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The expected exit time from $(\ell ,r)$ for a one-dimensional diffusion is ${\int }_{\ell }^{r}K(x,v)m(v)dv$ when finite. Differentiating this expression on each side of $v=x$ gives the equation $\mathcal{L}u=-1$ with zero endpoint values. For the SLE two-boundary-point ratio diffusion, $s(v)-s(1)$ behaves as $(v-1{)}^{1-4/\kappa }$ and the speed density behaves as $(v-1{)}^{4/\kappa }$, so the product is integrable at $1$ for $\kappa >4$.

Put $X_t=g_t(x)-W_t$, $Y_t=g_t(y)-W_t$ and $D_t=Y_t-X_t$. Before $T(x)$, order preservation gives $0<X_t<Y_t$ and $D_t>0$. The common noise cancels from their difference:Since $D$ has finite variation, applying the Itô formula to $Z=Y/D$ gives the SLE two-boundary-point ratio diffusionWith the strictly increasing clock $u(t)={\int }_0^t{D}_s^{-2}ds$, the Dambis-Dubins-Schwarz theorem yields a Brownian motion $B$ for which the time-changed process obeysThe sign of the new Brownian motion has absorbed the minus sign above.

The domain Markov property of a chordal Loewner chain says that, conditionally on the past before $T(x)$, the mapped future is a fresh SLE. Its boundary marked points are $X_t,Y_t$. HenceLocalize, for example, where $D_t\ge 1/n$, $Z_t\in [1+1/n,n]$ and $t\le n$. On each such stopped interval this is a bounded conditional-expectation martingale, and therefore a genuine martingale. No smoothness assumption on $f$ is needed for this argument. The exit calculation below identifies it explicitly.

For $\kappa \ge 8$, almost surely the positive-boundary swallowing times satisfy $T(x)<T(y)$ simultaneously for every $0<x<y$. The SLE two-boundary-point ratio diffusion has unbounded scale, ruling out escape before hitting $1$. For $4<\kappa <8$, its total scale $S_{\kappa }$ is finite, and the simultaneous-swallowing probability is $s_{\kappa }(y/(y-x))/S_{\kappa }$. At $\kappa =4$, fixed positive points have infinite swallowing times.