Corollary 18.34.9.7733E1B20E7B Let $v_ t=v+tq$ with $v,q\in C(X,\mathbb R)$. For $x\notin Z_ m$, the probability measure
F73FC2DA9CA6
18.34.10
\begin{equation} \mathrm{d}\sigma _{t,x}(y) =\frac{|k_{v_ t,x}(y)|_{h^ m}^2\mathrm{e}^{-mv_ t(y)}}{\| k_{v_ t,x}\| _{m,v_ t,\nu }^2}\, \mathrm{d}\nu (y) \end{equation}
satisfies
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18.34.11
\begin{equation} \frac{\mathrm{d}}{\mathrm{d}t}\log B_{v_ t}(x) =m\left(\int _ Xq\, \mathrm{d}\sigma _{t,x}-q(x)\right). \end{equation}
On $Z_ m$ the kernel and all its $t$-derivatives vanish. Moreover, $\int _ X\partial _ tB_{v_ t}\, \mathrm{d}\nu =0$.
Proof.
Since $\mathcal T_{t,1}u=mP_{v_ t}(qu)$, the case $k=1$ of (18.34.7) gives $\partial _ tk_{v_ t,x}=mP_{v_ t}(qk_{v_ t,x})$. Applying the fixed evaluation $\ell _ x$ and the Riesz identity gives
\[ \partial _ t\ell _ x(k_{v_ t,x}) =m\int _ Xq|k_{v_ t,x}|_{h^ m}^2 \mathrm{e}^{-mv_ t}\, \mathrm{d}\nu . \]
Dividing by the positive number $\ell _ x(k_{v_ t,x})=\| k_{v_ t,x}\| _{m,v_ t,\nu }^2$ and differentiating the factor $\mathrm{e}^{-mv_ t(x)}$ in (18.34.4) proves (18.34.11); the density in (18.34.10) has integral one by the definition of the norm. At points of $Z_ m$ the evaluations and Riesz vectors vanish for every weight. Finally, differentiating the identity $\int _ XB_{v_ t}\, \mathrm{d}\nu =N_ m$ in $C(X)$ proves the last assertion. For constant $q$ the right-hand side of (18.34.11) vanishes, in accordance with the invariance of the kernel under adding a constant to the weight.
$\square$
Pome
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