Causal convolution response, transfer function and dispersion relations

Fix $t\in \mathbb{R}$, $j\in \{1,\dots ,d\}$ and set $g_j(s):={\sum }_{k=1}^d{K}_{jk}(s)E_k(t-s)$ for $s>0$. Each summand is a product of the measurable function $K_{jk}$ with the continuous, hence Borel measurable, function $s\mapsto E_k(t-s)$; so $g_j$ is measurable. With $M:={\max }_k{\sup }_{\mathbb{R}}\mathrm{\mid }{E}_k\mathrm{\mid }<+\mathrm{\infty }$ one has $\mathrm{\mid }{g}_j(s)\mathrm{\mid }\le M{\sum }_{k=1}^d\mathrm{\mid }{K}_{jk}(s)\mathrm{\mid }$ for every $s>0$, and ${\sum }_{k=1}^d\mathrm{\mid }{K}_{jk}\mathrm{\mid }$ lies in $L^1(0,\mathrm{\infty })$; hence $g_j\in L^1(0,\mathrm{\infty })$ and $P_j(t):={\int }_0^{\mathrm{\infty }}{g}_j(s)ds$ converges absolutely. The termwise bound $\mathrm{\mid }{g}_j(s)\mathrm{\mid }\le {\sum }_{k=1}^d({\sup }_{\mathbb{R}}\mathrm{\mid }{E}_k\mathrm{\mid })\mathrm{\mid }{K}_{jk}(s)\mathrm{\mid }$ gives the exact stated estimate $\mathrm{\mid }{P}_j(t)\mathrm{\mid }\le {\sum }_{k=1}^d({\sup }_{\mathbb{R}}\mathrm{\mid }{E}_k\mathrm{\mid }){\int }_0^{\mathrm{\infty }}\mathrm{\mid }{K}_{jk}(s)\mathrm{\mid }ds\le M{\sum }_{k=1}^d{\int }_0^{\mathrm{\infty }}\mathrm{\mid }{K}_{jk}(s)\mathrm{\mid }ds$.

Fix $z=x+iy\in \mathbb{C}$. The map $s\mapsto e^{izs}$ is continuous, since it is the composition of the continuous linear map $s\mapsto izs$ with the continuous exponential, and $K_{jk}$ is measurable; hence $s\mapsto K_{jk}(s)e^{izs}$ is measurable as a product of measurable functions. On $\mathrm{supp}K\subseteq (0,b)$ one has $\mathrm{\mid }{e}^{izs}\mathrm{\mid }=\mathrm{\mid }{e}^{ixs}\mathrm{\mid }{e}^{-ys}=e^{-ys}\le e^{b\mathrm{\mid }y\mathrm{\mid }}$, so $\mathrm{\mid }{K}_{jk}(s)e^{izs}\mathrm{\mid }\le e^{b\mathrm{\mid }y\mathrm{\mid }}\mathrm{\mid }{K}_{jk}(s)\mathrm{\mid }$; the right-hand side is integrable, so the integrand is integrable and ${\chi }_{jk}(z):={\int }_0^{\mathrm{\infty }}{K}_{jk}(s)e^{izs}ds$ is well defined with $\mathrm{\mid }{\chi }_{jk}(z)\mathrm{\mid }\le e^{b\mathrm{\mid }y\mathrm{\mid }}\mathrm{\parallel }{K}_{jk}{\mathrm{\parallel }}_1$.

Clause (c) is exactly the supplied integrable-branch theorem [F14]: its hypotheses on $K$ are those stated in clause (c) (local absolute continuity entrywise in the sense of [F13], $K,K^{\mathrm{\prime }},sK\in L^1$), and it concludes the well-definedness and holomorphy of $\chi$ on the open upper half-plane, continuity on the closed upper half-plane and continuous first derivative on the real boundary, the entrywise principal-value formulas for every real $\omega$, and the bound $\mathrm{\mid }{\chi }_{jk}(z)\mathrm{\mid }\le \mathrm{\mid }{K}_{jk}(0)\mathrm{\mid }+\mathrm{\parallel }{K}_{jk}^{\mathrm{\prime }}{\mathrm{\parallel }}_1$ for $\mathrm{\mid }z\mathrm{\mid }\ge 1$ (which follows from its sharper $\mathrm{\mid }z{\mathrm{\mid }}^{-1}$ bound), together with the instantaneous-response subtraction for a real matrix ${\chi }_{\mathrm{\infty }}$ and the scalar case $d=1$. Continuity on the closed upper half-plane combines the lemma's holomorphy on the open half-plane, hence continuity there, with its continuity at every real boundary point.

