= Solution
Use $M_0=0$, as implicit in the specified initial value of $Y$. First let $[M]_t=t$. The <Lévy characterization of Brownian motion> makes $M$ <Brownian motion> relative to the given <filtration>. Set
$$
\varepsilon=\mathbb P(|G|>C+2)>0,\qquad G\sim N(0,1).
$$
On this event at time one, the drift has absolute size at most $C$, while $|Y_0|<1$, so $|Y_1|\ge|M_1|-|Y_0|-C>1$. Continuity forces exit by time one. Therefore
$$
\boxed{\mathbb P(T>1)\le1-\varepsilon.}
$$
The same estimate applies conditionally at each integer time without assuming anything about future drift independence. On $\{T>n\}$, $|Y_n|<1$; if $|M_{n+1}-M_n|>C+2$, the drift bound forces exit by $n+1$. That Brownian increment is independent of $\mathcal F_n$. Hence
$$
\mathbb P(T>n+1\mid\mathcal F_n)\le(1-\varepsilon)\mathbf1_{\{T>n\}},\qquad \mathbb P(T>n)\le(1-\varepsilon)^n.
$$
In particular $T<\infty$ almost surely.
For general $Q$, write $K_t=[M]_t$ and $D_t=\int_0^t A_s\,ds$. Continuity of $Q$ and monotonicity of its integral imply $Q_s\ge0$ at all times outside one null set. The drift condition gives
$$
|D_t-D_s|\le C(K_t-K_s)\qquad(s\le t).
$$
Let $\tau_u=\inf\{t:K_t>u\}$. The divergent clock makes $\tau_u$ finite for every $u$, and its continuity gives $K_{\tau_u}=u$. By the <Dambis-Dubins-Schwarz theorem>, $W_u=M_{\tau_u}$ is <Brownian motion> for $\mathcal G_u=\mathcal F_{\tau_u}$. On intervals where $K$ is constant, both $M$ and $D$ are constant. Thus $d_u=D_{\tau_u}$ is a continuous adapted function of clock time satisfying $|d_v-d_u|\le C(v-u)$. The time-changed process is
$$
\widetilde Y_u=Y_{\tau_u}=Y_0+W_u+d_u.
$$
The unit-clock argument uses only this increment bound on the drift, so its first exit time $S$ obeys $\mathbb P(S>n)\le(1-\varepsilon)^n$. It is finite almost surely, and a corresponding finite ordinary time has $|Y|\ge1$. This proves the <exit with drift controlled by quadratic variation> result
$$
\boxed{T<\infty\quad\text{almost surely}.}
$$
This proof needs only a nonnegative measurable bracket density, a continuous divergent integral clock, and the drift increment bound; pointwise continuity of the density is unnecessary. This observation permits the application below with measurable SDE coefficients.
For the multidimensional SDE, put $a_{ij}(x)=\sum_{k=1}^m\sigma_k^i(x)\sigma_k^j(x)$. The <generator of an SDE diffusion> is
$$
\boxed{Lf(x)=\sum_i b_i(x)\partial_if(x)+\frac12\sum_{i,j}a_{ij}(x)\partial_{ij}f(x).}
$$
For every $f\in C_b^2(\mathbb R^d)$, the <Itô formula> gives
$$
f(X_t^x)-f(x)-\int_0^tLf(X_s^x)\,ds=\sum_k\int_0^t\nabla f(X_s^x)\cdot\sigma_k(X_s^x)\,dB_s^k.
$$
Bounded derivatives and bounded noise fields make the right side square-integrable on each finite horizon, hence a true <martingale>. This is precisely the defining test-function property of an <L-diffusion>, in the martingale-problem sense.
Now fix $x$ and $\xi\ne0$. Its scalar projection has <martingale> part $M_t^\xi=\sum_k\int_0^t\langle\xi,\sigma_k(X_s^x)\rangle\,dB_s^k$ and drift $\langle\xi,b(X_s^x)\rangle$. The <uniform ellipticity> assumption gives
$$
\frac{d[M^\xi]_t}{dt}=\sum_k\langle\xi,\sigma_k(X_t^x)\rangle^2\ge\lambda|\xi|^2.
$$
Let $B_*=\sup_z|b(z)|$ and take $R>|\langle\xi,x\rangle|$. Rescale the projection by $R$. Its bracket density is at least $\lambda|\xi|^2/R^2$, its absolute drift is at most $B_*|\xi|/R$, and hence that drift is at most $C_R$ times its bracket density, with $C_R=B_*R/(\lambda|\xi|)$. Its clock diverges, so the exit lemma forces it out of $(-1,1)$ in finite time. Equivalently it leaves the slab $|\langle\xi,X_t^x\rangle|<R$. Taking a countable sequence of integer $R$ tending to infinity proves the <projection unboundedness of a uniformly elliptic diffusion>:
$$
\boxed{\sup_{t\ge0}|\langle\xi,X_t^x\rangle|=\infty\quad\text{almost surely, for each fixed }x,\ \xi\ne0.}
$$
For the final uniqueness result, let $\tau_D=\inf\{t:X_t^x\notin D\}$, with $x\in D$. Since $D$ is bounded, it lies in a finite slab; the preceding finite slab-exit time bounds $\tau_D$ from above. Thus $\tau_D<\infty$ almost surely. Continuity gives $X_{\tau_D}^x\in\partial D$.
Set $h=u-v$. It is bounded, has bounded derivatives, is zero on the boundary, and satisfies $Lh=0$ in $D$. The stopped test-function identity says $h(X_{t\wedge\tau_D}^x)$ is a <martingale>; equivalently it is a bounded <local martingale>, which suffices. Taking expectations and then using bounded convergence as $t\to\infty$ gives
$$
h(x)=\mathbb E_x h(X_{\tau_D})=0.
$$
Therefore the <bounded-domain harmonic uniqueness for a diffusion> conclusion is
$$
\boxed{u=v\text{ throughout }D.}
$$
In particular the argument supplies the representation $u(x)=\mathbb E_xu(X_{\tau_D})$ and establishes uniqueness without selecting a unique diffusion law.
Back to article page
