Bounded density for diffusions with diffusion coefficients bounded away from $0$ Consider a diffusion given by
$$X_t=\int_0^t a(s,X_s)\,dW_s$$
for $t\ge 0$, where $W_\cdot$ is a standard Wiener process/Brownian motion and $a$ is a smooth enough positive function bounded away from $0$.
Does then $X_1$ have a bounded pdf?

This interesting answer by James Martin shows that, without assuming that $a$ is bounded away from $0$ and replacing $a(s,X_s)$ by $a\big(s,(X_u\colon 0\le u\le s)\big)$, it is possible that $P(X_1=0)>0$. See also comments to that answer.
 A: The "yes" answer follows immediately from Theorem 2.5 in this paper by Kusuoka, which implies that $X_1$ has a normal-like pdf $p$, such that
$$c_1 e^{-b_1x^2}\le p(x)\le c_2 e^{-b_2x^2}$$
for all real $x$, where $c_1,b_1,c_2,b_2$ are positive real constants depending only on $\inf_{t,x} a(t,x)>0$, $\sup_{t,x} a(t,x)<\infty$, and $\sup_{t,x,y\ne x}|(a(t,x)-a(t,y)|/|x-y|<\infty$.
A: I believe so. The local martingale $S_t = \int_{0}^t a(s, X_s) \, dW_s$ is a time changed Brownian motion, so by the Dambis-Dubins-Schwartz theorem, it can be written as a time changed Brownian motion $B_{f(t)}$, where $f(t) := \int_{0}^t a(s, X_s)^2 ds$ is the quadratic variation of $X_t$.
Assuming $a$ is bounded away from $0$ we have that $f(1) > C$ almost surely for some $C > 0$. Now since $a$ is deterministic, we have that $X_t$ and hence $f(1)$ are supported on the Wiener probability space $\Omega$ of continuous functions starting at $0$. This is a Radon space, and so we can write, by the disintegration theorem
$$\mathbb P(X_t \in U) = \int_{\mathbb R} \int_{f(1)^{-1}(y)} \mathbf 1_{\{B_y \in U\}} \ \, d\mu_y \ d\eta(y)$$
where $\mu_y, \eta$ is the regular conditional distribution given $f(1)$.
Since $f(1) > C$ a.s., we have that for finite measure Borel sets $U$ that $\int \mathbf 1_{\{B_y \in U\}} \, \ d\mu_y$ is uniformly $O (\mathcal \nu(U))$ as $\mathcal \nu(U)\to 0$ for all $y$ on which $\eta$ is supported. Here $\mathcal \nu$ denotes the Lebesgue measure. (There is an issue here, see edit.)
Thus as $\mathcal \nu(U) \to 0$,
$$\mathbb P(X_1 \in U) = \int_{\mathbb R} O(\mathcal \nu(U)) \, d\eta = O(\mathcal \nu(U))$$
and we conclude that $X_1$ admits a density as claimed.
I think by looking at the $O(\nu(U))$ term we can show further that the density is bounded, but I have not worked out the details yet.
Remark: I believe this proof also shows why the case where $a$ isn’t bounded away from $0$ admits a counterexample of the form $P(X_1 = 0) > 0$. The $B_y$’s can concentrate at the origin, and so I think the absolute continuity can only ever fail at, or near $0$.
Edit: There is an issue with the given claim - $B_y$ is not necessarily mormal with respect to $\mu_y$ so more work has to be done here to justify the $O(\nu(U))$ claim.
