These are the conditions we impose on the Markov process and transition function.
Using this definition, we prove the backward Kolmogorov Equation.
Now it says that we have $u(x,t)=f(x)$ from the fact that $f \in C_b(\mathbb{R})$ and the arbitrariness of $\epsilon$. Taking $s \to t$, I can see that all we need is the middle term to $\to 0$ as $\epsilon \to 0$. But how can we ensure this here from the property of continuous boundedness of $f$?
Finally, in the use of the Taylor's theorem below, how do we get the remainder term for the second order Taylor as $\frac{1}{2}\frac{\partial^2 u(x,\rho)}{\partial x^2}(z-x)^2 \alpha_\epsilon$? I cannot find out which form of remainder gives this. I would greatly appreciate any help.
Taking $s\to t$, I can see that all we need is the middle term $\to 0$ as $\varepsilon \to 0$. But how can we ensure this here from the property of continuous boundedness of $f$?
Fix $x \in \mathbb R$. Since $f \in C_b(\mathbb R)$ we have that for any $a>0$ there exists $b>0$ (depending on $a$) s.t. $|f(x)-f(y)| < a$ for any $y$ s.t. $|x-y|<b$. Let $\omega_f(I) := \sup\{ |f(t)-f(u)|, u,t\in I\} $.
Claim: For any bounded interval $I$, $\omega_f(I)<\infty$, furthermore if $I_n$ is a shrinking system of intervals (i.e. $I_{n+1}\subset I_n$ for any $n$) s.t. $\bigcap_n I_ = \{x\}$, then $\omega_f(I_n) \searrow 0.$
Proof: Let $I$ be any bounded interval. Then for any $x,y \in I$ we have $|f(x)-f(y)| \leq |f(x)| + |f(y)| \leq 2\sup_{x\in I}|f(x)| <\infty $, due to $f\in C_b$. Hence $\omega_f(I) = \sup_{x,y\in I} |f(x)-f(y)| <\infty$.
Now suppose $I_n$ is a shrinking sequence and there exists $\varepsilon>0$ s.t. $\omega_f(I_n) \ge \varepsilon$ for all $n$. For this $\varepsilon$ due to continuity there is a $\delta>0$ s.t. $|f(x)-f(y)|<\varepsilon/4 $ for any y s.t. $|y-x|<\delta$. Well this means that $|f(y)-f(y')|<\varepsilon/2$ whenever $|y-x|<\delta$ and $|y'-x|<\delta$, furthermore $|y-y'|<2\delta$. Since $I_n\to \{x\}$, $diam(I_n)\to 0$, meaning that there is an index $n$ s.t. $diam(I_n)<2\delta$. This would mean that $$ \omega_f(I_n) \geq \varepsilon $$ but due to the previous argument $$ |f(y)-f(y')|<\varepsilon/2 $$ whenever $|y-y'|<2\delta$ and $\max\{|x-y|,|x-y'|\}<\delta$. This would mean that $$ \sup\{|f(y)-f(y')|: y,y'\in I_n\}\leq \varepsilon/2 $$ which contradicts our original assumption. $\Box$
Hence $$ \big|\int_{|x-y|\le \varepsilon} (f(y)-f(x)) P(dy,t|x,s)\big|\le \int_{|x-y|\le \varepsilon} |f(y)-f(x)| P(dy,t|x,s) \le \omega_f(\overline{B(x,\varepsilon)}) P(\overline{B(x,\varepsilon)},t|x,s)<\omega_f(\overline{B(x,\varepsilon)})$$ due to our previous claim this tends to zero as $\varepsilon \to 0$.
Finally, in the use of the Taylor's theorem below, how do we get the remainder term for the second order Taylor as $\frac12 \frac{\partial^2 u(x,ρ)}{\partial x^2}(z−x)^2(1+\alpha_\varepsilon)$?
According to Taylor's theorem $$ u(z,\rho)-u(x,\rho) = \partial_x u(x,\rho)(z-x) + \frac12 \partial^2_x u(\xi_z,\rho)(z-x)^2 $$ for some $\xi $ s.t. $|x-\xi_z|<|x-z|$. If $\partial_x^2 u$ is continuous then it can be shown that $\partial^2_x u(\xi_z, \rho) \to \partial_x^2 u(x,\rho)$ whenever $|z-x| \to 0$, i.e. $z\to x$. I don't think that the equality in (2.49) holds. It rather says that $ \partial^2u_x(z,\rho)$ is close to $\partial_x^2u(x,\rho)$ whenever $z$ is close to $x$, i.e. $$ |\partial_x^2 u(\xi_z,\rho) - \partial_x^2 u(x,\rho) | \le \sup_{t,|x-z|\le \varepsilon} \big| \partial_x^2u(x,t) - \partial_x^2 u(z,t) \big| $$