SpectralFactorizationTheorem2

Theorem [Spectral Factorization]: Let $$K : \mathbb{R} \to \mathbb{R}$$ be a positive definite stationary kernel function. By Bochner's theorem, there exists a non-negative spectral density function $$S : \mathbb{R} \to \mathbb{R}$$ such that:

$$K(t-s) = \frac{1}{2\pi} \int_{-\infty}^{\infty} S(\omega) e^{i\omega(t-s)} d\omega$$

Define $$h : \mathbb{R} \to \mathbb{R}$$ as:

$$h(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \sqrt{S(\omega)} e^{i\omega t} d\omega$$

Then:

$$K(t-s) = \int_{-\infty}^{\infty} h(t+\tau) h(s+\tau) d\tau$$

Proof: Starting with the right-hand side of the equation we want to prove:

$$\int_{-\infty}^{\infty} h(t+\tau) h(s+\tau) d\tau$$

Substituting the definition of $$h$$:

$$\int_{-\infty}^{\infty} \left(\frac{1}{2\pi} \int_{-\infty}^{\infty} \sqrt{S(\omega)} e^{i\omega(t+\tau)} d\omega\right) \left(\frac{1}{2\pi} \int_{-\infty}^{\infty} \sqrt{S(\nu)} e^{i\nu(s+\tau)} d\nu\right) d\tau$$

By Fubini's theorem, we can interchange the order of integration:

$$\frac{1}{4\pi^2} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \sqrt{S(\omega)S(\nu)} e^{i\omega t} e^{i\nu s} \left(\int_{-\infty}^{\infty} e^{i(\omega+\nu)\tau} d\tau\right) d\omega d\nu$$

The inner integral evaluates to:

$$\int_{-\infty}^{\infty} e^{i(\omega+\nu)\tau} d\tau = 2\pi\delta(\omega+\nu)$$

Substituting this back:

$$\frac{1}{2\pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \sqrt{S(\omega)S(\nu)} e^{i\omega t} e^{i\nu s} \delta(\omega+\nu) d\omega d\nu$$

For the $$\nu$$ integration:

$$\frac{1}{2\pi} \int_{-\infty}^{\infty} \sqrt{S(\omega)} e^{i\omega t} \left(\int_{-\infty}^{\infty} \sqrt{S(\nu)} e^{i\nu s} \delta(\omega+\nu) d\nu\right) d\omega$$

By the definition of the delta function:

$$\int_{-\infty}^{\infty} \sqrt{S(\nu)} e^{i\nu s} \delta(\omega+\nu) d\nu = \sqrt{S(-\omega)} e^{-i\omega s}$$

Substituting this result:

$$\frac{1}{2\pi} \int_{-\infty}^{\infty} \sqrt{S(\omega)} e^{i\omega t} \sqrt{S(-\omega)} e^{-i\omega s} d\omega$$

Since $$S(\omega)$$ is real and symmetric, $$S(\omega) = S(-\omega)$$, we have:

$$\sqrt{S(\omega)} \sqrt{S(-\omega)} = \sqrt{S(\omega)} \sqrt{S(\omega)} = S(\omega)$$

And for the exponentials:

$$e^{i\omega t} e^{-i\omega s} = e^{i\omega(t-s)}$$

Therefore:

$$\frac{1}{2\pi} \int_{-\infty}^{\infty} S(\omega) e^{i\omega(t-s)} d\omega$$

This is exactly the spectral representation of $$K(t-s)$$ that we started with, completing the proof.