Proposition 18.34.18.4920A08E02F1 For $q,v,w\in C(X,\mathbb R)$, write $\operatorname {osc}_ Xq=\sup _ Xq-\inf _ Xq$. Then
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18.34.19
\begin{equation} D\mu _ v[q]=m(\mathcal Q_ vq-q)\mu _ v, \qquad \| D\mu _ v[q]\| _{\mathrm{TV}} \leq \frac m2\operatorname {osc}_ Xq. \end{equation}
Consequently
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18.34.20
\begin{equation} \| \mu _ w-\mu _ v\| _{\mathrm{TV}} \leq 2\tanh \left(\frac m4\operatorname {osc}_ X(w-v)\right). \end{equation}
These bounds do not depend on $N_ m$. In particular, let $v_ m,w_ m\in C(X,\mathbb R)$ along a sequence of degrees with nonzero selected spaces. If $m\operatorname {osc}_ X(w_ m-v_ m)\to 0$, then the two sequences of normalized kernel probabilities have the same weak, or total variation, limit whenever one of them converges in that topology.
Proof.
The formula for $D\mu _ v[q]$ follows from corollary 18.34.9, since $N_ m$ does not depend on the weight. For the bound, put $g=q-\mathcal Q_ vq$, $A=\{ g>0\} $ and $p=\mu _ v(A)$. Since $\mathcal Q_ v$ preserves integrals, $g$ has mean zero, and selfadjointness gives
\[ \int _ Ag\, \mathrm{d}\mu _ v =\langle q,(I-\mathcal Q_ v)\mathbf1_ A\rangle _{L^2(\mu _ v)}. \]
The function $(I-\mathcal Q_ v)\mathbf1_ A$ is nonnegative on $A$ and nonpositive on its complement, almost everywhere, and its integral is zero. Subtracting $\inf _ Xq$ from $q$ does not change the pairing, which is therefore at most
\[ \operatorname {osc}_ Xq \bigl (p-\langle \mathbf1_ A,\mathcal Q_ v\mathbf1_ A\rangle \bigr ). \]
Positivity of $\mathcal Q_ v$, applied to $\mathbf1_ A-p$, gives $\langle \mathbf1_ A,\mathcal Q_ v\mathbf1_ A\rangle \geq p^2$: since $\mathcal Q_ v$ fixes constants and preserves integrals, $\langle \mathbf1_ A-p,\mathcal Q_ v(\mathbf1_ A-p)\rangle =\langle \mathbf1_ A,\mathcal Q_ v\mathbf1_ A\rangle -p^2$. As $g$ has mean zero, it follows that
\[ \| g\| _{L^1(\mu _ v)} =2\int _ Ag\, \mathrm{d}\mu _ v \leq 2\operatorname {osc}_ Xq\, p(1-p) \leq \tfrac 12\operatorname {osc}_ Xq. \]
The total variation norm of a signed measure is the integral of the absolute value of its density, so $\| D\mu _ v[q]\| _{\mathrm{TV}}=m\| g\| _{L^1(\mu _ v)}$, and (18.34.19) follows.
For the global estimate, take $q=w-v$, $v_ t=v+tq$, $\mu _ t=\mu _{v_ t}$ and $\mathcal Q_ t=\mathcal Q_{v_ t}$ for $0\leq t\leq 1$, and put $a=m\operatorname {osc}_ Xq$. For a Borel set $E$, the function $p_ t=\mu _ t(E)$ is differentiable, because integration against $\mathbf1_ E$ is a bounded linear functional for the total variation norm, and selfadjointness gives
\[ p'_ t=-m\langle q,(I-\mathcal Q_ t)\mathbf1_ E\rangle _{L^2(\mu _ t)}. \]
The function $(I-\mathcal Q_ t)\mathbf1_ E$ has positive and negative parts whose integrals both equal $p_ t-\langle \mathbf1_ E,\mathcal Q_ t\mathbf1_ E\rangle $. Bounding $q$ above and below on the supports of these parts, and applying positivity to $\mathbf1_ E-p_ t$ as before, we obtain
\[ |p'_ t|\leq a\bigl (p_ t-\langle \mathbf1_ E,\mathcal Q_ t\mathbf1_ E\rangle \bigr ) \leq ap_ t(1-p_ t). \]
All the kernels $B_{v_ t}$ are positive exactly off the common zero set $Z_ m$, so the probabilities $\mu _ t$ are mutually absolutely continuous. Hence, if $p_0$ is zero or one, then $p_ t=p_0$ for all $t$. Otherwise $0<p_ t<1$ for all $t$, and integration gives
\[ \left|\log \frac{p_1}{1-p_1}-\log \frac{p_0}{1-p_0}\right|\leq a. \]
Let $\sigma (z)=\mathrm{e}^ z/(1+\mathrm{e}^ z)$ be the logistic function. If $p_1\geq p_0$ and $z=\log (p_0/(1-p_0))$, then
\[ p_1-p_0\leq \sigma (z+a)-\sigma (z) =\frac{\sinh (a/2)}{\cosh (z+a/2)+\cosh (a/2)} \leq \tanh (a/4). \]
The case $p_1\leq p_0$ is symmetric. Since $\mu _ w-\mu _ v$ has total mass zero, its total variation norm is $2\sup _ E|\mu _ w(E)-\mu _ v(E)|$, which proves (18.34.20). Under the hypothesis of the last assertion the right-hand side tends to zero, which gives the final statement.
$\square$
Pome
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