52.4. EXTENSIONS OF MAXIMUM MODULUS THEOREM 1659

52.4.2 Hadamard Three Circles TheoremLet 0 < R1 < R2 and suppose f is analytic on {z ∈ C : R1 < |z|< R2} . Then letting R1 <a < b < R2, note that g(z)≡ exp(z) satisfies

g : {z ∈ C : lna < Rez < b}→ {z ∈ C : a < |z|< b}

is onto and that in fact, g maps the line lnr+ iy onto the circle reiθ . Now let M (x) be definedas above and m be defined by m(r)≡maxθ

∣∣ f (reiθ)∣∣ .Then for a < r < b, Corollary 52.4.4

implies

m(r) = supy

∣∣∣ f (elnr+iy)∣∣∣= M (lnr)≤M (lna)

lnb−lnrlnb−lna M (lnb)

lnr−lnalnb−lna

= m(a)ln(b/r)/ ln(b/a) m(b)ln(r/a)/ ln(b/a)

and so m(r)ln(b/a) ≤ m(a)ln(b/r) m(b)ln(r/a) . Taking logarithms, this yields

ln(

ba

)ln(m(r))≤ ln

(br

)ln(m(a))+ ln

( ra

)ln(m(b))

which says the same as r→ ln(m(r)) is a convex function of lnr.The next example, also in Rudin [113] is very dramatic. An unbelievably weak as-

sumption is made on the growth of the function and still you get a uniform bound in theconclusion.

Corollary 52.4.5 Let Ω ={

z ∈ C : |Im(z)|< π

2

}. Suppose f is analytic on Ω, continuous

on Ω, and there exist constants, α < 1 and A < ∞ such that

| f (z)| ≤ exp(Aexp(α |x|)) for z = x+ iy

and ∣∣∣ f (x± iπ

2

)∣∣∣≤ 1

for all x ∈ R. Then | f (z)| ≤ 1 on Ω.

Proof: This time let φ (z) = [exp(Aexp(β z))exp(Aexp(−β z))]−1 where α < β < 1.Then φ (z) ̸= 0 on Ω and for η > 0

|φ (z)|η =1

|exp(ηAexp(β z))exp(ηAexp(−β z))|Now

exp(ηAexp(β z))exp(ηAexp(−β z))

= exp(ηA(exp(β z)+ exp(−β z)))

= exp[ηA(

cos(βy)(

eβx + e−βx)+ isin(βy)

(eβx− e−βx

))]and so

|φ (z)|η =1

exp[ηA(cos(βy)

(eβx + e−βx

))]Now cosβy > 0 because β < 1 and |y|< π

2 . Therefore, limsupz→∞ | f (z)| |φ (z)|η ≤ 0 ≤ 1and so by Theorem 52.4.1, | f (z)| ≤ 1.