Skip to main content (access key 's'), Skip to navigation (access key 'n'), Accessibility information (access key '0')

Back to Simple Search

Advanced Search

Match of the following rules

Add Another Rule

Contains the following math formula (red border means the formula is incomplete)

Formula preview

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

Currently displaying 1 – 3 of 3

Order by Relevance | Title | Year of publication

A continuum of totally incomparable hereditarily indecomposable Banach spaces

I. Gasparis — 2002

Studia Mathematica

A family is constructed of cardinality equal to the continuum, whose members are totally incomparable hereditarily indecomposable Banach spaces.

More lr saturated L∞ spaces

Gasparis, I.; Papadiamantis, M. K.; Zisimopoulou, D. Z. — 2010

Serdica Mathematical Journal

2000 Mathematics Subject Classification: 05D10, 46B03. Given r ∈ (1, ∞), we construct a new L∞ separable Banach space which is lr saturated.

On the complemented subspaces of the Schreier spaces

I. Gasparis; D. Leung — 2000

Studia Mathematica

It is shown that for every 1 ≤ ξ < ω, two subspaces of the Schreier space $X^{ξ}$ generated by subsequences $(e_{l_{n}}^{ξ})$ and $(e_{m_{n}}^{ξ})$ , respectively, of the natural Schauder basis $(e_{n}^{ξ})$ of $X^{ξ}$ are isomorphic if and only if $(e_{l_{n}}^{ξ})$ and $(e_{m_{n}}^{ξ})$ are equivalent. Further, $X^{ξ}$ admits a continuum of mutually incomparable complemented subspaces spanned by subsequences of $(e_{n}^{ξ})$ . It is also shown that there exists a complemented subspace spanned by a block basis of $(e_{n}^{ξ})$ , which is not isomorphic to a subspace generated by a subsequence of $(e_{n}^{ζ})$ , for every $0 \leq ζ \leq ξ$ ....

Page 1

Download Results (CSV)