Convergent jump integrals and Hermitian Lamb operators

The frequency entries in [F1] are in $L^1(\mathbb{R})$, so [F4] with $\xi =t\mathrm{/}(2\pi )$ makes every time-kernel entry continuous. Its real and imaginary parts have open inverse images of open sets, hence are Borel and therefore Lebesgue measurable by [F7] and [F8]. Translations and reflections in [F3] preserve measurability of the stated $X_{\alpha }$ entries and of the kernel entries. Lebesgue measure on $\mathbb{R}$ is sigma-finite, using the finite-measure intervals $[-m,m]$, $m\ge 1$.

Write $\mathcal{L}$ for the Lebesgue sigma-algebra on $\mathbb{R}$. If $f:\mathbb{R}\to \mathbb{R}$ is Lebesgue measurable and $B\subseteq \mathbb{R}$ is Borel, then $\{(u,v):f(u)\in B\}=f^{-1}(B)\times \mathbb{R}$ and $\{(u,v):f(v)\in B\}=\mathbb{R}\times f^{-1}(B)$. These are rectangles in $\mathcal{L}\otimes \mathcal{L}$ by [F7], proving the coordinate-pullback assertion directly. Apply this to the real and imaginary parts of the translated and reflected entries from step 1.1; [F8] then gives their complex measurability on this product sigma-algebra.

In particular the coordinate functions $u,v$ are product-measurable, and [F8] makes $d(u,v)=u-v$ measurable. The sets $E_{+}=\{d>0\}$ and $E_{-}=\{d<0\}$ are measurable. For a measurable set $E$, every inverse image under its indicator ${\mathbf{1}}_E$ is one of $\mathrm{\varnothing },E,E^c,{\mathbb{R}}^2$, so [F7] makes the indicator measurable. Thus $\mathrm{sgn}(u-v)={\mathbf{1}}_{E_{+}}-{\mathbf{1}}_{E_{-}}$ is measurable by [F8]. The same fact supplies finite sums and products of the lifted complex entries. Nonnegative square roots are measurable as well: for measurable $r\ge 0$, $\{\sqrt{r}>c\}=\{r>c^2\}$ when $c\ge 0$, and this set is the whole space when $c<0$. Consequently the mixed norm $a(u)=\mathrm{\parallel }h(u){\mathrm{\parallel }}_{2,1}$, the majorants used below, every entry of the jump integrand, and every entry of the Lamb integrand are measurable on their stated one-variable or product spaces. Product-measurable functions are measurable for the completion, as the latter sigma-algebra contains the former. Fixing either variable gives the same finite arithmetic constructions on measurable one-variable functions, so every section is measurable too. This establishes the measurability hypotheses needed for [F2].

Put $a(u)=\mathrm{\parallel }h(u){\mathrm{\parallel }}_{2,1}$ and $S(A)={\sum }_{a,b}\mathrm{\mid }{A}_{ab}\mathrm{\mid }$. Apply [F6] in each column $b$ to the real vectors $x_a=\mathrm{\mid }{A}_{ab}\mathrm{\mid }$ and $y_a=1$, $1\le a\le n$. This gives ${\sum }_a\mathrm{\mid }{A}_{ab}\mathrm{\mid }\le \sqrt{n}({\sum }_a\mathrm{\mid }{A}_{ab}{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}$; summing over $b$ yields $S(A)\le \sqrt{n}\mathrm{\parallel }A{\mathrm{\parallel }}_{2,1}$. Hence ${\sum }_{\alpha }\mathrm{\mid }{h}_{\lambda \alpha }(t-s)\mathrm{\mid }\le \sqrt{n}a(t-s)$. By [F3] with $s\mapsto t-s$, $\int a(t-s)ds=G$. The jump integrand has norm at most $\sqrt{\gamma n}a(t-s)$, so every entry is integrable. For a matrix-valued integrable function $M$ and unit vectors $z,w$, [F5] gives $\mathrm{\mid }{w}^{†}(\int M)z\mathrm{\mid }=\mathrm{\mid }\int w^{†}Mz\mathrm{\mid }\le \int \mathrm{\parallel }M\mathrm{\parallel }$; taking the supremum over $z,w$ yields $\mathrm{\parallel }\int M\mathrm{\parallel }\le \int \mathrm{\parallel }M\mathrm{\parallel }$. This proves absolute convergence and $\mathrm{\parallel }{K}_{\lambda }(t)\mathrm{\parallel }\le \sqrt{\gamma n}G$.

The entrywise product estimate is $S(AB)\le S(A)S(B)$, by expanding each entry and applying the finite triangle inequality. Since $\mathrm{\parallel }{X}_{\alpha }{X}_{\beta }\mathrm{\parallel }\le 1$ directly from the operator-norm definition, the norm of the numerator integrand ${\sum }_{\alpha \beta }{X}_{\alpha }(t+u)X_{\beta }(t+v){\varphi }_{\alpha \beta }(u,v)$ is at most $S(h(u)h(-v))\le na(u)a(-v)$. Apply the nonnegative part of [F2] to this measurable majorant. Its iterated integral is $n(\int a(u)du)(\int a(-v)dv)=n{G}^2<\mathrm{\infty }$, with the second factor evaluated by reflection in [F3]. This proves absolute convergence before any complex Fubini interchange. Each section is also integrable, since fixing either variable leaves a finite value of one factor times an integrable function of the other.

By the integrable part of [F2], applied entrywise using step 5.1, either order of integration gives the same numerator matrix $N(t)$ of $Q(t)$. By [F5], conjugation commutes with an absolutely convergent scalar integral: write $f=p+iq$ with real integrable $p,q$, and use $\int \bar{f}=\int p-i\int q=\overline{\int f}$. Linearity handles finite matrix sums, so adjoints can be taken inside $N(t)$.

The adjoint symmetry in [F1] and the real antisymmetric sign function give $\varphi (v,u{)}^{†}=(h(v)h(-u){)}^{†}\mathrm{sgn}(v-u)=-h(u)h(-v)\mathrm{sgn}(u-v)=-\varphi (u,v)$. Taking the adjoint of $N(t)$ reverses the two Hermitian $X$ factors. Fubini as justified in step 6.1 permits exchanging integration order; then relabel $u,v$ and $\alpha ,\beta$. The resulting scalar coefficient is $\overline{{\varphi }_{\beta \alpha }(v,u)}=-{\varphi }_{\alpha \beta }(u,v)$, so $N(t{)}^{†}=-N(t)$. Since $\gamma$ is real, $(\gamma N(t)\mathrm{/}(2i){)}^{†}=\gamma N(t)\mathrm{/}(2i)$, proving that $Q(t)$ is Hermitian, with the asserted convergence and jump bound already established.