465 lines
32 KiB
TeX
465 lines
32 KiB
TeX
\documentclass[10pt,a4paper]{article}
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% }
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\newtheorem{theorem}{Theorem}
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\newtheorem{proposition}[theorem]{Proposition}
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\newtheorem{corollary}[theorem]{Corollary}
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\newtheorem{lemma}[theorem]{Lemma}
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\newtheorem{definition}[theorem]{Definition}
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\newtheorem{assumption}[theorem]{Assumption}
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\newtheorem{remark}[theorem]{Remark}
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\newtheorem{example}[theorem]{Example}
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\newtheorem{notation}{Notation}
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\def \R {\mathbb{R}}
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\def \N {\mathbb{N}}
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\def \d {\mathrm{d}}
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\newcommand{\red}[1]{\textcolor{red}{#1}}
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\begin{document}
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To start off we define the notation for the objects we will work on
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$$
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dX_t = B(t,X_t,\mu_{X_t})dt + \Sigma(t,X_t,\mu_{X_t})dW_t,\qquad X_0\sim\mu_0.
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$$
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Given the flow of marginals $\mu_t$ we can fix the coefficients and linearize the SDE with the linearized coefficients $B^\mu(t,x)$ and $\Sigma^\mu(t,x)$. Using this, we may define the infinitesimal generator
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$$
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\mathcal{A}_t^\mu=\frac{1}{2}\sum_{i,j=1}^N c^\mu_{ij}(t,x)\partial_{x_ix_j}+\sum_{i=1}^NB^\mu_i(t,x)\partial_{x_i}.
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$$
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Given this operator, under reasonable assumptions we have the existance of $p(s,x;t,y)$ fundamental solution of
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\begin{align*}
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(\partial_s+\mathcal{A}^\mu_s)p^\mu(s,x;t,y)&=0,\\
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(\partial_t-(\mathcal{A}^\mu_t)^*)p^\mu(s,x;t,y)&=0.
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\end{align*}
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Having the transition density $p^\mu$ we may define the forward translation operator
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$$
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U_\mu^{t,s}\phi(y)=\int p^\mu(s,x;t,y)\phi(x)dx,
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$$
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whose definition may be easily extended to $\mathcal{P}^2(\R^N)$ due to the gaussian estimates on $p$ (which are uniform over the choice of $\mu_t$):
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$$
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U_\mu^{t,s}u(y)=\int p^\mu(s,x;t,y)u(dx),\qquad u\in\mathcal{P}^2(\R^N).
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$$
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Via this operator we may define
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$$
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u^\mu_t(x)=U_\mu^{t,0}\mu_0,
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$$
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the density of the solution of the linearized SDE via the marginal flow $(\mu_t)_{t\in[0,T]}$ with initial law $\mu_0$.
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Via this density we are able to construct a new flow of marginals (the one of the solution of the linearized SDE via marginal flow $\mu_t$ and initial law $\mu_0$):
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$$
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\mathcal{L}^\mu_t(dy) = u^\mu_t(y)dy = \left(\int p^\mu(0,x;t,y)\mu_0(dx)\right)dy.
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$$
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\red{
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Here we will briefly state what \cite{Kolokoltsov} does to study the contraction properties on $L^1$ norm of $u^\mu$.
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To start we need the identity\footnote{In my calculations I get the adjoint operator $(\mathcal{A}_s)^*$ but in Kolokoltsov's paper there is the backward one.} (28) of \cite{Kolokoltsov}:
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\begin{equation}\label{e1}
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U^{t,0}_\mu - U^{t,0}_\nu = \int_0^t\frac{d}{ds}U^{t,s}_\nu U^{s,0}_\mu ds = \int_0^t U^{t,s}_\nu((\mathcal{A}^{\mu}_s)^*-(\mathcal{A}^{\nu}_s)^*)U^{s,0}_\mu ds.
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\end{equation}
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Then we need to observe that
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\begin{align}
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||U^{t,s} f||_{L^1} &= \int \left| \int p(s,x;t,y)f(x)dx\right|dy\nonumber\\
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&\stackrel{Gaussian\ estimates}{\leq}C\int\int |f(x+y)|\Gamma^+(y)dy dx = C||f||_{L^1}.\label{e2}
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\end{align}
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Also observe that
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\begin{align*}
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||(\mathcal{A}^\mu_s-\mathcal{A}^\nu_s)f||_{L^1}\leq C\sup_{t,x}\left(|c^\mu(t,x)-c^\nu(t,x)|+|B^\mu(t,x)-B^\nu(t,x)|\right)||f||_{W^{1,2}},
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\end{align*}
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and by the fact that the operator $U^{s,0}_\mu$ is a bounded operator in $W^{1,2}$, check footnote\footnote{I am unsure of this passage, we don't actually need the $0$-derivative and due to gaussian estimates we should get something similar to $||\partial U\ f||_{L^1}\leq s^{-1/2}||f||_{L^1}$ and $||\partial^2 U\ f||_{L^1}\leq s^{-1}||f||_{L^1}$ where the second derivative is no longer integrable wrt $s$. I think something similar to (24.2.7) in dispense is happening.} we get
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\begin{equation}\label{e3}
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||U^{s,0}_\mu f||_{W^{1,2}}\leq C\cdot s^{-1/2}||f||_{W^{1,2}}.
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\end{equation}
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By \eqref{e1}, \eqref{e2} and \eqref{e3} we get
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\begin{align}\nonumber
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||u^\mu_t-u^\nu_t||_{L^1}=||U^{t,0}_\mu\mu_0-U^{t,0}_\mu\mu_0||_{L^1}&\leq C\int_0^t s^{-1/2}ds ||\mu_0||_{L^1}\sup_{t,x}\left(|c^\mu(t,x)-c^\nu(t,x)|+|B^\mu(t,x)-B^\nu(t,x)|\right)\\
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&\leq C\sqrt{t} ||\mu_0||_{L^1}\sup_{t,x}\left(|c^\mu(t,x)-c^\nu(t,x)|+|B^\mu(t,x)-B^\nu(t,x)|\right).\label{e4}
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\end{align}
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Now that we briefly stated the ideas of \cite{Kolokoltsov} we can begin.}
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To leave as many doors open as possible we first define
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$$
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I(f) := \int f(x)(u^\mu_t(x)-u^\nu_t(x))dx,
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$$
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which at the moment may be seen as an indicator of closeness between the two densities, at a later moment we will take the $\sup$ for $f$ in some bounded functional space like the bounded Holder functions or the bounded functions.
