# Extending the analicity region of a function.

Prove or refute (by means of a counterexample) the following claim:

Let $\,f\,$ be an analytic function in $\,D:=\{z\;\;:\;\;|z-a|<r\,\}\,$ and continuous in $\,\overline{D}\,$ , then: there exists $\,\delta>0\,$ s.t. $\,f\,$ can be analitically continued to $\,\{z\;\;:\;\;|z-a|<r+\delta\}\,$

What I tried: At first I though this can be disproved by means of the example $$\frac{1}{1-z}=\sum_{n=0}^\infty z^n\,\,,\,|z|<1$$ but then I realized that it can't be said the function $\,\displaystyle{\frac{1}{1-z}}\,$ is continuous on $\,\overline D\,$ , so I began suspecting the claim is true.

We can assume $\,a=0\,$ to avoid cumbersome stuff.

Now, we have that $$\forall\,z\in D\,\,,\,\,f(z)=\sum_{n=0}a_nz^n$$ so I argued: suppose $\,z_0\in \partial D\,$ is s.t. $\,f\,$ cannot be continued through it. Let us choose a sequence $\,\{z_k\}\subset D\,\,\,s.t.\,\,\,z_k\xrightarrow [k\to\infty]{}z_0\,$ , then by continuity we get $$f(z_0)=\lim_{n\to\infty}f(z_k)=\lim_{k\to\infty}\sum_{n=0}^\infty a_nz_k^n\stackrel{\color{red}{attention!}}=\sum_{n=0}^\infty a_nz_0^n$$

Of course, the above is possible if we assume we can interchange the limit and the summation, which I'm not sure at all...I can't see here any monotone or dominated convergence theorem that'll allow this, so I'm pretty stuck.

Any hint, idea or solution will be duly appreciated.

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Abel's Theorem seems relevant. –  Ragib Zaman Jul 1 '12 at 16:00

Let $$f(z) = \sum_{n=1}^\infty \frac{z^n}{n^p}$$ where $p>1$. Then the radius of convergence is 1 and the series converges unformly (hence to a continuous function) on the closed unit disc.