DOI: 10.11606/t.45.2019.tde-26072019-151133
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Light groups of isometries and polyhedrality of Banach spaces

Abstract: Megrelishvili defines light groups of isomorphisms of a Banach space as the groups on which the weak and strong operator topologies coincide and proves that every bounded group of isomorphisms of Banach spaces with the point of continuity property (PCP) is light. We investigate this concept for isomorphism groups G of classical Banach spaces X without the PCP, especially isometry groups, and relate it to the existence of G-invariant LUR or strictly convex renormings of X. We give an example of a Banach space X… Show more

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Cited by 1 publication
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“…where the three vectors are disjointly supported (possibly y * i or z * i is 0), and y * i and z * i only have coordinates ±1 on their support. From the fact that for i = j, T (e * p i ± e * p j ) must be an extreme point and therefore does not have ± 1 2 coordinates, we deduce that the support of y * i is some finite set C independent of i and that z * i is disjointly supported from k, from y * i and from all other z * j . Since C is finite we find i = j such that y * i = y * j , and we compute The following example shows that a bounded operator that sends extreme points to extreme points and vectors of disjoint support to vectors of disjoint support is not necessarily given by a permutation of the basis.…”
Section: And For Anymentioning
confidence: 94%
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“…where the three vectors are disjointly supported (possibly y * i or z * i is 0), and y * i and z * i only have coordinates ±1 on their support. From the fact that for i = j, T (e * p i ± e * p j ) must be an extreme point and therefore does not have ± 1 2 coordinates, we deduce that the support of y * i is some finite set C independent of i and that z * i is disjointly supported from k, from y * i and from all other z * j . Since C is finite we find i = j such that y * i = y * j , and we compute The following example shows that a bounded operator that sends extreme points to extreme points and vectors of disjoint support to vectors of disjoint support is not necessarily given by a permutation of the basis.…”
Section: And For Anymentioning
confidence: 94%
“…In a similar manner, reversing the roles of n and m we obtain that α m k = ± 1 2 as well. Plugging these values in (1), and taking into account that all θ j α j k are non-negative, it follows that α j k = 0 for all j ∈ F \ {n, m}.…”
Section: And For Anymentioning
confidence: 99%
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