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%Now we will change a bit what \cite{Kolokoltsov} did to work in the Wasserstein framework. To start off we apply the duality formula for the Kantorovich–Rubinstein distance (Remark 6.5 of \cite{Villani}):
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%\begin{align*}
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%W^{(1)}(\mathcal{L}^\mu_t,\mathcal{L}^\nu_t) = \sup_{||f||_{Lip}\leq1}\left(\int f(x)(u^\mu_t(x)-u^\nu_t(x))dx\right)=:\sup_{||f||_{Lip}\leq1}I(f).
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%\end{align*}
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Now we apply this useful trick: if we write explicitly the definition of $u^\mu$ as the evaluation of $p^\mu$ on the distribution $\mu_0$ in $I(f)$ we can change the order of integration to evaluate $p^\mu$ on the regular distribution $f(x)dx$, this is useful because it switches the operators in Kolokoltsov's formula \eqref{e1} from being forward to being backwards while changing only marginally everything else. If we define the backward propagator operator
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\begin{equation}\label{e4-1}
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V^{s,t}_\mu g(y):=\int p^\mu(s,y;t,x)g(x)dx,
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\end{equation}
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we can expand $I(f)$ this way
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\begin{align*}
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I(f) &= \int\int f(x)(p^\mu-p^\nu)(0,y;t,x)\mu_0(dy)dx = \mu_0\left(\int f(x)(p^\mu-p^\nu)(0,\cdot;t,x)dx\right)\\
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&=\mu_0\left(\int_0^t\frac{d}{ds}\left(\int\int p^\mu(0,\cdot;s,z)p^\nu(s,z;t,x)f(x)dxdz\right)ds\right)\\
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&=\mu_0\left(\int_0^t \int\int\partial_{t_2}p^\mu(0,\cdot;s,z)p^\nu(s,z;t,x)f(x)dxdz + \int\int p^\mu(0,\cdot;s,z)\partial_{t_1}p^\nu(s,z;t,x)f(x)dxdz ds\right)
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\intertext{by the fact that $p$ is the fundamental solution for both the forward and backward PDEs}
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&=\mu_0\left(\int_0^t \int\int(\mathcal{A}_s^\mu)^* p^\mu(0,\cdot;s,z)p^\nu(s,z;t,x)f(x)dxdz - \int\int p^\mu(0,\cdot;s,z)\mathcal{A}_s^\nu p^\nu(s,z;t,x)f(x)dxdz ds\right)\\
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&=\mu_0\left(\int_0^t \int(\mathcal{A}_s^\mu)^* p^\mu(0,\cdot;s,z)V_\nu^{s,t}f(z)dz - \int (\mathcal{A}_s^\nu)^*p^\mu(0,\cdot;s,z) V_\nu^{s,t}f(z)dz ds\right)\\
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&=\mu_0\left(\int_0^t \int p^\mu(0,\cdot;s,z)\left(\mathcal{A}_s^\mu - \mathcal{A}_s^\nu\right)V_\nu^{s,t}f(z)dz ds\right)\\
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&=\mu_0\left(\int_0^t V_\mu^{0,s}\left(\mathcal{A}_s^\mu - \mathcal{A}_s^\nu\right)V_\nu^{s,t}f ds\right)=\int_0^t \mu_0\left(V_\mu^{0,s}\left(\mathcal{A}_s^\mu - \mathcal{A}_s^\nu\right)V_\nu^{s,t}f \right) ds.
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\end{align*}
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Using similar arguments it is possible to obtain also Kolokoltsov's formula \eqref{e1}:
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\begin{equation*}
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I(f) = \int_0^t f\left(U_\nu^{t,s}\left((\mathcal{A}_s^\mu)^* - (\mathcal{A}_s^\nu)^*\right)U_\mu^{s,0}\mu_0 \right) ds.
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\end{equation*}
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Now we can try to estimate $I(f)$:
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\begin{align}\nonumber
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I(f)&\leq \int_0^t \left|\mu_0\left( V^{0,s}_\mu\left( \mathcal{A}^\mu_s - \mathcal{A}^\nu_s \right)V^{s,t}_\nu f\right)\right| ds,
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\intertext{since $\mu_0$ is a probability measure we can bound $\mu_0(g)$ with the uniform bound of $g$: $\mu_0(g)\leq |g|_\infty$:}
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&\leq \int_0^t \sup_x\left(V^{0,s}_\mu\left( \mathcal{A}^\mu_s - \mathcal{A}^\nu_s \right)V^{s,t}_\nu f(x)\right) ds.\label{f1}
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\end{align}
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We observe that due to Holder's inequality and the fact that $p^\mu(0,x;s,y)dy$ is a probability measure for any fixed $x$ we have uniformly in $x$
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$$
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|V^{0,s}_\mu g(x)|=\left| \int p^\mu(0,x;s,y)g(y)dy \right|\leq ||g||_{L^\infty}.
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$$
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Thus continuing from \eqref{f1} we have
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\begin{align*}
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I(f)&\leq \int_0^t ||\left( \mathcal{A}^\mu_s - \mathcal{A}^\nu_s \right)V^{s,t}_\nu f(x)||_{L^\infty} ds\leq \\
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&\leq \sup_{s,x}\left(|B^\mu(s,x)-B^\nu(s,x)|+|c^\mu(s,x)-c^\nu(s,x)|\right)\int_0^t ||\bigtriangledown V^{s,t}_\nu f||_{L^\infty} + ||Hess\ V^{s,t}_\nu f||_{L^\infty} ds.
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\end{align*}
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\begin{theorem}\label{a1}
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\begin{comment}
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If $f$ is a Lipschitz function with $[f]_{Lip}\leq 1$ then
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\begin{enumerate}
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\item $||Hess\ V^{s,t}_\nu f||_{L^\infty}\leq \frac{C}{\sqrt{t-s}}$.
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\item $||\bigtriangledown V^{s,t}_\nu f||_{L^\infty}\leq C$.
