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Consider a disk in the complex plane. If one applies an affine-linear map to this disk, one obtains

For maps that are merely holomorphic instead of affine-linear, one has some variants of this assertion, which I am recording here mostly for my own reference:

Theorem 1 (Holomorphic images of disks)Let be a disk in the complex plane, and be a holomorphic function with .

- (i) (Open mapping theorem or inverse function theorem) contains a disk for some . (In fact there is even a holomorphic right inverse of from to .)
- (ii) (Bloch theorem) contains a disk for some absolute constant and some . (In fact there is even a holomorphic right inverse of from to .)
- (iii) (Koebe quarter theorem) If is injective, then contains the disk .
- (iv) If is a polynomial of degree , then contains the disk .
- (v) If one has a bound of the form for all and some , then contains the disk for some absolute constant . (In fact there is holomorphic right inverse of from to .)

Parts (i), (ii), (iii) of this theorem are standard, as indicated by the given links. I found part (iv) as (a consequence of) Theorem 2 of this paper of Degot, who remarks that it “seems not already known in spite of its simplicity”; an equivalent form of this result also appears in Lemma 4 of this paper of Miller. The proof is simple:

*Proof:* (Proof of (iv)) Let , then we have a lower bound for the log-derivative of at :

The constant in (iv) is completely sharp: if and is non-zero then contains the disk

but avoids the origin, thus does not contain any disk of the form . This example also shows that despite parts (ii), (iii) of the theorem, one cannot hope for a general inclusion of the form for an absolute constant .Part (v) is implicit in the standard proof of Bloch’s theorem (part (ii)), and is easy to establish:

*Proof:* (Proof of (v)) From the Cauchy inequalities one has for , hence by Taylor’s theorem with remainder for . By Rouche’s theorem, this implies that the function has a unique zero in for any , if is a sufficiently small absolute constant. The claim follows.

Note that part (v) implies part (i). A standard point picking argument also lets one deduce part (ii) from part (v):

*Proof:* (Proof of (ii)) By shrinking slightly if necessary we may assume that extends analytically to the closure of the disk . Let be the constant in (v) with ; we will prove (iii) with replaced by . If we have for all then we are done by (v), so we may assume without loss of generality that there is such that . If for all then by (v) we have

Here is another classical result stated by Alexander (and then proven by Kakeya and by Szego, but also implied to a classical theorem of Grace and Heawood) that is broadly compatible with parts (iii), (iv) of the above theorem:

Proposition 2Let be a disk in the complex plane, and be a polynomial of degree with for all . Then is injective on .

The radius is best possible, for the polynomial has non-vanishing on , but one has , and lie on the boundary of .

If one narrows slightly to then one can quickly prove this proposition as follows. Suppose for contradiction that there exist distinct with , thus if we let be the line segment contour from to then . However, by assumption we may factor where all the lie outside of . Elementary trigonometry then tells us that the argument of only varies by less than as traverses , hence the argument of only varies by less than . Thus takes values in an open half-plane avoiding the origin and so it is not possible for to vanish.

To recover the best constant of requires some effort. By taking contrapositives and applying an affine rescaling and some trigonometry, the proposition can be deduced from the following result, known variously as the Grace-Heawood theorem or the complex Rolle theorem.

Proposition 3 (Grace-Heawood theorem)Let be a polynomial of degree such that . Then contains a zero in the closure of .

This is in turn implied by a remarkable and powerful theorem of Grace (which we shall prove shortly). Given two polynomials of degree at most , define the *apolar* form by

Theorem 4 (Grace’s theorem)Let be a circle or line in , dividing into two open connected regions . Let be two polynomials of degree at most , with all the zeroes of lying in and all the zeroes of lying in . Then .

(Contrapositively: if , then the zeroes of cannot be separated from the zeroes of by a circle or line.)

Indeed, a brief calculation reveals the identity

where is the degree polynomial The zeroes of are for , so the Grace-Heawood theorem follows by applying Grace’s theorem with equal to the boundary of .The same method of proof gives the following nice consequence:

Theorem 5 (Perpendicular bisector theorem)Let be a polynomial such that for some distinct . Then the zeroes of cannot all lie on one side of the perpendicular bisector of . For instance, if , then the zeroes of cannot all lie in the halfplane or the halfplane .

I’d be interested in seeing a proof of this latter theorem that did not proceed via Grace’s theorem.

Now we give a proof of Grace’s theorem. The case can be established by direct computation, so suppose inductively that and that the claim has already been established for . Given the involvement of circles and lines it is natural to suspect that a Möbius transformation symmetry is involved. This is indeed the case and can be made precise as follows. Let denote the vector space of polynomials of degree at most , then the apolar form is a bilinear form . Each translation on the complex plane induces a corresponding map on , mapping each polynomial to its shift . We claim that the apolar form is invariant with respect to these translations:

Taking derivatives in , it suffices to establish the skew-adjointness relation but this is clear from the alternating form of (1).Next, we see that the inversion map also induces a corresponding map on , mapping each polynomial to its inversion . From (1) we see that this map also (projectively) preserves the apolar form:

More generally, the group of Möbius transformations on the Riemann sphere acts projectively on , with each Möbius transformation mapping each to , where is the unique (up to constants) rational function that maps this a map from to (its divisor is ). Since the Möbius transformations are generated by translations and inversion, we see that the action of Möbius transformations projectively preserves the apolar form; also, we see this action of on also moves the zeroes of each by (viewing polynomials of degree less than in as having zeroes at infinity). In particular, the hypotheses and conclusions of Grace’s theorem are preserved by this Möbius action. We can then apply such a transformation to move one of the zeroes of to infinity (thus making a polynomial of degree ), so that must now be a circle, with the zeroes of inside the circle and the remaining zeroes of outside the circle. But then By the Gauss-Lucas theorem, the zeroes of are also inside . The claim now follows from the induction hypothesis.
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