Convergence at border points

Stream: maths

[Filippo A. E. Nuccio]

I have a purely math question: suppose that $f(T)=\sum a_nT^n$ is a power series that converges on the closed disk $\overline{B(0,1)}$ centred at $0$ and radius $1$. Assume that $f(1/2)=0$, so I can divide the series by $(2T-1)$: I certainly get a holomorphic function on the open disk $B(0,1)$. Is it true/easy/obvious to check that $f(T)/(2T-1)$ still converges also on the border?

In my application, all $a_n\in\mathbb{Z}$, but I suspect that this is not crucial.

[Sebastien Gouezel]

What do you mean when you say that the power series converges on the closed disk? Pointwise convergence, or uniform convergence, or the holomorphic function can be extended beyond the open unit disk? Since you say that all your a_n are integers, I don't see in which sense you can have convergence.

[Filippo A. E. Nuccio]

I mean pointwise convergence at all points in the circle. The series is not (a priori) convergent beyond $\vert z\vert =1$, and it does not give raise to a holomorphic function beyond the unit disk, only on the open one.

[Sebastien Gouezel]

You can not have pointwise convergence at 1 if your coefficients are nonzero integers (as the general term of the series does not tend to zero).

[Filippo A. E. Nuccio]

Ah, I see the issue with integers: as a matter of fact, I should have been more precise, you are right. In my example, the radius is some $1/2<r<1$ (I thought it would simplify matters to assume $r=1$ but you are right it does not).

[Sebastien Gouezel]

The question definitely makes sense if you just don't assume that the a_n are integers.

[Filippo A. E. Nuccio]

You are right, but to stick to my example, it is better to work with some fixed $1/2<r<1$ and assume $f$ converges (uniformly) on the open $B(0,r)$, pointwise on $\overline{B(0,r)}$ and vanishes at $1/2$.

[Sebastien Gouezel]

Just to clarify the matter a little bit more: when you say pointwise convergence, you mean that $\sum_{i=0}^{n-1} a_i T^i$ converges to a limit when $n \to \infty$, but not that the series is converging in the mathlib sense (which would amoung to the convergence of the series of the absolute values). Right?

[Filippo A. E. Nuccio]

No, I mean in the absolute value sense; so $\sum a_i \vert z^i\vert$ converges to a limit when $n\to+\infty$, for all $\vert z \vert =r$: so $\sum a_i r^i$ converges to a limit.

[Johan Commelin]

I think we want absolute convergence

[Sebastien Gouezel]

ok. Then you're fine.

[Filippo A. E. Nuccio]

Ah, good!

[Filippo A. E. Nuccio]

Can you sketch the argument?

[Sebastien Gouezel]

Let me do it for r = 1.

[Filippo A. E. Nuccio]

Sure

[Sebastien Gouezel]

Let $c = 1/2$. Then $(f (T) - f (c)) / (T - c) = \sum a_n (T^n - c^n)/(T-c) = \sum a_n (T^{n-1} + c T^{n-2} + ... + c^{n-1})$. So, the series expansion for $(f(T) - f(c) / (T - c))$ is $\sum b_n T^n$ with $b_n = \sum_{i=1}^\infty a_{n+i} c^i$. If you sum the absolute values of the $b_n$, you get at most the one for the $a_n$ times the one for $c^i$, which is finite.

[Filippo A. E. Nuccio]

I agree, but this was my argument for convergency on the open disk. Does it give convergency at the boundary?

[Sebastien Gouezel]

If you assume that the series of the norms of $a_n$ is converging, then so is the series of the norms of $b_n$. Isn't it what you are asking?

[Filippo A. E. Nuccio]

Oh, right. Let me think for a second.

[Filippo A. E. Nuccio]

Can you help me with the final estimate? I agree that $b_n=\sum_{i\geq n+1}a_ic^i$; my previous idea was to bound these terms individually but you are suggesting that I use this estimate to directly bound their sum.

[Filippo A. E. Nuccio]

Perhaps I see: you are using the same argument used to study the radius of convergence of the Cauchy product of two series, here the $\sum a_nr^n$ and the $\sum c^n$?

[Sebastien Gouezel]

Yes. If you expand $\left(\sum a_i\right) \left(\sum c^i\right)$, you will see exactly all the terms that show up in $\sum b_n$.

[Filippo A. E. Nuccio]

Ah sure. Thanks! :pray:

[Notification Bot]

Filippo A. E. Nuccio has marked this topic as unresolved.