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\end{enumerate}
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\end{comment}
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If $f$ is a $C^\alpha_B$ function with $||f||_{C^\alpha_B}\leq 1$ then
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\begin{enumerate}
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\item $||Hess\ V^{s,t}_\nu f||_{L^\infty}\leq \frac{C}{|t-s|^{1-\frac{\alpha}{2}}}$.
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\end{enumerate}
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\end{theorem}
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\begin{proof}
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\begin{comment}
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Let's first tackle the case with $f$ Lipschitz. Let $x\in\R^N$. We have
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\begin{align*}
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\left| \partial_{x_ix_j}V^{s,t}_\nu f (x) \right|&=\left| \int\partial_{x_i x_j}p^\nu(s,x;t,y)f(y)dy \right|\leq \left| \int\partial_{x_i x_j}p^\nu(s,x;t,y)(f(y)-f(x))dy \right|+ \left| \int\partial_{x_i x_j}p^\nu(s,x;t,y)dyf(x) \right|\\
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&\leq \int |\partial_{x_ix_j}p^\nu(s,x;t,y)||x-y|dy + \left| \partial_{x_ix_j}\underbrace{\int p^\nu(s,x;t,y) dy}_{=1} \right||f(x)|\\
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&\leq \frac{C}{|t-s|}\int \Gamma^+(t-s,x-y)|x-y|dy + 0\leq \frac{C}{\sqrt{t-s}}.
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\end{align*}
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The second inequality for the Lipschitz case is done in a completely analogous manner. Let's consider the Holder case
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\end{comment}
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\begin{align*}
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\left| \partial_{x_ix_j}V^{s,t}_\nu f (x) \right|&=\left| \int\partial_{x_i x_j}p^\nu(s,x;t,y)f(y)dy \right|\leq \left| \int\partial_{x_i x_j}p^\nu(s,x;t,y)(f(y)-f(e^{(t-s)B}x))dy \right|+ \left| \int\partial_{x_i x_j}p^\nu(s,x;t,y)dyf(e^{(t-s)B}x) \right|\\
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&\leq \int |\partial_{x_ix_j}p^\nu(s,x;t,y)||e^{(t-s)B}x-y|^\alpha_B dy + \left| \partial_{x_ix_j}\underbrace{\int p^\nu(s,x;t,y) dy}_{=1} \right||f(e^{(t-s)B}x)|\\
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&\leq \frac{C_{B,\alpha}}{|t-s|}\int \Gamma^+(t-s,x-y)|x-e^{-(t-s)B}y|^\alpha_B dy + 0\leq \frac{C_{B,\alpha}}{|t-s|^{1-\frac{\alpha}{2}}}.
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\end{align*}
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by Lemma (A.5) of \cite{LucePagliaPascu}.
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\end{proof}
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We will now define
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$$
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d_{C^{\alpha}_B}(\mu,\nu)=\sup_{||f||_{C^{\alpha}_B}\leq 1}\left|\int f(x)\left(\mu(dx)-\nu(dx)\right) \right|
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$$
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the bounded anisotropic $\alpha$-Holder distance.
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\begin{theorem}
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The bounded anisotropic $\alpha$-Holder distance metrizes weak convergence of measures. More precisely given $(\mu_n)_{n\in\N}$ and $\mu$ probability measures
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$$
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d_{C^{\alpha}_B}(\mu_n,\mu)\rightarrow 0 \Leftrightarrow \mu_n\stackrel{d}{\rightarrow}\mu.
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$$
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\end{theorem}
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\begin{proof}
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The proof will be divided in two steps and is mostly taken from \url{https://sites.stat.washington.edu/jaw/COURSES/520s/522/HO.522.20/ch11c.pdf}
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1) First we prove that
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$$
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\mu_n\stackrel{d}{\rightarrow}\mu \Leftrightarrow \int f d\mu_n\rightarrow\int f d\mu,\ \forall f\in C^{\alpha}_B.
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$$
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If $\mu_n\stackrel{d}{\rightarrow}\mu$ then equivalently $\int f\mu_n\rightarrow\int f\mu$ for any function $f\in bC$ which in particular means that it is true for any $f\in C^{\alpha}_B$. The converse is true because if $\int f\mu_n\rightarrow\int f\mu$ for any function $f\in C^{\alpha}_B$ then in particular it is true for any $f\in bLip$ which by Portmanteau's theorem implies weak convergence.
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2) We will now prove that
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$$
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\int f d\mu_n\rightarrow\int fd\mu \ \forall f\in C^{\alpha}_B \Leftrightarrow d_{C^{\alpha}_B}(\mu_n,\mu)\rightarrow0.
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$$
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The easy implication is the right-to-left one: indeed by comparison theorem
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$$
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\lim_n \int f(x) \left( \mu_n(dx)-\mu(dx) \right)\leq \lim_n \sup_{||f||_{C^{\alpha}_B}\leq 1}\left|\int f(x) \left( \mu_n(dx)-\mu(dx) \right)\right| = \lim_n d_{C^{\alpha}_B}(\mu_n,\mu)\rightarrow0.
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$$
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The other way is more challenging, first by continuity from below of probability measures for any fixed $\epsilon>0$ there exists $K$ a compact set such that $\mu(K)>1-\epsilon$. Let $\mathcal{H}=\left\lbrace f\in C^{\alpha}_B\ |\ ||f||_{C^{\alpha}_B}\leq 1 \right\rbrace$, if we restrict each of these functions on $K$ we have that $\mathcal{H}\vert_K$ is totally bounded with respect to the $||\cdot||_\infty$ norm by Ascoli-Arzelà's theorem, in particular $\exists k$ finite and $f_1,\cdots f_k\in \mathcal{H}\vert_K$ such that for any $f\in\mathcal{H}$ $\exists j$ such that $\sup_K|f-f_j|\leq\epsilon$.
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Now if we consider $d_B(x,y)=|x-y|_B$ and $K^\epsilon=\left\lbrace x\in\R^N\ |\ d_B(x,K)\leq\epsilon \right\rbrace$ and $f,f_j$ as before we have
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$$
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\sup_{x\in K^\epsilon}|f(x)-f_j(x)|\leq \sup_{x\in K^\epsilon}\left( |f(x)-f(y_x)| + |f(y_x)-f_j(y_x)| + |f_j(y_x)-f_j(x)|\right)\leq \sup_{x\in K^\epsilon}\left( 2\epsilon^\alpha + \epsilon \right)\leq C_\alpha\epsilon^\alpha.
