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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.

8 votes
1 answer
330 views

Tokarev's theorem on Banach lattices which are Grothendieck spaces

When browsing the literature, I have found the following theorem of E. Tokarev: Let $X$ be a Banach lattice with weakly sequentially complete dual space. Then for any Banach space $Y$, every uncondit …
7 votes
1 answer
191 views

Projective tensor product of injective operators

I've seen claims that it is known that for a pair of bounded injective linear operators $T\colon X\to Y, S\colon W\to V$, their tensor product $T\otimes S\colon X \otimes_\pi W\to Y \otimes_\pi V$ nee …
15 votes
2 answers
923 views

Distinguishing topologically weak topologies of Banach spaces

Are the weak topologies of $\ell_1$ and $L_1$ homeomorphic? Strangely may it sound, the question seeks contrasts between norm and weak topologies of Banach spaces from the non-linear point of view. …
9 votes
0 answers
1k views

Weak compactness in $\mathcal{F}(X)$

Let $(X,0)$ be a pointed metric space and let $\mathcal{F}(X)$ be the natural predual of ${\rm Lip}_0(X)$, the space of Lipschitz functions on $X$ that map $0$ to $0$; here $\mathcal{F}(X)$ is really …
4 votes
1 answer
253 views

M-bases for $C(K)$-spaces, $K$ -scattered

Recall that a biorthogonal system $\{(x_i, x^*_i)\colon i\in I\}$ in a Banach space $E$ is a M-basis if $\{x_i\}_{i\in I}$ is linearly dense in $E$ and $\{x_i^*\}_{i\in I}$ separates points. Let me re …
6 votes
3 answers
2k views

Space of compact operators

I am interested in the Banach space $\mathcal{K}=\mathcal{K}(\ell^2)$ of compact operators on $\ell^2$, however my questions can be stated for any $\mathcal{K}(E)$, where $E$ is an arbitrary Banach sp …
7 votes
0 answers
123 views

The bidual of the space of divergence-free vector fields

Consider the Banach space $L_1(\mathbb R^n, \mathbb R^n)$ of integrable vector fields $(n>1$) together with its subspace $N$ formed by those vectors fields whose divergence (computed in the distributi …
6 votes
0 answers
99 views

Is every separable Banach space with the MAP 1-complemented in a space with a monotone basis?

The question, already phrased in the title, looks like a classical problem from Banach space theory from the 1970s. Hence, my question is more of a reference request in its nature. Can every separ …
15 votes
1 answer
679 views

Open bilinear maps that are not uniformly open

A map $f\colon X\to Y$ between metric spaces is uniformly open whenever for each $\varepsilon >0$ there is $\delta >0$ such that for any $x\in X$ one has $$B_Y\big(f(x),\delta\big)\subseteq f\big(B_X …
10 votes
0 answers
264 views

Are biduals of spaces of differentiable functions on hypercubes Grothendieck?

Consider the space $E_n = C^1([0,1]^n)$ of continuously differentiable functions with the usual norm $$\max\{ \|f\|_\infty, \|f^\prime_{x_1}\|_\infty, \ldots, \|f^\prime_{x_n}\|_\infty\}.$$ making it …
7 votes
0 answers
200 views

Equivalent strictly convex norms in spaces of small density

Can one construct in ZFC a Banach space of density character $\omega_1$ that does not have an equivalent strictly convex norm? Maybe one may apply some kind of a Löwenheim–Skolem-type argument to …
10 votes
0 answers
250 views

Do sufficiently large Banach spaces admit non-compact operators with not too large range?

As in the title, does there exist a cardinal number $\lambda$ such that for every Banach space $X$ of density/cardinality at least $\lambda$ there exists a non-compact bounded, linear operator $T\ …
5 votes
0 answers
115 views

An elementary inequality in normed spaces - the original reference sought

The following inequality is an elementary exercise in convexity: let $x,y$ be non-zero vectors in a normed space with $\|x\|, \|y\|\leqslant 1$. Suppose that $\|x-y\| \geqslant 1$. Then $$\left\|\fra …
5 votes
0 answers
137 views

Banach spaces complemented in their ultrapowers

By the principle of local reflexivity, the second dual $X^{**}$ of a Banach space $X$ is complemented in some ultrapower $X^U$ of $X$. Even when $X$ is separable, the index set of $U$ cannot be expect …
7 votes
1 answer
498 views

Is $L_q(X^*)$ complemented in $(L_p(X))^*$?

Let $X$ be a Banach space and let $p\in (1,\infty)$. If $q$ denotes the conjugate exponent to $p$, then $L_q(X^*)$ is easily seen to be isometric to a subspace of $(L_p(X))^*$ via the map $$f\mapsto \ …

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