For a SDE with smooth transition densities, if every point is "path-accessible", is every positive-measure set probabilistically accessible?

Question

Suppose we have a $C^\infty$ manifold $M$ and $C^\infty$ vector fields $b,\sigma_1,\ldots,\sigma_k$ on $M$, and for convenience define the set of vector fields $$ \mathcal{S} = \{b,\sigma_1,-\sigma_1,\sigma_2,-\sigma_2,\ldots,\sigma_k,-\sigma_k\}\text{.} $$ Assume also that the "parabolic Hörmander condition" is fulfilled: letting $\mathcal{L}_b(\sigma_1,\ldots,\sigma_k)$ be the smallest Lie algebra that both contains $\sigma_1,\ldots,\sigma_k$ and is closed under $v \mapsto [v,b]$, every $x \in M$ has $\{f(x):f \in \mathcal{L}_b(\sigma_1,\ldots,\sigma_k)\}=T_xM$.

Suppose that "$\mathcal{S}$ allows path-accessibility of every point from every open set"; that is:

For every non-empty open $X_1 \subset M$ and every point $x_2 \in M$ there exists a continuous path $\gamma \colon [0,T] \to M$ and values $0=t_0<t_1<\ldots<t_n=T$ such that

• $\gamma(0) \in X_1$ and $\gamma(T) = x_2$;
• for each $i=1,\ldots,n$ there exists $f_i \in \mathcal{S}$ such that $\left.\gamma\right|_{[t_{i-1},t_i]}$ is a finite-time orbit of the differential equation $\dot{x}=f_i(x)$.

Does it follow that for every point $x_1 \in M$ and every Lebesgue-positive measure Borel set $X_2 \in \mathcal{B}(M)$, there exists $t_{x_1,X_2} \geq 0$ such that the strong solution $Y_t$ of the Wiener-driven SDE \begin{align*} dY_t &= b(Y_t) dt + \sum_{j=1}^k \sigma_j(Y_t) \circ dW_t^j \\ Y_0 &= x_1 \end{align*} has $\mathbb{P}(Y_{t_{x_1,X_2}} \in X_2)>0$? If so, is this a known result (or easy consequence of a known result)?

Answer 1

The standard approach is to first use Hörmander’s theorem to prove that the law of $Y_t$ has a smooth density with respect to Lebesgue measure. To prove this density is strictly positive, it suffices to show: for all $x \in M$, $t>0$ and for some $s \in (0,t)$ the following set fills up $M$ $$ A(s,x):=\{ y_s(u) : u \in H^1((0,\infty), \mathbb{R}^k) \} $$ where $y_s(u)$ satisfies the associated control problem $$ \dot{y_s}=b(y_s)+\sum_{j=1}^k \sigma_j(y_s) \dot{u_j}(s)\;, \quad y_0=x\;. $$ These results (and much more) are neatly described (on flat spaces) in Section 3.3 of Michel and Pardoux 1990.

Answer 2

By adapting the arguments in Sec. 3.3.6.1 of the Michel & Pardoux notes linked to by Nawaf Bou-Rabee, I think I can prove the result. (I will assume for simplicity that the SDE has global existence of strong solutions.)

Lemma. Let $M$ be a $C^1$ manifold. Let $(I,\mathcal{I},\mu)$ and $(J,\mathcal{J},\nu)$ be probability spaces; and over these probability spaces respectively, let $(\Phi_\alpha)_{\alpha \in I}$ and $(\Psi_\beta)_{\beta \in J}$ be random $C^1$ self-embeddings of $M$. Fix $x_1,x_2 \in M$, let $p$ be the law over $\mu$ of $\alpha \mapsto \Phi_\alpha(x_1)$, and let $\tilde{p}$ be the law over $\mu \otimes \nu$ of $(\alpha,\beta) \mapsto \Psi_\beta(\Phi_\alpha(x_1))$. Suppose there exists $m>0$ and an open subset $U$ of a chart on $M$ such that

• for every $A \in \mathcal{B}(U)$, $p(A) \geq m\mathrm{Leb}(A)$;
• $\nu(\beta \in J : x_2 \in \Psi_\beta(U))>0$.

Then there exists $\tilde{m}>0$ and a neighbourhood $\tilde{U}$ of $x_2$ contained in a chart on $M$ such that for every $A \in \mathcal{B}(\tilde{U})$, $\tilde{p}(A) \geq \tilde{m}\mathrm{Leb}(A)$.

Proof. One can find a $\nu$-positive measure set $J' \subset J$, a neighbourhood $\tilde{U}$ of $x_2$ contained in a chart on $M$, and a value $r>0$, such that for all $\beta \in J'$ and $x \in \tilde{U}$, we have $\Psi_\beta^{-1}(x) \in U$ and $|\mathrm{det}(D(\Psi_\beta^{-1})(x))| \geq r$. Now take any $A \in \mathcal{B}(\tilde{U})$ and let $$ E := \{(\alpha,\beta) \in I \times J' : \Psi_\beta(\Phi_\alpha(x_1)) \in A \}\text{;} $$ then \begin{align*} \tilde{p}(A) \geq (\mu \otimes \nu)(E) &= \int_{J'} \mu(\alpha \in I : \Psi_\beta(\Phi_\alpha(x_1)) \in A) \, \nu(d\beta) \\ &= \int_{J'} p(\Psi_\beta^{-1}(A)) \, \nu(d\beta) \\ &\geq m\int_{J'} \mathrm{Leb}(\Psi_\beta^{-1}(A)) \, \nu(d\beta) \\ &\geq \underbrace{m\nu(J')r}_{\ \ \ =:\ \tilde{m}}\mathrm{Leb}(A). \quad\quad\quad\quad\quad \square \end{align*}

Now applying the Lemma to my problem: Fix $x_1$ and $X_2$. Now fix $x_2 \in M$ such that every neighbourhood of $x_2$ has positive-measure intersection with $X_2$. Due to Hörmander’s theorem, we can find $t'>0$, $m>0$ and a non-empty open subset $X_1$ of a chart on $M$ such that the law of $Y_{t'}$ has Lebesgue density of at least $m$ on $X_1$. Now let $T>0$ be as in the definition of path-accessibility applied to our set $X_1$ and our point $x_2$, and take $t_{x_1,X_2}:=t'+T$. Let $(I,\mathcal{I},\mu)$ be the Wiener space defined over $[0,t']$, let $(J,\mathcal{J},\nu)$ be the Wiener space defined over $[0,T]$, and let $\Phi$ and $\Psi$ be the respective time-$t'$ and time-$T$ mappings of the SDE. Since $X_1$ is open, applying the "support theorem" [Theorem 3.3.1(b) of the Michel & Pardoux notes] to the reverse-time SDE gives that $\nu(x_2 \in \Psi(X_1))>0$. Hence the Lemma gives the desired result.