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Let $f,g:(0,1) \to \mathbb{R}$ be analytic. It is well known that the product $h:=fg$ is then also analytic.

I would like to prove it by a using the following characterization of analyticity: $h$ is analytic if and only if, for any compact $K \subset (0,1)$, there exists $C>0$ such that $$|h^{(n)}(x)| \leq C^{n+1} n!$$ for any $x \in K$ and $n \geq 0$. My attempt: $$ |h^{(n)}|=\left|\sum_{k=0}^n \binom{n}{k} f^{(n-k)} g^{(k)}\right| \leq \sum_{k=0}^n \binom{n}{k} C^{n-k+1} (n-k)! C^{k+1} k! =C^{n+2} \sum_{k=0}^n \binom{n}{k} (n-k)! k! =C^{n+2} (n+1)! $$ I have n+1 instead of n...

K.defaoite
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1 Answers1

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Suggestion. The terms in the Cauchy product of two absolutely convergent series can be estimated by first normalizing the two series so they both sum to 1. (They are now essentially probability distributions.) Then you can see that each individual term in the Cauchy product (the convolution of the two distributions) must be uniformly bounded by 1.

In more detail, given functions $f$ and $g$ and their product $h=fg$, choose a slightly new constant $C$ so that

$a_n=\frac{f^n}{n! C^n}$ satisfies $a_n\leq .9^n$ and likewise

$b_n= \frac{g^n}{n! C^n}$ satisfies $|b_n|\leq .9^n$

Then you want to estimate the behavior of

$ c_n= \frac{ h^n}{n! C^n} $ which satisfies the algebraic identity $c_n=\sum_{k=0}^n a_{n-k} b_k$.

Now put absolute values across this expression and use the absolute summability suggestion to complete the proof.

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