In order to show that the infinitesimal generator of the Brownian motion is $\frac{1}{2}\Delta$, in this answer, first he writes equation $$ \frac{d}{dt} P_t f(x) = A P_tf(x), \tag{1} $$ then he derives the following approximation : $$ \mathbb{E}^x(f(B_t)) \approx f(x)+ \frac{t}{2} f''(x) $$ Then it is argued that " From (1) we see that $u(t,x) := \mathbb{E}^x(f(B_t))$ is the (unique) solution of the heat equation"
As discussed here, we cannot simply replace the approximation into the heat equation. If so,
- Why did the author of that post made this approximation? how did he use this approximation for the proof? if he didn't use it,
- Can anybody explain more his argument that : " From (1) we see that $u(t,x) := \mathbb{E}^x(f(B_t))$ is the (unique) solution of the heat equation..."?
Your confusion might arise because you think somehow the approximation serves to construct a solution to the heat equation. What is happening is that you start with a solution to some partial differential equation (PDE), and the approximation serves to identify this PDE as the heat equation. No proofs were given in either of the posts you linked. They are just formal arguments to help develop intuition.
Start with your second question. Equation (1) is $$\frac{d}{dt}P_t f(x) = A P_t f(x).$$ By definition, $P_t f(x) = \mathbb{E}^x [f(B_t)].$ Setting $u(t,x) = P_t f(x)$ in equation (1), we have $$\frac{d}{dt} u(t,x) = A u(t, x). \tag{$\spadesuit$}$$ Here, $A$ is a differential operator, so $u(t,x)$ solves some differential equation with some initial conditions. Which differential equation is it?
To guess which differential equation it is, the approximation $u(t,x) \approx f(x) + t f''(x)/2$ is used. Putting this directly into the left-hand side of ($\spadesuit$), you find $$\frac{d}{dt} u(t,x) = \frac{d}{dt}\left(f(x) + t\frac{f''(x)}{2}\right)=\frac{f''(x)}{2}=Au(t,x).$$ Based on this relation, can you guess what $A$ is?