Let $f(z)$ be analytic in the unit disc $D$. Suppose there is a constant $M$ such that $$\left|f^{(n)}(0)\right| \leq M^n$$ for all n sufficiently large. Show that $f(z)$ can be extended to all of $\mathbb{C}$ as an entire function.

My idea is to construct a function $g(z)$ whose restriction to $D$ is $f(z)$, and prove that its Taylor series has infinite radius of convergence, which means it is an entire function. But I do not know how to do that.

Do you have any idea?

Any help would be highly appreciated!

Thank you in advance!

  • $\begingroup$ What is the radius of convergence of the Taylor-MacLaurin series of $f$? $\endgroup$ – Daniel Fischer Dec 19 '15 at 14:56
  • 2
    $\begingroup$ Hint: define $g$ to be the formal power series for $f$. $\endgroup$ – Cameron Williams Dec 19 '15 at 14:59
  • $\begingroup$ @CameronWilliams I defined $g(z) = \sum f^{(k)}(0)z^k$ as your hint and used the method of vvnitram below. It worked. Thanks very much! $\endgroup$ – Phuoc Luan Duong Dec 20 '15 at 3:36

Let $m$ such that $|f^{(n)}(0)|\leq M^n$ for all $n\geq m$. Put $f(z)=\sum a_kz^k$ for all $z\in D$. Because $f^{(k)}(0)=k!a_k$, we have $|a_k|\leq \frac{M^k}{k!}$ for all $k\geq m$.

Let $z\in K,\text{ a compact set on } \mathbb{C}$ and let $r>0: |z|\leq r $ for all $z\in K$. $ |\sum a_kz^k|\leq |\sum_{k=0}^{m-1} a_kz^k| + |\sum_{k=m}^{\infty} a_kz^k|$ but $ |\sum_{k=m}^{\infty} a_kz^k|=|z^m|| |\sum_{k=0}^{\infty} a_{m+k}z^k|\leq |z^m|(\sum_{k=0}^{\infty} |a_{m+k}z^k|)\leq |z^m|(\sum_{k=0}^{\infty} \frac{M^{m+k}}{k!}|z^k|)\leq r^mM^me^{Mr}$.

Because $f$ converges on all compacts $K$,$f$ is holomorphic on $\mathbb{C}$.

  • $\begingroup$ Great! I used your idea with $g(z) = \sum f^{(k)}(0)z^k$ and it worked perfectly! Thanks very much! $\endgroup$ – Phuoc Luan Duong Dec 20 '15 at 3:31

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