# Showing an "obviously-optimal" control is optimal (without smoothness assumptions)

Let $$\mathcal{A}\subseteq\mathbb R$$ be a compact interval, $$T\in\mathbb R_+$$ be a finite horizon, and $$g:\mathbb R\to\mathbb R_+$$ be a continuous function with $$g\leq 1+|\cdot|$$. Consider an optimal control problem where, observing a standard Brownian motion $$\{B_t\}_t$$, I choose an $$\mathcal A$$-valued control $$\{a_t\}_t$$ to control the drift. I want to maximize $$\mathbb E g(X_T)$$, where $$X_t:=B_t + \int_0^t a_s \text{ d}s$$.

I have proven the following:

1) The optimal value function $$v:[0,T]\times\mathbb R\to\mathbb R_+$$ (taking a time $$t$$ and current level of $$X_t$$ to an optimal continuation value) is finite-valued.

2) There is a continuous cutoff function $$\kappa:[0,T]\to\mathbb R$$ such that, for every $$t\in[0,T)$$, the function $$v(t,\cdot)$$ is strictly increasing on $$[\kappa(t),\infty)$$ and strictly decreasing on $$(-\infty,\kappa(t)]$$.

I want to show that there is an optimal control where I choose $$\max \mathcal A$$ whenever $$X_t > \kappa(t)$$ and $$\min \mathcal A$$ whenever $$X_t < \kappa(t)$$. Ideally, I would show that any other optimal control almost surely agrees with this one at almost every time.

This seems, intuitively, like it must be true, but I don't see an easy proof (or know a reference), especially given that I've made no differentiability assumptions.

• "The optimal value function $v$ ... is finite-valued" you certainly need some assumptions about the growth of $g$. Apr 15, 2020 at 13:52
• Yes, sorry for being unclear. I meant that I had an instance of this problem in which I had already proven it was finite-valued. In my instance, $g\leq 1+|\cdot|$, so that finite-valuedness is easy. I'll edit to add this feature to the problem. Apr 15, 2020 at 15:05