A is selfadjoint: First, it must be shown that $A$ is a closed densely-defined selfadjoint operator on its domain $\mathcal{D}(A)$. To do this it suffices to show that
- $A$ is symmetric on its domain
- For $f \in L^{2}(\mathbb{R})$ there exist solutions $g_{\pm}\in\mathcal{D}(A)$ of $(A\pm iI)g_{\pm}=f$.
A problem has been posted with solution that shows the above conditions implie that $A$ is densely-defined and selfadjoint. ( Reference Why is this operator self-adoint )
Showing $A$ is symmetric: To show that $A$ is symmetric on its domain, suppose that $f,g\in L^{2}(\mathbb{R})$ are absolutely continuous with $f',g' \in L^{2}(\mathbb{R})$. Then
$$
\left.\int_{a}^{b}(Af)\overline{g}-f(\overline{Ag})\,dx=\int_{a}^{b}f'\overline{g}+f\overline{g}'\,dt=f\overline{g}\right|_{a}^{b}
$$
The integral on the left converges because $Af$,$f$,$Ag$,$g$ are square-integrable, and this integral converges as $b\rightarrow \infty$, $a\rightarrow -\infty$. That means that the evaluation terms on the right also have such limits. The evaluation limits must be $0$ because $f\overline{g}\in L^{1}(\mathbb{R})$. Hence,
$$
(Af,g)=(f,Ag),\;\;\; f,g \in\mathcal{D}(A).
$$
Deriving the Resolvent Operator:
Suppose $\lambda\notin\mathbb{R}$ and $f\in L^{2}(\mathbb{R})$. According to the plan, we find $g\in\mathcal{D}(A)$ such that
$$
(A-\lambda I)g = f \\
g'-i\lambda g = if \\
(e^{-i\lambda t}g)'= ie^{-i\lambda t}f
$$
Now the solution breaks into two cases: $\Im\lambda > 0$ and $\Im\lambda < 0$. If $\Im\lambda > 0$, then $e^{-i\lambda t}$ is integrable on $(-\infty,x]$. So, for this case, we choose the solution
$$
e^{-i\lambda t}g(t)= i\int_{-\infty}^{t}e^{-i\lambda u}f(u)\,du,\\
g(t) = i\int_{-\infty}^{t}e^{i\lambda(t-u)}f(u)\,du.
$$
In other words, the proposed solutions $g$ are
$$
g = R(\lambda)f = i\int_{-\infty}^{t}e^{i\lambda(t-x)}f(x)\,dx,\;\; \Im\lambda > 0,\\
g = R(\lambda)f = -i\int_{t}^{\infty}e^{i\lambda(t-x)}f(x)\,dx,\;\;\Im\lambda < 0.
$$
It can be shown that $g \in L^{2}(\mathbb{R})$ by careful application of Cauchy-Schwarz. Knowing this, it is apparent that $A$ is selfadjoint because $(A\pm iI)$ are surjective, which is enough to guarantee that $\mathcal{D}(A)$ is dense in $L^{2}(\mathbb{R})$ and $A=A^{\star}$ on this domain.
Applying Stone's Formula: Now consider
$$
R(s+i\epsilon)f-R(s-i\epsilon)f=i\int_{-\infty}^{\infty}e^{-\epsilon|t-x|}e^{is(t-x)}f(x)\,dx
$$
For convenience, suppose that $f$ is compactly supported in order to avoid convergence issues. This restriction is easily removed once the final result is obtained. Dividing by $2\pi i$ and integrating the above in $s$ over $[a,b]$ gives
$$
\frac{1}{2}\{E[a,b]+E(a,b)\}f = \lim_{\epsilon\downarrow 0}\frac{1}{2\pi}\int_{a}^{b}\int_{-\infty}^{\infty}e^{-\epsilon|t-x|}e^{-isx}f(x)\,dx\,ds.
$$
The convergence of this limit is guaranteed to take place in the norm of $L^{2}(\mathbb{R})$. It is easy to see that the spectral measure has no atoms so that $E[a,b]=E(a,b)$, and
$$
E[a,b]f = \frac{1}{2\pi}\int_{a}^{b}\left[\int_{-\infty}^{\infty}e^{-isx}f(x)\,dx\right]e^{ist}\,ds.
$$
The connection between the spectral measure for $A$ and the Fourier transform with its inverse is apparent.
