The Laplace transform
$$F(s) = \int^{∞}_{0}f(t)e^{-st} dt$$ can be understood much like the fourier transform, as a change of basis of an $L^2$ function to the eigen functions of the differential operator $\frac{d}{dt}$
Unlike the complex exponentials, which form an uncountable orthogonal basis, the $e^{-st}$ basis functions are not in general orthogonal to each other. Does this mean we are switching to a frame and not a minimal orthogonal basis? Does this at all imply were overcounting? Or is there just redundant information in our resulting function?
The operator $L_0f = -if'$ is symmetric on the domain $\mathcal{D}(L_0)$ of functions $f\in L^2[0,\infty)$ that are absolutely continuous on $[0,\infty)$, that vanish at $0$, and that satisfy $f'\in L^2[0,\infty)$. The operator $L_0$ is maximally symmetric, which means that it is closed and symmetric, but it is has no proper symmetric extension.
$L_0^*$ is not symmetric, but it is an extension of $L_0$. Indeed, $L_0^*f=-if'$ has the same domain except that the functions in the domain of $L_0^*$ do not necessarily vanish at $0$. The operator $L_0^*$ has spectrum equal to the upper half plane becase $e^{isx} \in L^2[0,\infty)$ for $\Im s > 0$ and $L_0^*e^{isx}=-i(-is e^{-isx})=se^{isx}$. The real axis is also in the spectrum because the spectrum is closed.
It is possible to extend the holomorphic functiona calculus to functions $F$ that are holomorphic on a neighborhood of the upper half plane: $$ F(L_0^*)f=\frac{1}{2\pi i}\int_{-i\epsilon-\infty}^{-i\epsilon+\infty}F(s)(s I -L_0^*)^{-1}fds. $$ This extends to the case where $\epsilon=0$ if $F$ is a bounded holomorphic function in the upper half plane with a.e. boundary function $F(s)$ on the real axis. If you turn everything clockwise in the complex plane by looking at $L_0 f = -f'$ instead of $-if'$, then the operator calculus is implemented by a Fourier transform for the outer integral and a Laplace transform for the inner integral. The functional calculus allows for the computation of $F(L_0^*)f$ if $F$ is a bounded holomorphic function in the right half plane.