Skip to main content

All Questions

10 questions with no upvoted or accepted answers
Filter by
Sorted by
Tagged with
11 votes
0 answers
388 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 ...
erz's user avatar
  • 5,529
4 votes
0 answers
152 views

Maximally fine topologies on $B(H)$ making the unit ball compact

Let $H$ be a Hilbert space, and $B(H)$ its algebra of bounded operators. One of the reasons the Ultraweak topology is (in a way) more useful than the weak operator topology is that the Ultraweak ...
Aareyan Manzoor's user avatar
4 votes
0 answers
185 views

A strongly open set which is not measurable in the weak operator topology

Let $H$ be a non-separable Hilbert space and $\{e_i\}_{i\in I}$ be an orthonormal basis for $H$. Let $J$ be a uncountable proper subset in $I$. Let us put $$E=\{x\in B(H): \lVert xe_j\rVert <1: \...
ABB's user avatar
  • 4,058
2 votes
0 answers
316 views

What are alternative or equivalent definitions of a positive-definite function on a group?

The standard definition of a positive-definite function on a group goes as follows: Let $\varphi : G \rightarrow L(H)$, where $G$ is a group (with an involution) and $H$ a Hilbert space. $L(H)$ is the ...
S-F's user avatar
  • 63
2 votes
0 answers
56 views

Existence of a suitable smooth kernel

Denote by $H=L^2([0,1])$ the Hilbert space of square integrable real valued functions on the interval $[0,1]$. Does there exist a nontrivial smooth real valued function $k$ on the unit interval such ...
Ali's user avatar
  • 4,135
2 votes
0 answers
116 views

Closable operators on Hilbert modules

For $T:{\frak{Dom}}(T) \to H$ a densely defined operator, with $H$ a (separable) Hilbert space, we know that $T$ is closable if its adjoint $T^*$ has dense domain in $H$. Does this extend to the (...
Max Schattman's user avatar
1 vote
0 answers
89 views

Let $T\in B(H)$. Then prove that $T$ is a contraction if and only if the closed unit disc is spectral set for $T$

Let's first recall the definition of Spectral set for a bounded operator $T$ on a hilbert space $H$. Let $X\subseteq \Bbb{C}$ be compact such that $\sigma(T)\subseteq X$. Define $$\mathcal{R}(X):=\...
DeltaEpsilon's user avatar
1 vote
0 answers
277 views

Adjoint for a non-densely defined unbounded operator on a Hilbert space

Let $\mathbf{H}$ be a Hilbert space, and $D$ an unbounded densely-defined operator on $\mathbf{H}$. As is well-known, every such operator admits an adjoint, with domain possibly different from that ...
Max Schattman's user avatar
1 vote
0 answers
74 views

If $f$ takes values in $L(H,L(H,\Bbb R))$ and $μ$ is a $H\hat ⊗_πH$-valued measure, how are $\int f\:dμ$ and $\int f⊗_π\text{id}_Hdμ$ related?

Let $H$ be a separable $\mathbb R$-Hilbert space $H\:\hat\otimes_\pi\:H$ denote the projective tensor product of $H$ and $H$ $(\Omega,\mathcal A)$ be a measurable space $\mu$ be a $H\:\hat\otimes_\pi\...
0xbadf00d's user avatar
  • 167
1 vote
0 answers
233 views

Bochner integrals with values in a Hilbert $A$-module

I'm wondering whether there exists a generalisation of Bochner integration with values in a Hilbert $A$-module $M$, where $A$ is a general $C^*$-algebra rather than $\mathbb{C}$ (and whether there are ...
geometricK's user avatar
  • 1,903