Principle of subordination

First we recall the statement of Schwarz’s lemma, which is a basic result in complex analysis. Let ${\Bbb D}$ be the unit disk in ${\Bbb C}$.

Schwarz’s lemma. Let $f \in H({\Bbb D})$. If $|f(z)| \leq 1$ for all $z \in {\Bbb D}$ and $f(0) = 0$, then $|f'(0)| \leq 1$, and $|f(z)| \leq |z|$ for all $z \in {\Bbb D}$. If $|f'(0)| = 1$ or $|f(z)| = |z|$ for some $z \in {\Bbb D}$, then $f(z) \equiv cz$ for some $c$ with $|c| = 1$.

To prove this, consider the holomorphic function $g(z) = \frac{f(z)}{z}$ and apply the maximum principle. (See any textbook for details.) Note that $|f'(0)|$ is maximized if and only if $f$ is a conformal self-map of ${\Bbb D}$ (with the normalization $f(0) = 0$). If we do not specify the value of $f(0)$, we get Pick’s lemma.

Pick’s lemma. Suppose $f: {\Bbb D} \rightarrow {\Bbb D}$ is holomorphic. Then

$|f'(z)| \leq \frac{1 - |f(z)|^2}{1 - |z|^2}$

for any $z \in {\Bbb D}$. Equality holds for some $z \in {\Bbb D}$ if and only if $f$ is a conformal self-map of ${\Bbb D}$ (and in that case equality holds everywhere).

Pick’s lemma leads naturally to the hyperbolic metric on ${\Bbb D}$.

Schwarz’s lemma has the following useful Corollary called the Principle of Subordination. It makes clear the role of conformal maps in Schwarz’s lemma. First we need a notation. If $f$ is holomorphic (or merely continuous) on ${\Bbb D}$ and $0 < r < 1$, we define

$M(r, f) = \max_{|z| = r} |f(z)|$.

Principle of Subordination. Let $f, g \in H({\Bbb D})$. Suppose that $f(0) = g(0)$, $g$ is one-to-one, and $f({\Bbb D}) \subset g({\Bbb D})$, then $|f'(0)| \leq |g'(0)|$ and $M(r, f) \leq M(r, g)$ for all $0 < r < 1$.

Proof. Since $g$ is one-to-one, $g$ is in fact a conformal map from ${\Bbb D}$ to $g({\Bbb D})$. Let $h = g^{-1}$ be the inverse of $g$. Then $h \circ f$ is holomorphic and maps ${\Bbb D}$ into ${\Bbb D}$ with $(h \circ f)(0) = 0$ since $f(0) = g(0)$. By Schwarz’s lemma, we have

$|(h \circ f)'(0)| \leq 1$ and $|(h \circ f)(z)| \leq |z|$

for all $z \in {\Bbb D}$. By the chain rule, $(h \circ f)'(0) = \frac{f'(0)}{g'(0)}$. So the first inequality gives $|f'(0)| \leq |g'(0)|$.

Fix $0 < r < 1$. The second inequality implies that

$f(\partial {\Bbb D}(0, r)) \subset g(\overline{{\Bbb D}}(0, r)$.

It follows that

$M(r, f) = \max_{|z| = r} |f(z)| \leq \max_{|z| \leq r} |g(z)| = \max_{|z| = r} |g(z)| = M(r, g)$.

Note that the second last equality follows from the maximum principle. $\Box$

As an exercise, formulate the condition for equality. Here we give a simple example.

Example. Let $R = \{z: Re(z) > 0\}$ be the right half plane. Let $f: {\Bbb D} \rightarrow R$ be holomorphic. We claim that $|f'(0)| \leq 2 Re(f(0))$.

