What is the meaning of the Fourier integral in a Hilbert space?
If I have a function (vector) $f(x)\in L^2(-L,L)$ I can expand it on an orthonormal basis: $$ f=\sum_{k=-\infty}^{+\infty}c_ku_k, $$ where, for instance, the basis is $$u_k=\frac{1}{\sqrt{2L}}e^{-2\pi i k x/(2L)}.$$
The coefficients $c_k$ are then given by $$ c_k=(u_k,f)=\frac{1}{\sqrt{2L}}\int_{-L}^Lf(x)e^{2\pi i k x/(2L)}dx, $$
in which I used the integral definition of the scalar product.
But now, if $k$ is in the continuum, and we extend the period of the function to infinity, we have that the function $f(x)$ is given by the inverse transform
$$ f(x)=\int_{-\infty}^\infty g(k)e^{-2 \pi i k x}dk, $$
where, now, the role of the $c_k$ coefficients (i.e. the contribution of each single $f(x)$) is played by the Fourier transform
$$ g(k)=\int_{-\infty}^\infty f(x)e^{2\pi i k x}dx. $$
What is the relation between the latter integral and my Hilbert space? Do we have a continuum of vectors in the basis of the vector space?
Edit: I do not know if my question makes sense. I am trying to understand the meaning of the Fourier transform in Hilbert spaces.
But, for instance, I read that the definition of a complete basis $\{e_i\}$ in an infinite dimensional space Hilbert space $H$ (the extension of the finite dimension definition) is given using a discrete sum. $$ \forall x\in H \text{ we have that }||x_n-x||\rightarrow 0, $$ where $x_n$ is the Fourier series of $x$ with respect to $\{e_i\}$.
Also $$ x=\sum_{i=1}^\infty a_i e_i, \;\;\; a_i=(e_i,x). $$
What kind of meaning could have something in the continuum? Are we dealing with $L^2(\mathbb{R})$?
Moreover, $e^{-2\pi i k x}$, with $k$ in the continuum, is not even orthonormal.