Does there exist an analytic function $f:\mathbb C \to \mathbb C$ such that $f(0)=1$ and for $z\in \mathbb C$,$|z|\ge1$ $|f(z)|\le e^{-|z|}$

Question

Does there exist an analytic function $f:\mathbb C \to \mathbb C$ such that

$f(0)=1$ and

$|f(z)|\le e^{-|z|}$ for $z\in \mathbb C$,$|z|\ge1$

I know existence of analytic function can be checked by C-R equations with having continuous partial derivatives of real and imaginary part of $f(z)$. I am not understanding how to check the existence of this kind of function. Any hint?

Answer 1

Since $f$ is holomorphic (so in particular continuous) on the closed unit disk $\{|z|\leq 1\}$, there is some $B>0$ such that $|f(z)|\leq B$ for all $|z|\leq 1$.

And by hypothesis $|f(z)|\leq e^{-|z|}\leq e^{-1}$ for all $|z|\geq 1$, so combining these inequalities we obtain $|f(z)|\leq \max\{B,e^{-1}\}$ for all $z\in\mathbb{C}$.

Thus $f$ is a bounded entire function, hence is constant by Liouville's theorem. Since $f(0)=1$ we must have $f(z)=1$ for all $z$, but this is impossible since $|f(z)|\leq e^{-|z|}$ for $|z|\geq 1$. So no such $f$ exists.

Answer 2

Since $|f(z)| \le e^{-|z|} = {1 \over e} < 1$ for $|z| =1$, the maximum modulus theorem implies that $|f(z)| \le {1 \over e}$ for $|z| <1$, which contradicts $f(0) = 1$.