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$$
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where $y_x$ is a point in $K$ such that $|x-y|_B<\epsilon$. $C_\alpha$ may be taken uniformly of $\epsilon$ as long as $\epsilon\leq 1$.
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Let $g(x)=\max\left( 0, 1-\frac{d_B(x,K)}{\epsilon} \right)$, evidently $g\in bLip\subseteq C^{\alpha}_B$ and $\mathds{1}_K\leq g\leq \mathds{1}_{K^\epsilon}$. Thus by taking $n$ big enough we have by convergence against $C^{\alpha}_B$ functions that
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$$
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\mu_n(K^\epsilon)\geq\int g(x) \mu_n(dx) > 1 - 2\epsilon.
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$$
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Thus by taking $f\in\mathcal{H}$ and the associated $f_j$ we have
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\begin{align*}
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\left| \int f(x)\left(\mu_n(dx)-\mu(dx)\right) \right|&= \left| \int (f(x)-f_j(x))\left(\mu_n(dx)-\mu(dx)\right) \right| + \left| \int f_j(x)\left(\mu_n(dx)-\mu(dx)\right) \right|\\
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&\leq \left| \int (f(x)-f_j(x))\mu_n(dx) \right| + \left| \int (f(x)-f_j(x))\mu(dx) \right| + \left| \int f_j(x)\left(\mu_n(dx)-\mu(dx)\right) \right|\\
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&\leq \left| \int_{K^\epsilon} (f(x)-f_j(x))\mu_n(dx) \right| + \left| \int_{(K^\epsilon)^c} (f(x)-f_j(x))\mu_n(dx) \right| + \left| \int_{K^\epsilon} (f(x)-f_j(x))\mu(dx) \right| + \\
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&\qquad + \left| \int_{(K^\epsilon)^c} (f(x)-f_j(x))\mu(dx) \right| + \left| \int f_j(x)\left(\mu_n(dx)-\mu(dx)\right) \right|\\
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&\leq C_\alpha\epsilon^\alpha + 4\epsilon + C_\alpha\epsilon^\alpha+2\epsilon+\epsilon\leq C_\alpha\epsilon^\alpha,
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\end{align*}
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where the last term gets bounded by taking $n$ big enough and by using convergence against $C^{\alpha}_B$ functions, this gives us the final result.
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\end{proof}
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\begin{theorem}
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For small values of $T$; if the coefficients of the SDE are $C^\alpha_B$ functions of $y$ uniformly in $(t,x)$ (the $C^\alpha_B$ norm is uniformly bounded in $(t,x)$) we have that the application $\mathcal{L}:C([0,T],\mathcal{P}(\R^N))\rightarrow C([0,T],\mathcal{P}(\R^N))$ that $\mathcal{L}((\mu_t)_{t\in[0,T]})=(\mathcal{L}^\mu_t)_{t\in[0,T]}$ is a contraction wrt the distance
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$$
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d((\mu_t)_{t\in[0,T]},(\nu_t)_{t\in[0,T]})=\sup_{t\in[0,T]}d_{C^\alpha_B}(\mu_t,\nu_t).
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$$
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\end{theorem}
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\begin{proof}
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We have
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\begin{align*}
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d_{C^{\alpha}_B}(\mathcal{L}^\mu_t,\mathcal{L}^\nu_t)&=\sup_{||f||_{C^{\alpha}_B}\leq 1}|I(f)|\leq\\
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&\stackrel{Th.\ \ref{a1}}{\leq} C\sup_{s,x} \left(|B^\mu(s,x)-B^\nu(s,x)|+|c^\mu(s,x)-c^\nu(s,x)|\right)\int_0^t \left(\frac{1}{|t-s|^{1-\frac{\alpha}{2}}} + \frac{1}{\sqrt{t-s}}\right)ds\\
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&\leq C_T|t|^\frac{\alpha}{2}\sup_{s,x} \left(|B^\mu(s,x)-B^\nu(s,x)|+|c^\mu(s,x)-c^\nu(s,x)|\right).
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\end{align*}
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Now we observe that since the coefficients are uniformly $C^\alpha_B$ we have
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||
$$
|
||
|B^\mu(s,x)-B^\nu(s,x)|=\left| \int b(s,x,y)\mu_s(dy) - \int b(s,x,y)\nu_s(dy) \right|\leq C d_{C^{\alpha}_B}(\mu_s,\nu_s),
|
||
$$
|
||
where $C=||b||_{C^{\alpha}_B}$, a priori it depends on $(s,x)$ but since $b$ uniformly $C^\alpha_B$ it can be taken uniformly in $(s,x)$. It is also possible to prove that
|
||
$$
|
||
|c^\mu(s,x)-c^\nu(s,x)|\leq C ||\sigma||_{\infty}d_{C^{\alpha}_B}(\mu_s,\nu_s).
|
||
$$
|
||
with these we can conclude that
|
||
$$
|
||
d_{C^{\alpha}_B}(\mathcal{L}^\mu_t,\mathcal{L}^\nu_t)\leq C|t|^\frac{\alpha}{2}\sup_{s\in[0,t]} d_{C^{\alpha}_B}(\mu_s,\nu_s),
|
||
$$
|
||
and thus
|
||
$$
|
||
\sup_{t\in[0,T]}d_{C^{\alpha}_B}(\mathcal{L}^\mu_t,\mathcal{L}^\nu_t)\leq C|T|^\frac{\alpha}{2}\sup_{t\in[0,T]}d_{C^{\alpha}_B}(\mu_t,\nu_t).
|
||
$$
|
||
which proves contraction for small valus of $T$.
