Brownian motion and Laplacian of a function

Question

Let $f:\mathbb{R}^d \to \mathbb{R}$ be continuous and $T:=\inf\{r>0,|B_r|=2\},$ where $|.|$ is the Euclidean norm.

Let $h(x)= E_x[\int_0^T f(B_r)dr], x\in B(0,1).$

Can we claim that for $x \in B(0,1),\Delta h(x)=f(x)?$ Why?

I tried the way used in Dirichlet's problem to do the proof with no sucess.

The question is related to: Solution for the Poisson problem

Answer 1

According to Problem 4.2.25 in [1] (thanks Kazuki OKAMURA for pointing that out):

Consider an open bounded subset $D$ of $\mathbb{R}^d$ and the bounded continuous functions $g:D\to\mathbb{R}$ and $f:\partial D\to\mathbb R$. Assume that $u:\overline{D}\to\mathbb R$ is continuous of class $C^2(D)\,,$ and solves the Poisson equation $$\tag{1} \textstyle\frac{1}{2}\Delta u=-g\,;\text{ in }D $$ subject to the boundary condition $u=f$ on $\partial D\,.$

Then the representation $$\tag{2} u(x)=\mathbb E^x\Big[f(W_{\tau_d})+\int_0^{\tau_D}g(W_t)\,dt\Big]\,;\quad x\in D $$ holds. (Here $W$ is a Brownian motion starting in $x$ and $\tau_D=\inf\{t\ge 0:W_t\in D^c\}$ is its exit time from $D\,.$)

• This is a modern formulation of a famous Theorem first proved by S. Kakutani.

• You want to prove the converse. That is: Does a function $h(x)$ defined by the RHS of (2) satisfy the Poisson equation $\frac{1}{2}\Delta h=-g$ and the boundary condition $h=f$ on $\partial D\,?$

• The answer is yes provided that it is properly formulated.

• First, it follows from the maximum principle that the solution $u$ in K&S' Problem 4.2.25 in [1] is unique. This implies

Corollary to Problem 4.2.25 in [1]: Let $$\tag{3} h(x):=\mathbb E^x\Big[f(W_{\tau_d})+\int_0^{\tau_D}g(W_t)\,dt\Big] $$ and assume that the BVP for $u$ above has a solution. Then $h(x)$ is that solution.

Remarks

• We can always define $h(x)$ by (3) but without making further assumptions on the regularity of $D$ and on the functions $f$ and $g$ we cannot guarantee that the BVP has a solution. For instance when $\partial D$ has the famous Lebesgue 'cusp' that is also discussed in [1] in their proof of Theorem 4.2.12 then we can modify $f$ at that cusp $x$ without changing $h$ because the BM never reaches that boundary point.

• In short: the formulation of Problem 4.2.25 is the most economic one to guarantee the converse in the sense of the above simple Corollary.

[1] I. Karatzas, S. Shreve, Brownian Motion and Stochastic Calculus.