I am reading about a topic on Kolmogorov equation and diffusion process. However, I am confused by the last using so called "regular function". I wonder what does this "regular function" mean here? And why do we even need to write the final equation in terms of $u$ not $F_{t}\left(x, y_{0}\right)$?
$$ \frac{\partial}{\partial t} F_{t}\left(x, y_{0}\right)=\rho(x) \frac{\partial}{\partial x} F_{t}\left(x, y_{0}\right)+\frac{1}{2} Q(x) \frac{\partial^{2}}{\partial x^{2}} F_{t}\left(x, y_{0}\right) \qquad(10.9.11) $$ which is the Kolmogorov backward equation for stationary diffusion processes. The initial condition is given by $$ \lim _{t \downarrow 0} F_{t}\left(x, y_{0}\right)= \begin{cases}1, & \text { if } x<y_{0} \\ 0, & \text { if } x>y_{0}\end{cases} $$ For a reasonably regular function $f(x)$ on $\mathbb{R}$, define $$ u(t, x)=\int_{\mathbb{R}} f(y) P_{t}(x, d y)=\int_{\mathbb{R}} f(y) d F_{t}(x, y) . $$ It follows from Equation (10.9.11) that $u$ satisfies the following equation: $$ \frac{\partial u}{\partial t}=\mathcal{A} u, \quad u(0, x)=f(x) $$