 Solution

Put $q(x)=\sigma (x{)}^{2}p(x)$. The explicit density formula and local integrability assumption giveHence its Fokker-Planck probability current $bp-q^{\prime }/2$ is zero. The stationary adjoint equation is $L^{\ast }p=-(bp{)}^{\prime }+q^{\prime \prime }/2=0$. We prove the required integral identity with cutoffs, avoiding an unstated boundary condition on derivatives of $u$.

Choose smooth $0\le {\chi }_R\le 1$, equal to one on $[-R,R]$, supported in $[-2R,2R]$, with $|{\chi }_R^{\prime }|\le C_1/R$ and $|{\chi }_R^{\prime \prime }|\le C_2/R^2$. Since $pLu=(q{u}_x{)}^{\prime }/2$, integrate the backward equation in time and integrate by parts twice in space:These integrations have compact support, as permitted in the question. Bounded coefficients and normalized $p$ giveThe right side is bounded in absolute value bywhich tends to zero. Dominated convergence on the left therefore provesThis is the cutoff proof of invariance for a zero-flux diffusion density. In particular, if an independent initial state has density $p$, part (a) and Fubini give a constant expectation of this test function at every time. This consequence alone is sufficient for the final coupling argument.