An analogue of Uhl's theorem on the convexity of the image of vector measures (Q2825583)

Let \(U\) be a closed absolutely convex absorbing subset of a Fréchet space \(E\). By \(p_U\) denote the Minkowski functional of \(U\). Assume that \(p_U\) is a norm on \(E\). Denote by \(E_U\) the completion of \((E, p_U)\) and denote by \(J_U\) the natural embedding of \(E\) into \(E_U\).NEWLINENEWLINEDue to the author's definition, such a subset \(U\) is said to be anti-compact if, for every bounded subset \(A \subset E\), its image \(J_U(A)\) is pre-compact in \(E_U\). The author demonstrates that every separable Fréchet space \(E\) contains an anti-compact subset. The main result of the paper says that, for every separable Fréchet space \(E\), every \(\sigma\)-algebra of subsets \(\Sigma \subset 2^\Omega\), and every non-atomic countably-additive measure \(\mu : \Sigma \to E\) of bounded variation, there is an anti-compact subset \(U \subset E\) such that the closure of \(J_U\mu(\Sigma)\) is convex and compact in \(E_U\).NEWLINENEWLINEReviewer's remark. The results are correct in the case of Banach spaces. In the case of general Fréchet spaces, the demonstrations are unclear for me, and the validity of the results appears questionable. For example, in the separable Fréchet space \(\mathbb R^{\mathbb N}\) of all sequences \(x = (x_1, x_2, \ldots)\) of reals, equipped with the standard product topology, there is no anti-compact subset. Indeed, denote by \(\{e_n \}_{n \in \mathbb N}\) the canonical basis of \(\mathbb R^{\mathbb N}\), i.e., \(e_1 = (1, 0, 0, \ldots)\), \(e_2 = (0, 1, 0, \ldots)\), etc., and let \(U \subset \mathbb R^{\mathbb N}\) be an absolutely convex absorbing subset. Assume that \(p_U\) is a norm on \(\mathbb R^{\mathbb N}\). Then the set \(A = \left\{\frac{ne_n}{p_U(e_n)}\right\}_{n \in \mathbb N}\) is bounded in the original topology of \(\mathbb R^{\mathbb N}\), but is unbounded in the norm \(p_U\). This means that \(U\) is not anti-compact. Taking into account that every Fréchet space \(E\) is barreled, every anti-compact subset \(U\) of \(E\) is a neighborhood of zero, and consequently the norm \(p_U\) is continuous. This means that the results of the paper can be correct only in those Fréchet spaces that possess continuous norms.