|
||
\end{proof}
|
||
\begin{comment}
|
||
\red{
|
||
We will now define\footnote{to be precise the Wasserstein metric is defined differently, still by the duality formula for the Wasserstein $1$-distance (Remark 6.5 of \cite{Villani}) the two definitions are equivalent}
|
||
\begin{align*}
|
||
d_{bL}(\mu,\nu)&=\sup_{||f||_{bLip}\leq 1} \left|\int f(x)\left(\mu(dx)-\nu(dx)\right)\right|,\\
|
||
W^{(1)}(\mu,\nu)&=\sup_{||f||_{Lip}\leq 1} \left|\int f(x)\left(\mu(dx)-\nu(dx)\right)\right|,
|
||
\end{align*}
|
||
respectively the bounded Lipschitz distance and the Wasserstein $1$-distance. It is possible to prove that convergence in these distances implies weak convergence of measures.
|
||
First let's concentrate on the bounded Lipschitz case, consider the coefficients of the SDE to be bounded and globally Lipschitz in the $y$ variable ($b(t,x,y)$), then we have
|
||
\begin{align*}
|
||
d_{bL}(\mathcal{L}^\mu_t,\mathcal{L}^\nu_t)&= \sup_{||f||_{bLip}\leq 1}|I(f)|\leq\\
|
||
&\stackrel{Assumption\ 1}{\leq} C\sup_{s,x} \left(|B^\mu(s,x)-B^\nu(s,x)|+|c^\mu(s,x)-c^\nu(s,x)|\right)\int_0^t \frac{1}{\sqrt{t-s}}\left(||f||_{L^\infty}+||f||_{Lip}\right)ds\\
|
||
\intertext{but the sum of these norms of $f$ is equal to $||f||_{bLip}$ which is less than $1$,}
|
||
&\leq C\sqrt{t}\sup_{s,x} \left(|B^\mu(s,x)-B^\nu(s,x)|+|c^\mu(s,x)-c^\nu(s,x)|\right).
|
||
\end{align*}
|
||
Now we most importantly observe that since the coefficients are bounded and Lipschitz we have
|
||
$$
|
||
|B^\mu(s,x)-B^\nu(s,x)|=\left| \int b(s,x,y)\mu_s(dy) - \int b(s,x,y)\nu_s(dy) \right|\leq C d_{bL}(\mu_s,\nu_s),
|
||
$$
|
||
where $C=||b||_{bLip}$, a priori it depends on $(s,x)$ but since $b$ is globally Lipschitz and bounded it can be taken uniformly in $(s,x)$. It is also possible to prove that
|
||
$$
|
||
|c^\mu(s,x)-c^\nu(s,x)|\leq C ||\sigma||_{\infty}d_{bL}(\mu_s,\nu_s).
|
||
$$
|
||
with these we can conclude that
|
||
$$
|
||
d_{bL}(\mathcal{L}^\mu_t,\mathcal{L}^\nu_t)\leq C\sqrt{t}\sup_{s\in[0,t]} d_{bL}(\mu_s,\nu_s),
|
||
$$
|
||
and thus
|
||
$$
|
||
\sup_{t\in[0,T]}d_{bL}(\mathcal{L}^\mu_t,\mathcal{L}^\nu_t)\leq C\sqrt{T}\sup_{t\in[0,T]}d_{bL}(\mu_t,\nu_t).
|
||
$$
|
||
which proves contraction for small valus of $T$.
|
||
If we consider the case with coefficients globally Lipschitz in $y$ but with $b$ possibly unbounded we have
|
||
\begin{align*}
|
||
W^{(1)}(\mathcal{L}^\mu_t,\mathcal{L}^\nu_t)&=\sup_{||f||_{Lip}\leq 1}|I(f)|\leq\\
|
||
&\stackrel{Assumption\ 1}{\leq} C\sqrt{t} \sup_{s,x} \left(|B^\mu(s,x)-B^\nu(s,x)|+|c^\mu(s,x)-c^\nu(s,x)|\right)
|
||
\intertext{using the estimates on $B$ and $c$ as above but with the Wasserstein distance since now $b$ is not bounded}
|
||
&\leq C\sqrt{t}\sup_{s\in[0,t]}W^{(1)}(\mu_s,\nu_s)
|
||
\end{align*}
|
||
and thus
|
||
$$
|
||
\sup_{t\in[0,T]}W^{(1)}(\mathcal{L}^\mu_t,\mathcal{L}^\nu_t)\leq C\sqrt{T}\sup_{t\in[0,T]}W^{(1)}(\mu_t,\nu_t).
|
||
$$
|
||
which proves contraction for small valus of $T$.}
|
||
\end{comment}
|
||
\begin{remark}
|
||
This approach of having the $\sup$ in the distance over the space of functions of the same regularity of the coefficients of the SDE seems quite natural ($b$ and $f$ in the same bounded space). It doesn't seem impossible to use these types of techniques for even broader classes of coefficients as long as there are gaussian estimates.
|
||
\end{remark}
|
||
%At this point we can try to bound $I(f)$:
|
||
%\begin{equation}\label{e6}
|
||
%\mu_0(g)=\int g(y)\mu_0(dy)\leq ||g||_{L^1(\mu_0)}.
|
||
%\end{equation}
|
||
%By Gaussian estimates we can observe that the backward propagation operator $V^{0,s}$ is a bounded operator\footnote{Here there is an error in the last equality since $\mu_0$ is not translation invariant. We could try to work with the $W^{(2)}$ distance or make this last inequality work.} in $L^1(\mu_0)$:
|
||
%\begin{align}\label{e7}
|
||
%||V^{0,s}_\mu g||_{L^1(\mu_0)}\leq \int\left|\int p^\mu(0,x;s,y)g(y)dy\right|\mu_0(dx)\leq C\int\int |g(z+x)|\Gamma^+(|s|,z)dz\mu_0(dx)=C||g||_{L^1(\mu_0)}.
|
||
%\end{align}
|
||
%Similarly to \cite{Kolokoltsov}, but without using the adjoint operator so without the need to ask for regularity of the coefficients, we can show the following bound:
|
||
%\begin{align}\nonumber
|
||
%||(\mathcal{A}^\mu_s-\mathcal{A}^\nu_s)g||_{L^1(\mu_0)}&=||(B^\mu_s-B^\nu_s)\bigtriangledown g +\frac{1}{2}\left\langle (c^\mu_s-c^\nu_s)\bigtriangledown, \bigtriangledown\right\rangle g||_{L^1(\mu_0)}\\ \label{e8}
|
||
%&\leq \sup_{s,x}\left( |B^\mu_s(x)-B^\nu_s(x)| + |c^\mu_s(x)-c^\nu_s(x)| \right)||g||_{W^{2,1}(\mu_0)}.
