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Growth and Distortion Theorems

Once a map is in the class S, its size and its stretch are no longer free — they are trapped between two explicit curves. The growth and distortion theorems put hard sharp walls on |f(z)| and on |f'(z)|, and the Koebe function touches every wall.

From one coefficient bound to control everywhere

By now you have earned two prized facts about the normalized class. Guide 1 set the stage: a function f is in the class S when it is univalent (injective and holomorphic) on the unit disk and normalized so that f(0) = 0 and f'(0) = 1, so its Taylor series reads f(z) = z + a_2 z^2 + a_3 z^3 + .... Guide 2 then squeezed real juice from the area theorem, giving the famous Bieberbach bound |a_2| <= 2 and, as its geometric shadow, the Koebe one-quarter theorem: the image f(D) always contains the disk of radius 1/4 about the origin.

Those two results are about a single number (the second coefficient) and a single point (the origin). The growth and distortion theorems are the moment that local control spreads out to cover the whole disk. They answer two very physical questions about any f in S. First: if I walk out to a point z at distance r = |z| from the centre, how far from the origin can f(z) be, and how close to it must it stay? Second: how much can the map stretch or compress lengths there — how big or small can the derivative |f'(z)| be? In both cases the answer is not 'it depends'; it is a pair of explicit, best-possible inequalities.

The distortion theorem: how much the derivative may stretch

Let us state the sharper of the two first, because growth follows from it by integration. The distortion theorem says that for every f in S and every z with |z| = r < 1, the derivative is pinned between two walls that depend only on r: (1 - r) / (1 + r)^3 <= |f'(z)| <= (1 + r) / (1 - r)^3. Read it slowly. The modulus of the derivative is a stretch factor — it tells you how much a tiny circle at z is magnified by f. So the theorem promises that no S-map can stretch lengths by more than (1 + r)/(1 - r)^3 at radius r, nor shrink them below (1 - r)/(1 + r)^3, no matter how clever the map is.

for every f in S and every |z| = r < 1:

  growth:      r / (1 + r)^2  <=  |f(z)|   <=  r / (1 - r)^2

  distortion:  (1 - r)/(1 + r)^3  <=  |f'(z)|  <=  (1 + r)/(1 - r)^3

as r -> 1 the upper walls blow up; the lower walls go to 0.
the Koebe function K(z) = z / (1 - z)^2 hits every wall.
The two theorems side by side: |f(z)| and |f'(z)| are each caged between an explicit lower and upper curve in r.

Here is the engine of the proof, and it is wonderfully reusable. We already know |a_2| <= 2 for every S-function. The trick is that this bound does not just apply to f itself; it applies to a whole family of disguised copies of f. Fix a point z_0 in the disk and pre-compose f with the disk automorphism that slides z_0 to the centre, then re-normalize the result back into S. The new function is still in S, so its own second coefficient also obeys the bound |a_2| <= 2 — but when you unwind what that second coefficient is in terms of f at z_0, the inequality turns into a statement about f'(z_0) and f''(z_0). Out drops a sharp bound on the logarithmic derivative, and integrating it gives the distortion walls.

The growth theorem: how far the image can travel

Now integrate. To find |f(z)| you can walk out along the straight radius from 0 to z and accumulate the stretch |f'| along the way. Feeding the distortion walls into that integral collapses neatly, and the growth theorem falls out: r / (1 + r)^2 <= |f(z)| <= r / (1 - r)^2 for every f in S at |z| = r. The lower wall is the more startling one. It says an S-map can never collapse a point of modulus r too close to the origin — its image stays at distance at least r/(1 + r)^2. Injectivity, again, is forbidding the map from folding the disk in on itself.

And here the one-quarter theorem reappears, for free, as a limit. Let r climb toward 1 in the lower bound r/(1 + r)^2. Its smallest possible value over the whole disk is the limit as r -> 1, which is 1/(1 + 1)^2 = 1/4. So every point of the image has modulus at least 1/4 in the worst case — exactly the radius of the Koebe disk you met last guide. The one-quarter theorem is not a separate miracle; it is the boundary case of the growth theorem's lower wall, read off at the rim of the disk.

  1. Start from the distortion theorem's lower bound |f'(t e^(i theta))| >= (1 - t)/(1 + t)^3 along the radius from 0 to z, where t runs from 0 to r.
  2. Since |f(z)| equals the length of the image of that radius is at least the integral of |f'| along it, integrate the lower bound from t = 0 to t = r.
  3. The elementary integral of (1 - t)/(1 + t)^3 from 0 to r is exactly r/(1 + r)^2 — that is the growth theorem's lower wall.
  4. Let r -> 1: the wall tends to 1/4, so every boundary direction's image stays at distance at least 1/4, recovering the Koebe one-quarter theorem.

The Koebe function presses against every wall

A bound is only as impressive as its sharpness, so the natural worry is: are these walls reachable, or are they generously loose? They are perfectly tight, and one single function touches all of them at once. The Koebe function K(z) = z / (1 - z)^2 is the recurring villain-hero of this whole rung. Its Taylor expansion is z + 2 z^2 + 3 z^3 + 4 z^4 + ..., so a_2 = 2 hits the Bieberbach bound exactly, and in general a_n = n. It maps the disk univalently onto the entire plane with a single slit removed — the ray from -1/4 out to -infinity along the negative real axis — which is why its image just barely fails to cover the point -1/4.

Watch it press the walls. Put z = r on the positive real axis. Then K(r) = r/(1 - r)^2, which is exactly the upper growth wall — so the upper bound on |f(z)| is achieved. Put z = -r instead and K(-r) = -r/(1 + r)^2, whose modulus r/(1 + r)^2 is exactly the lower growth wall. Differentiate to get K'(z) = (1 + z)/(1 - z)^3; at z = r this gives (1 + r)/(1 - r)^3, the upper distortion wall, and at z = -r it gives (1 - r)/(1 + r)^3, the lower one. Every inequality of this guide is touched, with equality, by rotations of this one map. That is what 'sharp' means.

Why this matters, and where the road heads next

Step back and feel the shape of what we have. The class S is an infinite-dimensional space of maps, wildly varied — yet the moment you ask about size or stretch, all that variety is corralled into a thin band between two universal curves, with the Koebe function riding the edge. This is the same flavour of rigidity you first tasted in the Schwarz lemma, where holding the origin fixed forced |f(z)| <= |z|. Geometric function theory is, at heart, the art of converting one normalization (here f(0) = 0, f'(0) = 1) into global, quantitative, best-possible control.

There is a working payoff, too. The growth theorem's upper wall r/(1 - r)^2 gives you an instant, hypothesis-free a priori bound on any univalent map of the disk once you renormalize it into S — useful whenever you build conformal maps numerically and need to know they cannot run off to infinity too fast. The lower walls guarantee a map does not degenerate. These are exactly the compactness estimates that make the class S a well-behaved, closed family, which is the technical bedrock under the existence theorems of the subject.

And the Koebe function leaves us with a tantalizing clue for what comes next. Its coefficients are a_n = n, dead on the integers, and its second coefficient sits exactly at the Bieberbach bound a_2 = 2. That coincidence is no accident: it whispers the boldest guess in the whole field — that |a_n| <= n for every f in S, with the Koebe function as the lone extremal. That guess is the Bieberbach conjecture, which stood open for sixty-nine years and is the subject of the next guide, where de Branges's theorem and Loewner chains finally close it.