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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.
3
votes
1
answer
161
views
Approximating continuous functions from $K\times L$ into $[0,1]$
Let $K$ and $L$ be compact Hausdorff spaces, let $f:K\times L\to [0,1]$ be continuous and let $\varepsilon>0$. Can we find continuous $g_{1},...,g_{n}:K\to[0,+\infty)$ and $h_{1},...,h_{n}:L\to[0,+\in …
3
votes
1
answer
293
views
Pointwise convergence and disjoint sequences in $C(K)$
Let $K$ be a Hausdorff compact space and let $C(K)$ be the space of continuous real-valued functions on $K$. A sequence $(h_n)$ in $C(K)$ is called almost disjoint if there is a sequence $(g_n)$ with …
6
votes
2
answers
234
views
Continuity of a differential of a Banach-valued holomorphic map
Originally posted on MSE.
Let $U$ be an open set in $\mathbb{C}^{n}$ let $F$ be a Banach space (in my case even a dual Banach space), and let $\varphi:U\to F$ be a holomorphic map. I seem to be able …
2
votes
1
answer
189
views
Biorthogonal weakly null basic sequences
Let $E$ be a Banach space, let $e_{n}\in E$ and $g_{n}\in E^{*}$ be biorthogonal basic sequences (i.e. $\left<e_n,g_m\right>=\delta_{mn}$ ). Moreover, both of these sequences are weakly null. (note th …
1
vote
Accepted
Biorthogonal weakly null basic sequences
I think I've found a counterexample in the literature. I would really appreciate if somebody verified that I didn't get confused about the terminology of Orlicz spaces.
Recall that a Banach space $E$ …
2
votes
1
answer
245
views
Is the union of good equivalence relations on a compact space good?
Let $X$, $Y_1$ and $Y_2$ be a compact Hausdorff spaces and let $\varphi_i:X\to Y_i$ be a continuous surjection (and so a quotient map).
Let $\sim$ be the minimal closed equivalence relation on $X$ tha …
5
votes
2
answers
245
views
Is there a topology that makes every basic sequence null?
Let $E$ be a Banach space. Let $F$ be the collection of all $f\in E^*$ such that $\left<f,e_n\right>\to 0$, for every normalized basic sequence $\{e_n\}$. It is easy to see that $F$ is a closed subspa …
0
votes
Is there a topology that makes every basic sequence null?
This answer is supplementary to the one of Bill Johnson, to fill in some details.
A sequence $\{e_n\}$ in a Banach space $E$ is called a basic sequence of type P* if (among other equivalent definitio …
1
vote
1
answer
225
views
Is a topology sandwiched between two norms compactly generated?
Recall that a Hausdorf topological space $X$ is called compactly generated if any set whose intersections with compacts are compact is closed. Locally compact and first countable spaces are compactly …
5
votes
1
answer
205
views
If a subspace $F$ is contained in a subspace $G$, and $H$ is close to $G$, can we choose a s...
Let $E$ be a Banach space. Recall that the collection of all closed linear subspaces of $E$ can be turned into a metric space in a number of ways. In particular, consider the notion of a gap: if $G$ a …
9
votes
1
answer
271
views
Is a Banach lattice isomorphic to a Hilbert space in fact a Hilbert lattice?
The title says it all:
Let $E$ be a Banach lattice, which is isomorphic to a Hilbert space (as normed spaces). Is there an equivalent Hilbert norm on $E$, which still makes it a Banach lattice with r …
2
votes
1
answer
69
views
Equicontinuity-like property of a convex compact set
Let $X$ be a Tychonoff topological space and let $x\in X$. Let $B\subset C(X)$ be convex and compact in the topology of pointwise convergence, and such that $f(x)=1$, for every $f\in B$.
Is there an …
4
votes
1
answer
289
views
Supremum over which sets makes $H^{\infty}$ non-separable?
It is known that the space $H^{\infty}$ of bounded holomorphic functions on the unit disk $D$ is non-separable with respect to the supremum norm $\|\cdot\|_{\infty}^{D}$. Let $E\subset D$ be connected …
6
votes
Accepted
Supremum over which sets makes $H^{\infty}$ non-separable?
Jochen Wengenroth suggested to look at Carleson's interpolation theorem, and it seems like it completely answers my question. Namely, the following is true.
Let $E$ be a subset of $D$. Then $H^\infty …
11
votes
0
answers
381
views
Von Neumann Inequality in Banach spaces
It is known that the only Banach space that satisfies the von-Neumann inequality is the Hilbert space:
Theorem (see e.g. Pisier, "Similarity Problems and Completely Bounded Maps", p 27) For a Banach …