Let $f(z)=\sum_{j=0}^\infty a_j z^j $ denote an entire function satisfying the estimate$$ |f(z)|\leq M e^{|z|}$$ for all $z\in \mathbb{C}$ for some constant $M$. Prove that the coefficient $a_j$ satisfy $$|a_j|\leq M \big(\frac{e}{j}\big)^j ,j=0,1,2,3.....$$

I was trying to apply the Cauchy estimate:

$$ |f^j(z_0) |\leq \frac{M j!}{R^j}$$ where are is radius of large disk. $$\frac{|f^j(z_0)|}{j!} =|a_j| \leq \frac{M e^{|z|}}{R^j}$$ Now I have problem, $ e^{|z|}$ . Can I write a Taylor series of $e^{|z|}$ on disk $D(0,\epsilon)$

I am not sure I should I do further , any idea would be appreciated!


1 Answer 1


The Cauchy estimate reads $$ \frac{|f^{(j)}(0)|}{j!}\le\frac{\sup_{|z|=R}|f(z)|}{R^j},\quad R>0. $$ Since $|f(z)|\le M\,e^{|z|}$, we have $\sup_{|z|=R}|f(z)|\le M\,e^R$. Now take $R=j$: $$ \frac{|f^{(j)}(0)|}{j!}\le\frac{M\,e^j}{j^j}=M\,\Bigl(\frac{e}{j}\Bigr)^j. $$


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