# Comparison principle of classical solution for a Dirichlet problem

The goal of this note is to show the uniqueness of ${C^{2}(O) \cap C(\bar O)}$ solution of $\displaystyle \mathcal L u (x) - q u(x) + \ell (x) = 0, \ x\in O \ \ \ \ \ (1)$

with boundary value $\displaystyle u(x) = g(x), \ x\in O^{c}. \ \ \ \ \ (2)$

In this below, we assume that

1. ${O}$ is bounded open set in ${\mathbb R^{d}}$;
2. For ${\phi\in C^{2}}$ and ${x\in \mathbb R^{d}}$, ${\mathcal L \phi(x) = b(x) \cdot D \phi(x) + \frac 1 2 tr (A(x) D^{2} \phi(x) ) + \mathcal I \phi(x)}$;
3. For ${\phi\in C^{2}}$ and ${x\in \mathbb R^{d}}$, ${\mathcal I \phi(x) = \int_{\mathbb R} [ \phi(x+y) - \phi(x) - \phi'(x) y I_{B_{1}} (y)] \nu (dy)}$ for a Levy measure ${\nu(\cdot)}$.

Recall that a Levy measure ${\nu}$ on ${\mathbb R^{d}}$ is a non-negative measure satisfying $\displaystyle \nu(\{0\}) = 0; \ \int_{\mathbb R^{d}} (1 \wedge |x|^{2}) \nu(dx) <\infty.$

Now we observe that $\displaystyle \mathcal I \phi(x) = \mathcal I_{1,1} \phi(x) + \mathcal I_{2,1} \phi(x) = \mathcal I_{1,r} \phi(x) + \mathcal I_{2,r} \phi(x) - D\phi(x) \cdot \int_{B_{1}\setminus B_{r}} y \nu(dy) ,$

where $\displaystyle \mathcal I_{1,r} \phi(x) = \int_{B_{r}} (\phi(x+y) - \phi(x) - D \phi(x) \cdot y) \nu(dy)$

and $\displaystyle \mathcal I_{2,r} \phi(x) = \int_{\mathbb R \setminus B_{r}} (\phi(x+y) - \phi(x)) \nu(dy).$

Proposition 1 For any ${r\in (0,1]}$, ${C^{2}_{b} \subset \mathcal D(I_{1,r})}$ and ${C_{b} \subset \mathcal D(I_{2,r})}$.

Proof: It’s the consequence of Taylor expansion. $\Box$

Proposition 2 ${\int_{B_{\epsilon}} |y|^{2} \nu(dy) \rightarrow 0}$ as ${\epsilon \rightarrow 0}$.

Proof: Let ${f(r) = \int_{B_{r}} |y|^{2} \nu(dy)}$. If ${f(1) = 0}$, then it’s done. If ${f(1) >0}$, then the function ${g(r) = f(r)/f(1)}$ is CDF of the probability distribution on the sample space ${[0,1]}$ given by $\displaystyle \forall 0\le a \le b \le 1, \ \mathbb P\{[a, b]\} = g(b) - g(a).$

Therefore, ${g}$ is RCLL, so is ${f}$. Since ${f(0) = 0}$, we conclude ${\lim_{\epsilon \rightarrow 0^{+}} f(\epsilon) = f(0) = 0.}$ $\Box$

Proposition 3 Suppose ${\phi \in C^{2}_{b}}$ and ${\hat x = \arg\max_{\mathbb R^{d}} \phi}$ is the global maximum of ${\phi}$, then $\displaystyle \lim\sup_{r \rightarrow 0^{+}} (\mathcal I_{1,r} + \mathcal I_{2,r}) \phi(\hat x) \le 0.$

Proof: Since ${\phi(\hat x + y) - \phi(\hat x) \le 0}$, we have ${\mathcal I_{2,r} \phi(\hat x) \le 0}$ for all ${r\in (0,1]}$. By mean value theorem, there exists ${\xi_{r}(y) \in B_{r}}$ such that $\displaystyle \mathcal I_{1,r} \phi (\hat x) = \int_{B_{r}} \frac 1 2 y^{t} D^{2} \phi (\xi_{r}(y)) y \ \nu (dy) \le \frac 1 2 \|D^{2} \phi\|_{\infty, B_{r}(\hat x)} \int_{B_{r}} |y|^{2} \nu (dy).$

Finally, we can take ${\lim\sup_{r\rightarrow 0}}$ and apply Proposition 2 to the last inequality to obtain the proof. $\Box$

Next, if we set $\displaystyle b_{r}(x) = b(x) - \int_{B_{1}\setminus B_{r}} y \nu(dy), \ \forall r\in (0,1],$

then ${b= b_{1}}$ and $\displaystyle \mathcal L \phi = b_{r} \cdot D\phi + \frac 1 2 tr(A \ D^{2} \phi) + \mathcal I_{1,r} \phi + \mathcal I_{2,r} \phi.$

If we define $\displaystyle F_{r}(x, p, X) = - b_{r} (x) \cdot p - \frac 1 2 tr( A(x) X) - \ell(x),$

then (1) can be rewritten by $\displaystyle F[u](x) := F_{r}(x, Du(x), D^{2} u(x)) - \mathcal I_{1,r} u(x) - \mathcal I_{2,r} u(x) + qu(x) = 0$

for any ${r\in (0,1]}$.

We say, ${u\in C^{2}(O) \cap C(\bar O)}$ is a subsolution if ${F[u] \le 0}$ on ${O}$, and is a supersolution if ${F[u] \ge 0}$. Next, we will show the comparison principle. The proof given below may be not the simplest one, but may be easy to be adapted to the proof of comparison principle in the context of the viscosity solution.

Theorem 4 Let ${q>0}$, and also let ${u}$ and ${v}$ be sub and super solutions, respectively. If ${u\le v}$ on ${\partial O}$, then ${u\le v}$ on ${\bar O}$.

Proof: Because ${w: = u -v \in C(\bar O)}$, there exists ${\hat x = \arg\max_{\bar O} w \in \bar O}$. Assuming ${w(\hat x) = \delta >0}$, then ${\hat x}$ shall satisfy

1. ${\hat x \in O}$;
2. ${u(\hat x) = v(\hat x) + \delta}$;
3. ${Du(\hat x) = D v(\hat x)}$;
4. ${D^{2} u(\hat x) \le D^{2} v(\hat x)}$;

and we try to find a contradiction below. From the sub and super solution property of ${F[u](\hat x) \le 0 \le F[v](\hat x)}$, we can write $\displaystyle \begin{array} {ll} & F_{r}(\hat x, Du(\hat x), D^{2} u(\hat x)) - \mathcal I_{1,r} u(\hat x) - \mathcal I_{2,r} u(\hat x) + qu(\hat x) \le 0 \\ \le & F_{r}(\hat x, Dv(\hat x), D^{2} v(\hat x)) - \mathcal I_{1,r} v(\hat x) - \mathcal I_{2,r} v(\hat x) + qv(\hat x) \\ \le & F_{r}(\hat x, Du(\hat x), D^{2} u(\hat x)) - \mathcal I_{1,r} v(\hat x) - \mathcal I_{2,r} v(\hat x) + qu(\hat x) - q\delta \end{array}$

The last inequality is the consequence of monotonicity of ${F_{r}}$ on the third variable together with properties of ${\hat x}$. This leads to an inequality $\displaystyle 0 < q\delta \le (\mathcal I_{1,r} + \mathcal I_{2,r}) w(\hat x),$

which makes a contradiction to Proposition 3. $\Box$