# A characterization of the solution of 1-d obstacle equation

The value function of optimal stopping problem is related to Bellman equation of the type $\displaystyle F[u](x) = \min\{(\alpha - \mathcal{A})u(x), u(x) - g(x)\} = 0. \ \ \ \ \ (1)$

where the operator ${\mathcal{A}}$ is defined by $\displaystyle \mathcal{A} \varphi = b \varphi' + \frac 1 2 \sigma^2 \varphi'', \forall \varphi \in C^2(\mathbb{R}).$

Suppose comparison principle [CP] holds for the viscosity solution, then the unique solution can be characterized by the pointwise infimum of the piecewise smooth continuous viscosity supersolutions.

To proceed, we first define the space of supersolution, for any open set ${\mathcal{O}}$ $\displaystyle \mathcal{U}^+(\mathcal{O}) = \{u\in C^2(\mathcal{\bar O}): F[u](x) \ge 0, \forall x\in \mathcal{\bar O}. \}.$

We extend a function of ${u:\mathcal{O} \rightarrow \mathbb{R}}$ into ${\bar u: \mathbb{R}\mapsto \mathbb{R}}$ by ${\bar u(x) = u(x) I_{\mathcal{\bar O}}+ \infty I_{\mathbb{R} \setminus \mathcal{O}}}$. If there is no confuse, we shall use ${u}$ in replace of ${\bar u}$. Also define the space of continuous piecsewise smooth supersolution by $\displaystyle \mathcal{U}^+ = \{ u\in C(\mathbb{R}): u = \min_{i=1}^n u_i \hbox{ for some } u_i \in \mathcal{U}^+(\mathcal{O}_i) , \ \cup_{i=1}^n \mathcal{O}_i = \mathbb{R}, \ n<\infty. \}$

Consider $\displaystyle u = \inf\{w : w\in \mathcal{U}^+\} \ \ \ \ \ (2)$

Proposition 1 Suppose [CP] holds for (1) and ${u}$ is continuous, then ${u}$ is the unique viscosity solution of (1).

Proof: First, one can show an arbitrary ${w \in \mathcal{U}^+}$ is a supersoluiton by the similar to the proof of Proposition 2.8 of [CC95]. Since ${u}$ is continuous, Lemma 4.5 of [CIL92] implies ${u}$ is again a supersolution.

Suppose ${u}$ is not the subsolution. Then, there exists ${x_0\in \mathbb{R}}$ and a parabola $\displaystyle P(x) = \frac{a_2}{2} (x-x_0)^2 + a_1 (x-x_0) + u(x_0)$

that touches ${u}$ at ${x_0}$ from the above, satisfying $\displaystyle F[P](x_0) = \min\{(\alpha - \mathcal{A})P(x_0), P(x_0) - g(x_0)\} \ge \varepsilon > 0.$

By continuity of ${x \mapsto F[P](x)}$, there exists ${\delta>0}$ such that $\displaystyle \min\{(\alpha - \mathcal{A})P(x), P(x) - g(x)\} \ge \frac{3\varepsilon}{4}>0, \ \forall |x-x_0| <\delta.$

Consider $\displaystyle \bar P(x) = P(x) - \delta^2 \varepsilon_1 + 2 \varepsilon_1 (x-x_0)^2$

where $\displaystyle \varepsilon_1 = \frac{\varepsilon}{4 \delta^2 (\alpha+1) + 4 |\sigma|_{\infty, B_\delta(x_0)}^2 + 8 \delta |b|_{\infty, B_\delta(x_0)}}.$

Then, we conclude

1. ${\bar P(x)> P(x) \ge u(x)}$ on ${B_\delta(x_0) \setminus \overline{B_{\delta/\sqrt 2}(x_0)}}$, since ${\bar P = P}$ when ${|x-x_0| = \delta/\sqrt 2}$. In particular, we have $\displaystyle \bar P(x) > u(x) \hbox{ if } x = x_0 \pm \delta. \ \ \ \ \ (3)$

2. ${\bar P \in \mathcal{U}^+(B_{\delta}(x_0)),}$ since $\displaystyle \bar P(x) - g(x) \ge \varepsilon/2 >0, \ \hbox{ and } (\alpha - \mathcal{A}) \bar P(x) \ge 0, \ \forall x\in B_{\delta}(x_0). \ \ \ \ \ (4)$

By strict inequality (3) and the definition (2) of ${u}$, there exist two functions ${w_+, w_- \in \mathcal{U}^+}$ such that $\displaystyle u(x_0 \pm \delta) \le w_{\pm} (x_0 \pm \delta) < \bar P(x_0 \pm \delta).$

Note that, $\displaystyle \bar w := \min\{ w_+, w_-, \bar P\} \in \mathcal{U}^+,$

satisfies $\displaystyle \bar w (x_0) = \bar P(x_0) = u(x_0) - \delta^2 \varepsilon_1 < u(x_0),$

which leads to a contradiction to the definition (2) of ${u}$. $\Box$

1. Chao Zhu says:|
So if the CP holds, the value of the dual is the unique viscosity solution to the equation $F[u](x)=0$?
• kenneth says:|