Proof. Write $f(0) = a + bi$, where $a = Re(f(0)) > 0$. Define $g(z) = a \frac{z + 1}{z - 1} + bi$. Then $g$ is a conformal map from ${\Bbb D}$ to $R$ with $f(0) = g(0)$. Clearly $f({\Bbb D}) \subset R = g({\Bbb D})$. By the principle of subordination, we have $|f'(0)| \leq |g'(0)|$. By direct calculation, $g'(0) = 2 Re(f(0))$ and the result follows. $\Box$

Using this estimate, we can prove the following interesting result.

Example (a prelim problem). Let $f: {\Bbb D} \rightarrow {\Bbb D} \setminus \{0\}$ be holomorphic. Then $|f'(0)| \leq \frac{2}{e}$.

Proof. Since $f$ omits $0$, there exists $g \in H({\Bbb D})$ such that $f = e^g$. Note that

$|f(z)| = e^{Re(g(z))}$ and $f'(z) = f(z)g'(z)$.

Since $e^{Re(g(z))} = |f(z)| < 1$, we have $Re(g(z)) < 0$. It follows that $z \mapsto -g(z)$ maps ${\Bbb D}$ into $R$. Let $x = -Re(g(0)) > 0$. By the above result, we have

$|g'(0)| \leq 2x$.

It follows that

$|f'(0)| \leq |f(0)||g'(0)| \leq 2xe^{-x}$.

Maximizing the last expression over $x > 0$, we have

$|f'(0)| \leq \frac{2}{e}$. $\Box$

More examples will be added later.

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2 Responses to Principle of subordination

1. KKK says:

Nice post. Admittedly I have never thought about hyperbolic metric this way ( I am making my remark here for my own benefit). Let me try to figure out what you mean by the Pick’s lemma gives rise to the hyperbolic metric. Perhaps this is somewhat trivial and is well-known to many of you.

In most textbooks, the hyperbolic metric $g$ is given first and then it is deduced that the isometry group is exactly the group of all conformal self maps of the disk i.e. disk-preserving holomorphic and antiholomorphic functions. Suppose we want to reverse this process and want to find a (hyperbolic) metric which is preserved by the conformal self maps. We observe from the equality case of the Pick’s lemma that for $w=w(z)$ to be a conformal self map,

$|\frac{dw}{dz}|= \frac{1-|w|^2}{1-|z|^2}.$
i.e. $\frac{|dw|}{1-|w|^2}= \frac{|dz|}{1-|z|^2}.$

Thus we find that the hyperbolic metric is given by (up to a positive factor)

$g= \frac{|dz|^2}{(1-|z|^2)^2}= \frac{dx^2+dy^2}{(1-x^2-y^2)^2}.$

Interestingly, the other form of Pick’s lemma is given by $|\phi(f(\alpha),f(w))|\leq |\phi(\alpha,w)|$ where $\phi(\alpha,z)=\phi_{\alpha}(z)=\frac{z-\alpha}{1-\overline \alpha z}$, with equality holds if and only if it is a conformal self map. (This can be seen by $|\phi_\beta\circ f\circ \phi_{-\alpha} (z)|\leq|z|$, where $\beta=f(\alpha)$ and putting $z=\phi_\alpha (w)$. ) Writing it out, it is

$|\frac{f(w)-f(z)}{1-\overline{f(w)} f(z)} |\leq |\frac{w-z}{1-\overline w z}|,$

with equality hold iff $f$ is a conformal self map. It suggests (without integrating) that the hyperbolic distance is actually given by (up a scaling)
$d(z,w)=|\frac{z-w}{1-\overline w z}|$. HOWEVER, it is different from the usual definition $d(z,w)= \tanh^{-1}(|\frac{z-w}{1-\overline w z}|)$, I can’t think if any way to “see” (except integrating) that the correct definition is the later one instead of the former one (actually I haven’t checked if the former one really satisfies the triangle inequality). Can anyone help?

2. The former one really satisfies the triangle inequality and we call it the pseudo-hyperbolic distance. The pseudo-hyperbolic distance is not additive along hyperbolic geodesics.