Suppose an $n$-dimensional process $(X_t)_{0 \leq t \leq 1}$ satisfies an SDE of the form:

$$dX_t = u_t(X_t) \,dt + dB_t, ~~X_0 = 0$$

where $(B_t)_{t\geq 0}$ is a Brownian motion with $B_1 \sim N(0,K)$, and $K$ is positive definite.

Does anyone have a reference for simple conditions on $u_t$ that will ensure that $\operatorname{law}(X_t)$ has full support on $\mathbb{R}^n$?

For example, for $t>0$, if there exists a finite constant $C_t$ such that $|u_s(x)| \leq C_t(|x|+1)$ for all $0 \leq s\leq t$, does this ensure $\operatorname{law}(X_t)$ has full support?

Of course, some regularity on $u_t$ is needed; consider the standard Brownian bridge, which has $X_1 = 0$ a.s., but $\operatorname{law}(X_t)$ has full support for each $0< t<1$.

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    $\begingroup$ At least in the case where there exists a density, one can use the Hormander-Malliavin theory to get conditions for strict positivity $p(x,y)>0$ for all $x\in \mathbb{R}^{n}$. $\endgroup$ Commented Apr 15, 2023 at 19:28
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    $\begingroup$ As mentioned here mathoverflow.net/questions/424273/…, one source is in Michel-Pardoux "Introduction Malliavin calculus" contains a general result involving the vector fields in 3.3.1 for positivity in the Stratonivich form (and so you have to use the conversion to Ito). $\endgroup$ Commented Apr 15, 2023 at 19:29
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    $\begingroup$ I think this question is actually closer to the OP's: mathoverflow.net/q/438425 Full support is much weaker than regularity of the density, so the conditions on $u_t$ are much weaker as well. $\endgroup$
    – Pierre PC
    Commented Apr 15, 2023 at 21:28
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    $\begingroup$ These theorems are usually called Stroock-Varadhan type, if that makes it easier to look it up. $\endgroup$
    – Pierre PC
    Commented Apr 15, 2023 at 21:32
  • $\begingroup$ @PierrePC and ThomasKojar, thanks for the pointers. This gives me quite a bit to go on. The Stroock-Varadhan-type support theorems seem pretty complicated in terms of their conditions, so I'm still hopeful there's something simpler that would ensure full support, even if it's not as generally stated as possible (the constant diffusion coefficient might help here also). $\endgroup$
    – Tom
    Commented Apr 15, 2023 at 21:59

1 Answer 1


I am writing down a very similar answer to this one, that I wrote for the similar question SDE with non-degenerate diffusion visits every point. I am not sure how closely it answers your question, since it is not a reference for the exact result you quote, but it does follow as a corollary.

The support of the whole trajectory $X$ in $\mathcal C([0,\infty),\mathbb R^n)$ is described by the so-called Stroock-Varadhan support theorem. In On the support of diffusion processes with applications to the strong maximum principle, their Theorem 3.1 ensures that the (unique) solution to the martingale problem associated with $u_t+\frac12\Delta_K$ has full support provided $u:[0,\infty)\times\mathbb R^d\to\mathbb R^d$ is bounded measurable.

A strong solution to the equation $\mathrm dX_t=u_t(X_t)\mathrm dt + \mathrm dB_t$ will always be a solution to the martingale problem, so this answers your question in the case where (1) you have conditions guaranteeing existence of a strong solution and (2) $u$ is bounded measurable (you can also define the solution as a solution to the martingale problem, in which case (2) is sufficient for existence and uniqueness).

A natural case that doesn't quite fit the above is when $u$ is unbounded but you have a unique strong and weak solution up to some explosion time using some other argument $\tau$. We can still rely on this theorem in more general situations, and to illustrate my point I will consider the classical case where $u$ is continuous in $(t,x)$, locally Lipschitz in $x$ (for instance $u$ is $\mathcal C^1$ in $(t,x)$). Then one can define the solution $X^R>0$ to the equation $$ \mathrm dX^R_t = \big(0\vee(2R-|X^R_t|)\wedge1\big)u_t(X^R_t)\mathrm dt + \mathrm dB_t, $$ which is defined for all times and coincides with $X$ until one of the two exists the ball of radius $R$. This means that $$\mathbb P(X_t\in U)\geq\mathbb P(X_t\in U\text{ and }\forall s\leq t,|X_s|<R)=\mathbb P(X^R_t\in U\text{ and }\forall s\leq t,|X^R_s|<R)$$ for $R$ large enough. By the theorem above, we know that this last probability must be positive, and so is the first one.

A similar argument shows that your condition $|u_s(x)|\leq C_t(1+|x|)$ is enough, provided $u$ is measurable, and your notion of solution coincides with the unique martingale problem solution.

I should add that these results rely on applying a Girsanov argument, making a change of probability with density (if I am not mistaken with my use of $K$ and $K^{-1}$) $$\exp\left(\int_0^tu_s(X_s)K^{-1}\mathrm dB_s-\frac12\int_0^tu_s\mathrm ds\right).$$


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