1787

Then from 57.0.1 and 57.0.2,

∣∣∣∣∣P(z)−p(n)

∏k=1

(1+unk (z)

)∣∣∣∣∣≤

∣∣∣∣∣P(z)−n

∏k=1

(1+uk (z))

∣∣∣∣∣+∣∣∣∣∣ n

∏k=1

(1+uk (z))−p(n)

∏k=1

(1+unk (z)

)∣∣∣∣∣≤ ε +

∣∣∣∣∣ n

∏k=1

(1+uk (z))−p(n)

∏k=1

(1+unk (z)

)∣∣∣∣∣≤ ε +

∣∣∣∣∣ n

∏k=1

(1+ |uk (z)|)

∣∣∣∣∣∣∣∣∣∣1− ∏

nk>n

(1+unk (z)

)∣∣∣∣∣≤ ε +

∣∣∣∣∣ N0

∏k=1

(1+ |uk (z)|)

∣∣∣∣∣∣∣∣∣∣ n

∏k=N0+1

(1+ |uk (z)|)

∣∣∣∣∣∣∣∣∣∣1− ∏

nk>n

(1+unk (z)

)∣∣∣∣∣≤ ε +Ce

∣∣∣∣∣∏nk>n

(1+∣∣unk (z)

∣∣)−1

∣∣∣∣∣≤ ε +Ce

∣∣∣∣∣M(p(n))

∏k=n+1

(1+∣∣unk (z)

∣∣)−1

∣∣∣∣∣where M (p(n)) is the largest index in the permuted list,

{n1,n2, · · · ,np(n)

}. then from

57.0.1, this last term is dominated by

ε +Ce2

∣∣∣∣∣ln(

M(p(n))

∏k=n+1

(1+∣∣unk (z)

∣∣))− ln1

∣∣∣∣∣≤ ε +Ce2

∞

∑k=n+1

ln(1+∣∣unk

∣∣)≤ ε +Ce2∞

∑k=n+1

∣∣unk

∣∣< 2ε

for all n large enough uniformly in z ∈ H. Therefore,∣∣∣P(z)−∏

p(n)k=1

(1+unk (z)

)∣∣∣ < 2ε

whenever n is large enough. This proves the part about the permutation.

It remains to verify the assertion about the points, z0, where P(z0) = 0. Obviously, ifun (z0) =−1, then P(z0) = 0. Suppose then that P(z0) = 0 and M > N0. Then

∣∣∣∣∣ M

∏k=1

(1+uk (z0))

∣∣∣∣∣=