# Prove there exists $c\in\mathbb{C}$ such that $|c| \leq 1$ and $f(z)=ce^{z}$

complex-analysis

Let $$(a_k)_{k\geq 0}$$ be a real sequence such that $$\lim_{k\rightarrow \infty} a_k =+\infty$$. Let $$f:\mathbb{C}\rightarrow\mathbb{C}$$ be a holomorphic function such that:
$$\forall k \in \mathbb{N}, \forall n\in\mathbb{N},~~|f^{(n)}(a_k)|\leq e^{-a_k}$$

Prove there exists $$c\in\mathbb{C}$$ such that $$|c| \leq 1$$ and,
$$\forall z \in\mathbb{C}, ~~f(z)=ce^{-z}$$

Do you have an idea about how to approach this problem?

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 $$|c| \le 1$$.