All Questions
8 questions with no upvoted or accepted answers
11
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388
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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 ...
- 5,529
4
votes
0
answers
160
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Solution without using any k-theory tools
Let $A$ be the UHF-algebra of type $2^{\infty}$. Suppose that $p$ and $q$ are two projections in $A$ and $\tau(p) = \tau(q)$, where $\tau$ is the unique normalized trace. Then there is a partail ...
- 581
2
votes
0
answers
317
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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 ...
- 63
2
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0
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56
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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 ...
- 4,135
1
vote
0
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89
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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):=\...
- 111
1
vote
0
answers
70
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Is $F: B(H)\to B(H\otimes K)$, $X\mapsto X\otimes \mathbb{1}_K$ completely contractive?
Take Hilbert spaces $H$ and $K$. Consider a linear map $F: B(H)\to B(H\otimes K)$, $X\mapsto X\otimes \mathbb{1}_K$.
Is it true that $F$ is completely contractive? If it is, I would be very grateful ...
1
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0
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233
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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 ...
- 1,903
0
votes
0
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35
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Operator-form correspondence without lower semiboundedness
When dealing with a normal unbounded operator $A$, it is often useful to frame questions about the operator in terms of questions about the associated form $\omega,$ which has domain $D(|A|^{1/2})$ ...
- 109