If $f:D\to \mathbb C$, is holomorphic on $D$ does $f(z)=\sum_{k=0}^\infty \frac{f^{(n)}(0)}{n!}z^n$ for all $z\in D$?

Question

Let $D\subset \mathbb C$ an domain. If $f:D\to \mathbb C$, is holomorphic on $D$ does $$f(z)=\sum_{k=0}^\infty \frac{f^{(n)}(0)}{n!}z^n\tag{E}$$ for all $z\in D$ ?

I know that $(E)$ hold on a small $B_r(0)\subset D$, but does it also hold on $D$ ?

Answer 1

Not necessarily. If $D=\mathbb C\setminus\{1\}$ and $f(z)=\frac1{1-z}$, then you have$$f(z)=\sum_{n=0}^\infty z^n$$if and only if $\lvert z\rvert<1$.

However, for every open ball $B_r(0)$ contained in $D$ (not just for small ones), it is true that$$\bigl(\forall z\in B_r(0)\bigr):f(z)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}z^n.$$

Answer 2

If $f$ is holomorphic on $D$ and $0 \in D$, then in fact $f(z)=\sum_{k=0}^\infty \frac{f^{(n)}(0)}{n!}z^n$ in some open disk $B_r(0) \subset D$. More precisely, let $$r(D) = \sup \{r \mid B_r(0) \subset U \} .$$ Then $B_{r(D)}(0) \subset D$, therefore the radius of convergence of the Taylor series is $R(f) \ge r(D)$. See for example Radius of convergence of Taylor series of holomorphic function. By the identity theorem for holomorphic functions, on $B_{r(D)}(0)$ we have $f(z)=\sum_{k=0}^\infty \frac{f^{(n)}(0)}{n!}z^n$.

If $D \subset B_{R(f)}(0)$ (which is possible only when $R(f) > r(D)$ or $R(f) = r(D)$ and $D = B_{r(D)}(0) = B_{R(f)}(0)$) , we have good luck. But if $D \subsetneqq B_{R(f)}(0)$, we find $z \in D$ with $\lvert z \rvert > R(f)$ and it is definitely wrong since the Taylor series diverges in $z$.

Note that we have $D \subset B_{r(D)}(0)$ if and only if $D = B_{r(D)}(0)$, i.e. $D$ is an open disk with center $0$. Moreover, we can always find a holomorphic $f : D \to \mathbb C$ whose Taylor series has radius of convergence $R(f) = r(D)$. To see this, note that there must exist $s \in \mathbb C \setminus D$ such that $\lvert s \rvert = r(D)$. Then José Carlos Santos's example generalizes by taking $f : D \to \mathbb C, f(z) = 1/(z-s)$.

Therefore $f(z)=\sum_{k=0}^\infty \frac{f^{(n)}(0)}{n!}z^n$ on $D$ for all holomorphic $f : D \to \mathbb C$ if and only if $D$ is an open disk with center $0$.