Starlike, Convex, and the Schwarzian

From bounding maps to building them

Everything in this rung so far has been a one-way street: you hand me a function that you already know lives in the class S, and I tell you what it cannot do — its second coefficient cannot exceed 2, its image must cover the disk of radius 1/4, every |a_n| obeys |a_n| <= n by de Branges. Beautiful, but notice the awkward catch hiding inside every one of those statements. They all begin 'let f be univalent'. They presume the very thing — injectivity — that is the hardest property in the whole subject to actually check. A univalent function is global and subtle; you cannot read injectivity off a power series by inspection.

This last guide closes the loop by going the other way. Suppose you only have a formula, and you want a certificate that it is univalent — without ever testing f(z_1) = f(z_2) by hand over an infinite domain. Geometric function theory offers two very different kinds of certificate. The first is geometric: if the image is a shape so well-behaved that overlap is visibly impossible — a star you can see across, or a region with no dents — then injectivity follows, and the shape is detectable by a single inequality on f. The second is analytic and almost magical: the Schwarzian derivative, a number attached to f, whose smallness forces univalence outright. Two roads, both ending at the word you have chased all rung: injective.

Starlike: see every point along a ray

Picture standing at the origin inside the image region. The region is starlike (with respect to 0) if, from that one vantage point, you can see every other point of the region along a straight line that never leaves it — exactly like standing at the heart of a star and looking out along each spike. Formally, for every w in the image, the whole segment from 0 to w stays inside. A disk centered at 0 is starlike; so is the Koebe slit image, since the only missing direction is a single ray and you can still see along every other. A crescent is not — part of it is hidden behind the bite.

Here is the lovely part: that visual condition has an exact algebraic translation. A starlike function is an f in S whose image is starlike about 0, and the test is one inequality, Re( z f'(z) / f(z) ) > 0 for all z in the disk. The intuition is direct. As you march z counterclockwise around a circle |z| = r, the quantity z f'/f measures how fast the argument of the image point f(z) turns as seen from the origin. Demanding its real part stay positive says that turning never reverses — arg f(z) climbs steadily and never backtracks. A point seen from the origin whose direction only ever rotates forward is a point you can always reach along an unobstructed ray. That monotone winding of the argument is starlikeness.

Convex: no dents, and a smaller bound

Tighten the shape one notch further. A region is convex if, whenever you pick any two points inside it — not just the origin and one other — the straight segment joining them stays entirely inside. No dents, no notches, no missing bites. A disk, a half-plane, the inside of an ellipse: convex. A star or a crescent: not. A convex univalent function is an f in S whose image is a convex set, and once more there is a clean test: Re( 1 + z f''(z) / f'(z) ) > 0 for all z in the disk. The quantity 1 + z f''/f' tracks how the tangent direction along the image boundary turns; keeping its real part positive says the boundary always bends the same way and never reverses its curving — which is exactly what 'no dents' means.

Two tidy facts make this family feel inevitable. First, a hierarchy: every convex map is starlike, every starlike map is close-to-convex, and all three are univalent. (Convex needs every chord inside; starlike needs only the chords from 0; so convex is the stronger demand and sits on top.) Second, Alexander's theorem links the two tests as derivatives of one another: f is convex if and only if z f'(z) is starlike. So the convex world is just the starlike world differentiated, a satisfying piece of bookkeeping that lets you trade one positivity condition for the other.

And here is the payoff that ties this guide back to the whole rung. Because convex maps are so heavily constrained, their coefficients are far smaller than the general class allows: for a convex f, |a_n| <= 1 for every n — not |a_n| <= n. The extremal is the half-plane map f(z) = z/(1 - z) = z + z^2 + z^3 + ... , whose coefficients are all exactly 1, mapping the disk onto a half-plane. Compare that to the Koebe function z/(1 - z)^2 with a_n = n, the extremal of the unconstrained class. The convex bound 1 sits far below the universal bound n — geometry that is more rigid pays you back with sharper arithmetic.

Lay the three classes side by side and the pattern is clean. A convex map passes Re( 1 + z f''/f' ) > 0 and obeys |a_n| <= 1, with the half-plane map z/(1 - z) as extremal. A starlike map passes the looser Re( z f'/f ) > 0 and obeys |a_n| <= n, with the Koebe slit map z/(1 - z)^2 extremal. The full class S only knows univalence and meets the same |a_n| <= n with the very same Koebe extremal. The hierarchy convex implies starlike implies close-to-convex implies univalent runs from tightest geometry and smallest bound at the top down to the widest, most permissive class at the bottom — each step out loosens the shape and, where it can, loosens the arithmetic.

