Consider a function $f: (0,1) \to \mathbb R$ defined as the following power series: $$ f(x) = \sum_{k=0}^\infty a_k (1-x)^k. $$ This series has $a_k > 0$ and is assumed to converge point-wise for all $x \in (0,1)$.

My question is:

If $x$ in the above expression is replaced by a complex variable $z$, can it be assured that $\sum_{k=0}^\infty a_k (1-z)^k$ converges for all $z$ with $|z-1|<1$?

Here is my attempt at a proof, but I'm not sure it is correct.

The series for $x \in (0,1)$ is absolutely convergent in $(0,1)$ by the assumptions. That is, $\sum_{k=0}^\infty a_k (1-x)^k = \sum_{k=0}^\infty a_k |1-x|^k$ is finite for all such $x$. Replacing $x$ by any complex $z$ such that $|1-z| = |1-x|$, the resulting $\sum_{k=0}^\infty a_k |1-z|^k$ converges to the same value. This implies that the complex series $\sum_{k=0}^\infty a_k (1-z)^k$ (absolutely) converges. Since this holds for any $x \in (0,1)$, the complex series converges for any $z$ in the disk $|z-1|<1$.

Are all my steps correct? If so, is there any shorter proof? If not, is there a counterexample?


The steps are correct. Perhaps you can simplify notation by considering the series $\Sigma_{k=0}^\infty a_ky^k$. It has a radius of convergence $R$; this means that $\Sigma_{k=0}^\infty a_ky^k$ converges absolutely for $\lvert y \rvert < R$ and diverges for $\lvert y \rvert > R$. You know that it converges for $0 < y < 1$; this is possible only when $1 \le R$. The complex series $\Sigma_{k=0}^\infty a_kz^k$ has the same radius $R$ of convergence as the real series.


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