A power series is real analytic on its radius of convergence.

Question

Currently stuck on the last part of 15.2.8(e) of this problem:

I don't know how to apply Fubini's theorem since one index relies on the other.

Having slept on it, I've almost got it figured out except for one thing. Part of Fubini's theorem states (in my book) that if

$$\sum_{(n,m) \in \mathbb{N} \times \mathbb{N}} f(n,m)$$

converges absolutely to some limit $L$, then

$$\sum_{n=0}^{\infty} \left ( \sum_{m=0}^{\infty} f(n, m)\right )$$

also converges absolutely to $L$.

What I'm trying to figure out is if the converse of this statement is true. Because it seems that in order to apply Fubini's theorem to this problem it needs to be.

Answer 1

Note that if the terms are non-negative or the double series converges absolutely then $$ \begin{align*} \sum_{m\geqslant 0}d_m(x-b)^m&=\sum_{m\geqslant 0}\sum_{n\geqslant m}\binom{n}{m}(b-a)^{n-m}(x-b)^mc_n\\ &=\sum_{n\geqslant m\geqslant 0}\binom{n}{m}(b-a)^{n-m}(x-b)^mc_n\\ &=\sum_{n\geqslant 0}\sum_{m=0}^n\binom{n}{m}(b-a)^{n-m}(x-b)^mc_n\\ &=\sum_{n\geqslant 0}c_n(b-a+x-b)^n\\ &=\sum_{n\geqslant 0}(x-a)^nc_n \end{align*} $$

To "free" one variable of the other in the summation signs you also can use some indicator function as follows: $$ \begin{align*} \sum_{m\geqslant 0}d_m(x-b)^m&=\sum_{m\geqslant 0}\sum_{n\geqslant m}\binom{n}{m}(b-a)^{n-m}(x-b)^mc_n\\ &=\sum_{m\geqslant 0}\sum_{n\geqslant 0}\binom{n}{m}(b-a)^{n-m}(x-b)^mc_n\,\chi _{\Bbb N \cap [m,\infty )}(n) \end{align*} $$ and writing the last expression as a double integral with respect to the counting measure $\delta $ we have $$ \int_{\Bbb N }\int_{\Bbb N }\binom{n}{m}(b-a)^{n-m}(x-b)^mc_n\,\chi _{\Bbb N \cap [m,\infty )}(n)\,\delta (n)\, \delta (m) $$

Answer 2

I think Rudin's approach is very clear. Following him then, without loss of generality $a=0,$ so $(b-s,b+s)\subseteq (-r,r).$ Then, by the binomial theorem,

$\sum^{\infty}_{n=0}c_nx^n=\sum^{\infty}_{n=0}c_n((x-b)+b)^n=\sum^{\infty}_{n=0}c_n\left[\sum^{n}_{m=0}\binom{n}{m}b^{n-m}(x-b)^m\right].$

In this double sum, we are summing the (infinite number of) rows of a lower triangular matrix. We want to change this to a sum over the columns, so that the above becomes

$\sum^{\infty}_{m=0}\left[\sum^{\infty}_{n=m}\binom{n}{m}c_nb^{n-m}\right](x-b)^m$, and then we may set the bracketed term equal to $d_n.$

So all that remains is to see that

$\sum^{\infty}_{n=0}\left|c_n\left[\sum^{n}_{m=0}\binom{n}{m}b^{n-m}(x-b)^m\right]\right|=\sum^{\infty}_{n=0}\sum^{n}_{m=0}\left|c_n\binom{n}{m}b^{n-m}(x-b)^m\right|=$

$\sum^{\infty}_{n=0}|c_n|\cdot (|(x-b)|+|b|)^n$ converges by the comparison test, once $|(x-b)|+|b|<r.$

But if $x\in (b-s,b+s),$ then $|x-b|<s<r-|b|\Rightarrow |(x-b)|+|b|<r,$ which completes the proof.