Dispersion relations for an absolutely continuous integrable causal kernel

Fix an entry $k=K_{jk}$. On each $[0,N]$ the FTC representation of [L1] gives $k(s)=k(0)+{\int }_0^s{k}^{\mathrm{\prime }}(u)du$ for every $s\in [0,N]$. Since $k^{\mathrm{\prime }}\in L^1(0,\mathrm{\infty })$, the limit $\mathrm{\ell }:=k(0)+{\int }_0^{\mathrm{\infty }}{k}^{\mathrm{\prime }}(u)du$ exists and $k(s)\to \mathrm{\ell }$ as $s\to \mathrm{\infty }$, because $\mathrm{\mid }{k}^{\mathrm{\prime }}(u)\mathrm{\mid }{\mathbf{1}}_{[0,s]}(u)\le \mathrm{\mid }{k}^{\mathrm{\prime }}(u)\mathrm{\mid }$ and dominated convergence [L2] applies. If $\mathrm{\ell }\mathrm{\ne }0$ there is $s_0$ with $\mathrm{\mid }k(s)\mathrm{\mid }\ge \mathrm{\mid }\mathrm{\ell }\mathrm{\mid }\mathrm{/}2$ for all $s\ge s_0$, so monotonicity of the nonnegative integral [L2] gives ${\int }_0^N\mathrm{\mid }k\mathrm{\mid }\ge (\mathrm{\mid }\mathrm{\ell }\mathrm{\mid }\mathrm{/}2)(N-s_0)\to \mathrm{\infty }$, contradicting $k\in L^1$; hence $\mathrm{\ell }=0$ and $k(s)\to 0$.

For $\mathrm{Im}z\ge 0$ the integral defining the entry ${\chi }_{jk}(z)={\int }_0^{\mathrm{\infty }}k(s)e^{izs}ds$ converges absolutely with $\mathrm{\mid }{\chi }_{jk}(z)\mathrm{\mid }\le {\int }_0^{\mathrm{\infty }}\mathrm{\mid }k(s)\mathrm{\mid }{e}^{-\mathrm{Im}z\cdot s}ds\le \mathrm{\parallel }k{\mathrm{\parallel }}_1$, and it is continuous on the closed upper half-plane: if $z_n\to z$ in $\{\mathrm{Im}\ge 0\}$, then $k(s)e^{i{z}_{n}s}\to k(s)e^{izs}$ pointwise with majorant $\mathrm{\mid }k(s)\mathrm{\mid }$, so [L2] gives ${\chi }_{jk}(z_n)\to {\chi }_{jk}(z)$.

Write $z=x+iy$ on $H$. The parameter derivatives of the integrand $k(s)e^{izs}$ in $x$ and $y$ are $isk(s)e^{izs}$ and $-sk(s)e^{izs}$, whose absolute values are bounded by the measurable function $s\mathrm{\mid }k(s)\mathrm{\mid }$, integrable by hypothesis; Differentiation under the integral sign applied on a bounded rectangle around any point of $H$ therefore gives ${\mathrm{\partial }}_x{\chi }_{jk}(z)={\int }_0^{\mathrm{\infty }}isk(s)e^{izs}ds$ and ${\mathrm{\partial }}_y{\chi }_{jk}(z)=-{\int }_0^{\mathrm{\infty }}sk(s)e^{izs}ds$, and both partials are continuous on $H$ by [L2]. Since ${\mathrm{\partial }}_y{\chi }_{jk}=-{\int }_0^{\mathrm{\infty }}sk(s)e^{izs}ds=i{\mathrm{\partial }}_x{\chi }_{jk}$, the Cauchy-Riemann equations hold with continuous partials, and Continuous first partial derivatives and the Cauchy–Riemann equations imply complex differentiability pointwise, and holomorphy when they hold throughout an open set makes ${\chi }_{jk}$ holomorphic on the upper half-plane.

Along the real boundary, fix $\omega \in \mathbb{R}$: for $t$ in a bounded interval around $\omega$ the integrand $k(s)e^{its}$ has $t$-derivative $isk(s)e^{its}$, bounded by the integrable $s\mathrm{\mid }k(s)\mathrm{\mid }$, so Differentiation under the integral sign gives ${\chi }_{jk}^{\mathrm{\prime }}(t)={\int }_0^{\mathrm{\infty }}isk(s)e^{its}ds$ for every such $t$, and this derivative is continuous in $t$ by dominated convergence [L2]. Thus ${\chi }_{jk}$ is continuous on the closed upper half-plane (step 1.2) and $C^1$ on its real boundary.

