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Consider a disk in the complex plane. If one applies an affine-linear map
to this disk, one obtains
Theorem 1 (Holomorphic images of disks) Letbe 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
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 2 Letbe 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) Letbe 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) Letbe 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
The same method of proof gives the following nice consequence:
Theorem 5 (Perpendicular bisector theorem) Letbe 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:
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:
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