|
||
%\end{align}
|
||
%The following inequality still needs to be properly proved but is stated as (13) in \cite{Kolokoltsov} and seems reasonable
|
||
%\begin{equation}\label{e9}
|
||
%||V_\nu^{s,t}f||_{W^{2,1}(\mu_0)}\leq Cs^{-\frac{1}{2}}||f||_{W^{1,1}(\mu_0)}.
|
||
%\end{equation}
|
||
%Now we employ a couple of observations. Firstly $I(f)$ is invariant with respect to translations: $I(f+c)=I(f),\ \forall c\in\R$; for this reason without loss of generality $f(0)=0$ and thus
|
||
%\begin{align*}
|
||
%||f||_{L^1(\mu_0)}&=\int|f(x)|\mu_0(dx)\leq \int|f(x)-f(0)|\mu_0(dx) + \int|f(0)|\mu_0(dx)\leq ||f||_{Lip}\int |x|\mu_0(dx) = C,\\
|
||
%||\bigtriangledown f||_{L^1(\mu_0)}&\leq \int ||f||_{Lip}\mu_0(dx) = 1.
|
||
%\end{align*}
|
||
%Thus putting everything together in \eqref{e6}-\eqref{e9} and noticing that the estimate is uniform for any $f$ with Lipschitz constant bounded by $1$ we obtain
|
||
%$$
|
||
%W^{1}(\mathcal{L}^\mu_t,\mathcal{L}^\nu_t)\leq \int_0^t C\sup\left( |B^\mu-B^\nu| + |a^\mu-a^\nu| \right)s^{-\frac{1}{2}}ds = Ct^{\frac{1}{2}}\sup\left( |B^\mu-B^\nu| + |c^\mu-c^\nu| \right).
|
||
%$$
|
||
%Now, if we are able to prove an inequality in the form $\sup\left( |B^\mu-B^\nu| + |c^\mu-c^\nu|\right)\leq \sup_t W^{(1)}(\mu_t,\nu_t)$ we have a contraction. If the coefficients are Lipschitz in the third variable this is mostly trivial
|
||
%\begin{gather*}
|
||
%\sup\left( |B^\mu-B^\nu|\right)\leq \sup_{t,x}\left(\left|\int b(t,x,y_1)\mu_t(dy_1)-\int b(t,x,y_2)\nu_t(dy_2)\right| \right)\leq L\sup_{t,x}\left( \int|y_1-y_2|\gamma(dy_1,dy_2) \right)\\
|
||
%\stackrel{\inf}{\Rightarrow} \sup\left( |B^\mu-B^\nu|\right)\leq L\sup_{t,x}W^{(1)}(\mu_t,\nu_t),
|
||
%\end{gather*}
|
||
%this should also work for $c$ but the matrix form is a bit more challenging. If everything goes according to plan we should achieve that for small $T$ we have the contraction
|
||
%$$
|
||
%\sup_tW^{(1)}(\mathcal{L}^\mu_t,\mathcal{L}^\nu_t)\leq C\sqrt{T} \sup_tW^{(1)}(\mu_t,\nu_t).
|
||
%$$
|
||
|
||
|
||
|
||
%\begin{small}
|
||
%\red{
|
||
%Let's concentrate for a moment on $B$:
|
||
%\begin{align*}
|
||
%|B^\mu(t,x)-B^\nu(t,x)|&=|B(t,x,\mu_t)-B(t,x,\nu_t)|=\left|\int b(t,x,y)\mu_t(dy)-\int b(t,x,z)\nu_t(dz)\right|
|
||
%\intertext{Given $\gamma$ a law of marginals $\mu_t$ and $\nu_t$}
|
||
%&\leq \left|\int b(t,x,y)-b(t,x,z)\gamma(dy,dz)\right|
|
||
%\stackrel{\alpha-Hold}{\leq} ||b||_{C^{0,\alpha}}\int |y-z|^\alpha\gamma(dy,dz)
|
||
%\intertext{This is uniform over $\gamma$, by passing to the inf we arrive to the Wasserstein distance}
|
||
%|B^\mu(t,x)-B^\nu(t,x)|&\leq ||b||_{C^{0,\alpha}} W^{(\alpha)}(\mu_t,\nu_t)^\alpha\stackrel{Jensen}{\leq} ||b||_{C^{0,\alpha}}W^{(2)}(\mu_t,\nu_t)^\alpha.
|
||
%\end{align*}
|
||
%Probably something similar may be proven for $c$, effectively $c=\Sigma\Sigma^*$ and $\Sigma$ is $\alpha$-Holderian wrt the Wasserstein distance.\\
|
||
%This should almost prove what we need, we just need an inequality to bound\footnote{to accomplish such a feat the "$\inf$" part of the Wasserstein is fundamental since if for simplicity we consider the product of $Unif_{[0,1]}$ with itself it is very far from $0$ while the Wasserstein distance between a law and itself is $0$. More heuristically with the $\inf$ the Wasserstein considers heavily correlated random variables pair which drastically lower the value of the distance.} $W^{(2)}(\mathcal{L}^\mu_t,\mathcal{L}^\nu_t)$ with $||u^\mu_t-u^\nu_t||_{L^1}$. Alternatively we could bound $\sup_tW^{(2)}(\mu_t,\nu_t)$ with the $L^1$ norm of the difference of the densities associated with $\mu_t$ and $\nu_t$. Depending on the case the contraction is applied on $C([0,T]\times\mathcal{P}^2(\R^n))$ or in $C([0,T]\times L^1(\R^N))$.\\
|
||
%A little observation with the $W^{(1)}$ distance: due to Kantarovich's duality theorem we have (\cite{Villani} remark 6.5)
|
||
%\begin{align*}
|
||
%W^{(1)}(\mu,\nu)&=\sup_{||f||_{Lip}\leq1}\left(\int f\mu-\int f\nu\right)\stackrel{AC}{=}\sup_{||f||_{Lip}\leq1}\left(\int f(x)(\gamma_\mu(x)-\gamma_\nu(x))dx\right)\\
|
||
%&\stackrel{Lipschitz}{\leq}\int|x-x_0|\cdot|\gamma_\mu(x)-\gamma_\nu(x)|dx
|
||
%\end{align*}
|
||
%In some cases the inequality seems an equality. For simplicity let's consider the $1$-d case and define $\beta(dt)=|\gamma_\mu(t)-\gamma_\nu(t)|dt$:
|
||
%\begin{align*}
|
||
%W^{(1)}(\mu,\nu)&\leq\int_{-\infty}^{+\infty}|t|\beta(dt)=\int_0^{+\infty}t\beta(dt)-\int_{-\infty}^0t\beta(dt)\\
|
||
%&=\int_0^{+\infty}\int_0^tdx\beta(dt)+\int_{-\infty}^0\int_t^0dx\beta(dt)=\int_0^{+\infty}\beta([x,+\infty[)dx+\int_{-\infty}^0\beta(]-\infty,x])dx\\
|
||
%&=\int_0^{+\infty}\beta(]-\infty,-x]\cup[x,+\infty[)dx=\int_0^{+\infty}\int_{\R\setminus[-x,x]}|\gamma_\mu(t)-\gamma_\nu(t)|dtdx.
