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Let $\theta :{\mathbb{R}}^d\to (0,\Theta ]$ be a Borel measurable function bounded away from zero on every compact set. If $B$ is standard $d$-dimensional Brownian motion, the displayed clock is finite on compact intervals, continuous, strictly increasing, and unbounded. Its inverse ${\tau }_t=A^{-1}(t)$ is a Brownian stopping time, with ${\tau }_t\le 2\Theta t$. The optional sampling theorem and the change of variables $d{\tau }_t=2\theta (B_{{\tau }_t})dt$ show that $X_t=B_{{\tau }_t}$ solves the martingale problem for $\mathcal{L}=\theta \Delta$ on test functions $C_c^{\infty }({\mathbb{R}}^d)$. The filtration is ${\mathcal{F}}_{{\tau }_t}^B$. These are true martingales: on a finite new-time horizon the sampling times have a deterministic bound. Continuous strictly positive $\theta$ satisfies the compact-set lower bound. Pointwise positivity alone does not; pointwise positivity does not ensure weak existence gives a counterexample.

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 $\Theta =1$: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 withThe criterion for a continuous local martingale with bounded quadratic variation makes it a square-integrable martingale, and the Itô isometry gives $\mathbb{E}{X}_t^2=\mathbb{E}[X{]}_t$.

Apply the occupation-times formula with the indicator of the singleton $\{0\}$. Since that singleton has zero Lebesgue measure,Thus $X$ spends zero ordinary time at zero. Away from zero, $\theta (x)\le x^2$, soThe Gronwall inequality forces $\mathbb{E}{X}_t^2=0$ for every $t$. Continuity then forces $X\equiv 0$ almost surely, contradicting either the zero occupation time just proved or $[X{]}_t=2t$ for that constant path. Consequently

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 byThe 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 $A_u\ge u/(2\Theta )$, so $A_u\to \infty$. Its inverse is continuous and satisfies ${\tau }_t\le 2\Theta t$. 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 thatis a true martingale in the filtration ${\mathcal{G}}_t={\mathcal{F}}_{{\tau }_t}^B$. The inverse-clock substitution $d{\tau }_t=2\theta (X_t)dt$, valid almost everywhere, rewrites it asThis proves the Brownian time change for a variable diffusion speed:The factor $2$ in the clock is necessary because standard Brownian motion has infinitesimal generator $\Delta /2$.