my question may be no well posed. Consider a probability space $(\Omega,\mathcal{F},\mathbb{P})$, $\sigma$-subalgebra $\mathcal{G}$ of $\mathcal{F}$ and a function $\pi:L^{0}(\Omega,\mathcal{F},\mathbb{P})\to L^0(\Omega,\mathcal{G},\mathbb{P})$. Suppose that $\pi$ is an $L^0(\Omega,\mathcal{G},\mathbb{P})-sublinear\ function$ i.e.
- $\forall a\in L^0(\Omega,\mathcal{G},\mathbb{P})_+, \forall x\in L^0(\Omega,\mathcal{F},\mathbb{P}),\ \pi(ax)=a\pi(x)$;
- $\forall x,y\in L^0(\Omega,\mathcal{F},\mathbb{P}), \pi(x+y)\leq \pi(x)+\pi(y)$;
Then assume that $L^0(\Omega,\mathcal{F},\mathbb{P}),\ L^0(\Omega,\mathcal{G},\mathbb{P})$ are endowed with the metric of convergence in probability and that $\pi$ is continuous. Then is it true that the following set: $$\left\lbrace f:L^0(\Omega,\mathcal{F},\mathbb{P})\to L^0(\Omega,\mathcal{G},\mathbb{P}): f\ is\ L^0-linear\ and\ f\leq \pi\right\rbrace$$ is compact w.r.t the metric of convergence of probability? The definition of the $L^0$-linearity is the following: - $\forall a\in L^0(\Omega,\mathcal{G},\mathbb{P}), \forall x\in L^0(\Omega,\mathcal{F},\mathbb{P}),\ f(ax)=af(x)$; - $\forall x,y\in L^0(\Omega,\mathcal{F},\mathbb{P}),\ f(x+y)=f(x)+f(y)$;
Notation: $ L^0(\Omega,\mathcal{G},\mathbb{P})_+$ denotes the function which almostu surely non-negative, the orders $\leq$ holds almost surely and the order between functions is the usual pointwise ordering.