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Holomorphic images of disks

28 November, 2020 in 246A - complex analysis, expository, math.CV | Tags: , , | by | 14 comments

Consider a disk {D(z_0,r) := \{ z: |z-z_0| < r \}} in the complex plane. If one applies an affine-linear map {f(z) = az+b} to this disk, one obtains

\displaystyle f(D(z_0,r)) = D(f(z_0), |f'(z_0)| r).

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 {D(z_0,r)} be a disk in the complex plane, and {f: D(z_0,r) \rightarrow {\bf C}} be a holomorphic function with {f'(z_0) \neq 0}.
  • (i) (Open mapping theorem or inverse function theorem) {f(D(z_0,r))} contains a disk {D(f(z_0),\varepsilon)} for some {\varepsilon>0}. (In fact there is even a holomorphic right inverse of {f} from {D(f(z_0), \varepsilon)} to {D(z_0,r)}.)
  • (ii) (Bloch theorem) {f(D(z_0,r))} contains a disk {D(w, c |f'(z_0)| r)} for some absolute constant {c>0} and some {w \in {\bf C}}. (In fact there is even a holomorphic right inverse of {f} from {D(w, c |f'(z_0)| r)} to {D(z_0,r)}.)
  • (iii) (Koebe quarter theorem) If {f} is injective, then {f(D(z_0,r))} contains the disk {D(f(z_0), \frac{1}{4} |f'(z_0)| r)}.
  • (iv) If {f} is a polynomial of degree {n}, then {f(D(z_0,r))} contains the disk {D(f(z_0), \frac{1}{n} |f'(z_0)| r)}.
  • (v) If one has a bound of the form {|f'(z)| \leq A |f'(z_0)|} for all {z \in D(z_0,r)} and some {A>1}, then {f(D(z_0,r))} contains the disk {D(f(z_0), \frac{c}{A} |f'(z_0)| r)} for some absolute constant {c>0}. (In fact there is holomorphic right inverse of {f} from {D(f(z_0), \frac{c}{A} |f'(z_0)| r)} to {D(z_0,r)}.)

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 {w \in D(f(z_0), \frac{1}{n} |f'(z_0)| r)}, then we have a lower bound for the log-derivative of {f(z)-w} at {z_0}:

\displaystyle \frac{|f'(z_0)|}{|f(z_0)-w|} > \frac{n}{r}

(with the convention that the left-hand side is infinite when {f(z_0)=w}). But by the fundamental theorem of algebra we have

\displaystyle \frac{f'(z_0)}{f(z_0)-w} = \sum_{j=1}^n \frac{1}{z_0-\zeta_j}

where {\zeta_1,\dots,\zeta_n} are the roots of the polynomial {f(z)-w} (counting multiplicity). By the pigeonhole principle, there must therefore exist a root {\zeta_j} of {f(z) - w} such that

\displaystyle \frac{1}{|z_0-\zeta_j|} > \frac{1}{r}

and hence {\zeta_j \in D(z_0,r)}. Thus {f(D(z_0,r))} contains {w}, and the claim follows. \Box

The constant {\frac{1}{n}} in (iv) is completely sharp: if {f(z) = z^n} and {z_0} is non-zero then {f(D(z_0,|z_0|))} contains the disk

\displaystyle D(f(z_0), \frac{1}{n} |f'(z_0)| r) = D( z_0^n, |z_0|^n)

but avoids the origin, thus does not contain any disk of the form {D( z_0^n, |z_0|^n+\varepsilon)}. This example also shows that despite parts (ii), (iii) of the theorem, one cannot hope for a general inclusion of the form

\displaystyle f(D(z_0,r)) \supset D(f(z_0), c |f'(z_0)| r )

for an absolute constant {c>0}.

