Philip Brooker

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Name Philip Brooker
Member for 3 years
Seen 8 hours ago
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Location Canberra, Australia
Age 31
My research interests are primarily in Banach space theory and related areas.
May
2
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Mar
24
awarded  Citizen Patrol
Mar
13
comment B(H) as a direct sum of a closed two sided ideal and a subalgebra
$ ^\ast$ - I'll show myself out...
Mar
13
comment B(H) as a direct sum of a closed two sided ideal and a subalgebra
For $K_\lambda(H)$, you can of course just take the uniform closure in the case I mention above to get the desired closed ideal; I guess this special case is why Luft (and, later, Daws) used a generalised notion of total boundedness rather than density character to define the operator ideals $K_\lambda$.
Mar
13
comment B(H) as a direct sum of a closed two sided ideal and a subalgebra
Nice answer, Tomek. For the sake of completeness$^\ast$, I mention that, with the definitions given in your answer, $K_\lambda(H)$ is not closed in $B(H)$ when $\lambda< dens(H)$ is of cofinality $\omega$; similarly, $\ell_\infty^\lambda (dens(H))$ is not closed in $\ell_\infty(dens(H))$ when $\lambda< dens(H)$ is of cofinality $\omega$ (these spaces are otherwise closed in their respective overspaces). The definition of $\ell_\infty^\lambda (dens(H))$ should presumably be those $f\in \ell_\infty(dens(H))$ with $\vert\{ \alpha\in dens(H)\mid\vert f(\alpha)\vert>\epsilon\}\vert <\lambda$.
Feb
21
comment Strictly singular operators and their adjoints
Shortly after the discussion above it occurred to me that one of the main features of Pelczynski's example is that it is the embedding of a Banach space into its second dual, and such embeddings never have strictly singular adjoint because the adjoint is the canonical projection of the third dual onto the first dual (in hindsight, all very obvious!). So to find a counterexample one could try to find $X$ with $X^{\ast\ast}$ separable and the canonical embedding of $X$ into $X^{\ast\ast}$ strictly cosingular; I (naively) wonder if a James-Lindenstrauss type of construction is worth considering?
Feb
21
comment Weak*-closed and complemented subspaces of dual Banach spaces
I removed the incorrect James tree space example from my answer, so hopefully my answer is now correct.
Feb
21
revised Weak*-closed and complemented subspaces of dual Banach spaces
Removed incorrect example from my answer.
Jan
5
comment Non-super reflexive space
@Bojan Kwitek: examples of the kind of space you asked about in your second question do exist; the Pisier-Xu paper is Random series in the real interpolation spaces between the spaces $v_p$, see link.springer.com/chapter/10.1007%2FBFb0078146 .
Nov
30
comment A basis of the space of continuous function of countable ordinals $C(\alpha) = C [0, \alpha]$
The best way to understand this consists in studying Bourgain's representation of the spaces $C(\alpha)$ as tree spaces; this is from Section 2 of Bourgain's paper The Szlenk index and operators on $C(K)$ spaces. Bourgain doesn't mention bases there, however an exposition of Bourgain's ideas, along with an observation of Odell about bases of separable $C(\alpha)$ spaces as per your question, can be found in Section 4.B of Rosenthal's survey article on $C(K)$ spaces in Volume 2 of the Handbook of the Geometry of Banach Spaces (in particular, see page 1583 onwards).