 Solution

Relative to a filtration $({\mathcal{F}}_n)$, a martingale is an adapted integrable process with $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$; a supermartingale has the corresponding inequality $\le$. A stopping time $T$ has $\{T\le n\}\in {\mathcal{F}}_n$ for every $n$.

For $T\le N$, the pathwise identityhas ${\mathcal{F}}_{k-1}$-measurable indicators. Conditioning each summand gives zero expected increment for a martingale, and a nonpositive one for a supermartingale. All terms are integrable because the sum is finite. Thus $\mathbb{E}{M}_T=\mathbb{E}{M}_0$ and $\mathbb{E}{\hat{M}}_T\le \mathbb{E}{\hat{M}}_0$, proving the requested inequality when their initial values agree.

A self-financing strategy holds ${\pi }_n$ shares during period $n$, chosen using information at time $n-1$. Its cash investment is $X_{n-1}-{\pi }_n{S}_{n-1}$, which earns the risk-free return. HenceUse the terminal condition $V(N,x,s)=U(x)$ from the original PDF. Conditional on the current state and past, the next return has the law of ${\xi }_1$ and is independent of the past. Therefore the conditional expectation of $V(n+1,X_{n+1},S_{n+1})$ for the chosen holding ${\pi }_{n+1}$ is at most its supremum over all holdings, namely $V(n,X_n,S_n)$. Assuming the finite values and integrability implicit in this stochastic-control formulation, the value process is a supermartingale.

It follows that $\mathbb{E}U(X_N)\le V(0,X_0,S_0)$ for every admissible strategy. If the strategy ${\pi }^{\ast }$ makes this process a martingale, equality holds for it. ${\pi }^{\ast }$ attains the upper bound and is optimal. No existence of an optimizer or finiteness of an unconstrained supremum is being assumed without the stated admissibility conditions.