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 {f''(z) = O(\frac{A}{r} |f'(z_0)|)} for {z \in D(z_0,r/2)}, hence by Taylor’s theorem with remainder {f(z) = f(z_0) + f'(z_0) (z-z_0) (1 + O( A \frac{|z-z_0|}{r} ) )} for {z \in D(z_0, r/2)}. By Rouche’s theorem, this implies that the function {f(z)-w} has a unique zero in {D(z_0, 2cr/A)} for any {w \in D(f(z_0), cr|f'(z_0)|/A)}, if {c>0} is a sufficiently small absolute constant. The claim follows. \Box

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 {r} slightly if necessary we may assume that {f} extends analytically to the closure of the disk {D(z_0,r)}. Let {c} be the constant in (v) with {A=2}; we will prove (iii) with {c} replaced by {c/2}. If we have {|f'(z)| \leq 2 |f'(z_0)|} for all {z \in D(z_0,r/2)} then we are done by (v), so we may assume without loss of generality that there is {z_1 \in D(z_0,r/2)} such that {|f'(z_1)| > 2 |f'(z_0)|}. If {|f'(z)| \leq 2 |f'(z_1)|} for all {z \in D(z_1,r/4)} then by (v) we have

\displaystyle f( D(z_0, r) ) \supset f( D(z_1,r/2) ) \supset D( f(z_1), \frac{c}{2} |f'(z_1)| \frac{r}{2} )

\displaystyle \supset D( f(z_1), \frac{c}{2} |f'(z_0)| r )

and we are again done. Hence we may assume without loss of generality that there is {z_2 \in D(z_1,r/4)} such that {|f'(z_2)| > 2 |f'(z_1)|}. Iterating this procedure in the obvious fashion we either are done, or obtain a Cauchy sequence {z_0, z_1, \dots} in {D(z_0,r)} such that {f'(z_j)} goes to infinity as {j \rightarrow \infty}, which contradicts the analytic nature of {f} (and hence continuous nature of {f'}) on the closure of {D(z_0,r)}. This gives the claim. \Box

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 Let {D(z_0,r)} be a disk in the complex plane, and {f: D(z_0,r) \rightarrow {\bf C}} be a polynomial of degree {n \geq 1} with {f'(z) \neq 0} for all {z \in D(z_0,r)}. Then {f} is injective on {D(z_0, \sin\frac{\pi}{n})}.

The radius {\sin \frac{\pi}{n}} is best possible, for the polynomial {f(z) = z^n} has {f'} non-vanishing on {D(1,1)}, but one has {f(\cos(\pi/n) e^{i \pi/n}) = f(\cos(\pi/n) e^{-i\pi/n})}, and {\cos(\pi/n) e^{i \pi/n}, \cos(\pi/n) e^{-i\pi/n}} lie on the boundary of {D(1,\sin \frac{\pi}{n})}.

If one narrows {\sin \frac{\pi}{n}} slightly to {\sin \frac{\pi}{2n}} then one can quickly prove this proposition as follows. Suppose for contradiction that there exist distinct {z_1, z_2 \in D(z_0, \sin\frac{\pi}{n})} with {f(z_1)=f(z_2)}, thus if we let {\gamma} be the line segment contour from {z_1} to {z_2} then {\int_\gamma f'(z)\ dz}. However, by assumption we may factor {f'(z) = c (z-\zeta_1) \dots (z-\zeta_{n-1})} where all the {\zeta_j} lie outside of {D(z_0,r)}. Elementary trigonometry then tells us that the argument of {z-\zeta_j} only varies by less than {\frac{\pi}{n}} as {z} traverses {\gamma}, hence the argument of {f'(z)} only varies by less than {\pi}. Thus {f'(z)} takes values in an open half-plane avoiding the origin and so it is not possible for {\int_\gamma f'(z)\ dz} to vanish.

To recover the best constant of {\sin \frac{\pi}{n}} 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 {f: {\bf C} \rightarrow {\bf C}} be a polynomial of degree {n \geq 1} such that {f(1)=f(-1)}. Then {f'} contains a zero in the closure of {D( 0, \cot \frac{\pi}{n} )}.

