Let us consider the Poisson problem
\begin{cases} \frac{1}{2}u''=-f\qquad\text{in}\,\,(a,b)\\u(a)=u(b)=0 \end{cases}
where $f:(a,b)\to\mathbb{R}$ is continuous and bounded. We have obtained two solutions, one analytical, the other probabilistic:
- $u(x)=\int^{a}_bg(x,y)f(y)dy$, where $g$ is the Green function
- $u(x) =\mathbb{E}^x[\int^\tau_0 f(B_t)]$ where $\tau=\tau_{(a,b)}$ is the first exit time from $(a,b)$ and $(B_t)_t$ is a standard Brownian motion. Making equal this two solutions, we obtain, after some passages: $$g(x,y)=\int^{\infty}_0\mathbb{P}^x(B_t\in dy,t<\tau)dt$$
Our professor give us the following interpretation.
"Does there exist $f:\mathbb{R}^+\to \mathbb{R}$ continuous such that for each level $a\in\mathbb{R}$ we have that $$\int_0^{\infty}\mathbb{1}_{(f(x)=a)}(x)dx>0?$$
Yes, Brownian Motion has this property, infact $\forall a\in\mathbb{R}$ $$\mathbb{E^x}[\int^{\tau}_0\mathbb{1}_{(f(B_t)=a)dt}]>0$$."
Can someone explain this interpretation, please?
There are so many misprints in the question that it is difficult to know what is really asked. In any case, the identity $$ g(x,y)=\int^{\infty}_0\mathbb{P}^x(B_t\in dy,t<\tau)\,\mathrm dt $$ should read $$ g(x,y)dy=\int^{\infty}_0\mathbb{P}^x(B_t\in dy,t<\tau)\,\mathrm dt. $$ Later on, the assertion that for all $a$, $$\mathbb{E^x}[\int^{\tau}_0\mathbb{1}_{(f(B_t)=a)dt}]>0$$ is just absurd and should probably read $$\mathbb{E^x}\left[\int^{\tau}_0\mathbb{1}_{B_t=a}\,\mathrm dt\right]>0. $$ Unfortunately this is wrong as well, since for every $a$, $$ \int^{\infty}_0\mathbb{1}_{B_t=a}\,\mathrm dt=0, $$ with full probability. The result your professor is alluding to here might be the existence of a local time, that is, some family $(L_t^a)$ of random variables such that, for every measurable function $f$, the so-called occupation formula holds, that is, $$ \int^{\tau}_0f(B_t)\,\mathrm dt=\int_\mathbb R f(a)L_\tau^a\,\mathrm da. $$