Uniformly convergent series and products of entire functions

[ForMyThunder]

If the sum of a sequence of functions $$a_n$$ converges uniformly, how is it that the product of $$1+a_n$$ converges uniformly? I know that this is true if the $$a_n$$ are constants but how does this translate to functions?

 

[HallsofIvy]

It seems to me almost trivial. If \(\displaystyle a_n(x)\) converges uniformly to a(x) then:

Given any $\epsilon> 0$, there exist N such that if n> N, [/itex]|a_n(x)- a(x)|< \epsilon[/itex] for all x.

Well, [/itex]|(a_n(x)+ 1)- (a(x)+ 1)|= |a_n(x)- a(x)|[/itex] for all x! So it immediately follows that $a_n(x)+ 1$ converges uniformly to a(x)+ 1.

(Oh, and it really does not make sense to talk about a sequence of constants converging "uniformly" since different x values will make no difference.)

 

[ForMyThunder]

I asked if $$\sum |a_n(z)|$$ converges uniformly, does this imply $$\prod(1+a_n(z))$$ converges uniformly?