The question is this: Does there exist an entire function $f$ such that $\lim_{z\rightarrow \infty}f(z)=0$ and $f(0)=1$.
I immediately would point to $f(z)=e^{-z}$. It is entire and satisfies the above. However my professor gives this answer:
No. Since entire functions are bounded on bounded sets, the condition $\lim_{z\rightarrow \infty}f(z)=0$ implies that $f$ is bounded on the whole complex plane. Then by Liouville's Theorem, $f(z)=f(0)$ and $\lim_{z\rightarrow \infty}f(z)=1\neq 0.$
Who is correct? Maybe the question is wrong and it is supposed to have the limit for $\pm \infty$?
The function $z \mapsto e^{-z}$ doesn't satisfy the conditions. Actually the limit $$\lim_{z\to \infty} e^{-z}$$ doesn't exist. (For exemple let $z$ go to infinity on $i\mathbb{R}$) Hence it is not a counterexample.
The proposition is correct, by Liouville's theorem the function must by $1$ but then the limit is not $0$ so there are no such functions.