57.5. BLASCHKE PRODUCTS 1805

and so the assumption on the sum gives uniform convergence of the product on B(0,r) toan analytic function. Since r < 1 is arbitrary, this shows B(z) is analytic on B(0,1) and hasthe specified zeros because the only place the factors equal zero are at the αn or 0.

Now consider the factors in the product. The claim is that they are all no larger inabsolute value than 1. This is very easy to see from the maximum modulus theorem. Let|α| < 1 and φ (z) = α−z

1−αz . Then φ is analytic near B(0,1) because its only pole is 1/α .Consider z = eiθ . Then ∣∣∣φ (eiθ

)∣∣∣= ∣∣∣∣ α− eiθ

1−αeiθ

∣∣∣∣= ∣∣∣∣1−αe−iθ

1−αeiθ

∣∣∣∣= 1.

Thus the modulus of φ (z) equals 1 on ∂B(0,1) . Therefore, by the maximum modulustheorem, |φ (z)| < 1 if |z| < 1. This proves the claim that the terms in the product are nolarger than 1 and shows the function determined by the Blaschke product is bounded. Thisproves the theorem.

Note in the conditions for this theorem the one for the sum, ∑∞n=1 (1−|αn|) < ∞. The

Blaschke product gives an analytic function, whose absolute value is bounded by 1 andwhich has the αn as zeros. What if you had a bounded function, analytic on B(0,1) whichhad zeros at {αk}? Could you conclude the condition on the sum? The answer is yes. Infact, you can get by with less than the assumption that f is bounded but this will not bepresented here. See Rudin [113]. This theorem is an exciting use of Jensen’s equation.

Theorem 57.5.2 Suppose f is an analytic function on B(0,1) , f (0) ̸= 0, and | f (z)| ≤M for all z ∈ B(0,1) . Suppose also that the zeros of f are {αk}∞

k=1 , listed according tomultiplicity. Then ∑

∞k=1 (1−|αk|)< ∞.

Proof: If there are only finitely many zeros, there is nothing to prove so assume thereare infinitely many. Also let the zeros be listed such that |αn| ≤ |αn+1| · · · Let n(r) denotethe number of zeros in B(0,r) . By Jensen’s formula,

ln | f (0)|+n(r)

∑i=1

lnr− ln |α i|=1

2π

∫ 2π

0ln∣∣∣ f (reiθ

)∣∣∣dθ ≤ ln(M) .

Therefore, by the mean value theorem,

n(r)

∑i=1

1r(r−|α i|)≤

n(r)

∑i=1

lnr− ln |α i| ≤ ln(M)− ln | f (0)|

As r→ 1−,n(r)→ ∞, and so an application of Fatous lemma yields

∞

∑i=1

(1−|α i|)≤ lim infr→1−

n(r)

∑i=1

1r(r−|α i|)≤ ln(M)− ln | f (0)| .

This proves the theorem.You don’t need the assumption that f (0) ̸= 0.