}\varphi}{(1-\varphi)^{2}}\frac{d}{dt}\left(\sum\limits_{k=1}^{\infty}\left(\frac{4}{k}-k\left|d_{k}(t)\right|^{2}\right)\varphi^{k}\right)dt.\]

Using the relations (3) and (5) gives after some further computations ($\zeta:=re^{i\theta}$, $r\in(0,1)$)

\[\omega(z)=\int\limits_{0}^{\infty}\frac{e^{t}\varphi}{1-\varphi^{2}}\sum\limits_{k=1}^{\infty}A_{k}(t)\varphi^{k}dt,\]

where \[A_{k}(t):=\lim\limits_{r\to 1}\frac{1}{2\pi}\int\limits_{0}^{2\pi}\operatorname{Re}p(\zeta,t)\;\left|1+2\sum\limits_{j=1}^{k}jd_{j}(t)\zeta^{j}-kd_{k}(t)\zeta^{k}\right|^{2}\;d\theta.\]

If we now write

\[W_{k}(z,t)=\frac{e^{t}\varphi^{k+1}}{1-\varphi^{2}}=:\sum\limits_{n=k}^{\infty}\Lambda_{k}^{n}(t)z^{n+1}, \]

then finally

\[\omega(z)=\sum\limits_{n=1}^{\infty}\left(\int\limits_{0}^{2\pi}\sum\limits_{k=1}^{\infty}\Lambda_{k}^{n}(t)A_{k}(t)dt\right)z^{n+1}. \]

Remember that the Milin conjecture is equivalent to the fact that $\omega$ has non-negative coefficients. On the other hand Loewner's equation (2), $\operatorname{Re}p(\zeta,t)>0$, yields

\[A_{k}(t)\geq 0\qquad(t\geq 0),\]

so that the Milin conjecture follows if

\[\Lambda_{k}^{n}(t)\geq 0\qquad(t\geq 0,\quad k,n\in\mathbb{N}). \tag{6}\]

This, again, is a positivity statement of a system of special real functions. These functions, however, are intimately related with the Koebe function and therefore with geometric function theory.

The range of $\varphi=K^{-1}(e^{-t}K)$ is the unit disk with a slit on the real axis. Since the mapping ($\gamma\in\mathbb{R}$)

\[h_{\gamma}(z):=\frac{z}{1-2\cos\gamma\cdot z+z^{2}}\]

maps the unit disk for $\gamma\neq 0\pmod{\pi}$ onto a domain which is slit on the real axis twice, we can also interpret $\varphi$ as the composition $\varphi=h_{\gamma}^{-1}(e^{-t}h_{\gamma})$ for a suitable pair $(\theta,\gamma)$, and a computation shows the relation

\[\cos\gamma=(1-e^{-t})+e^{-t}\cos\theta. \tag{7}\]

We therefore get

\begin{align*} h_{\gamma}(z) &= e^{t}\cdot h_{\theta}(\varphi)=\frac{e^{t}\varphi}{1-\varphi^{2}}\left(\frac{1-\varphi^{2}}{1-2\cos\theta\cdot\varphi+\varphi^{2}}\right) \\ &= \frac{e^{t}\varphi}{1-\varphi^{2}}\operatorname{Re}\frac{1+e^{i\theta}\varphi}{1-e^{i\theta}\varphi}=\frac{e^{t}\varphi}{1-\varphi^{2}}\left(1+2\sum\limits_{k=1}^{\infty}\varphi^{k}\cos k\theta\right) \\&= \frac{e^{t}\varphi}{1-\varphi^{2}}+2\sum\limits_{k=1}^{\infty}\left(\sum\limits_{n=1}^{\infty}\Lambda_{k}^{n}(t)z^{n+1}\right)\cos k\theta. \tag{8} \end{align*}

From here there are two ways to conclude. 

