Search Results

If $\|\cdot\|_1$ is a continuous strictly convex norm on $(X,\|\cdot\|_0)$, then $\|x\|_2=\|x\|_0 + \|x\|_1$ defines an strictly convex norm on $X$ equivalent to $\|\cdot\|_0$. Therefore, a space like …

Suppose that you have two Banach spaces $X$ and $Y$, and a (bounded) operator $TX:\to Y$. The operator $T$ is called semi-embedding if $T$ is injective and $T(B_X)$ is closed in $Y$. So your are aski …

[J. Lindenstrauss and L. Tzafriri. Classical Banach spaces I. Sequence spaces. Springer 1977]. In page 76, after Prop. 2.c.3, it says that the proof of 2.c.3 shows that an operator $T:\ell_p\to\ell_p$ …

Several geometric properties equivalent to non-reflexivity for a Banach space were studied by R.C. James in "Some self-dual properties of normed linear spaces". Ann. of Math. Studies 69 (1972), 159-17 …

The answer is no: you can even have $T_1=T_3=0$ and $T_2$ equal to the identity $id$ on an infinite dimensional Banach space. Indeed, consider the following commutative diagram with exact rows: $$\ …

The question has a negative answer: Following the idea in Bill Johnson's comment, I looked at the work of Argyros. In this paper (see the reference below), there are several examples of hereditarily i …

This is a detailed version of a previous comment that gives a partial answer. If $K$ is a compact space and $C(K)$ is isometrically isomorphic to $X^{**}$, then $X$ is a $\mathcal{L}_{\infty}$-space. …

A (non-trivial) ultrafilter $\mathcal{U}$ on $\mathbb{N}$ is a p-point if and only if every bounded sequence $(a_n)$ of real numbers contains a convergent subsequence $(a_{n_k})$ such that $\{n_k\}\in …

You can find this result in the book of T. Kato. Perturbation theory for linear operators. Springer 1980, 1995. In Theorem IV.4.2, page 219.

In this reference (Definition 2.47) a Banach space $X$ is called separably automorphic if given a separable space $Y$ and isomorphic copies $A$ and $B$ of $Y$ in $X$, every bijective operator $T:A\to …

An operator $T:X\to Y$ is Banach-Saks if for every bounded sequence $(x_n)$ in $X$ there is a subsequence $(Tx_{n_k})$ such that the Cesàro means $N^{-1}(\sum_{k=1}^N Tx_k)$ from a norm-convergent sub …

Let $\mathcal{K}$, $\mathcal{W}$ and $\mathcal{C}$ denote the compact, weakly compact and completely continuous operators, respectively, and let $\mathcal{W}^{-1}\circ \mathcal{K}$ denote the operator …

There exists a separable reflexive Banach space $\hat X^*$ which is indecomposable but not HI, and admits a subspace $Y$ such that both $Y$ and $\hat X^*/Y$ are HI. This example provides a negative an …

The answer is yes. Indeed, let us denote $Y_h=\ell_\infty(B_{Y^*})$. For every $L\in\mathcal{K}(X,Y_h)$, $\|T\|_m=\|JT\|_m=\|JT-L\|_m\leq \|JT-L\|$. Hence $\|T\|_m\leq \|JT\|_e$. Conversely, since …

You can proceed as follows. Let $(x_n^{**})\in \ell_2(X^{**})$. Then $$(x_n^{**})\in T^{**-1}(\ell_2(X))\Rightarrow T^{**}(x_n^{**})= (\frac{x_n^{**}}{n})\in \ell_2(X),$$ hence $(x_n^{**})\in \ell_2 …