Strict and uniform convexity of the measurable section spaces (Q2910526)

Let \((\Omega,\Sigma,\mu)\), \(E\) and \(\mathcal X\) be a complete \(\sigma\)-finite measure space, a Köthe space over \((\Omega,\Sigma,\mu)\) and a measurable bundle of Banach spaces over \((\Omega,\Sigma,\mu)\), respectively. Then NEWLINE\[NEWLINE E(\mathcal X):=\{f\in L^{0}(\mu,\mathcal{X}):\|f(\cdot)\|_{\mathcal X(\cdot)}\in E\},NEWLINE\]NEWLINE equipped with the norm \(|\|f\||:=\|\|f(\cdot)\|_{\mathcal X(\cdot)}\|_{E}\), is a Banach space.NEWLINENEWLINERecall that a Banach space \(X\) is said to be uniformly convex (uniformly rotund) if, for any \(\varepsilon\in(0,2]\), there exists \(\delta\in(0,1)\) such that, for any \(x,y\in B(X)\) (the unit ball of \(X\)), we have \(\|\frac{x+y}{2}\|_{X}\leq1-\delta\) whenever \(\|x-y\|_{X}\geq\varepsilon\). Analogously, a Banach space \(X\) is said to be strictly convex (rotund) if, for any \(x,y\in S(X)\) (the unit sphere of \(X\)), we have \(\|\frac{x+y}{2}\|_{X}<1\).NEWLINENEWLINEThe main result of the paper is the following theorem: If \(E\) is uniformly (strictly) convex and almost all fibers \({\mathcal X}_{t}\) are uniformly (resp. strictly) convex, then \(E({\mathcal X})\) is uniformly (resp. strictly) convex. Moreover, if \(E({\mathcal X})\) is uniformly (resp. strictly) convex, then \(E\) is uniformly (resp. strictly) convex and every fiber \({\mathcal X}_{t}\) such that \(\rho({\mathcal X}_{t}):=\mu(\{s\in\Omega\colon{\mathcal X}_{s}\text{ is isomorphic to }{\mathcal X}_{t}\})>0\) is uniformly (resp. strictly) convex.