asplund space
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2021 ◽  
Vol 392 ◽  
pp. 108041 ◽  
Author(s):  
Petr Hájek ◽  
Tommaso Russo ◽  
Jacopo Somaglia ◽  
Stevo Todorčević


Mathematics ◽  
2020 ◽  
Vol 8 (12) ◽  
pp. 2243
Author(s):  
Yaroslav Bazaykin ◽  
Dušan Bednařík ◽  
Veronika Borůvková ◽  
Tomáš Zuščák

The aim of the paper is to generalize results by Sikorska on some functional equations for set-valued functions. In the paper, a tool is described for solving a generalized type of an integral-functional equation for a set-valued function F:X→cc(Y), where X is a real vector space and Y is a locally convex real linear metric space with an invariant metric. Most general results are described in the case of a compact topological group G equipped with the right-invariant Haar measure acting on X. Further results are found if the group G is finite or Y is Asplund space. The main results are applied to an example where X=R2 and Y=Rn, n∈N, and G is the unitary group U(1).



2019 ◽  
Vol 2019 ◽  
pp. 1-12
Author(s):  
Shaoqiang Shang ◽  
Yunan Cui

In this paper, we prove that if C⁎⁎ is a ε-separable bounded subset of X⁎⁎, then every convex function g≤σC is Ga^teaux differentiable at a dense Gδ subset G of X⁎ if and only if every subset of ∂σC(0)∩X is weakly dentable. Moreover, we also prove that if C is a closed convex set, then dσC(x⁎)=x if and only if x is a weakly exposed point of C exposed by x⁎. Finally, we prove that X is an Asplund space if and only if, for every bounded closed convex set C⁎ of X⁎, there exists a dense subset G of X⁎⁎ such that σC⁎ is Ga^teaux differentiable on G and dσC⁎(G)⊂C⁎. We also prove that X is an Asplund space if and only if, for every w⁎-lower semicontinuous convex function f, there exists a dense subset G of X⁎⁎ such that f is Ga^teaux differentiable on G and df(G)⊂X⁎.



2014 ◽  
Vol 2014 ◽  
pp. 1-7
Author(s):  
J. J. Wang ◽  
W. Song

We mainly present several equivalent characterizations of the strong metric subregularity of the Mordukhovich subdifferential for an extended-real-valued lower semicontinuous, prox-regular, and subdifferentially continuous function acting on an Asplund space.



Author(s):  
Joram Lindenstrauss ◽  
David Preiss ◽  
Tiˇser Jaroslav

This book deals with the existence of Fréchet derivatives of Lipschitz functions from X to Y, where X is an Asplund space and Y has the Radon-Nikodým property (RNP). It considers whether every countable collection of real-valued Lipschitz functions on an Asplund space has a common point of Fréchet differentiability. It also examines the conditions under which all Lipschitz mapping of X to finite dimensional spaces not only possess points of Fréchet differentiability, but possess so many of them that even the multidimensional mean value estimate holds. Other topics include the notion of the Radon-Nikodým property and main results on Gâteaux differentiability of Lipschitz functions and related notions of null sets; separable determination and variational principles; and differentiability of Lipschitz maps on Hilbert spaces.



2011 ◽  
Vol 83 (3) ◽  
pp. 450-455
Author(s):  
J. R. GILES

AbstractA Banach space is an Asplund space if every continuous gauge has a point where the subdifferential mapping is Hausdorff weak upper semi-continuous with weakly compact image. This contributes towards the solution of a problem posed by Godefroy, Montesinos and Zizler.



2009 ◽  
Vol 79 (2) ◽  
pp. 309-317 ◽  
Author(s):  
J. R. GILES

AbstractThe deep Preiss theorem states that a Lipschitz function on a nonempty open subset of an Asplund space is densely Fréchet differentiable. However, the simpler Fabian–Preiss lemma implies that it is Fréchet intermediately differentiable on a dense subset and that for a large class of Lipschitz functions this dense subset is residual. Results are presented for Asplund generated spaces.



2007 ◽  
Vol 82 (1-2) ◽  
pp. 104-109
Author(s):  
V. I. Rybakov
Keyword(s):  


2002 ◽  
Vol 130 (7) ◽  
pp. 2139-2143 ◽  
Author(s):  
Ondřej F. K. Kalenda


2001 ◽  
Vol 129 (12) ◽  
pp. 3741-3747 ◽  
Author(s):  
Petar S. Kenderov ◽  
Warren B. Moors ◽  
Scott Sciffer


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