Consider a stochastic control problem, $$v^C(0,x) = \mathbb{E} \Big[\int_0^\tau f(X_t,C_t) d t + (T-\tau)|X_\tau|\Big] $$ where $X_t$ is a weak solution to the SDE $$dX_t = C_t dB_t, \quad X_0 = x \in (-1,1).$$ $\tau:= \inf\{t: X_t \notin (-1,1)\} \wedge T$.
$C$ is a $\mathbb{R}_+$ valued progressively measurable stochastic process for some space $(\Omega, \mathbb{F}=(\mathcal{F}_t)_{t\ge 0}, \mathbb{P})$ for which the SDE above has a solution for some standard brownian motion $B_t$.
$f(\cdot,\cdot)$ is a measurable function. Also, $f(\cdot, c)$ is continuous for any $c$. I would like to claim the following:
Claim: Given any admissible control $C$, $\exists$ a Markovian control $C'$ (i.e. $C'_t = g(t,X_t)$ )(possibly on a different probability space) such that $v^{C'}(0,x) \ge v^C(0,x)$.
Does this sound true? There are various results in the vicinity of this that I have found in a survey by Borkar but, not being a practitioner myself, they are a bit hard to follow. Any help would be appreciated. Thanks.
