Quantitative switched coefficient and inertial rate

Assume the conditional mathematical spectra and switched-coefficient bound of Exact defined Markov thermometer and switched coefficient bound, with $a>0$, $c>0$, and a real nonzero $h\in C_c^{\mathrm{\infty }}(\mathbb{R})$. For each proper switching duration $\mathsf{T}>0$ and excitation frequency $\nu >0$, define the normalized accelerated coefficient $Q_{\mathsf{T}}^{(a)}(\nu )=\frac{R_{\mathsf{T}}^{(C_a)}(\nu )}{\mathsf{T}\mathrm{\parallel }h{\mathrm{\parallel }}_2^2}\mathrm{.}$ With $K_h=\frac{1}{2\pi \mathrm{\parallel }h{\mathrm{\parallel }}_2^2}{\int }_{\mathbb{R}}\mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }\widehat{h}(z){\mathrm{\mid }}^{2}dz<\mathrm{\infty },$ the exact leading-coefficient estimate is $\mathrm{\mid }{Q}_{\mathsf{T}}^{(a)}(\nu )-S_a(\nu )\mathrm{\mid }\le \frac{\mathrm{Lip}(S_a)K_h}{\mathsf{T}}\mathrm{.}$ For the inertial covariance, the same definition gives $0\le \frac{R_{\mathsf{T}}^{(C_0)}(\nu )}{\mathsf{T}\mathrm{\parallel }h{\mathrm{\parallel }}_2^2}\le \frac{K_h}{2\pi c^2\mathsf{T}},$ because $S_0(\nu )=0$ for $\nu >0$ and $\mathrm{Lip}(S_0)=1\mathrm{/}(2\pi c^2)$. Thus the inertial excitation coefficient divided by duration tends to zero, while the accelerated one tends to its conditional spectrum. These are finite-switch coefficient statements; they do not claim an all-order detector probability, strict nonzero response for every finite switch, or a measured outcome.