3
$\begingroup$

Let $(a_n)_{n\in\mathbb N}$ and $(b_n)_{n\in\mathbb N}$ be sequences of $\mathbb Q_p$ such that the function $f:z\in\mathbb Q_p\to\sum_{n\ge0}a_nz^n+b_nz^{n+1}$converges in $\{|z|_p<1\}$. Assume that the series $\sum_{n\ge0}a_n+b_n$ converges in $\mathbb Q_p$. Can the function $f$ be continued in a larger disk in an analytic function?

Thanks in advance for for any hint or answer.

$\endgroup$
3
  • $\begingroup$ You are not assuming that $\sum a_nz^n$ and/or $\sum b_nz^n$ converge, right? $\endgroup$
    – efs
    Mar 9, 2019 at 4:20
  • $\begingroup$ $\sum_{n\ge}a_nz^n$ and $\sum_{n\ge}b_nz^n$ converge for $|z|_p<1$ but not $\sum_{n\ge}a_n$ and $\sum_{n\ge0}b_n$. $\endgroup$
    – joaopa
    Mar 9, 2019 at 4:33
  • $\begingroup$ Ok, I understand your question now. I deleted my anwer. $\endgroup$
    – efs
    Mar 9, 2019 at 6:06

1 Answer 1

2
$\begingroup$

I think the answer is "not necessarily", by the following (counter)example. First, let $b_n=-a_n$ for $n\ge0$. Then $\sum(a_n+b_n)=0$ and $$f(z)=\sum_{n\ge0} a_n(1-z)z^n=\sum_{n\ge0} (a_n-a_{n-1})z^n,$$ where we set $a_{-1}=0$. Now, define $$a_n:=\frac{-1}{n+1}.$$ Then, $\sum a_nz^n$ converges if and only if $|z|_p<1$. Also, $$f(z)=\sum_{n\ge0}\frac{1}{n(n+1)}z^n,$$ which converges if and only if $|z|_p<1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.