Showing $\sum c_n = \sum a_n \sum b_n$ if $\sum c_n x^n=\sum a_n x^n \sum b_n x^n$ on $[0, 1)$

Question

Given that $\sum c_n x^n=\sum a_n x^n \sum b_n x^n$ on $[0, 1)$ and all partial sums $\sum c_n x^n, \sum a_n x^n, \sum b_n x^n$ converges uniformly on $[0, 1]$, is it true that $\sum c_n = \sum a_n \sum b_n$?

Answer 1

Yes. In the conditions you mentioned the sum $\sum c_n$ is equal to the Abel sum which is $\lim_{x\to{1^{-}}}\sum c_n x^n$, and same can be said about the other two sums. So then:

$\sum c_n=\lim_{x\to{1^{-}}}\sum c_n x^n=\lim_{x\to{1^{-}}}\sum a_n x^n\sum b_n x^n=\lim_{x\to{1^{-}}}\sum a_n x^n\lim_{x\to{1^{-}}}\sum b_n x^n=$

$=\sum a_n\sum b_n$

And all we used is arithmetic of limits.