= Solution
Let $B=\|b\|_\infty$ and define the <scale function of a one-dimensional diffusion>
$$
\boxed{
g(x)=\int_0^x
\exp\left(-2\int_0^u b(v)\,dv\right)\,du.}
$$
Since $b$ is continuous, $g\in C^2$, and
$$
g'(x)=\exp\left(-2\int_0^x b(v)\,dv\right)>0,\qquad
g''(x)=-2b(x)g'(x).
$$
Thus $g$ is strictly increasing. The <Itô formula> for the additive-noise equation gives
$$
d\,g(X_t)
=\left(b(X_t)g'(X_t)+\frac12g''(X_t)\right)dt+g'(X_t)dW_t
=g'(X_t)dW_t.
$$
The integrand is locally square-integrable: a continuous path $X$ has compact range on every finite interval, where $g'$ is bounded. Hence
$$
\boxed{Y_t=g(X_t)\text{ is a continuous local martingale}.}
$$
This is the <scale transform for an additive-noise diffusion>. Its range is the open interval $I=g(\mathbb R)$, which need not be all of $\mathbb R$.
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