|
||
%\end{align*}
|
||
%Unfortunately here there is no space for bounding the Wasserstein distance with the $L^1$ distance: if we consider the densities uniform in $[n,n+1]$ and $[n+1/2,n+3/2]$ we get that the constant $C$ in $W^{(1)}(\mu,\nu)\leq C||\gamma_\mu-\gamma_\nu||_{L^1}$ needs to be arbitrarily big. The following theorem should shelter us from this type of problems.}
|
||
%\end{small}
|
||
|
||
There is the property of tightness for the family of measures that are solution of an SDE with $\alpha$-Holder coefficients and with initial law with finite $p$-moment:
|
||
\begin{theorem}\label{t1}
|
||
Let $\mu_0\in\mathcal{P}^p(\R^N)$. Let $p(s,x;t,y)$ be a fundamental solution of a forward Kolmogorov equation with $\alpha$-Holderian coefficients so that Gaussian estimates exist. Then for any $\epsilon>0$ there exists $K>0$ such that
|
||
$$
|
||
\int_{B_K^c}u_t(x)dx<\epsilon,\qquad \int_{B_K^c}|x|^pu_t(x)dx<\epsilon.
|
||
$$
|
||
Where $u_t(x)=\int p(0,y;t,x)\mu_0(dy)$.
|
||
\end{theorem}
|
||
\begin{proof}
|
||
The proof is a little variation of (3.2) in \cite{Kolokoltsov}; indeed the first inequality is proved there. Fix $\epsilon>0$. Let $\tilde{\epsilon}>0$ that we will fix later. Since $\mu_0$ is a measure with finite $p$-moment there exists $K>0$ such that
|
||
$$
|
||
\mu_0(B_K^c)<\tilde{\epsilon},\qquad \int_{B_K^c}|x|^p\mu_0(dx)<\tilde{\epsilon}.
|
||
$$
|
||
Let $\tilde{K}>0$ that we will fix later.
|
||
\begin{align*}
|
||
\int_{|x|\geq K+\tilde{K}}|x|^pu_t(x)dx&\stackrel{Gaussian\ estimates}{\leq} C\int_{|x|\geq K+\tilde{K}}|x|^p\int\Gamma^+(x-\xi,t)\mu_0(d\xi)dx\\
|
||
&\leq C\int_{|\xi|\geq K,y\in\R^N}|y+\xi|^p\Gamma^+(y,t)\mu_0(d\xi)dy + C\int_{|\xi|\leq K,y\geq\tilde{K}}|y+\xi|^p\Gamma^+(y,t)\mu_0(d\xi)dy\\
|
||
&\leq C_p\mu_0(B_K^c)\int |y|^p\Gamma^+(y,t)dy+C_p\int_{B_K^c}|\xi|^p\mu_0(d\xi)\\
|
||
&\qquad+C_p\mu_0(B_k)\int_{B_{\tilde{K}}^c}|y|^p\Gamma^+(y,t)dy + C_p\int|\xi|^p\mu_0(d\xi)\int_{B_{\tilde{K}}^c}\Gamma^+(y,t)dy
|
||
\intertext{The first two terms get bounded by the preliminary inequalities and the fact that the Gaussian has finite $p$ moment. The last two terms get bounded by a constant $\tilde{C}_{p,\tilde{K},T}$ that goes to $0$ as $\tilde{K}$ goes to $+\infty$.}
|
||
&\leq C_{T,p}\tilde{\epsilon} + C_p\tilde{\epsilon} + C_p\tilde{C}_{p,\tilde{K},T} + C_{p,\mu_0}\tilde{C}_{p,\tilde{K},T}.
|
||
\end{align*}
|
||
if we choose $\tilde{\epsilon}$ small enough and $\tilde{K}$ big enough the final result will be smaller than $\epsilon$. We must also notice that all the estimates and the constant do not depend directly on $u_t(x)$ but on the Gaussian estimates and so they hold uniformly for the whole family of solutions.
|
||
\end{proof}
|
||
We may notice that the family of the marginals is bounded in $\mathcal{P}^p$ with the Wasserstein metric.
|
||
\begin{theorem}
|
||
Let $(\mu_i)_{i\in\mathcal{I}}$ be the family of the marginals of solutions to SDEs with the same initial datum and $\alpha$-Holderian coefficients. (Written like this is not very rigorous but for example given $(\mu_t)_{t\in[0,T]}$ the flow of marginals of a solution to an SDE as in the hypothesis we have that $\mu_t$ is an element of the family for every $t\in[0,T]$).
|
||
|
||
Then the family is bounded as a subset of $\mathcal{P}^p(\R^N)$ equipped with the Wasserstein metric .
|
||
\end{theorem}
|
||
\begin{proof}
|
||
Let $\mu_1$ and $\mu_2$ be elements of the family. Fix $\epsilon>0$. By theorem \ref{t1} we know that exists $K>0$ such that
|
||
$$
|
||
\int_{B^c_K}\mu_i(dx)<\epsilon,\qquad \int_{B^c_K}|x|^p\mu_i(dx)<\epsilon,\qquad i=1,2.