The Schwarzian: distance from a Mobius map

The geometric tests are wonderful when the image happens to be a nice shape, but most univalent maps are neither starlike nor convex. For a universal certificate we need something blind to the trivial maps and sensitive only to genuine bending. That something is the Schwarzian derivative. Recall from the conformal rung that a Mobius transformation (az + b)/(cz + d) is the simplest possible non-trivial holomorphic map — it preserves circles and angles and is determined by three points. The Schwarzian is engineered to ignore exactly these maps: it vanishes precisely on Mobius transformations, so it measures how far f departs from being one.

  Schwarzian derivative of f  (where f' is nonzero):

        Sf  =  (f''/f')'  -  (1/2) (f''/f')^2
            =  f'''/f'  -  (3/2) (f''/f')^2

  key properties:
    Sf = 0           exactly when f is Mobius, (a z + b)/(c z + d)
    S(M of f) = Sf   post-composing with any Mobius M leaves Sf unchanged

That second property — invariance under post-composition with a Mobius map, S(M of f) = Sf — is the deep one. It says the Schwarzian has already quotiented out the Mobius freedom, the same freedom we quotiented out by normalizing f(0) = 0 and f'(0) = 1 back in guide 1. So the Schwarzian sees only the part of f that the Schwarz lemma and the normalizations could not legislate away: the intrinsic, higher-order bending. This is exactly the kind of quantity that ought to detect univalence, because univalence too is invariant under following f by an injective Mobius map.

Nehari's criterion: one inequality certifies injectivity

Now the punchline that makes the Schwarzian indispensable. Nehari's criterion says: if f is holomorphic on the unit disk and its Schwarzian stays small enough — precisely |Sf(z)| <= 2/(1 - |z|^2)^2 for every z in the disk — then f is univalent on the whole disk. One inequality, checked pointwise, and global injectivity drops out, with no need to wrestle with f(z_1) = f(z_2) directly. The weight 2/(1 - |z|^2)^2 is not arbitrary: it is the natural hyperbolic measuring stick on the disk, exactly the gauge in which 'staying small' has Mobius-invariant meaning. And the constant 2 is sharp — you cannot replace it with anything larger.

1. Start from your holomorphic f with f' never zero on the disk, and compute the ratio f''/f' (its logarithmic derivative).
2. Form the Schwarzian Sf = (f''/f')' - (1/2)(f''/f')^2. By construction this kills any Mobius part of f, so it reports only genuine bending.
3. Check the single pointwise inequality |Sf(z)| <= 2/(1 - |z|^2)^2 across the disk, comparing your Schwarzian against the hyperbolic weight.
4. If it holds everywhere, conclude f is univalent on the disk — a global certificate earned from a local check, never having to test injectivity by hand.

Closing the rung

Step back and see the shape of the whole rung. You began by isolating the single extra word, injective, that turns a holomorphic map into a univalent one, and you normalized to the class S. You watched injectivity leave its fingerprint as |a_2| <= 2, proved through the area theorem, and you met the Koebe slit map sitting on every edge. You saw the one-quarter theorem, the growth and distortion theorems, and finally the sixty-nine-year siege of the Bieberbach conjecture |a_n| <= n, settled by de Branges. Every one of those was a bound on maps already known to be univalent.

This guide supplied the missing direction: how to certify univalence in the first place. Geometry hands you the starlike and convex sub-classes, where a visible shape — a star you see across, a region with no dents — turns into a single positivity test on f, and the tighter shape buys the smaller coefficient bound (|a_n| <= 1 for convex, against the universal n). Analysis hands you the Schwarzian and Nehari's inequality, a Mobius-blind gauge whose smallness forces injectivity outright. Together they answer the question the whole rung quietly presupposed: not just what can a univalent function do, but how do I know I have one.

Carry forward the two honest caveats that keep these tools sharp. The geometric tests are tied to a center and a normalization: starlikeness is always 'with respect to 0', and the clean numbers 1, 2, 1/4, n only mean what they mean on the normalized class S of the disk. And the certificates run one way: convex implies starlike implies close-to-convex implies univalent, and Nehari's bound implies univalent — but none of these arrows reverse, so a univalent map may be none of these shapes and may flunk Nehari's test. They are sufficient certificates, not characterizations. Held that honestly, you now command both halves of the theory: the bounds that constrain univalent maps, and the tests that build them.