Theorem ($L^{2}$ Fourier Transform): Let $f\in L^{2}(\mathbb{R})$. Then the following limit exists in $L^{2}(\mathbb{R})$
$$
\mathscr{F}f=\lim_{r\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\int_{-r}^{r}f(x)e^{-isx}\,dx
$$
Furthermore, $\mathscr{F}$ is a unitary operator on $L^{2}(\mathbb{R})$ whose inverse is given by
$$
\mathscr{F}^{-1} f= \lim_{r\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\int_{-r}^{r}f(s)e^{ist}\,ds.
$$
The spectral measure $E$ for the operator $A$ on an interval $[a,b]$ is given by
$$
E[a,b]f = \mathscr{F}^{-1}\chi_{[a,b]}\mathscr{F}f,
$$
where $\chi_{[a,b]}$ is the characteristic function of $[a,b]$. Therefore,
$$
Af = \mathscr{F}^{-1}M \mathscr{F}f
$$
where $M$ is the multiplication operator on $L^{2}(\mathbb{R})$ defined by $(Mf)(s)=sf(s)$. Thus, $f \in \mathcal{D}(A)$ iff $s\hat{f}(s) \in L^{2}(\mathbb{R})$ where $\hat{f}(s)=\mathscr{F}f$ is the Fourier transform of $f$. In other words, a function $f\in L^{2}$ is absolutely continuous with $L^{2}$ derivative iff $s\hat{f}(s)$ is in $L^{2}$!
Proof:
Define $\mathscr{F}$ as in the theorem. For compactly supported $f\in L^{2}$, the operator $\mathscr{F}f$ is clearly defined pointwise, and it has been shown that
$$
E[a,b]f = \frac{1}{\sqrt{2\pi}}\int_{a}^{b}(\mathscr{F}f)(s)e^{ist}\,ds.
$$
The above is guaranteed to be in $L^{2}$. In fact, because $E$ is the spectral measure for $A$,
$$
\|E[a,b]f\|^{2}=(E[a,b]f,f)=\frac{1}{\sqrt{2\pi}}\int_{a}^{b}(\mathscr{F}f)(s)\int_{-\infty}^{\infty}\overline{f(t)e^{-ist}}\,dt=\int_{a}^{b}|\mathscr{F}f|^{2}\,ds.
$$
By properties of the spectral measure, $\lim_{r\rightarrow\infty}E[-r,r]f=f$, where the convergence is in $L^{2}$. Therefore, the following limit exists and is finite:
$$
\|f\|^{2} = \int_{-\infty}^{\infty}|\mathscr{F}f|^{2}\,dx.
$$
In particular, $\mathscr{F} : L^{2}\rightarrow L^{2}$ is an isometry, which is not yet known to be unitary. This is Parseval's Equality for the $L^{2}$ Fourier transform $\mathscr{F}$. Furthermore, from the first equation of the proof, it is seen that the following limit also exists in $L^{2}$:
$$
f = \lim_{r\rightarrow\infty}E[-r,r]f = \lim_{r\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\int_{-r}^{r}(\mathscr{F}f)(s)e^{ist}\,ds
$$
So this is the inverse $\mathscr{F}^{-1}$ of the Fourier transform $\mathscr{F}$. By the symmetry of these expressions, it is easy to see that the two operators invert each other in either direction. So they are inverses, which means that $\mathscr{F}$ is a unitary map from $L^{2}$ to $L^{2}$, and the spectral measure of $A$ for an interval $[a,b]$ is
$$
E[a,b]f = \mathscr{F}^{-1}\chi_{[a,b]}\mathscr{F}f.
$$
So differentiation is unitarily equivalent to the multiplication operator $Mf=sf(s)$ on $L^{2}(\mathbb{R})$. The Borel functional calculus is
$$
F(A)f = \int F(\lambda)dE(\lambda)f = \mathscr{F}^{-1}M_{F}\mathscr{F}f
$$
where $M_{F}$ is multiplication by the Borel function $F$. Part of the statement of the Spectral Theorem is that the domain of $A$ consists of the set of all $f\in L^{2}$ such that
$$
\int \lambda^{2}d\|E(\lambda)f\|^{2} = \int \lambda^{2}|\mathscr{F}f|^{2}\,d\lambda < \infty.
$$
This alone is a remarkable result: A function $f \in L^{2}$ is absolutely continuous on $\mathbb{R}$ with $f'\in L^{2}$ iff $s\hat{f}(s) \in L^{2}$, where $\hat{f}$ is the Fourier transform of $f$. $\;\;\Box$
Best Answer
No, the amplitude spectrum is a nonnegative function (as you can see in the paper too), which does not have an imaginary part. One can represent a complex number, such as $S(\omega)$, as $re^{i\phi}$ with $r\ge 0$ and $\phi\in [0,2\pi)$. Here $r$ is the amplitude and $\phi$ is the phase. The amplitude spectrum is the plot of the absolute value of Fourier transform, while the phase spectrum is the plot of the phase (argument). The prevailing idea is that the amplitude spectrum is the useful part, and one would like to recover the signal from it as much as possible.