Write $a_j={\alpha }_j+i{\beta }_j$ and $e_j=p_j+i{q}_j$ for the components of $a={\epsilon }_0\chi (\omega )e$ and $e$. By [F7], $e^{i\omega t}=\cos (\omega t)+i\sin (\omega t)$, and $\mathrm{Re}(w)=\mathrm{Re}(\bar{w})$ for every $w\in \mathbb{C}$; therefore $E_j(t)=\mathrm{Re}(e_j{e}^{-i\omega t})=\mathrm{Re}(\overline{e_j}{e}^{i\omega t})=p_j\cos (\omega t)+q_j\sin (\omega t)$ and $P_j(t)=\mathrm{Re}(a_j{e}^{-i\omega t})=\mathrm{Re}(\overline{a_j}{e}^{i\omega t})={\alpha }_j\cos (\omega t)+{\beta }_j\sin (\omega t)$, using the component computation $\mathrm{Re}((u+iv)(\cos +i\sin ))=u\cos -v\sin$.

The estimate of step 1.1 is uniform in $t$, so ${\sup }_{t\in \mathbb{R}}\mathrm{\mid }{P}_j(t)\mathrm{\mid }\le M{\sum }_{k=1}^d\mathrm{\parallel }{K}_{jk}{\mathrm{\parallel }}_1$ and $P=(P_1,\dots ,P_d)$ is bounded. The map $E\mapsto P$ is linear: for $\alpha ,\beta \in \mathbb{R}$ and bounded continuous $E,F$ the integrand defining $P_{\alpha E+\beta F}$ equals $\alpha$ times that of $P_E$ plus $\beta$ times that of $P_F$ pointwise, so the linearity of the scalar integral gives $P_{\alpha E+\beta F}=\alpha P_E+\beta P_F$.

Fix $j$ and apply [F5] with $X=(0,\mathrm{\infty })$ carrying Lebesgue measure, $I=\mathbb{R}$ and $f(s,t):={\sum }_{k=1}^d{K}_{jk}(s)E_k(t-s)$: for every $t$ the function $f(\cdot ,t)$ is integrable by step 1.1; for every $s$ the map $t\mapsto f(s,t)$ is continuous, being a finite sum of products of constants with the continuous functions $t\mapsto E_k(t-s)$; and $\mathrm{\mid }f(s,t)\mathrm{\mid }\le M{\sum }_{k=1}^d\mathrm{\mid }{K}_{jk}(s)\mathrm{\mid }$ for the single integrable function named in step 1.1. Therefore $t\mapsto P_j(t)$ is continuous on $\mathbb{R}$, and $P$ is continuous componentwise.

Causality: if $E_1,E_2$ are bounded continuous and agree on $(-\mathrm{\infty },t_0]$, and if $t\le t_0$, then for every $s>0$ the argument $t-s$ lies in $(-\mathrm{\infty },t_0]$, so the integrands of step 1.1 agree pointwise and hence $P_{E_1}(t)=P_{E_2}(t)$; thus the response at time $t$ depends only on the history up to time $t$.

Fix $y$ and apply [F6] in the parameter $x$ to $f(s,x):=K_{jk}(s)e^{i(x+iy)s}$ on a bounded interval of $x$-values: for each $x$ the function $f(\cdot ,x)$ is integrable by step 1.2, the map $x\mapsto K_{jk}(s)e^{i(x+iy)s}$ is differentiable for every $s$ with derivative $is{K}_{jk}(s)e^{i(x+iy)s}$, and on the bounded parameter set this derivative is dominated by $(1+b)e^{b\mathrm{\mid }y\mathrm{\mid }}\mathrm{\mid }{K}_{jk}(s)\mathrm{\mid }$, an integrable function. Hence $x\mapsto {\chi }_{jk}(x+iy)$ is differentiable with ${\mathrm{\partial }}_x{\chi }_{jk}(z)={\int }_0^{\mathrm{\infty }}is{K}_{jk}(s)e^{izs}ds$; exchanging the roles of $x$ and $y$ and repeating the same argument gives ${\mathrm{\partial }}_y{\chi }_{jk}(z)={\int }_0^{\mathrm{\infty }}(-s)K_{jk}(s)e^{izs}ds$. Both partials are continuous in $(x,y)$: if $(x_n,y_n)\to (x,y)$, the integrands $is{K}_{jk}(s)e^{i(x_n+i{y}_n)s}$ converge pointwise to $is{K}_{jk}(s)e^{izs}$ and are dominated on a bounded parameter neighbourhood by $(1+b)e^{bY}\mathrm{\mid }{K}_{jk}(s)\mathrm{\mid }$ for a suitable $Y$, so [F4] gives convergence of the integrals.

