Borel summation

Question

If $f(z)=\sum_{n=0}^\infty a_n z^n$ is a formal power series with complex coefficients, then its Borel transform is defined by $$B(f)(z)=\sum_{n=0}^\infty a_n \tfrac{z^n}{n!}.$$ Suppose that $f$ and $g\neq 0$ are formal power series such that $B(g)$ and $B(fg)$ are entire functions, i.e., analytic for all $z\in \mathbb{C}$. Is it true that $B(f)$ is an entire function?

Answer 1

I think that I was able to prove that it is true. It is enough to prove that if $g(z)=1-(a_1z+a_2 z^2+\cdots)$ and $B(g)$ is entire, then $B(1/g)$ is entire. Using a geometric series expansion for $1/g$ we get that $B(1/g)$ can be written as $$ 1+\sum \frac{(a_1 z/1!)^{m_1} (a_2 z^2/2!)^{m_2}\cdots }{m_1! m_2! \cdots} \, \frac{k!\ (1!)^{m_1} (2!)^{m_2}\cdots }{(m_1+2m_2+\cdots)!} $$ where the sum is over all integers $k\geq 1$ and all sequences $(m_1,m_2,\dots)$ of non-negative integers, such that, $m_1+m_2+\cdots =k$. If we prove that the second fraction in the sum is $\leq 1$, then the entire sum can be bounded by $$ \exp\Big( \sum_{k=1}^\infty \frac{|a_k|}{k!}\, |z|^k \Big) $$ which is clearly a finite number for all $z\in \mathbb{C}$, because $B(g)$ is entire. The estimate that it remains to be proved is equivalent to the following fact. Let $n\geq 1$ be an integer and $n=n_1+\cdots +n_k$ be a partition (with $n_i\geq 1$), then $$ n!\geq k! \, n_1!\cdots n_k!. $$ This is easy to prove for each $k$ by induction on $n$. Note that the starting point of the induction is $n=k$, in which case $n_1=\cdots=n_k=1$.