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I was asked to show that if $f(z)$ is holomorphic at a neighbourhood of $a$, and $\sum_{n\geq0}f^{(n)}(a)$ converges, then $f$ is entire and $\sum_{n\geq0}f^{(n)}(z)$ converges for all z in the complex plane.It is easy to prove that $f$ is entire since $f^{(n)}(a)$ is bounded and $\frac{|z-a|^{n}}{n!}$ is smaller than a geometric sequence for sufficiently large $n$. I have trouble with the second assertion. I can only get the result when the convergence at $a$ is absolute. The method was shown Show that the series of derivatives of a certain entire function converges everywhere. But for the non-absolute convergent case, I don't know how to proceed.

Jiang Tianshu
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1 Answers1

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$a=0$ $$\sum_{m=0}^M f^{(m)}(z) = \sum_{m=0}^M\sum_{k\ge 0} f^{(k+m)}(0) \frac{z^k}{k!} = \sum_{k\ge 0}\frac{z^k}{k!} \sum_{m=0}^M f^{(k+m)}(0)= \sum_{k\ge 0}\frac{z^k}{k!}(b_k-b_{k+M+1}) $$

where $b_k = \sum_{m=0}^\infty f^{(k+m)}(0)$ converges and $\lim_{k\to \infty} b_k=0$.

Whence $$\sum_{m=0}^\infty f^{(m)}(z)=\sum_{k\ge 0}\frac{z^k}{k!}b_k- \lim_{M\to \infty} \sum_{k\ge 0}\frac{z^k}{k!} b_{k+M+1}$$

$$=\sum_{k\ge 0}\frac{z^k}{k!}b_k- \lim_{M\to \infty} O( \exp(|z|) \sup_{k>M}|b_k|) =\sum_{k\ge 0}\frac{z^k}{k!}b_k$$

reuns
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