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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 …

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 …

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 …

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 …

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^{* …

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 …

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^{ …

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 …

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 …

Let $K$ be a subset of the positive integers $\mathbb{N}$. For each $n\in \mathbb{N}$, $K_{n}$ denotes the set $\{k\in K: k\leq n\}$ and $|K_{n}|$ denotes the number of the elements in $K_{n}$. The na …

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 …

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 …

$c_{0}$, 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, …

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 …

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 $(x_{n})_{n=1}^\infty$. …