1

Let $D = \left\{ z \in \mathbf{C} \colon \left| z \right| < 1 \right\}$. Let $f \colon D \to \mathbf{C}$ be analytic such that $$\sum_{n = 0}^\infty \left| f^{(n)}(0) \right| < \infty. $$ One can show that $$ \sum_{n = 0}^\infty \frac{f^{(n)}(0)}{n!} z^n \text{ converges for all } z, $$ i.e. $f$ extends to an entire function defined by this power series.

Question: Show that $$ \sum_{n = 0}^\infty f^{(n)}(z) \text{ converges for all } z. $$

This problem appeared on a qualifying exam. Am I just supposed to differentiate the power series term by term to calculate $f^{(n)}(z)$ and examine the "series of series" that results? (Edit: This is exactly what I'm supposed to do.) This does not seem right... I would appreciate any help.

Doug
  • 2,797

1 Answers1

3

Hint: We have $f(z) = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}z^n.$ This implies

$$|f^{(m)}(z)| = \left|\sum_{n=m}^{\infty}\frac{f^{(n)}(0)}{(n-m)!}z^{n-m}\right| \le \sum_{n=m}^{\infty}\frac{|f^{(n)}(0)|}{(n-m)!}\,|z|^{n-m}.$$

Now sum on $m$ and reverse the order of summation.

zhw.
  • 105,693