Timeline for Boundedness of nonlinear continuous functionals

10 events

when toggle format	what		by	license	comment
S Nov 18, 2014 at 11:37	history	suggested	user50278	CC BY-SA 3.0	$ missing everywhere
Nov 18, 2014 at 11:28	review	Suggested edits
S Nov 18, 2014 at 11:37
Feb 7, 2010 at 16:53	vote	accept	Ady
Jan 13, 2010 at 17:03	vote	accept	Ady
Jan 13, 2010 at 17:03
Dec 29, 2009 at 17:49	comment	added	Bill Johnson		Nice construction, Anon. A simpler answer to Ady's question about getting e in E s.t. ... is that the functions in C[0,1] that vanish at t_n for all n<i is a finite codimensional subspace of C[0,1].
Dec 29, 2009 at 16:32	comment	added	Anonymous		For every n < i fix some t_n in V_n. Since E is infinite-dimensional you can find a vector e in E such that e has norm one and satisfies e(t_n) < 1/5 for all n < i [indeed, you may select a basic sequence (e_k) in E with basis constant, say, 2 and such that lim_k e_k(t_n) exists for all n < i; so, if k is large enough you have that e_k(t_n)-e_{k+1}(t_n) is almost zero for all n < i; so, for a sufficiently large k, the vector e=(e_k-e_{k+1})/\\|e_k-e_{k+1}\\| is as desired]. By definition, the vector e is not in U_n for all n < i.
Dec 29, 2009 at 13:19	comment	added	Ady		Why "for every integer i and every infinte-dimensional subspace E of C[0,1] there is a norm-one vector e in E such e is NOT in U_n for every n < i" ?
Dec 28, 2009 at 16:53	history	edited	Anonymous	CC BY-SA 2.5	added 2 characters in body
Dec 28, 2009 at 16:32	history	edited	Anonymous	CC BY-SA 2.5	added 18 characters in body
Dec 28, 2009 at 16:06	history	answered	Anonymous	CC BY-SA 2.5