This result proves the conjecture is true. Consider a stationary complex random function $\zeta(t)$. Stationarity requires that the autocorrelation function depend on the difference $t –t’$ only and that the mean value must be a constant (not a member of a distribution function nor depend on $t$). Assume the power spectral density (hereafter called the spectrum) is absolutely continuous everywhere. The Fourier transform of $\zeta(t)$ itself does not exist in general, because stationary random functions are generally neither absolutely integrable nor square integrable, i.e. they do not satisfy Dirichlet’s condition. The correct representation of the random variable $\zeta(t)$ is either as a Fourier series or as a Fourier-Stieltjes integral [see, for example, Akira Ishimaru, Wave Propagation and Scattering in Random Media, (IEEE Press, 1997) Appendix A] Here, the Fourier-Stieltjes integral will be used
$$\zeta(t) = \int_{-\infty}^{+\infty} e^{i\omega t}dA(\omega)$$
where $dA(\omega)$ is the random amplitude. Typically, the mean value and the autocorrelation function are found using ensemble averages, but this method provides little insight into the behavior of $dA(\omega)$. Hence, the average value and autocorrelation function for $\zeta(t)$ will be found using temporal averages. The ergodic hypothesis states that for a stationary process the temporal average and the ensemble average are equivalent.
$\mathbf Temporal Averages -$ The temporal average $\overline{\zeta(t)}$ is calculated over a finite interval $T$, and then the limit is taken as $T$ goes to infinity
$$\overline {\zeta(t)} = \lim_{T\to \infty} \frac 1T \int_{-T/2}^{+T/2}\zeta(t) dt$$
Substitute the Fourier-Stieltjes integral into the equation and perform the $t$ integration to get
$$\overline {\zeta(t)} = \lim_{T\to \infty} \frac 1T \int_{-\infty}^{+\infty} dA(\omega) \int_{-T/2}^{+T/2} e^{i\omega t}dt = \lim_{T\to \infty} \frac 1T \int_{-\infty}^{+\infty} dA(\omega) \frac {2sin(\omega T/2)} {\omega} $$
Use the following expressions to take the limit
$$ \lim_{T\to \infty} \frac {2\sin(\omega T/2)}{\omega} \rightarrow 2\pi\delta(\omega) \quad \mathrm{and} \quad \lim_{T\to \infty} \frac 1T = \lim_{T\to \infty} \frac {\Delta\omega} {2\pi}\rightarrow \frac {d\omega}{2\pi} $$
which gives for the temporal average
$$\overline {\zeta(t)} = \int_{-\infty}^{+\infty} dA(\omega) \ \delta(\omega) \ d\omega = dA(0) $$
Finally, for a zero-mean process
$$\overline {\zeta(t)} = \langle \zeta \rangle = dA(0) = 0 $$
Thus, the mean value of $\zeta(t)$ obtained via temporal averaging is $dA(0)$ which is the random amplitude evaluated at the origin.