For $z\mathrm{\ne }0$ with $\mathrm{Im}z\ge 0$ and $N>0$, apply Integration by parts for absolutely continuous functions on $[0,N]$ to the absolutely continuous pair $(k,\mathrm{Re}{e}^{izs})$ and to $(k,\mathrm{Im}{e}^{izs})$; the function $s\mapsto e^{izs}$ is $C^1$ hence absolutely continuous by [L1], and combining the two real identities gives ${\int }_0^{N}k(s)e^{izs}ds=\frac{k(N)e^{izN}-k(0)}{iz}-\frac{1}{iz}{\int }_0^N{k}^{\mathrm{\prime }}(s)e^{izs}ds$. Since $\mathrm{\mid }{e}^{izN}\mathrm{\mid }\le 1$ for $\mathrm{Im}z\ge 0$ and $k(N)\to 0$ by step 1.1, dominated convergence [L2] with majorant $\mathrm{\mid }{k}^{\mathrm{\prime }}\mathrm{\mid }$ gives, in the limit $N\to \mathrm{\infty }$, ${\chi }_{jk}(z)=-\frac{k(0)}{iz}-\frac{1}{iz}{\int }_0^{\mathrm{\infty }}{k}^{\mathrm{\prime }}(s)e^{izs}ds$; hence $\mathrm{\mid }{\chi }_{jk}(z)\mathrm{\mid }\le \frac{\mathrm{\mid }k(0)\mathrm{\mid }+\mathrm{\parallel }{k}^{\mathrm{\prime }}{\mathrm{\parallel }}_1}{\mathrm{\mid }z\mathrm{\mid }}$ for every nonzero $z$ with $\mathrm{Im}z\ge 0$, in particular $\mathrm{\mid }{\chi }_{jk}(z)\mathrm{\mid }\le \mathrm{\mid }k(0)\mathrm{\mid }+\mathrm{\parallel }{k}^{\mathrm{\prime }}{\mathrm{\parallel }}_1$ for $\mathrm{\mid }z\mathrm{\mid }\ge 1$, and ${\chi }_{jk}(z)=O(1\mathrm{/}\mathrm{\mid }z\mathrm{\mid })$ uniformly on the closed upper half-plane.

Let $F={\chi }_{jk}$, which is holomorphic on $H$, continuous on the closed upper half-plane and $O(1\mathrm{/}\mathrm{\mid }z\mathrm{\mid })$ there by steps 1.3, 1.2 and 2.2. Fix $\omega \in \mathbb{R}$, choose $R>\mathrm{\mid }\omega \mathrm{\mid }+2$ and $0<\eta <1$, and let $\gamma$ be the closed contour consisting of the real segments $[\omega -R,\omega -\eta ]$ and $[\omega +\eta ,\omega +R]$ traversed from left to right, the clockwise small upper semicircle $z=\omega +\eta e^{i\theta }$ with $\theta$ decreasing from $\pi$ to $0$, and the counterclockwise large upper semicircle $z=\omega +R{e}^{i\theta }$ with $\theta$ increasing from $0$ to $\pi$; every piece is rectifiable and the contour is closed. For $\delta >0$ put $f_{\delta }(z)=F(z+i\delta )\mathrm{/}(z-\omega )$ on the open set $D_{\delta }=\{z:\mathrm{Im}z>-\delta \mathrm{/}2\}\setminus \{\omega -iv:v\ge 0\}$: this set is star-shaped about $\omega +iA$ for $A>1$, because the removed ray lies in $\{\mathrm{Im}\le 0\}$ while every point of $\gamma$ has $\mathrm{Im}\ge 0$ and different from $\omega$, so the segment from $\omega +iA$ to it stays in $D_{\delta }$. On $D_{\delta }$ the function $F(\cdot +i\delta )$ is holomorphic (translate of a holomorphic function) and $1\mathrm{/}(z-\omega )$ is holomorphic off $\omega$, and $\gamma$ lies in $D_{\delta }$; Cauchy's theorem on a star-shaped domain: every closed rectifiable contour integral of a holomorphic function is zero therefore gives ${\int }_{\gamma }{f}_{\delta }=0$. As $\delta \downarrow 0$ the integrands $f_{\delta }$ converge uniformly on the trace of $\gamma$ to $F(z)\mathrm{/}(z-\omega )$: the trace is compact, all shifted points lie in a compact subset of the closed upper half-plane, and $F$ is uniformly continuous there by Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous; so A uniformly convergent sequence of continuous integrands on a fixed contour permits passage of the limit through the complex line integral gives ${\int }_{\gamma }F(z)\mathrm{/}(z-\omega )dz=0$.

