Proof of Backward Kolmogorov Equation. Question on remainder of Taylor's theorem and limit of integrals on decreasing balls 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.

 A: 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.
$$ |\partial_x^2 u(\xi_z,\rho) - \partial_x^2 u(x,\rho) | \le \sup_{t,|x-z|\le \varepsilon} \big| \partial_x^2u(x,t) - \partial_x^2 u(z,t) \big| $$