In the following analysis the random variable $\zeta(t)$ will be decomposed into the sum of zero-mean random variable $\zeta_0(t)$ and the mean $\langle\zeta\rangle$ so that the mean value is everywhere explicit. Now the temporal average for the autocorrelation function of $\overline{\zeta(t)\zeta^*(t’)}$ is given by
$$ \overline{\zeta(t)\zeta^*(t+\tau)} = \overline{\zeta_0(t)\zeta_0^*(t+\tau)} + \langle \zeta \rangle^2 = \sigma^2C(\tau) + \langle \zeta \rangle^2 $$
$$ \lim_{T\to \infty} \frac 1T \iint_{-\infty}^{+\infty} e^{-i\omega\tau}dA(\omega)dA(\omega’) \int_{-T/2}^{+T/2} e^{i(\omega-\omega’)t}dt $$
Performing the $t$ integration yields
$$ \sigma^2C(\tau) + \langle \zeta \rangle^2 = \iint_{-\infty}^{+\infty} e^{-i\omega\tau}dA(\omega)dA(\omega’) \lim_{T\to \infty} \frac 1T \frac {2sin[(\omega-\omega’) T/2]} {(\omega-\omega’)} $$
The following expressions are used in the limit process
$$ \lim_{T\to \infty} \frac {2\sin[(\omega-\omega’) T/2]}{(\omega-\omega’)} \rightarrow 2\pi\delta(\omega-\omega’) \quad \mathrm{and} \quad \lim_{T\to \infty} \frac 1T = \lim_{T\to \infty} \frac {\Delta\omega} {2\pi}\rightarrow \frac {d\omega}{2\pi} $$
Substituting these expressions into the autocorrelation gives
$$ \sigma^2C(\tau) + \langle \zeta \rangle^2 = \iint_{-\infty}^{+\infty} e^{-i\omega\tau}dA(\omega)dA(\omega’) \delta(\omega-\omega’) \ d\omega $$
Finally, performing the $\omega$ integration yields
$$ \sigma^2C(\tau) + \langle \zeta \rangle^2 = \int_{-\infty}^{+\infty} e^{-i\omega'\tau}|dA(\omega’)|^2 \qquad \mathrm {Equation \ (1)}$$
The quantity $dA(\omega’)$ can be related to the mean value $\langle\zeta\rangle$ and spectrum $S(\omega)$ by taking the inverse Fourier transform of the autocorrelation function
$$ \int_{-\infty}^{+\infty}\sigma^2C(\tau)e^{i\omega\tau} \ d\tau + \int_{-\infty}^{+\infty} \langle\zeta \rangle^2 e^{i\omega\tau} \ d\tau = \int_{-\infty}^{+\infty}|dA(\omega’)|^2\int_{-\infty}^{+\infty}e^{i(\omega-\omega’)\tau}\ d\tau$$
Performing the $\tau$ integration yields
$$ S(\omega)+2\pi \langle \zeta \rangle^2 \delta(\omega)= 2\pi \int_{-\infty}^{+\infty} |dA(\omega’)|^2 \delta(\omega-\omega’) $$
Consider the case of $\omega = 0$ for a zero-mean process, i.e. $\langle\zeta\rangle = 0$; both sides of the expression reduce to
$$ S(0)+0 = 2\pi \int_{-\infty}^{+\infty} |dA(\omega’)|^2 \delta(\omega’) $$
The Dirac delta function forces the terms on the right side to be evaluated at $\omega’ = 0$ to produce
$$ S(0) = 2\pi |dA(0)|^2\delta(\omega’) = 2\pi \langle \zeta \rangle ^2 \delta(\omega’) =0 $$
where the previous result $dA(0) = \langle\zeta\rangle$ has been used. Thus, it is shown that the spectrum for a zero-mean process is zero when the frequency is zero, i.e. $S(0) = 0$. It can be seen in Equation (1) for the general case that the required value of $|dA(\omega’)|^2$ must be given by
$$|dA(\omega’)|^2 = \frac 1{2\pi}S(\omega’) \ d\omega’+|dA(0)|^2 \delta(\omega’) \ d\omega’ $$
$\mathbf Discussion -$ The vanishing of the spectrum at $\omega = 0$ establishes some new requirements on the autocorrelation function and spectrum. For a zero-mean random variable, when $\tau = 0$ the autocorrelation form of the Wiener-Khinchin theorem reduces to
$$ \sigma^2 = \frac 1{2\pi}\int_{-\infty}^{+\infty}S(\omega)\ d\omega $$
This expression has proven useful and has wide application. Now consider the spectrum form of the Wiener-Khinchin theorem when $\omega = 0$. This expression becomes
$$ S(0) = \int_{-\infty}^{+\infty}\sigma^2C(\tau)\ d\tau =0 $$
Besides the information on the behavior of the spectrum, this gives useful information on the autocorrelation function. For example, the autocovariance function must have both positive and negative excursions, and the area under these excursions must be equal otherwise the integral cannot be zero. This means that the simple Gaussian or exponential forms often used for the autocorrelation function are not valid for a zero mean process – they have no negative excursions anywhere.
