Let $\mathbb{L}$ be the set of all entire functions, so every $f \in \mathbb{L}$ can be rewritten as a Maclaurin series: $$f(x)=\sum_{n\geq0} \frac{f^{(n}(0)}{n!}x^n.$$
I'm tring to come out with a definition of an inner product such that this serie can be interpreted as an expansion of an orthonormal basis of a Hilbert space:
$$f=\sum_{n\geq 0} <f,x^n> x^n.$$
The set $\{x^n\}_{n\geq 0}$ is a great candidate to play the role of the orthonormal basis, because it can recover $\mathbb{L}$ with its linear span.
In order to define the inner product, it seems natural to ask for the following property: $$<f,x^n> \,\,:= \frac{1}{n!}\frac{d^nf}{d^nx}(0)$$
which also have desirable results over the selected base: $$<x^m, x^n> = \delta_{m,n}.$$
Linearity on the first component can be deduced from derivative properties: \begin{equation}\tag{1}<c_1f+c_2g\,,\,x^n> \,\,= c_1<f,x^n>+\,c_2<g,x^n>.\end{equation}
To overpass the definition of $<f,g>$ for all $f,g\in \mathbb{L}$, while keeping the properties of a good inner product, I think it is sufficient to define: $$<x^n,f> \,:= \,\overline{<f,x^n>} $$
with this and (1), the inner product of $f$ vs $g$ can be defined as: \begin{align} <f,g> &:= \,<\sum_{m\geq0} \frac{f^{(m}(0)}{m!}x^m, \sum_{n\geq0} \frac{g^{(n}(0)}{n!}x^n>= \sum_{m,n\geq0}\Big(\frac{f^{(m}(0) \cdot \overline{g^{(n}(0)}}{m!n!}<x^m,x^n >\Big)=\\ &= \sum_{n \geq0}\frac{f^{(n}(0) \cdot \overline{g^{(n}(0)}}{n!^2} \end{align}
hence the norm of any entire funtion will be: $$||f||^2 = <f,f> =\sum_{n\geq 0} \frac{||f^{(n}(0)||^2}{n!^2}$$ For example $||e^x|| = \sqrt{\sum {1/n!^2}} \approx 1.509...$
Questions:
- Is this a good definition of an inner product?
- Is $\mathbb{L}$ complete respect the distance $d(f,g):=||f-g||$?
- Can this definition be extended for $<*,\frac{1}{x^n}>$ so it can cover more than $\mathbb{L}$ (for example functions with a Laurent series)?