Reality identity: for real $\omega$, conjugating the partial sums of the absolutely convergent series defining $e^{i\omega s}$ term by term gives ${\sum }_{n\le m}\overline{(i\omega s{)}^n\mathrm{/}n!}={\sum }_{n\le m}(-i\omega s{)}^n\mathrm{/}n!$, because conjugation is a continuous real-field automorphism with $\bar{i}=-i$ and fixes the real numbers $1\mathrm{/}n!$; passing to the limit with continuity of conjugation gives $\overline{e^{i\omega s}}=e^{-i\omega s}$. Since $K_{jk}$ is real, $\overline{K_{jk}(s)e^{i\omega s}}=K_{jk}(s)e^{-i\omega s}$. For $h:=K_{jk}(\cdot )e^{i\omega \cdot }\in L^1(0,\mathrm{\infty })$ one has $\overline{\int h}=(\int \mathrm{Re}h)-i(\int \mathrm{Im}h)=\int (\mathrm{Re}h-i\mathrm{Im}h)=\int \bar{h}$ by [F1] and the complex linearity of the integral; hence ${\chi }_{jk}(-\omega )={\int }_0^{\mathrm{\infty }}{K}_{jk}(s)e^{-i\omega s}ds=\int \bar{h}=\overline{\int h}=\overline{{\chi }_{jk}(\omega )}$.

By-parts identity: fix $z\mathrm{\ne }0$. On $[0,b]$ the real and imaginary parts of $s\mapsto K_{jk}(s)$ and of $v(s):=e^{izs}\mathrm{/}(iz)$ are differentiable with continuous derivatives, and $K_{jk}(0)=K_{jk}(b)=K_{jk}^{\mathrm{\prime }}(0)=K_{jk}^{\mathrm{\prime }}(b)=0$ because $\mathrm{supp}K\subseteq (0,b)$; applying [F10] first with $u=K_{jk}$ and then with $u=K_{jk}^{\mathrm{\prime }}$ gives ${\int }_0^b{K}_{jk}(s)e^{izs}ds=-\frac{1}{iz}{\int }_0^b{K}_{jk}^{\mathrm{\prime }}(s)e^{izs}ds=\frac{1}{(iz{)}^2}{\int }_0^b{K}_{jk}^{\mathrm{\prime }\mathrm{\prime }}(s)e^{izs}ds=-z^{-2}{\int }_0^b{K}_{jk}^{\mathrm{\prime }\mathrm{\prime }}(s)e^{izs}ds$, the boundary terms vanishing at both steps. All three integrands are bounded and Riemann integrable on $[0,b]$, so [F10] transfers the identity to Lebesgue integrals; and since $K_{jk}$, $K_{jk}^{\mathrm{\prime }}$ and $K_{jk}^{\mathrm{\prime }\mathrm{\prime }}$ all vanish off $(0,b)$, the a.e.-equality clause of [F10] replaces each integral over $[0,b]$ by the integral over $(0,\mathrm{\infty })$. Hence ${\chi }_{jk}(z)=-z^{-2}{\int }_0^{\mathrm{\infty }}{K}_{jk}^{\mathrm{\prime }\mathrm{\prime }}(s)e^{izs}ds$ for every $z\mathrm{\ne }0$.

Differentiating the expressions of step 1.4 and using [F11] together with the algebra of derivatives gives $P_j^{\mathrm{\prime }}(t)=\omega (-{\alpha }_j\sin (\omega t)+{\beta }_j\cos (\omega t))$; on the other hand, expanding $(-i\omega )({\alpha }_j+i{\beta }_j)=\omega {\beta }_j-i\omega {\alpha }_j$ and multiplying by $e^{-i\omega t}=\cos (\omega t)-i\sin (\omega t)$ shows that the real part of $(-i\omega )a_j{e}^{-i\omega t}$ equals $\omega {\beta }_j\cos (\omega t)-\omega {\alpha }_j\sin (\omega t)$, the same number. Hence $P^{\mathrm{\prime }}(t)=\mathrm{Re}((-i\omega )a{e}^{-i\omega t})=\mathrm{Re}((-i\omega ){\epsilon }_0\chi (\omega )e{e}^{-i\omega t})$ componentwise, and $P^{\mathrm{\prime }}$ is the real part of a monochromatic complex expression with vector amplitude $d:=(-i\omega ){\epsilon }_0\chi (\omega )e$.

