= Solution
Use the <random walk> interpretation and <optimal stopping value function> recursion specified in part (b). If $h$ is <convex>, translating and integrating its <convexity> inequality gives, for $0<\theta<1$,
$$
\begin{aligned}
(Ph)(\theta x+(1-\theta)y)
&=\int h\big(\theta(x+z)+(1-\theta)(y+z)\big)\,\nu(dz)\\
&\leq\theta(Ph)(x)+(1-\theta)(Ph)(y).
\end{aligned}
$$
Thus the <transition operator> preserves <convex functions>. Also the <pointwise maximum of convex functions> is <convex>: each of two <convex functions> at an intermediate point is bounded by the same convex combination of their pointwise maximum at the endpoints. Starting with $V(T,\cdot)=f$, backward induction in
$$
V(t,\cdot)=\max\{f,PV(t+1,\cdot)\}
$$
therefore proves \b[the asserted convexity in the intended model]:
$$
\boxed{V(t,\cdot)\text{ is convex for every }0\leq t\leq T.}
$$
The statewise integrability convention in part (b) gives finite <convex functions> on all real states. The same inequality holds for the canonical extended value function wherever the expectations are well-defined. Integrability only along one started process need not make that canonical function finite at unused states: for $T=1$, $S_0=0$, $f(s)=e^{s^2}$ and increment density proportional to $e^{-y^2-|y|}$, $\mathbb E f(S_1)<\infty$, but $\mathbb E f(s+Y_1)=\infty$ for $s>1/2$. This concerns the canonical recursion away from visited states, rather than the almost-sure identity for $U$. Arbitrary off-state versions of $V$ need not be <convex>; the recursive version is the one meant here. No zero-mean assumption on the increments was used, and <convexity> alone does not make $f(S_t)$ a <submartingale> for an arbitrary drift.
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