Exact defined Markov thermometer and switched coefficient bound

Put $g_{\mathsf{T}}(t)=h(t\mathrm{/}\mathsf{T})$ and $A_{\mathsf{T}}(s)=\int g_{\mathsf{T}}(t)g_{\mathsf{T}}(t-s)dt$. Compact support [F6] and [F5] permit the change of variables and interchange of the two ordinary integrals. Setting $u=t$ and $v=t-s$, and using the reality of $h$, gives the negative-sign angular transform $\widehat{A_{\mathsf{T}}}(\lambda )={\int }_{{\mathbb{R}}^2}{g}_{\mathsf{T}}(u)g_{\mathsf{T}}(v)e^{-i\lambda (u-v)}dudv=\widehat{g_{\mathsf{T}}}(\lambda )\widehat{g_{\mathsf{T}}}(-\lambda )=\mathrm{\mid }\widehat{g_{\mathsf{T}}}(\lambda ){\mathrm{\mid }}^2={\mathsf{T}}^2\mathrm{\mid }\widehat{h}(\mathsf{T}\lambda ){\mathrm{\mid }}^2\mathrm{.}$ By [F2] and [F3], the inverse angular-frequency relation is $C(s)=(2\pi {)}^{-1}\int S(\omega )e^{i\omega s}d\omega$ in the distributional sense: it is the negative-sign $2\pi$ inverse after the change $\xi =\omega \mathrm{/}(2\pi )$. Pairing with $e^{-i\nu s}{A}_{\mathsf{T}}(s)$ gives the absolutely convergent formula $R_{\mathsf{T}}^{(C)}(\nu )=\frac{{\mathsf{T}}^2}{2\pi }{\int }_{\mathbb{R}}S(\omega )\mathrm{\mid }\widehat{h}(\mathsf{T}(\nu -\omega )){\mathrm{\mid }}^{2}d\omega \mathrm{.}$ Here $S$ has at most linear growth and $\widehat{h}$ is rapidly decreasing, so the integral is finite.

Since $h\in C_c^{\mathrm{\infty }}$ by [F6], each derivative of $\widehat{h}$ is obtained by transforming a compactly supported smooth function $(-iu{)}^{j}h(u)$; repeated integration by parts shows every such derivative decreases faster than any inverse power. Hence $\int \mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }\widehat{h}(z){\mathrm{\mid }}^{2}dz<\mathrm{\infty }$. Plancherel in [F4], followed by $z=2\pi \xi$, yields ${\int }_{\mathbb{R}}\mathrm{\mid }\widehat{h}(z){\mathrm{\mid }}^{2}dz=2\pi \mathrm{\parallel }h{\mathrm{\parallel }}_2^2\mathrm{.}$ Changing variables $z=\mathsf{T}(\nu -\omega )$ in 1.1 therefore gives $\frac{R_{\mathsf{T}}^{(C)}(\nu )}{\mathsf{T}\mathrm{\parallel }h{\mathrm{\parallel }}_2^2}=\frac{1}{2\pi \mathrm{\parallel }h{\mathrm{\parallel }}_2^2}{\int }_{\mathbb{R}}S(\nu -z\mathrm{/}\mathsf{T})\mathrm{\mid }\widehat{h}(z){\mathrm{\mid }}^{2}dz\mathrm{.}$

For $C=C_0$, [F1] gives $S_0(\nu )=\max (-\nu ,0)\mathrm{/}(2\pi c^2)$. On $\nu \le 0$ this is linear with slope $-1\mathrm{/}(2\pi c^2)$, on $\nu \ge 0$ it is zero, and for $x\le 0\le y$, $\mathrm{\mid }{S}_0(x)-S_0(y)\mathrm{\mid }=-x\mathrm{/}(2\pi c^2)\le (y-x)\mathrm{/}(2\pi c^2)$. Thus its global Lipschitz constant is $1\mathrm{/}(2\pi c^2)$. For $C=C_a$, the conditional formula in [F1] is smooth at zero by its Taylor expansion, has bounded derivative on every compact interval, has bounded derivative as $\nu \to +\mathrm{\infty }$ by exponential decay. Its derivative is $S_a^{\mathrm{\prime }}(\nu )=\frac{1}{2\pi c^2}\frac{e^{b\nu }-1-b\nu e^{b\nu }}{(e^{b\nu }-1{)}^2},b=2\pi c\mathrm{/}a\mathrm{.}$ The Taylor expansion at zero makes this derivative finite there; as $\nu \to +\mathrm{\infty }$ it decays exponentially times a polynomial, and as $\nu \to -\mathrm{\infty }$ it tends $-1\mathrm{/}(2\pi c^2)$. Thus it is bounded on both tails and on compact intervals, so both spectra are globally Lipschitz (the accelerated assertion is conditional on [L1]). Subtract $S(\nu )$ in step 2.1 and use $\mathrm{\mid }S(\nu -z\mathrm{/}\mathsf{T})-S(\nu )\mathrm{\mid }\le \mathrm{Lip}(S)\mathrm{\mid }z\mathrm{\mid }\mathrm{/}\mathsf{T}$. The definition of $K_h$ gives the stated error bound and its $1\mathrm{/}\mathsf{T}$ limit.

The accelerated spectrum is positive: for $\nu >0$ numerator and denominator in [F1] are positive, for $\nu <0$ they are both negative, and the zero limit is positive. Thus $r_{+},r_{-}>0$. By [A2], the second component of ${\pi }^{\mathrm{\prime }}=\pi Q$ gives $p^{\mathrm{\prime }}(t)=r_{+}(1-p(t))-r_{-}p(t)=r_{+}-(r_{+}+r_{-})p(t)$. Solving this scalar linear ODE gives $p(t)=p_{\ast }+(p(0)-p_{\ast })e^{-(r_{+}+r_{-})t}$ with $p_{\ast }=r_{+}\mathrm{/}(r_{+}+r_{-})$. For $t\ge 0$ seconds, this is a convex combination of $p(0)$ and $p_{\ast }\in (0,1)$, so it remains a probability. Apply [L1] at $\nu =E\mathrm{/}\mathrm{\hslash }>0$ to get $r_{-}\mathrm{/}{r}_{+}=S_a(-\nu )\mathrm{/}{S}_a(\nu )=e^{2\pi c\nu \mathrm{/}a}$, so $p_{\ast }=1\mathrm{/}(1+e^{2\pi cE\mathrm{/}(a\mathrm{\hslash })})$. The ratio $p_{\ast }\mathrm{/}(1-p_{\ast })$ is therefore $e^{-2\pi cE\mathrm{/}(a\mathrm{\hslash })}$. Comparing with its defining Boltzmann form gives $T_{\mathrm{e}\mathrm{f}\mathrm{f}}=\mathrm{\hslash }a\mathrm{/}(2\pi c{k}_B)$. This calculation concerns only the specified Markov model and its adopted rates.