ULE kernel adjoint symmetry and rate identities

The positive square root and Fourier integral defining $h=g$ are those supplied by the kernel contract. Its hypotheses give $G>0$ and finite $G,H$. Thus $\tau =H\mathrm{/}G$ and $\mathrm{\Gamma }=4\gamma G^2$ are well-defined and finite; direct cancellation gives $\mathrm{\Gamma }\tau =4\gamma GH$.

Let $f=u+iv$ be any complex integrable scalar function. By [F4], its real and imaginary parts $u,v$ are measurable and integrable. The function $-v$ is measurable by [F3], and $(-v{)}^{+}=v^{-}$ and $(-v{)}^{-}=v^{+}$. Thus the real-integral definition gives $\int (-v)=\int v^{-}-\int v^{+}=-\int v$. Applying the complex measurability criterion in [F4] to $u$ and $-v$ shows that $\bar{f}=u-iv$ is measurable. It has the same modulus as $f$, so it is integrable. The complex-integral definition now gives $\int \bar{f}=\int u+i\int (-v)=\int u-i\int v=\overline{\int f}$. This proves the conjugation rule from the integral definitions and the declared measurability criterion.

Fix real $t$ and indices $a,b$. The scalar Fourier integrand $f(\omega )=h_{ba}(\omega )e^{-i\omega t}$ is integrable: its modulus is $\mathrm{\mid }{h}_{ba}(\omega )\mathrm{\mid }$, which is integrable by [F1]. Here $A^{†}$ means the conjugate transpose, so $(A^{†}{)}_{ab}=\overline{A_{ba}}$. Apply step 1.2 to this integrand. Since $h(\omega )$ is Hermitian, $\overline{h_{ba}(\omega )}=h_{ab}(\omega )$, and hence $(h(t{)}^{†}{)}_{ab}=\overline{\int h_{ba}(\omega )e^{-i\omega t}d\omega }=\int h_{ab}(\omega )e^{i\omega t}d\omega =h_{ab}(-t)$. This holds for every entry, proving $h(-t)=h(t{)}^{†}$; the rate identities were proved in step 1.1.