I am struggling with computing the functional of a diffusion process. Assume that $(X_t)_{t \in [0,\infty)}$ is a one-dimensional, time-homogeneous, diffusion process such that the infinitesimal coefficients\begin{equation}\lim_{h \downarrow 0} \frac{1}{h} \mathbb{E} \Big[ \big( X(t+h) - X(t) \big)^p \mid \{ X(h) = x \} \Big] \quad p \in \{1,2\}\end{equation}depend only on $x$ and not on $t$, and such that\begin{equation}\mathbb{P} \big( \tau_z < \infty \mid \{X(0) = x\} \big) > 0 \quad \forall x,z \in \mathbb{R}\end{equation}where $\tau_z$ is the first hitting time of $z$.
The problem at hand is the derivation of\begin{equation}u(x) = \mathbb{P} \big( \{ \tau_b < \tau_a \} \mid \{ X(0) ) = x \} \big) \quad x \in (a,b) \subset \mathbb{R}\end{equation}At the beginning of the derivation, my textbook (Karlin-Taylor, A second course in stochastic processes, 1975, page 193) writes the following:
I tried to obtain this result using the law of total probabilities in the text of this question, but I don't know if it is correct. As an alternative, I just thought that the result may simply follow from\begin{align}\mathbb{E} \big[ u(X(h)) \mid \{ X(0) = x \} \big]& = u(x) + u^{\prime}(x) \cdot \mathbb{E} \big[ \Delta_h X_0 \mid \{ X(0) = x \} \big] + \frac{1}{2} u^{\prime \prime} (x) \cdot \big[ (\Delta_h X_0)^2 \mid \{ X(0) = x \} \big] + o([\Delta_h X_0]^2) \\& = u(x) + o(h)\end{align}where $[\Delta_h X_0]^p = \big( X(h) - X(0) \big)^p$.
But, first, the additional terms on the right are not supposed to be $o(h)$, and then I wouldn't see why the textbook invoked the law of total probabilities.
If the law of total probabilities must play a role, I would observe that, with $h$ small (which shold allow to restrict the extremes of the integral), one has\begin{align}u(x) & \approx \int_{a}^{b} \mathbb{P} \big( \{ \tau_b < \tau_a \} \mid \{ X(h) = y \} \big) \cdot \mathbb{P} \big( \{ X(h) \in d y \} \mid \{ X(0) = x \} \big) \\\\\text{and } \\\\\mathbb{E} \big[ u(X(h)) \mid \{ X(0) = x \} \big] & \approx \int_{a}^{b} \mathbb{P} \big( \{ \tau_b < \tau_a \} \mid \{ X(0) = y \} \big) \cdot \mathbb{P} \big( \{ X(h) \in d y \} \mid \{ X(0) = x \} \big)\end{align}
and then try to prove that\begin{equation}\mathbb{P} \big( \{ \tau_b < \tau_a \} \mid \{ X(h) = y \} \big) \underset{h \to 0}{\longrightarrow} \mathbb{P} \big( \{ \tau_b < \tau_a \} \mid \{ X(0) = y \} \big) \quad \forall y \in (a,b)\end{equation}Perhaps time-homogeneity could help here?