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:
14 comments
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28 November, 2020 at 1:21 pm
Anonymous
What are the best possible values of the absolute constants in (ii) and (v)?
28 November, 2020 at 3:57 pm
Terence Tao
For (ii) the best constant is known as Bloch’s constant; it is somewhere between 0.433 and 0.4719 but the precise value is not known. The best constant in (v) is of roughly a similar order of magnitude but I don’t think it’s value has been intensively studied.
28 November, 2020 at 7:07 pm
Anonym
Dear Professor,
Except the contribution of Dégot, I suggest to you the recent Ph.D Thesis of Taboka Prince Chalebgwa on the geometry of complex polynomials, and his paper entitled “Sendov conjecture: On a paper of Dégot” which has been published in Analysis Mathematica in 2018
28 November, 2020 at 8:04 pm
Anonymous
Your present post seems in fact related to the famous Ilief-Sendov conjecture.
Here is an interesting reference on this topic.
Zhaizhao Meng. “Proof of the Sendov conjecture for polynomials of degre ni ne.” Preprint uploaded to the ArXiv in 2017. Last revised on April 17, 2018.
29 November, 2020 at 4:05 am
Alan
Following the previous messages, could you precise if your very interesting contribution in the present post has an eventual strong or weak connection with the well-known Iliev-Sendov conjecture?
30 November, 2020 at 10:58 am
Anonymous
…by (v) we have
Part of this long line is missing in the display:
[Corrected, thanks – T.]
30 November, 2020 at 2:03 pm
Anonymous
Are there similar theorems for meromorphic images of disks?
1 December, 2020 at 9:13 am
Jacob Manaker
What is the intuition for the apolar form?
3 December, 2020 at 8:54 am
Jacob Manaker
Plaumann (1410.5935) notes that the apolar form coincides, “up to a conjugation in the second argument…with Fischer’s inner product…well known for identifying the adjoint of complex differentiation with the multipli-cation by the variable”; and has “geometric implications…surprising and multifold [8].” Following up on note 8 (Algebra of Invariants, Grace & Young, 1903), it appears that the apolar form is sensitive to multiple zeros, so that it can be used to detect tangency and (dually) collinearity.
For the purposes of this theorem, I think it’s probably best to black-box the form: it is (Plaumann again) “the only joint invariant under affine substitutions which is linear in the coefficients” and the “zero location property [of the Grace-Heawood theorem] is equivalent to apolarity.” So if we seek a tool to detect zeros separated by clines, and want a proof by Möbius transforms, the apolar form is the only game in town.
(Yes, I’m answering a question I posed. I don’t care.)
1 December, 2020 at 7:36 pm
HuaFoFeng
In order to answer to the question of Alan, the current post of Pr. Tao has a fortiori nothing to do with Iliev -Sendov conjecture, even if the same research subfield of geometrical aspects of complex polynomials is here considered. To the best of my knowledge, this conjecture remains an open problem today. It was only proven in particular cases and up to n = 9 nowadays by Meng (see the corresponding aforementioned reference in a previous commentary).
2 December, 2020 at 7:44 pm
Anonymous
is it true that the Stein Shakarchi Complex analysis book doesn’t have the standard definition of holomorphic functions?
2 December, 2020 at 9:12 pm
Rob
Should it be “Let V_n denote the vector.space” instead of “V_n denote the vector space”?
[Corrected, thanks – T.]
3 December, 2020 at 9:39 am
Anonym 2
In my opinion, I see rather “If the V_n denote (…)”
8 December, 2020 at 7:07 pm
Sendov’s conjecture for sufficiently high degree polynomials | What's new
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