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

\displaystyle (f,g)_n := \sum_{k=0}^n (-1)^k f^{(k)}(0) g^{(n-k)}(0). \ \ \ \ \ (1)

Theorem 4 (Grace’s theorem) Let {C} be a circle or line in {{\bf C}}, dividing {{\bf C} \backslash C} into two open connected regions {\Omega_1, \Omega_2}. Let {f,g} be two polynomials of degree at most {n \geq 1}, with all the zeroes of {f} lying in {\Omega_1} and all the zeroes of {g} lying in {\Omega_2}. Then {(f,g)_n \neq 0}.

(Contrapositively: if {(f,g)_n=0}, then the zeroes of {f} cannot be separated from the zeroes of {g} by a circle or line.)

Indeed, a brief calculation reveals the identity

\displaystyle f(1) - f(-1) = (f', g)_{n-1}

where {g} is the degree {n-1} polynomial

\displaystyle g(z) := \frac{1}{n!} ((z+1)^n - (z-1)^n).

The zeroes of {g} are {i \cot \frac{\pi j}{n}} for {j=1,\dots,n-1}, so the Grace-Heawood theorem follows by applying Grace’s theorem with {C} equal to the boundary of {D(0, \cot \frac{\pi}{n})}.

The same method of proof gives the following nice consequence:

Theorem 5 (Perpendicular bisector theorem) Let {f: {\bf C} \rightarrow C} be a polynomial such that {f(z_1)=f(z_2)} for some distinct {z_1,z_2}. Then the zeroes of {f'} cannot all lie on one side of the perpendicular bisector of {z_1,z_2}. For instance, if {f(1)=f(-1)}, then the zeroes of {f'} cannot all lie in the halfplane {\{ z: \mathrm{Re} z > 0 \}} or the halfplane {\{ z: \mathrm{Re} z < 0 \}}.

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 {n=1} can be established by direct computation, so suppose inductively that {n>1} and that the claim has already been established for {n-1}. 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 {V_n} denote the vector space of polynomials {f} of degree at most {n}, then the apolar form is a bilinear form {(,)_n: V_n \times V_n \rightarrow {\bf C}}. Each translation {z \mapsto z+a} on the complex plane induces a corresponding map on {V_n}, mapping each polynomial {f} to its shift {\tau_a f(z) := f(z-a)}. We claim that the apolar form is invariant with respect to these translations:

\displaystyle ( \tau_a f, \tau_a g )_n = (f,g)_n.

Taking derivatives in {a}, it suffices to establish the skew-adjointness relation

\displaystyle (f', g)_n + (f,g')_n = 0

but this is clear from the alternating form of (1).

Next, we see that the inversion map {z \mapsto 1/z} also induces a corresponding map on {V_n}, mapping each polynomial {f \in V_n} to its inversion {\iota f(z) := z^n f(1/z)}. From (1) we see that this map also (projectively) preserves the apolar form:

\displaystyle (\iota f, \iota g)_n = (-1)^n (f,g)_n.

More generally, the group of Möbius transformations on the Riemann sphere acts projectively on {V_n}, with each Möbius transformation {T: {\bf C} \rightarrow {\bf C}} mapping each {f \in V_n} to {Tf(z) := g_T(z) f(T^{-1} z)}, where {g_T} is the unique (up to constants) rational function that maps this a map from {V_n} to {V_n} (its divisor is {n(T \infty) - n(\infty)}). 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 {T} on {V_n} also moves the zeroes of each {f \in V_n} by {T} (viewing polynomials of degree less than {n} in {V_n} 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 {f} to infinity (thus making {f} a polynomial of degree {n-1}), so that {C} must now be a circle, with the zeroes of {g} inside the circle and the remaining zeroes of {f} outside the circle. But then

\displaystyle (f,g)_n = (f, g')_{n-1}.

By the Gauss-Lucas theorem, the zeroes of {g'} are also inside {C}. The claim now follows from the induction hypothesis.

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