1786 CHAPTER 57. INFINITE PRODUCTS

Now having picked N0, the assumption the un are bounded on H implies there exists aconstant, C, independent of z ∈ H such that for all z ∈ H,

N0

∏k=1

(1+ |uk (z)|)<C. (57.0.2)

Let N0 < M < N. Then∣∣∣∣∣ N

∏k=1

(1+uk (z))−M

∏k=1

(1+uk (z))

∣∣∣∣∣≤

N0

∏k=1

(1+ |uk (z)|)

∣∣∣∣∣ N

∏k=N0+1

(1+uk (z))−M

∏k=N0+1

(1+uk (z))

∣∣∣∣∣≤ C

∣∣∣∣∣ N

∏k=N0+1

(1+uk (z))−M

∏k=N0+1

(1+uk (z))

∣∣∣∣∣≤ C

(M

∏k=N0+1

(1+ |uk (z)|)

)∣∣∣∣∣ N

∏k=M+1

(1+uk (z))−1

∣∣∣∣∣≤ Ce

∣∣∣∣∣ N

∏k=M+1

(1+ |uk (z)|)−1

∣∣∣∣∣ .Since 1≤∏

Nk=M+1 (1+ |uk (z)|)≤ e, it follows the term on the far right is dominated by

Ce2

∣∣∣∣∣ln(

N

∏k=M+1

(1+ |uk (z)|)

)− ln1

∣∣∣∣∣≤ Ce2

N

∑k=M+1

ln(1+ |uk (z)|)

≤ Ce2N

∑k=M+1

|uk (z)|< ε

uniformly in z ∈ H provided M is large enough. This follows from the simple observationthat if 1< x< e, then x−1≤ e(lnx− ln1). Therefore, {∏m

k=1 (1+uk (z))}∞

m=1 is uniformlyCauchy on H and therefore, converges uniformly on H. Let P(z) denote the function itconverges to.

What about the permutations? Let {n1,n2, · · ·} be a permutation of the indices. Letε > 0 be given and let N0 be such that if n > N0,∣∣∣∣∣ n

∏k=1

(1+uk (z))−P(z)

∣∣∣∣∣< ε

for all z ∈ H. Let {1,2, · · · ,n} ⊆{

n1,n2, · · · ,np(n)}

where p(n) is an increasing sequence.