Section 6 9 October 2026

Damped temporal response with a noncommuting physical form

Jeremy Rodgers · Independent Researcher
doi:10.5281/zenodo.23259571

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6 Damped temporal response with a noncommuting physical form

The physical derivative form generally does not commute with the interacting complement. It may therefore be placed outside an actual resolvent, as below, but cannot be inserted into a scalar spectral measure as though it commuted with every spectral projection. This use of a resolvent and a time integral is related to the Kato smoothing framework [6, 7]. The statement here is at one earned $\eta>0$ ; it does not assert the uniform imaginary-part bounds required by a global smoothing theorem.

Theorem 6.1 (Ordered damped response)

Let $A$ be fixed self-adjoint, $K=A+c\ge I$ , and $\mathcal L=LF^{-1}:\mathcal U\to\mathcal V^*$ bounded. Let $T_{\rm ph}\ge I$ be a positive closed form whose domain contains $\mathcal V$ continuously. For fixed $\eta>0$ suppose the actual ordered sandwich satisfies the measurable bound

\|T_{\rm ph}^{1/2}(A-E-i\eta)^{-1}\mathcal L\|^2 \le S_\eta(E).

(6.1)

For $a\in L^2(0,T;\mathcal U)$ , extend the forcing by zero after $T$ , and let $q$ be its causal zero-initial form-dual response. Define $a_\eta(s)=e^{-\eta s}a(s)1_{[0,T]}(s)$ and use the unitary Fourier convention $\widehat u(E)=(2\pi)^{-1/2}\int_{\mathbb R}e^{iEt}u(t)\,dt$ . If the right side is finite, then

\int_0^\infty e^{-2\eta t} \|T_{\rm ph}^{1/2}q(t)\|^2\,dt \le \int_{\mathbb R}S_\eta(E) \|\widehat a_\eta(E)\|^2\,dE .

(6.2)

In particular a uniform $S_\eta(E)\le S_\eta$ gives the upper bound $S_\eta\|a_\eta\|_2^2$ . A bound $S_\eta(E)\le S_0+S_1E^2$ instead gives

S_0\|a_\eta\|_2^2+S_1\|\dot a_\eta\|_2^2,

(6.3)

provided the actual zero extension $a_\eta$ belongs to $H^1(\mathbb R;\mathcal U)$ .

Proof

The fixed group $e^{-iAt}$ is unitary also on the $K$ form dual. The causal integral

q(t)=-i\int_0^{\min(t,T)} e^{-iA(t-s)}\mathcal L a(s)\,ds

is therefore a continuous $\mathcal V^*$ vector. It is zero for negative times and bounded in that dual norm for all positive times. Its damped extension $q_\eta=e^{-\eta t}q\,1_{[0,\infty)}$ belongs to $L^2(\mathbb R;\mathcal V^*)$ . Its value at zero is zero, so differentiation creates no entrance delta distribution. Fourier transformation of the causal equation, or Fubini applied to its damped convolution, gives the exact form-dual identity

\widehat q_\eta(E) =-(A-E-i\eta)^{-1}\mathcal L\widehat a_\eta(E).

(6.4)

For almost every $E$ the right side lies in $\mathcal V$ , hence in $\operatorname{Dom}T_{\rm ph}^{1/2}$ . The assumed finite integral and (6.1) place it in the $L^2$ graph of $T_{\rm ph}^{1/2}$ . Coercivity $T_{\rm ph}\ge I$ also places it in physical $L^2(\mathcal H)$ . Fourier transformation commutes with this closed spatial operator: to see this, replace it by its bounded spectral truncations, use Hilbert-valued Plancherel there, and pass by its closed graph and monotone convergence of the graph norms. Thus inverse Fourier transformation identifies this same dual solution as a physical $L^2(\operatorname{Dom}T_{\rm ph}^{1/2})$ response. Plancherel and (6.1) prove (6.2). In this argument the truncations are of $T_{\rm ph}$ for the Fourier graph identity; none is commuted through $A$ or its resolvent.

For a uniform coefficient use Plancherel on $a_\eta$ . For the quadratic coefficient use $\|E\widehat a_\eta\|_2=\|\dot a_\eta\|_2$ . That equality requires the stated whole-line $H^1$ condition and proves (6.3).

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The theorem gives an integrated first-form response, not an all-times pointwise form bound. If the input is $H^1$ inside $[0,T]$ but has nonzero endpoint values, its zero extension has delta derivatives and is not $H^1(\mathbb R)$ . One must keep the endpoint dressings, as in (4.9) or Theorem 3.3, or prove a physical pulse collar that removes those jumps. The endpoint problem cannot be repaired by simply omitting the positive frequency-weighted term.

The physical finite-interval graph norm obeys

\|T_{\rm ph}^{1/2}q\|_{L^2(0,T)} \le e^{\eta T} \left(\int_{\mathbb R}S_\eta(E) \|\widehat a_\eta(E)\|^2\,dE\right)^{1/2}.

(6.5)

If $q_0\ne0$ , its homogeneous response must be added separately. For example, an earned $T_{\rm ph}\le\kappa K$ and $q_0\in\mathcal V$ give

\left(\int_0^\infty e^{-2\eta t} \|T_{\rm ph}^{1/2}e^{-iAt}q_0\|^2dt\right)^{1/2} \le\sqrt{\frac{\kappa}{2\eta}}\|K^{1/2}q_0\|.

Add this to the square root of the forced bound; the square of the sum, not the sum of squares, is a generally valid full-response bound. For a noncoercive derivative form, replace it by $I+T_{\rm ph}$ or establish a separate Hilbert-norm sandwich. Temporal damping has neither projected out a bound pole nor supplied a missing source graph.