By step 2.4 and the linearity of the integral, ${\mathrm{\partial }}_y{\chi }_{jk}(z)={\int }_0^{\mathrm{\infty }}(-s)K_{jk}(s)e^{izs}ds={\int }_0^{\mathrm{\infty }}i(is{K}_{jk}(s)e^{izs})ds=i{\mathrm{\partial }}_x{\chi }_{jk}(z)$ for every $z$. Writing ${\chi }_{jk}=u+iv$, this says $u_y+i{v}_y=i(u_x+i{v}_x)=-v_x+i{u}_x$, that is $u_y=-v_x$ and $v_y=u_x$; the four partials exist everywhere and are continuous by step 2.4, so [F8] makes ${\chi }_{jk}$ complex differentiable at every point of $\mathbb{C}$, that is, entire.

Bounds and boundary continuity: for $z\mathrm{\ne }0$ with $\mathrm{Im}z\ge 0$, step 2.6 and $\mathrm{\mid }{e}^{izs}\mathrm{\mid }\le 1$ for $s\ge 0$ give $\mathrm{\mid }{\chi }_{jk}(z)\mathrm{\mid }\le \mathrm{\mid }z{\mathrm{\mid }}^{-2}\mathrm{\parallel }{K}_{jk}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_1$; for arbitrary $z\mathrm{\ne }0$ one has $\mathrm{\mid }{e}^{izs}\mathrm{\mid }=e^{-s\mathrm{Im}z}\le e^{b\max (0,-\mathrm{Im}z)}$ on $\mathrm{supp}K\subseteq (0,b)$, whence $\mathrm{\mid }{\chi }_{jk}(z)\mathrm{\mid }\le e^{b\max (0,-\mathrm{Im}z)}\mathrm{\parallel }{K}_{jk}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_1\mathrm{\mid }z{\mathrm{\mid }}^{-2}$. Also ${\chi }_{jk}$ is continuous on the closed upper half-plane: if $z_n\to z$ with $\mathrm{Im}{z}_n\ge 0$, the integrands of step 1.2 converge pointwise and are dominated by $e^{bY}\mathrm{\mid }{K}_{jk}\mathrm{\mid }$ once all $\mathrm{Im}{z}_n$ lie below a bound $Y$, so [F4] gives ${\chi }_{jk}(z_n)\to {\chi }_{jk}(z)$.

Applying the period-average identity of [F15] with amplitude vectors $a=e$ and $d=(-i\omega ){\epsilon }_0\chi (\omega )e$ componentwise, and using step 2.7, $\frac{1}{T}{\int }_0^{T}E(t)\cdot P^{\mathrm{\prime }}(t)dt={\sum }_{j=1}^d\frac{1}{T}{\int }_0^T\mathrm{Re}(e_j{e}^{-i\omega t})\mathrm{Re}(d_j{e}^{-i\omega t})dt={\sum }_{j=1}^d\frac{1}{2}\mathrm{Re}\left(e_j\overline{d_j}\right)=\frac{1}{2}\mathrm{Re}(e\cdot \bar{d})$. Since $\bar{d}=i\omega {\epsilon }_0\overline{\chi (\omega )e}$ and $e\cdot \overline{\chi (\omega )e}={\sum }_{j,k}{e}_j\overline{{\chi }_{jk}(\omega )}\overline{e_k}=\overline{e^{\ast }\chi (\omega )e}$, this equals $\frac{{\epsilon }_0\omega }{2}\mathrm{Re}\left(i\overline{e^{\ast }\chi (\omega )e}\right)=\frac{{\epsilon }_0\omega }{2}\mathrm{Im}(e^{\ast }\chi (\omega )e)$.

