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Given the set of all power series with radius of convergence ($r$ in the definition) equal to one:

$$A:=\{\sum a_kz^k | r =1\}$$

Does $A$ form a vector space?

The radius of convergence doesn't change when you multiply a scalar inside, however if you add them the radius will be at least $1$ or bigger than $1$. Also the zero vector would be $a_k=0$ with $k\in \mathbb{N}$ and that would mean the series converges everywhere, so $r=\infty$.

Can we change the conditions of the set so that it becomes a vector space?

Yes, if we set $A:=\{\sum a_k z^k|r\ge 1\}$.

Are these thoughts correct?

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Looks right to me. –  Alex Becker Sep 10 '12 at 20:12
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A tiny detail: the scalar you multiply with shouldn't be zero if you don't want to change the radius of convergence. –  t.b. Sep 10 '12 at 20:24
    
Maybe I should have said explicitly that $r=1$ excludes the zero series to be in $A$, so that already rules out that $A$ is a vector space... –  t.b. Sep 11 '12 at 11:54

1 Answer 1

up vote 2 down vote accepted

Yes, that works. The resulting vector space is exactly the space of holomorphic functions on the unit disk.

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