# Is there a “monotonicity” property for analytic continuation?

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?

• 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. – David C. Ullrich Jan 15 at 18:15
• Better example: $b_n=-2^n$, $a_n=1$, so both series have positive radius of convergence. – David C. Ullrich Jan 15 at 18:40
• 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. – Dylan Jan 15 at 18:57

## 1 Answer

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.)

• thank you for the examples! They are both very good. – Dylan Jan 15 at 18:58