In the following formula
$$Af(x) := \lim_{t \downarrow 0} \frac{\mathbb{E}^x(f(B_t))-f(x)}{t} $$
- If $B_t$ is the Brownian motion, what are $f(B_t)$ and $f(x)$? here some explanation was given. Can anybody please provide with more intuitive explanation?
- In the way to prove that $A=\frac{1}{2}\Delta$ here, and after $P_t f(x) = \mathbb{E}^x[f(X_t)]$, he defines $u(t,x) := P_t f(x)$. Which interpretations/properties of $P_t$ led to the latter definition? In other words, why the expected value of f(x) is a deterministic function depending on $x$ and $t$?
(1) If $f$ is a function then $f(B_t)$ and $f(x)$ are simply the evaluation of that function at the points $B_t$ and $x$ respectively. Here, since $x$ is simply the starting point of your Brownian motion $f(x)$ needs no further interpretation. For fixed $t$, $f(B_t)$ is itself a random variable since for each $\omega \in \Omega$ (the underlying probability space), you should set $(f(B_t))(\omega) = f(B_t(\omega))$ where the right hand side is just evaluating the function $f$ at the number $B_t(\omega)$.
(2) Note that the claim is not that $\mathbb{E}^x[f(x)]$ is a function that depends on $x$ and $t$ but rather that $$u(t,x) := \mathbb{E}^x[f(X_t)]$$ is a function depending on $x$ and $t$. If you are specifically interested in Brownian motion, you should replace the general Markov process $X_t$ with $B_t$. This is a function of $t$ since the expectation depends on the time at which we look at the process (e.g. our Brownian motion). It is also a function of a space point $x$ because we have to decide which point our Brownian motion should start from; in the definition of $u$ this is denoted by writing $\mathbb{E}^x$ which is the expectation with respect to the law of the Brownian motion started from the point $x$. Finally, this is a deterministic function since we took an expectation so there is no $\omega$ dependence (randomness) left on the right hand side of the definition.