D. Freeman
Moving Parseval frames for vector bundles.
Thu, 03/08/2012 - 15:01 — SomNambulAuthors: D. Freeman, D. Poore, A. R. Wei, M. Wyse
Subjects: Geometric Topology
AbstractParseval frames can be thought of as redundant or linearly dependent
coordinate systems for Hilbert spaces, and have important applications in such
areas as signal processing, data compression, and sampling theory. We extend
the notion of a Parseval frame for a fixed Hilbert space to that of a moving
Parseval frame for a vector bundle over a manifold. Many vector bundles do not
have a moving basis, but in contrast to this every vector bundle over a
paracompact manifold has a moving Parseval frame.- Read more
- 5 comments
- login or register to post comments
Greedy bases for Besov spaces.
Wed, 10/21/2009 - 14:15 — SomNambulAuthors: S. J. Dilworth, D. Freeman, E. Odell, Th. Schlumprecht
Subjects: Functional Analysis
AbstractWe prove thatthe Banach space $(\oplus_{n=1}^\infty \ell_p^n)_{\ell_q}$,
which is isomorphic to certain Besov spaces, has a greedy basis whenever $1\leq
p \leq\infty$ and $1<q<\infty$. Furthermore, the Banach spaces
$(\oplus_{n=1}^\infty \ell_p^n)_{\ell_1}$, with $1<p\le \infty$, and
$(\oplus_{n=1}^\infty \ell_p^n)_{c_0}$, with $1\le p<\infty$ do not have a
greedy bases. We prove as well that the space $(\oplus_{n=1}^\infty
\ell_p^n)_{\ell_q}$ has a 1-greedy basis if and only if $1\leq p=q\le \infty$.Greedy bases for Besov spaces.
Wed, 10/21/2009 - 14:15 — SomNambulAuthors: S. J. Dilworth, D. Freeman, E. Odell, Th. Schlumprecht
Subjects: Functional Analysis
AbstractWe prove thatthe Banach space $(\oplus_{n=1}^\infty \ell_p^n)_{\ell_q}$,
which is isomorphic to certain Besov spaces, has a greedy basis whenever $1\leq
p \leq\infty$ and $1<q<\infty$. Furthermore, the Banach spaces
$(\oplus_{n=1}^\infty \ell_p^n)_{\ell_1}$, with $1<p\le \infty$, and
$(\oplus_{n=1}^\infty \ell_p^n)_{c_0}$, with $1\le p<\infty$ do not have a
greedy bases. We prove as well that the space $(\oplus_{n=1}^\infty
\ell_p^n)_{\ell_q}$ has a 1-greedy basis if and only if $1\leq p=q\le \infty$.
User login