|
||
$$
|
||
In particular given a measure $\gamma$ on $\R^{2N}$ with marginals $\mu_1$ and $\mu_2$ we have that exists $\tilde{K}$ (uniformly in $\gamma$) such that
|
||
\begin{equation*}
|
||
\int_{B^c_{\tilde{K}}}\gamma(dy,dz)\leq \int\int_{B^c_{K}\times B^c_{K}}\gamma(dy,dz)\leq \int\int_{B^c_{K}\times \R^N}\gamma(dy,dz)=\int_{B^c_K}\mu_1(dy)<\epsilon.
|
||
\end{equation*}
|
||
This also works for the $p$-moment and we get
|
||
\begin{equation*}
|
||
\int_{B^c_{\tilde{K}}}|y|^p\gamma(dy,dz)\leq \int\int_{B^c_{K}\times B^c_{K}}|y|^p\gamma(dy,dz)\leq \int\int_{B^c_{K}\times \R^N}|y|^p\gamma(dy,dz)=\int_{B^c_K}|y|^p\mu_1(dy)<\epsilon.
|
||
\end{equation*}
|
||
This means that we can bound the Wasserstein distance between the two in the following way:
|
||
\begin{align*}
|
||
W^{(p)}(\mu_1,\mu_2)^p&=\inf_{\gamma}\int\int(y-z)^p\gamma(dy,dz)\leq\inf_\gamma \int_{B^c_{\tilde{K}}}(y-z)^p\gamma(dy,dz) + \int_{B_{\tilde{K}}}(y-z)^p\gamma(dy,dz)\\
|
||
&\leq C_p\left(\int_{B^c_{\tilde{K}}}|y|^p\gamma(dy,dz) + \int_{B^c_{\tilde{K}}}|z|^p\gamma(dy,dz)\right) + \int_{B_{\tilde{K}}}diam(B_{\tilde{K}})^p\gamma(dy,dz)\\
|
||
&\leq 2C_p\epsilon + diam(B_{\tilde{K}})^p.
|
||
\end{align*}
|
||
\end{proof}
|
||
The preceding theorem in particular proves that the $p$-moments are uniformly bounded, for this reason the following theorem is valid in our case.
|
||
\begin{theorem}
|
||
Let $(\mu_i)_{i\in\mathcal{I}}\subset\mathcal{P}^p(\R^N)$ be a family of probability measures with tightness property as of Theorem \ref{t1} and such that the $p$-moments are uniformly bounded. Then for any sequence $(\mu_n)_{n\in\N}$ there exists a subsequence $(\mu_{n_m})_{m\in\N}$ and a measure $\mu\in\mathcal{P}^p(\R^N)$ such that $$\mu_{n_m}\stackrel{Wasserstein}{\rightarrow}\mu.$$
|
||
\end{theorem}
|
||
\begin{proof}
|
||
By theorem (6.9) of \cite{Villani} we have that Wasserstein convergence in $\mathcal{P}^p$ is equivalent to weak convergence and convergence of the $p$-moment. We know that
|
||
$$
|
||
\mu_i(B_K^c)<\epsilon,\qquad \int_{B_K^c}|x|^p\mu_i(dx)<\epsilon.
|
||
$$
|
||
Thus if we define $P_i(dx)=|x|^p\mu_i(dx)$ we have that $(P_i)_{i\in\mathcal{I}}$ is a tight family of uniformly finite measures, in particular we may use a generalization of Prokhorov's theorem (Theorem 8.6.2 of \cite{Bogachev}) and get that for any sequence there exists a subsequence $P_{n_m}$ that converges weakly to a finite measure $P$.
|
||
\begin{equation}\label{e10}
|
||
P_{n_m}(\phi)\rightarrow P(\phi),\qquad\forall \phi\in C_0^{\infty}(\R^N).
|
||
\end{equation}
|
||
By using the same theorem on $\mu_{n_m}$ we can find a new subsequence (that will still be witten as $\mu_{n_m}$) such that $\mu_{n_m}$ converges weakly to $\mu$.
|
||
\begin{equation}\label{e11}
|
||
\mu_{n_m}(\phi)\rightarrow \mu(\phi),\qquad\forall \phi\in C_0^{\infty}(\R^N).
|
||
\end{equation}
|
||
Consider now $\phi\in C_0^{\infty}(\R^N)$.
|
||
\begin{align*}
|
||
&\int |x|^p\phi(x)\mu_{n_m}(dx)\stackrel{\eqref{e11}}{\rightarrow} \int |x|^p\phi(x)\mu(dx)\\
|
||
&=P_{n_m}(\phi)\stackrel{\eqref{e10}}{\rightarrow} P(\phi).
|
||
\end{align*}
|
||
By uniqueness of the limit we have for any $\phi\in C_0^{\infty}$ that $P(\phi)=\int|x|^p\phi(x)\mu(dx)$. This proves convergence of the $p$-moment and thus with weak convergence we have Wasserstein convergence.
|
||
\end{proof}
|
||
%Most probably some of these $L^p$ spaces will need to be changed, for example it could be helpful to only work with $W^{(1)}$ and $L^1$ to follow more closely Kolokoltsov.
|
||
\begin{thebibliography}{90}
|
||
\bibitem{Bogachev} Vladimir I. Bogachev - Measure Theory (2007)
|
||
\bibitem{Kolokoltsov} Vassili N. Kolokoltsov - Nonlinear Diffusions and Stable-Like Processes
|
||
with Coefficients Depending on the Median or VaR (2013)
|
||
\bibitem{LucePagliaPascu} G. Lucertini, A. Pagliarani, A. Pascucci - Optimal regularity for degenerate Kolmogorov equations in non-divergence form with rough-in-time coefficients (2024), https://doi.org/10.1007/s00028-023-00916-9
|
||
\bibitem{YAOZHONG} Yaozhong Hu, Michael A. Kouritzin, Jiayu Zheng - Nonlinear McKean-Vlasov diffusions under the weak Hormander condition with quantile-dependent coefficients (2021), https://arxiv.org/abs/2101.04080
|
||
\bibitem{Villani} Cédric Villani - Optimal Transport Old and New (2009)
|
||
\end{thebibliography}
|
||
\end{document} |