Convex operators and convex integrands (Q750872)

Let \(\Omega\) be an arbitrary set, let \(\Sigma\) be a \(\sigma\)-field of subsets of \(\Omega\) (the measurable sets), and let \(\mu\) denote a nonnegative \(\sigma\)-finite measure on \(\Sigma\). Let S(\(\Omega\)) be the space of all finite valued measurable functions on \(\Omega\). We identify f and \(g\in S(\Omega)\) if they differ only on a set of \(\mu\)-measure zero. With the usual ordering S(\(\Omega\)) is an order complete vector lattice. A mapping \(F:\;R^ d\supset D(F)\to S(\Omega)\) is called a convex operator if D(F), the domain of F, is a convex set of the d- dimensional Euclidean space \(R^ d\) and \(F(\lambda x+(1-\lambda)y)\leq \lambda F(x)+(1-\lambda)F(y)\) holds for every x,y\(\in D(F)\) and \(\lambda\in [0,1]\). Many kinds of ordered vector spaces can be regarded as subspaces of S(\(\Omega\)), and hence this class of convex operators covers many cases. A function \(f:\;R^ d\times \Omega \to R\cup \{+\infty \}\) is said to be a convex integrand if f(\(\cdot,t)\) is a convex function for each \(t\in \Omega\). We say that a convex integrand f is a representation of a convex operator F if f(x,t) is measurable in \(t\in \Omega\) and \(f(x,\cdot)=F(x)\) holds for every \(x\in D(F)\). Our main result asserts that every convex operator \(F:\;R^ d\supset D(F)\to S(\Omega)\) has at least a representation of F. For the proof of this result in the one dimensional case see \textit{S. Koshi}, \textit{H. C. Lai} and the author [Hokkaido Math. J. 14, 75-84 (1985; Zbl 0569.49008)]. In the general case, the proof is more complicated. We consider the relations between convex operators and their representations. Further, we generalize the Fenchel-Moreau theorem by using representations, and give some conditions under which a convex operator can be represented by a normal convex integrand.