Carlson’s Theorem

Srivatsan Rajagopal

May 3, 2015

The starting point is the Maximum modulus principle :

1 Maximum Modulus Principle

If, f(z) is an analytic function regular in a region D and on its boundary C, and |f(z)| ≤ M on C, then f(z) < M in theinterior of D and f(z) ≤M throughout D

1.1 Proof

We need the following lemma : For any continous function φ(x)

φ(x) ≤ k∫ b

a

φ(x)dx ≥ k(b− a) =⇒ φ(x) = k

This is shown by noting that, if, φ(ξ) < k for some ξ ∈ [a, b], then, since φ(x) is continous,there is an interval [ξ − δ, ξ + δ]for which φ(x) ≤ k − ε (ε needn’t be small). Then,∫ b

a

φ(x)dx ≤ 2δ(k − ε) + (b− a− 2δ)k = (b− a)k − 2δε

which contradicts the assumption made on the integral. This shows φ(x) = kNow, suppose that at an interior point z0 of D, |f(z0)| ≥ |f(z)| for any z on C or in D.Let Γ be a circle with centre z0 contained entirely in D. By Cauchy’s theorem, we know

f(z0) =1

2πi

∫Γ

f(z)

z − z0dz (1)

Put z − z0 = reiθ, f(z)f(z0) = ρeiφ. ρ and φ are functions of θ on Γ and Eq.(1) is written as

1 =1

2π

∫ 2π

0

ρeiφdθ

By hypothesis ρ ≤ 1. But

1 = | 1

2π

∫ 2π

0

ρeiφdθ| ≤ 1

2π

∫ 2π

0

ρdθ

ρ is continous on Γ. Therefore, from the lemma, we immediately know

ρ = 1

This gives

1 =1

2π

∫ 2π

0

eiφdθ

The real part then gives

1 =1

2π

∫ 2π

0

cos(φ)dθ

Exactly similar arguments give φ = 0. This shows f(z) = f(z0) on Γ. Since an identical argument can be made for circlescentred on each point of Γ, we get f(z) is constant if it attains a maximum in the interior. Hence, for a non constant analyticfunction, the conclusion of the theorem holds.

We now consider the following extension of the Maximum Modulus Principle

1

2 Phragmen-Lindelof Theorem

Let C be a simple closed contour, and let f(z) be regular inside and on C except at one point P of C. Let |f(z)| ≤ Mon C except at P .

Further let there be a function ω(z), regular and not zero inside C such that |ω(z)| ≤ 1 inside C, and such that, if ε isany given positive number, we can find a system of curves, arbitrarily near to P and connecting the two sides of C aroundP , on which

|ω(z)εf(z)| ≤M

Then, |f(z)| ≤M at all points inside C

2.1 Proof

Choose an arbitrary point z0 inside C. Consider the function

F (z) = ω(z)εf(z)

By the hypothesis about ω(z), we can find a contour surrounding z0 on which F (z) is regular and satisfies

|F (z)| ≤M

By the Maximum Modulus principle, we get

|F (z0)| ≤M =⇒ |f(z0)| ≤M |ω(z0)|−ε

Taking ε→ 0, we obtain

|f(z0)| ≤M

3 Comments

The exceptional point may be replaced by any finite, or even an infinite number of points, provided that suitable interpolatingfunctions ωi(z) for each such point maybe found.

In practice, the exceptional point is at infinity.

4 Similar Results

A bunch of results similar to the above can be proved which would be useful in understanding Carlson’s theorem.

4.1

Let f(z) be an analytic function of z = reiθ, regular in the region D between two straight lines making an angle π/α at theorigin, and on the lines themselves. Suppose that

f(z) ≤M (2)

on the lines and that, as r →∞,

|f(z)| = O(erβ

)

, where β < α, uniformly in the angle. Then, actually, Eq.(2) holds throughout D

4.1.1 Proof

Assume the two lines are θ = ± 12π/α. let

F (z) = e−εzγ

f(z)

