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A Banach space is a complete normed vector space: A vector space equipped with a norm such that every Cauchy sequence converges.
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Quantifications of unconditional bases
Let $(z_{n})_{n=1}^{\infty}$ be a sequence in a Banach space $X$. We set $$ \textrm{ca}((z_{n})_{n=1}^{\infty})=\inf_{n}\sup_{k,l\geq n}\|z_{k}-z_{l}\|.$$ Clearly, $(z_{n})_{n=1}^{\infty}$ is norm-bou …
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vote
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Bases and reflexivity in Banach spaces
R. C. James's classical paper in the early 1950s introduced and investigated two special classes of bases, shrinking bases and boundedly complete bases. These two special classes of bases are used to …
3
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answers
60
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A measure of non-reflexivity of Banach spaces
Let $X$ be a Banach space and define
$\gamma(X)=\sup\{|\lim\limits_{n}\lim\limits_{m}\langle x^{*}_{m},x_{n}\rangle-\lim\limits_{m}\lim\limits_{n}\langle x^{*}_{m},x_{n}\rangle|:(x_{n})_{n}$ is a sequ …
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1
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322
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Weak compactness of the closed unit ball of $L_{\infty}(\mu,X)$ in $L_{1}(\mu,X)$
It is known that the closed unit ball of $L_{\infty}(\mu)$ is weakly compact in $L_{1}(\mu)$. A natural question arises in the case of spaces of Bochner integral functions:
Question. Let $X$ be a Bana …
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3
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160
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Absolutely summing operators from $l_{p}$ to $l_{q}$
Recall that an operator $T:X\rightarrow Y$ is called absolutely summing if there exists a constant $C>0$ such that
$$\sum_{i=1}^{n}\|Tx_{i}\|\leq C \sup_{x^{*}\in B_{X^{*}}}\sum_{i=1}^{n}|\langle x^{* …
2
votes
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answers
86
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The solid hull and weak*-closure in Banach lattices
Let $E$ be a Banach lattice. A subset $A$ of $E$ is called solid if $|x|\leq |y|$ for some $y\in A$ implies that $x\in A$. For a subset $A$ of $E$, the solid hull of $A$ is the smallest solid set incl …
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3
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140
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Weak*-null sequences in dual spaces
Let $(x_{n})_{n}$ be a sequence in a Banach space $X$. Assume that the set $\{x_{n}:n=1,2,\cdots\}$ is finite. Let $(f_{m})_{m}$ be a weak*-null sequence in $X^{*}$ satisfying the following conditions …
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1
answer
345
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Weakly null sequences in Banach spaces
Every weakly null sequence in a Banach space, as a subset, is clearly relatively weakly compact. To quantify the elementary fact, we need the following quantities:
$$\delta_{0}((x_{n})_{n}):=\sup_{x^{ …
4
votes
1
answer
237
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Boundedly complete bases
Let us recall that a basis $(x_{n})_{n}$ for a Banach space $X$ is boundedly complete if for every scalar sequence $(a_{n})_{n}$ with $\sup\limits_{n}\|\sum\limits_{i=1}^{n}a_{i}x_{i}\|<\infty$, the s …
4
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$c_{0}$ has no boundedly complete basis
Recall that a basis $(x_{n})_{n}$ for a Banach space $X$ is called boundedly complete if for every scalar sequence $(a_{n})_{n}$ with $\sup_{n}\|\sum_{i=1}^{n}a_{i}x_{i}\|<\infty$, the series $\sum_{n …
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1
answer
399
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Bases in $c_{0}$
$c_{0}$, the space of the scalar sequence that converges to $0$ endowed with the sup norm, has two well-known bases: the unit vector basis $(e_{n})_{n}$, where $e_{n}(k)=1$ if $k=n$ and $0$ otherwise, …
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102
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A question on the Haar basis for $L_{1}[0,1]$
Let $(x_{n})_{n=1}^\infty$ be a basis for a Banach space $X$. It is important to know the exact expression of the norm of $\|\sum_{i=1}^{n}a_{i}x_{i}\|$ for all $n$ and all scalars $a_{1},a_{2},\ldots …
1
vote
1
answer
126
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Quantifications of boundedly complete bases
Let $(x_{n})_{n=1}^\infty$ be a bounded sequence in a Banach space $X$. We set
$$\textrm{ca}((x_{n})_{n=1}^\infty)=\inf_{n}\sup_{k,l\geq n}\|x_{k}-x_{l}\|.$$
Then $(x_{n})_{n=1}^\infty$ is norm-Cauchy …
3
votes
1
answer
140
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Quantifying shrinking bases
Let $X$ be a Banach space and let $(x_{n})_{n=1}^\infty$ be a (Schauder) basis for $X$. Let $(x^{*}_{n})_{n=1}^{\infty}$ be the biorthogonal functionals associated to the basis $(x_{n})_{n=1}^\infty$. …
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answers
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(1+)-complemented, (1+)-isomorphic copies of $L_{1}$ in dual Banach spaces
I am thinking about the following question:
Question. Suppose that $X$ is a Banach space such that for every $\epsilon>0$, the dual $X^{*}$ contains a subspace $(1+\epsilon)$-isomorphic to $L_{1}[0,1] …