1658 CHAPTER 52. THE OPEN MAPPING THEOREM

and so by Theorem 52.4.1 | f (z)| ≤M.

The following is another interesting case. This case is presented in Rudin [113]

Corollary 52.4.3 Let Ω be the open set consisting of {z ∈ C : a < Rez < b} and supposef is analytic on Ω , continuous on Ω, and bounded on Ω. Suppose also that f (z) ≤ 1 onthe two lines Rez = a and Rez = b. Then | f (z)| ≤ 1 for all z ∈Ω.

Proof: This time let φ (z)= 11+z−a . Thus |φ (z)| ≤ 1 because Re(z−a)> 0 and φ (z) ̸= 0

for all z ∈ Ω. Also, limsupz→∞ |φ (z)|η = 0 for every η > 0. Therefore, if a is a point ofthe sides of Ω, limsupz→a | f (z)| ≤ 1 while limsupz→∞ | f (z)| |φ (z)|η = 0 ≤ 1 and so byTheorem 52.4.1, | f (z)| ≤ 1 on Ω.

This corollary yields an interesting conclusion.

Corollary 52.4.4 Let Ω be the open set consisting of {z ∈ C : a < Rez < b} and supposef is analytic on Ω , continuous on Ω, and bounded on Ω. Define

M (x)≡ sup{| f (z)| : Rez = x}

Then for x ∈ (a,b).

M (x)≤M (a)b−xb−a M (b)

x−ab−a .

Proof: Let ε > 0 and define

g(z)≡ (M (a)+ ε)b−zb−a (M (b)+ ε)

z−ab−a

where for M > 0 and z ∈ C, Mz ≡ exp(z ln(M)) . Thus g ̸= 0 and so f/g is analytic on Ω

and continuous on Ω. Also on the left side,∣∣∣∣ f (a+ iy)g(a+ iy)

∣∣∣∣=∣∣∣∣∣ f (a+ iy)

(M (a)+ ε)b−a−iy

b−a

∣∣∣∣∣=∣∣∣∣∣ f (a+ iy)

(M (a)+ ε)b−ab−a

∣∣∣∣∣≤ 1

while on the right side a similar computation shows∣∣∣ f

g

∣∣∣≤ 1 also. Therefore, by Corollary52.4.3 | f/g| ≤ 1 on Ω. Therefore, letting x+ iy = z,

| f (z)| ≤∣∣∣(M (a)+ ε)

b−zb−a (M (b)+ ε)

z−ab−a

∣∣∣= ∣∣∣(M (a)+ ε)b−xb−a (M (b)+ ε)

x−ab−a

∣∣∣and so

M (x)≤ (M (a)+ ε)b−xb−a (M (b)+ ε)

x−ab−a .

Since ε > 0 is arbitrary, it yields the conclusion of the corollary.Another way of saying this is that x→ ln(M (x)) is a convex function.This corollary has an interesting application known as the Hadamard three circles the-

orem.