Uniformly convergent series and products of entire functions

  • #1
ForMyThunder
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If the sum of a sequence of functions [tex]a_n[/tex] converges uniformly, how is it that the product of [tex]1+a_n[/tex] converges uniformly? I know that this is true if the [tex]a_n[/tex] are constants but how does this translate to functions?
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
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It seems to me almost trivial. If \(\displaystyle a_n(x)\) converges uniformly to a(x) then:

Given any [itex]\epsilon> 0[/itex], 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 [itex]a_n(x)+ 1[/itex] 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.)
 
  • #3
ForMyThunder
149
0
I asked if [tex]\sum |a_n(z)|[/tex] converges uniformly, does this imply [tex]\prod(1+a_n(z))[/tex] converges uniformly?
 

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