Dispersion relations: fix real $\omega$ and $j,k$, and put $\phi (z):={\chi }_{jk}(z)\mathrm{/}(z-\omega )$ for $z\mathrm{\ne }\omega$. For $R>\mathrm{\mid }\omega \mathrm{\mid }+1$ and $0<\delta <1$ let $\gamma$ be the closed piecewise $C^1$ contour running along $[\omega -R,\omega -\delta ]$, then along the clockwise semicircle $z=\omega +\delta e^{i\theta }$, $\pi \ge \theta \ge 0$, then along $[\omega +\delta ,\omega +R]$, and then back along the counterclockwise large semicircle $z=\omega +R{e}^{i\theta }$, $0\le \theta \le \pi$. The contour lies in the slit plane $D:=\mathbb{C}\setminus \{\omega -iv:v\ge 0\}$, where $\phi$ is holomorphic because ${\chi }_{jk}$ is entire and $z\mapsto (z-\omega {)}^{-1}$ is holomorphic off $\omega$. This domain is star-shaped about $\omega +iA$ for every $A>0$: along a segment from this centre to a point off the removed ray, either its imaginary part stays positive or its real part differs from $\omega$ except at the centre. The straight contraction to the centre, with the base-point connector and its reverse, therefore gives a null homotopy relative to the base point. Hence [F9] gives ${\int }_{\gamma }\phi dz=0$. The large arc contributes at most $\pi \mathrm{\parallel }{K}_{jk}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_1\mathrm{/}(R-\mathrm{\mid }\omega \mathrm{\mid }{)}^2\to 0$ by step 3.2; the small arc contributes $-i\pi {\chi }_{jk}(\omega )+o(1)$ as $\delta \downarrow 0$ because ${\chi }_{jk}(\omega +\delta e^{i\theta })\to {\chi }_{jk}(\omega )$ uniformly in $\theta$; and the two real segments converge to the principal value of ${\int }_{\mathbb{R}}{\chi }_{jk}(\xi )\mathrm{/}(\xi -\omega )d\xi$ as first $\delta \downarrow 0$ and then $R\to \mathrm{\infty }$: near $\omega$ the term ${\chi }_{jk}(\omega )\mathrm{/}(\xi -\omega )$ cancels between the symmetric punctured intervals while $({\chi }_{jk}(\xi )-{\chi }_{jk}(\omega ))\mathrm{/}(\xi -\omega )$ stays bounded, and for $\mathrm{\mid }\xi \mathrm{\mid }\ge \max (1,2\mathrm{\mid }\omega \mathrm{\mid })$ the estimate $\mathrm{\mid }{\chi }_{jk}(\xi )\mathrm{/}(\xi -\omega )\mathrm{\mid }\le 2\mathrm{\parallel }{K}_{jk}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_1\mathrm{\mid }\xi {\mathrm{\mid }}^{-2}$ makes the tails absolutely integrable. Passing to the limits in the identity $0={\int }_{\gamma }\phi dz$ therefore yields $\mathrm{PV}{\int }_{\mathbb{R}}{\chi }_{jk}(\xi )\mathrm{/}(\xi -\omega )d\xi =i\pi {\chi }_{jk}(\omega )$; taking real and imaginary parts gives the two displayed relations, since $-\pi \mathrm{Im}{\chi }_{jk}(\omega )$ and $\pi \mathrm{Re}{\chi }_{jk}(\omega )$ are respectively the real and imaginary parts of $i\pi {\chi }_{jk}(\omega )$.

Finally $M(\omega {)}^{\ast }=(\chi (\omega {)}^{\ast }-\chi (\omega ))\mathrm{/}(-2i)=M(\omega )$, so $M(\omega )$ is Hermitian, and $M(\omega )=(\chi (\omega )-\chi (\omega {)}^{\ast })\mathrm{/}(2i)$ has quadratic form $e^{\ast }M(\omega )e=(e^{\ast }\chi (\omega )e-e^{\ast }\chi (\omega {)}^{\ast }e)\mathrm{/}(2i)$, and $e^{\ast }\chi (\omega {)}^{\ast }e={\sum }_{j,k}\overline{e_j}\overline{{\chi }_{kj}(\omega )}{e}_k=\overline{e^{\ast }\chi (\omega )e}$ because $\chi (\omega {)}^{\ast }$ is the conjugate transpose; hence $e^{\ast }M(\omega )e=\mathrm{Im}(e^{\ast }\chi (\omega )e)$ for every $e\in {\mathbb{C}}^d$. By step 3.3 the period-averaged work equals $\frac{{\epsilon }_0\omega }{2}{e}^{\ast }M(\omega )e$ with $\frac{{\epsilon }_0\omega }{2}>0$, so it is nonnegative for every $e$ if and only if $e^{\ast }M(\omega )e\ge 0$ for every $e$; clause (a) establishes causality without assuming this positivity condition.