Why in this case $f$ should be entire?

Question

Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be a function.

Assume $f$ satisfies follows:

$f$ is analytic at a point $z_0$.
$\limsup\limits_{n\to\infty} \left|\frac{f^{(n)}(z_0)}{n!}\right|^{1/n}=0$.

Why in this case $f$ should be analytic everywhere?

That is, define $g(z)=\sum \frac{f^{(n)}(z_0)}{n!} (z-z_0)^n$.

Why $g=f$ on $\mathbb{C}$?

What gurantees that they coincide even outside of a given neighborhood of $z_0$?

That's a good question. Indeed, the given conditions don't imply that $f\equiv g$, only that it holds in some neighbourhood of $z_0$. — Daniel Fischer, Jun 16 '14 at 18:29
Isn't it possible to find the maximum radius $r$ to have $f\equiv g$ in $D_r(z_0)$ ? — Fardad Pouran, Jun 16 '14 at 18:49

score 1 · Answer 1 · answered Jun 16 '14 at 20:52

1

As you note yourself, this is not true without additional assumptions on $f$. Take for example $z_0=0$, $$g(z)=e^z=\sum_{n=0}^\infty \frac{z^n}{n!}$$ and let $f$ be a function that is equal to $g$ on some (arbitrarily small) neighbourhood of $0$ and everywhere discontinuous elsewhere.

answered Jun 16 '14 at 20:52

mrf

43,639

Being analytic is not still sufficient, am I right ? – Fardad Pouran Jun 16 '14 at 21:07
@FardadPouran Actually analytic and holomorphic are the same. – Daniele A Jun 16 '14 at 21:25
@DanieleA, surely, I meant can a function be analytic in $\Omega$, and $g$ be analytic in $\Omega$ and agrees with $f$ on a small neighborhood of $z_0$, but $f\neq g$ in $\Omega-D_r(Z_0)$ ? – Fardad Pouran Jun 16 '14 at 21:43
As $f(z)=g(z)$ in a sequence of ${z_n}$ without accumulation point, $f=g$ everywhere in $\Omega$ :) – Fardad Pouran Jun 16 '14 at 21:45
Yes, as long as $\Omega$ is connected. – Daniele A Jun 16 '14 at 21:47
It's really good : Unique extension :) – Fardad Pouran Jun 16 '14 at 22:18

score 1 · Answer 2 · answered Jun 16 '14 at 22:02

The fact that your function is analytic in $z_0$ means that there exists a radius $R>0$ s.t. $$ f(z)=\sum_{n=0}^{+\infty}c_n(z-z_0)^n $$ holds for every $z\in B(z_0,R[$.

But... who is $R$? By definition $R$ is the largest possible in which $f$ is analytic; but in this way we don't say nothing useful.

The fact is that your second hypotesis guarantees the convergence of the power series on the ball $B(z_0,T[$ where $1/T=\limsup_n|c_n|^{1/n}=0$ so $T=+\infty$ and you have a power series that defines a function that is analytic on the whole complex plane (in fact $B(z_0,+\infty[=\mathbb C$). Call this function $g$. You now have two functions, $f$ and $g$, both analytic on a nhbrd on $z_0$ in which they coincide. Hence by the identity principle, they must coincide everywhere. But $g$ is defined on the whole complex plane, hence it will be THE (unique possible) analytic extension of $f$.

Why in this case $f$ should be entire?

2 Answers2