Estimate the pieces of the identity of step 3.1, using $\mathrm{\mid }F(z)\mathrm{\mid }\le C\mathrm{/}\mathrm{\mid }z\mathrm{\mid }$ from step 2.2. On the large arc $\mathrm{\mid }z-\omega \mathrm{\mid }=R$ and $\mathrm{\mid }z\mathrm{\mid }\ge R-\mathrm{\mid }\omega \mathrm{\mid }$, so the integrand is at most $C\mathrm{/}(R(R-\mathrm{\mid }\omega \mathrm{\mid }))$ and ML estimate: a contour integral is bounded by a supremum bound times path length bounds the contribution by $\pi RC\mathrm{/}(R(R-\mathrm{\mid }\omega \mathrm{\mid }))\to 0$ as $R\to \mathrm{\infty }$. On the small arc, parametrizing $z=\omega +\eta e^{i\theta }$ with $\theta$ decreasing from $\pi$ to $0$ gives $\int =-i{\int }_0^{\pi }F(\omega +\eta e^{i\theta })d\theta \to -i\pi F(\omega )$ as $\eta \downarrow 0$ by continuity of $F$. The two real segments contribute ${\int }_{\eta <\mathrm{\mid }t\mathrm{\mid }<R}F(\omega +t)\mathrm{/}tdt$ with $t$ real; on $\eta <\mathrm{\mid }t\mathrm{\mid }\le 1$ subtract $F(\omega )$, whose quotient has zero integral on this symmetric punctured interval. The difference quotient $[F(\omega +t)-F(\omega )]\mathrm{/}t$ is bounded there and tends to $F^{\mathrm{\prime }}(\omega )$ by the boundary $C^1$ property (step 2.1). On the tails keep the ORIGINAL integrand $F(\omega +t)\mathrm{/}t$: for $\mathrm{\mid }t\mathrm{\mid }\ge \max (1,2\mathrm{\mid }\omega \mathrm{\mid })$ its modulus is at most $2C\mathrm{/}\mathrm{\mid }t{\mathrm{\mid }}^2$ by step 2.2, while the intervening compact intervals are regular. Thus the principal value exists by local cancellation and its two tails converge absolutely. Passing to the limit first in $\eta$ and then in $R$ in the identity ${\int }_{\gamma }F\mathrm{/}(z-\omega )=0$ therefore yields $\mathrm{PV}{\int }_{\mathbb{R}}\frac{F(\xi )}{\xi -\omega }d\xi =i\pi F(\omega )$, and the tails converge absolutely, so the centering at infinity does not affect the limit.

Write $F=u+iv$ with $u=\mathrm{Re}F$, $v=\mathrm{Im}F$. Splitting the identity of step 4.1 into real and imaginary parts gives $\mathrm{PV}{\int }_{\mathbb{R}}\frac{u(\xi )}{\xi -\omega }d\xi +i\mathrm{PV}{\int }_{\mathbb{R}}\frac{v(\xi )}{\xi -\omega }d\xi =i\pi (u(\omega )+iv(\omega ))=-\pi v(\omega )+i\pi u(\omega )$; comparing parts, $u(\omega )=\frac{1}{\pi }\mathrm{PV}{\int }_{\mathbb{R}}\frac{v(\xi )}{\xi -\omega }d\xi$ and $v(\omega )=-\frac{1}{\pi }\mathrm{PV}{\int }_{\mathbb{R}}\frac{u(\xi )}{\xi -\omega }d\xi$.

Each entry $k=K_{jk}$ is locally absolutely continuous with $k,k^{\mathrm{\prime }}\in L^1(0,\mathrm{\infty })$ and $sk(s)\in L^1(0,\mathrm{\infty })$ by the hypotheses on $K$, so steps 1.3, 2.1, 2.2 and 5.1 apply to it: the entries of $\chi$ are holomorphic on the upper half-plane, continuous on the closed upper half-plane and $C^1$ on the real boundary, obey $\mathrm{\mid }{\chi }_{jk}(z)\mathrm{\mid }\le (\mathrm{\mid }k(0)\mathrm{\mid }+\mathrm{\parallel }{k}^{\mathrm{\prime }}{\mathrm{\parallel }}_1)\mathrm{/}\mathrm{\mid }z\mathrm{\mid }$ for nonzero $z$ with $\mathrm{Im}z\ge 0$ (hence $\mathrm{\mid }{\chi }_{jk}(z)\mathrm{\mid }\le \mathrm{\mid }k(0)\mathrm{\mid }+\mathrm{\parallel }{k}^{\mathrm{\prime }}{\mathrm{\parallel }}_1$ for $\mathrm{\mid }z\mathrm{\mid }\ge 1$), and satisfy the two displayed principal-value dispersion formulas. For an instantaneous real matrix ${\chi }_{\mathrm{\infty }}$ the identities applied to $\chi ={\chi }_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}-{\chi }_{\mathrm{\infty }}$ give the stated form of the total transfer function; the scalar case is $d=1$. Nothing in the argument uses or produces a passivity quadratic-form condition, and no growth assumption beyond the stated integrability of $K,K^{\mathrm{\prime }},sK$ is made.