Kramers–Kronig relations

Remarks

Proof idea: integrate $\chi(z)/(z - \omega)$ over the boundary of the half-disc of radius $R$ in $H$, with a small half-circle of radius $\varepsilon$ around $\omega$ removed. By Cauchy's theorem the integral is $0$. The big half-circle contributes at most $\pi \max \abs{\chi} \cdot R/(R - \abs{\omega}) \to 0$, and the small one tends to $-i\pi \chi(\omega)$. So $\mathrm{p.v.} \int \frac{\chi(\omega')}{\omega' - \omega} \, d\omega' = i\pi \chi(\omega)$, and the real and imaginary parts give the two relations. Example: $\chi(z) = \frac{1}{z + i}$ gives $\chi_1(\omega) = \frac{\omega}{\omega^2 + 1}$ and $\chi_2(\omega) = -\frac{1}{\omega^2 + 1}$, and these satisfy both relations. In terms of the Hilbert transform, each of $\chi_1$ and $\chi_2$ determines the other.

Going deeper

For a student

The relations are a direct consequence of Cauchy's integral formula on a half-plane. They say that the real and imaginary parts of the boundary values of a holomorphic function on $H$ that decays at infinity are Hilbert transforms of each other. For square-integrable boundary values, Titchmarsh's theorem makes this an equivalence with the Hardy space $H^2$ of the upper half-plane. They are used to check measured spectra, to recover phases from amplitudes, and as dispersion relations in particle physics.