I have two related questions on basic properties of stationary random functions. I'm following Uriel Frisch's book on Turbulence, Chapter 4, Section 4.5 (Spectrum of stationary random functions). Let $v(t)$ be a continuous stationary centered random function and define its standard and low-pass filtered Fourier transforms as: $ v(t) = \int_\mathbb{R} e^{ift} \hat{v}(f) \mathrm{d}f$ and $v_F(t)= \int_{|f|\leq F} e^{ift} \hat{v}(f) \mathrm{d}f$. I will be happy to work with distributions whenever necessary. Since $v(t)$ is stationary, then $v_F(t)$ and $[v_F(t)]^2$ are also stationary, hence they satisfy an ergodic theorem, i.e. their time average is equal to their ensemble average (denoted in the following by $\langle \cdot \rangle$). Now the author defines the cumulative energy spectrum as $$ \mathcal{E}(F) = \frac{1}{2} \langle [v_F(t)]^2\rangle. $$
First, the author claims that using Parceval's theorem, one can show that the cumulative energy spectrum is non-decreasing in $F$. I have tried to see this but I don't. What I tried so far is to plug in the definition of $v_F$ and its ergodic property:
$$\mathcal{E}(F) = \frac{1}{2} \lim_{T\to \infty} \frac{1}{2T} \int_{-T}^T \mathrm{d}t\int_{-F}^F \mathrm{d}f \int_{-F}^F \mathrm{d}f' e^{i(f+f')t}\hat{v}(f) \hat{v}(f')$$ Now, it seems tempting to use that $\int_{-\infty}^\infty e^{i(f+f')t} \mathrm{d}t= 2\pi \delta(f+f')$, but because of the $1/T$ factor, I don't see how.
Second, he defines what he calls the energy spectrum as $E(f) \equiv \frac{\mathrm{d}\mathcal{E}(f)}{\mathrm{d}f}\geq 0$ (it is the power-spectral-density if I am not mistaken). I am not so worried about the existence of the derivative, my question is rather: how to show that this is equivalent to other definitions of the power spectral density? As done here, one can alternatively define the power spectrum to be $$E(f) = \lim_{T\to\infty} \frac{1}{2\pi T} |\hat{v}_T(f)|^2,$$ where $\hat{v}_T(f)=\int_{-T}^T v(t) \exp(-ift) \mathrm{d}t$ (different definitions of Fourier trf from Frisch's book). My question here is: how to show that these two definitions are equivalent?
Since noone has answered my question, I will post a non-rigorous derivation that I came up with. It is easier to first answer the second question, the positivity of the resulting derivative implies the non-increasing property.
Integrate the relation $E(F)=\frac{\mathrm{d}\mathcal{E}(F)}{\mathrm{d}F}$ in $t$ from $-T/2$ to $T/2$ and divide by $T$. Then we get
$$E(F) = \frac{1}{2T} \frac{\mathrm{d}}{\mathrm{d}F}\left\langle \int_{-F}^F\int_{-F}^F \underbrace{\int_{-T/2}^{T/2} \exp(i(f+f')t) \mathrm{d}t}_{= 2T/2 \,\mathrm{sinc}[(f+f')T/2]} \,\hat{v}(f)\hat{v}(f') \,\mathrm{d}f \mathrm{d}f' \right\rangle $$ Now, for large T, we may replace $T/2\,\mathrm{sinc}[(f+f')T/2]\to \delta(f+f')$. This then allows us to compute one frequency integral, the differentiation eliminates another. At the same time, for large $T$, we may approximate the Fourier transforms by those to be based on a finite time interval $[-T/2,T/2]$. Formally $$ E(F) = \lim_{T\to\infty} \frac{2\pi}{2T} \frac{\mathrm{d}}{\mathrm{d}F} \int_{-F}^F \left\langle |\hat{v}_T(f)|^2 \right\rangle \mathrm{d}f = \lim_{T\to\infty} \frac{2\pi}{T} \left\langle |\hat{v}_T(F)|^2 \right\rangle, $$ where $v_T(f)=\int_{-T/2}^{T/2} v(t)\exp(ift)\mathrm{d}t$. This is the correct result for our Fourier transform convention. It is manifestly positive. Also, from it, one can derive the Wiener-Khinchin theorem.