If I have two complex functions defined by power series $A(z) = \sum a_n z^n $, $B(z) = \sum b_n z^n$ with $|a_n| \ge |b_n|$ for all $n$, and I know that $A$ converges in some set $U_1$ and defines a function of $z$ with an analytic extension to set $U_2$, then of course $B$ converges in $U_1$ also, but does this imply that $B$ has an analytic extension to $U_2$?

It would be nice if this were true, but I am a bit concerned because it's not hard to imagine a situation in which $B(z)$ that slightly shifts the region where $A(z)$ fails to have an analytic continuation. Can anyone help me think of a concrete counter example, or give an idea towards a proof?

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    $\begingroup$ As stated, what you say about convergence is false, never mind analytic continuation. (Consider $b_n=-n!$, $a_n=1/n!$.) Surely you meant $a_n\ge b_n\ge 0$, or maybe just $|a_n|\ge |b_n|$. With that change it seems like a reasonable question - my guess is the answer is "of course not", but I don't have a counterexample handy. $\endgroup$ – David C. Ullrich Jan 15 '19 at 18:15
  • $\begingroup$ Better example: $b_n=-2^n$, $a_n=1$, so both series have positive radius of convergence. $\endgroup$ – David C. Ullrich Jan 15 '19 at 18:40
  • $\begingroup$ Good catch, I did indeed mean to have absolute values; edited that. My guess is also "of course not" but I am having trouble thinking of an example. $\endgroup$ – DJA Jan 15 '19 at 18:57

We need to modify the hypotheses to get a reasonable question; if $b_n<0$ and $a_n>0$ then $b_n\le a_n$ but there's obviously no inclusion that follows even for the region of convergence.

Two reasonable questions might be the above assuming $a_n\ge b_n\ge0$ or assuming just $|a_n|\ge |b_n|$. The answer is no for the second version: If $A(z)=1/(z-1)$ and $B(z)=1/(z+1)$ then $|b_n|=|a_n|$ but there is no inclusion between the two regions of continuation.

Ah, a more interesting counterexample with $a_n\ge b_n\ge0$: $$A(z)=\sum z^n,$$ $$B(z)=\sum z^{2^n}.$$Then $A$ extends to a function holomorphic in $\Bbb C\setminus\{1\}$, while it's well known that $B$ does not extend past any point of the unit circle.

(Heh: Replacing $B$ by $(A+B)/3$ gives a counterexample with $a_n>b_n>0$, the strongest version of the hypothesis I can think of. So it's really no, with no way to fix it.)

  • $\begingroup$ thank you for the examples! They are both very good. $\endgroup$ – DJA Jan 15 '19 at 18:58

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