Prove there exists $c\in\mathbb{C}$ such that $|c| \leq 1$ and $f(z)=ce^{z}$
Solution 1:
For $k \ge 0$ consider the power series expansion of $f$ in $a_k$: $$ f(z)=\sum_{j=0}^\infty c_j^{(k)}(z-a_k)^j. $$ Now $|f^{(n)}(a_k)| = |n!c_n^{(k)}| \le \exp(-a_k)$ $(n \ge 0)$. Thus $$ |f(z)| \le \sum_{j=0}^\infty \frac{\exp(-a_k)}{j!}|z-a_k|^j = \exp(-a_k+|z-a_k|). $$ Thus $$ |\exp(z)f(z)| \le \exp(\Re(z-a_k)+|z-a_k|) \quad (k \ge 0). $$ Now $a_k \to \infty$ implies $\Re(z-a_k)+|z-a_k| \to 0$ $(k \to \infty)$ for each $z \in \mathbb{C}$. Thus $|\exp(z)f(z)|\le 1$ on $\mathbb{C}$. By Liouville's Theorem $f(z)=c \exp(-z)$ for some $c$ with $|c| \le 1$.