, where β < γ < α and ε > 0. Then

|F (z)| = e−εrγcosγθ|f(z)|

2

. On θ = ± 12π/α, cosγθ > 0, since γ < α. Hence, on these lines

|F (z)| ≤ |f(z)| ≤M

Also, on the arc |θ| ≤ 12π/α of the circle |z| = R,

|F (z)| ≤ e−εRγcos 1

2γπ/α|f(z)| < AeRβ−εRγcos 1

2γπ/α

which tends to zero as R → ∞. Hence if R is sufficiently large, |F (z)| ≤ M on the arc also. By the Maximum modulusprinciple, we get

|F (z)| ≤M

throughout the interior of the region |θ| ≤ 12π/α, r ≤ R. Since R is arbitrarily large, this holds throughout D. Thus

|f(z)| ≤Meεrγ

. Taking ε→ 0, we get the stated result.

4.2

Define the function

h(θ) = limr→∞log|f(reiθ)|

V (r)

Here, V (r) is chosen so that the limit is purely a function of θWe have the following result.Assume V (r) = rρ in the above definition. Also, let α < θ1 < θ2 < β and θ1 − θ2 < π/ρ and

h(θ1) ≤ h1

h(θ2) ≤ h2

Let H(θ) be the function of form acosρθ + bsinρθ which has the values h1, h2 at θ1, θ2 respectively. Then,

h(θ) ≤ H(θ) (θ1 ≤ θ ≤ θ2)

4.2.1 Proof

Let Hδ(θ) be the function aδcosρθ + bδsinρθ that is equal to h1 + δ, h2 + δ at θ = θ1, θ = θ2 respectively. Let

F (z) = f(z)e−(aδ−ibδ)zρ

Then,

|F (z)| = |f(z)|e−Hδ(θ)rρ

and so, if r is large enough,

|F (reiθ1)| = O(e(h1+δ)rρ−Hδ(θ1)rρ) = O(1)

with a similar statement to be made about |F (reiθ2)|.By the theorem proved in the earlier subsection, we get that F (z) is bounded in the angle (θ1, θ2). Therefore, we get

f(z) = O(eHδ(θ)rρ)

uniformly in the angle. Thus, h(θ) ≤ Hδ(θ) for θ1 ≤ θ ≤ θ2. Taking δ → 0 the result follows.

5 Carlson’s theorem 1

Let f(z) be regular and of form O(ek|z|) for Re(z) ≥ 0 and let f(z) = O(e−a|z|), where a > 0, on the imaginary axis. Thenf(z) = 0 identically.

3

5.1 Proof

In the theorem of section 4.2, choose ρ = 1, θ1 = 0, θ2 = 12π, h1 = k, h2 = −a and take δ = 0 throughout the argument. The

result implies

f(z) = O(ekcosθ−a|sinθ|r) (3)

for 0 ≤ θ ≤ 12π A similar argument shows that this also holds for − 1

2π ≤ θ ≤ 0 Now, let

F (z) = ezωf(z)

where ω is a large positive number. Then Eq.(3) implies there is a constant M , independent of ω, such that

|F (z)| ≤Me(k+ω)cosθ−a|sinθ|r(−1

2π ≤ θ ≤ 1

2π)

In particular, we have

|F (z)| ≤M (4)

for θ = ± 12π and θ = ±α where α = arc tan(k+ω

a )Now, use the theorem of section 4.1 to conclude that Eq.(4) holds in each of the three regions (− 1

2π, α), (−α, α), (α, 12π).

It then follows that Eq.(4) holds throughout − 12π ≤ θ ≤

12π.

Hence,

|f(z)| ≤Me−rωcosθ

Taking ω →∞ gives the required result.

6 Carlson’s Theorem 2

If f(z) is regular and of the form O(ek|z|), where k < π, for Re(z) ≥ 0, and f(z) = 0 for z = 0, 1, 2, ..., then f(z) = 0identically.

6.1 Proof

Choose

F (z) = f(z)cosecπz

This is regular in the region Re(z) ≥ 0. Moreover, except at the points z = n, cosecπz is bounded and hence F (z) is of form

F (z) = O(ek|z|)

throughout Re(z) ≥ 0. Also,

F (z) = O(e(k−π)|z|)

on the imaginary axis. The result follows from the previous theorem.

References:

E.C. Titchmarsh, The Theory of Functions (2nd Edition)

4