= Solution
\b[The existence assertion is false under the hypotheses printed in the source.] No measurability of $\theta$ is specified. Even requiring $\theta$ to be a <Borel measurable function> does not suffice: a coefficient can be positive at every point while approaching zero too rapidly near the starting point. This is the obstruction described by <pointwise positivity does not ensure weak existence>.
Here is a counterexample in dimension one, with $\Theta=1$:
$$
\theta(0)=1,\qquad \theta(x)=\min(x^2,1)\quad(x\ne0).
$$
Suppose that a solution to the <martingale problem> with domain $C_c^\infty(\mathbb R)$ existed. The cutoff argument from part (a)(ii), now with $\mathcal L=\theta\Delta$, makes $X$ a <continuous local martingale> with
$$
[X]_t=2\int_0^t\theta(X_s)\,ds\leq2t.
$$
The criterion for a <continuous local martingale with bounded quadratic variation> makes it a <square-integrable> <martingale>, and the <Itô isometry> gives $\mathbb EX_t^2=\mathbb E[X]_t$.
Apply the <occupation-times formula> with the indicator of the singleton $\{0\}$. Since that singleton has zero <Lebesgue measure>,
$$
0=\int_0^t\mathbf1_{\{X_s=0\}}\,d[X]_s
=2\int_0^t\mathbf1_{\{X_s=0\}}\,ds.
$$
Thus $X$ spends zero ordinary time at zero. Away from zero, $\theta(x)\leq x^2$, so
$$
0\leq\mathbb EX_t^2
=2\mathbb E\int_0^t\theta(X_s)\,ds
\leq2\int_0^t\mathbb EX_s^2\,ds.
$$
The <Gronwall inequality> forces $\mathbb EX_t^2=0$ for every $t$. Continuity then forces $X\equiv0$ <almost surely>, contradicting either the zero occupation time just proved or $[X]_t=2t$ for that constant path. Consequently
$$
\boxed{0<\theta\leq\Theta\text{, even Borel measurable, does not ensure existence}.}
$$
For completeness, the intended construction works if $\theta$ is a <Borel measurable function> and bounded away from zero on every compact set. In particular, it works for continuous strictly positive $\theta$. Let $B$ be standard $d$-dimensional <Brownian motion>, and define a clock and its inverse by
$$
A_u=\int_0^u\frac{dr}{2\theta(B_r)},\qquad
\tau_t=\inf\{u\geq0:A_u>t\},\qquad X_t=B_{\tau_t}.
$$
The Brownian path on each finite interval has compact range, so the added hypothesis makes $A_u$ finite. The clock is continuous and strictly increasing, and $A_u\geq u/(2\Theta)$, so $A_u\to\infty$. Its inverse is continuous and satisfies $\tau_t\leq2\Theta t$. Each $\tau_t$ is a <stopping time> for the Brownian <filtration>.
For $f\in C_c^\infty(\mathbb R^d)$, the <optional sampling theorem> applied at these bounded random times shows that
$$
f(B_{\tau_t})-\frac12\int_0^{\tau_t}\Delta f(B_r)\,dr
$$
is a true <martingale> in the <filtration> $\mathcal G_t=\mathcal F^B_{\tau_t}$. The inverse-clock substitution $d\tau_t=2\theta(X_t)dt$, valid almost everywhere, rewrites it as
$$
f(X_t)-\int_0^t\theta(X_s)\Delta f(X_s)\,ds.
$$
This proves the <Brownian time change for a variable diffusion speed>:
$$
\boxed{X_t=B_{A^{-1}(t)}\text{ solves the corrected martingale problem}.}
$$
The factor $2$ in the clock is necessary because standard <Brownian motion> has <infinitesimal generator> $\Delta/2$.
Back to article page