Approach 1: The addition theorem

The first uses the following: 

Spherical harmonic addition theorem/addition theorem of the Legendre polynomials: Let $P_\ell(x)=\frac{1}{2^\ell\ell!}\frac{\partial^\ell}{\partial x^\ell}(x^2-1)^\ell$ be the Legendre polynomials, and for $m>0$ let

$P_\ell^m(x)=(-1)^m(x^2-1)^{m/2}\frac{\partial^m}{\partial x^m}P_\ell(x)$ be the associated Legendre polynomial.

Define the spherical harmonics $Y_{\ell m}(\theta, \phi)=\sqrt{\frac{2\ell+1}{4\pi}\frac{(\ell+m)!}{(\ell-m)!}}P_\ell^m(\cos\theta)e^{im\phi}$ and for real numbers $\theta, \theta', \phi, \phi'$ define $\gamma$ be such that the following holds

$\cos\gamma=\cos\theta\cos\theta'+\sin\theta\sin\theta'\cos(\phi'-\phi)$.

Then we have $P_\ell(\cos\gamma)=\frac{4\pi}{2\ell+1}\sum_{m=-\ell}^\ell Y_{\ell m}^{*}(\theta', \phi')Y_{\ell m}(\theta, \phi)$

I don't really know a good proof of this, but I will keep looking and blog about it if/when I find one. For now we take this as a blackbox.

Since

\[\eta(z) := h_\gamma(z) = \frac{1}{z} \left(1 - 2\cos\gamma \cdot z + z^2\right)\]

is the generating function of the Legendre polynomials, it follows from (7) and the addition theorem of the Legendre polynomials, an explicit representation of the form

\[\eta(z) = \sum_{n=0}^\infty \nu_n^2 z^n + 2 \sum_{k=1}^\infty \sum_{n=k}^\infty \frac{(n-k)!}{(n+k)!} \mu_{kn}^2 z^n \cos k\theta \quad (\nu_n, \mu_{kn} \in \mathbb{R} \ (k, n \in \mathbb{N}))\]

with obviously non-negative coefficients. After squaring this result, it follows from (8) that (6) is fulfilled, and therefore the Milin conjecture is true.

Approach 2: WZ method

The second approach goes via the previously discussed Wilf-Zeilberger algorithm, which doesn't require the messy addition theorem but instead requires a bit of familiarity with computers to find.

The coefficients of the function $\eta(z)$ given by

\[\frac{1}{1 - z(2x^2 + (1-z^2)(w+1/w)) + z^2} = \sum_{n=0}^\infty \sum_{k=0}^n \frac{(n-k)!}{(n+k)!} (1 - x^2)^k C_{k,n}(x) (w^k + w^{-k}) z^n\]

are polynomials $C_{k,n}(x) \in \mathbb{Z}[x]$ with rational coefficients. Again, it will suffice to prove that the coefficients are non-negative, and we will again try to show that they form the squares of another system of polynomials

\[D_{k,n}(x)^2 = C_{k,n}(x) \tag{9}\]

 To do this, one should calculate the first polynomials $D_{k,n}(x)$ for $0 \leq k \leq n \leq 20$, and then "guess" a holonomic recurrence equation w.r.t. $n$, i.e., a homogeneous linear differential equation with polynomial coefficients $\sigma_j \in \mathbb{Z}[k, n]$,

\[\sigma_2 D_{k,n+2}(x) + \sigma_1 D_{k,n+1}(x) + \sigma_0 D_{k,n}(x) = 0,\]

and use linear algebra to find $\sigma_0, \sigma_1, \sigma_2$. It turns out that

\[(n - k + 2)D_{k,n+2}(x) - (2n + 3)x D_{k,n+1}(x) + (n + k + 1)D_{k,n}(x) = 0 \]

and only the initial values for $0 \leq k \leq n \leq 6$ were necessary to find this equation.

Assume $E_{k,n}(x)$ is the solution of (30) with the appropriate initial values. Then by another application of linear algebra this recurrence equation can be "squared", i.e., it is possible to calculate the recurrence equation $R$ of third order valid for $E_{k,n}(x)^2$. Then by the WZ-method one can show that $R$ is valid for $C_{k,n}(x)$. This gives an a posteriori proof of the guessed (9).