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Spectrum, resolvent, numerical range, functional calculus, operator semigroups. Special classes of operators: compact, Fredholm, dissipative, differential, integral, pseudodifferential, etc.

0 votes
0 answers
140 views

A possible generalization of Pitt's theorem

Inspired by Pitt's theorem and this post we ask the following question: First we remind the Pitt's theorem and also introduce a particular Banach space as averaging of $\ell^p$ spaces for $1\leq p …
Ali Taghavi's user avatar
1 vote
1 answer
125 views

Trace class operators in the unit ball of a finite dimensional subvector space of $B(H)$

Let $F\subset B(H)$ be a finite dimensional subvector space of the space of all bounded operators on a Hilbert space. Question: Is there an upper bound for $$\{|tr(T)| \text{where} \quad T\in …
Ali Taghavi's user avatar
2 votes
1 answer
181 views

A Possible characterization of F.D or AF commutative $C^{*}$ algebras

By F.D or AF $C^{*}$ algebra,we mean finite dimensional or approximately finite dimensional $C^{*}$ algebra. Let $A$ be a unital commutative $C^{*}$ algebra with the property that for every …
Ali Taghavi's user avatar
7 votes
Accepted

A matrix norm inequality II

No the norm of the left side can be very large. For example $\left\| X^{-1}AXB \right\| $ is an unbounded function in $(x,y)$ where $A,X,B$ are the following matrices: Put $A=B= \begi …
Ali Taghavi's user avatar
2 votes
0 answers
684 views

Universal $C^*$ algebra generated by two self adjoint elements with $x^2+y^2=1+{(xy-yx)}^2$

Is there universal $C^*$ algebra generated by self adjoint elements $x,y$ subject to relation $x^2+y^2=1+{(xy-yx)}^2$? What is a precise description of such algebra?
Ali Taghavi's user avatar
4 votes
0 answers
100 views

A bounded operator associated with a principal bundle

Assume that $(X, B, G)$ is a $G$- principal bundle where $G$ is a compact topological or Lie group. The normalized Haar measure of (each fiber of ) $X$ is denoted by $\mu$. The space of continuos comp …
Ali Taghavi's user avatar
3 votes
1 answer
173 views

Obstructions for $C^\star$ algebras to contain a $Z^\star$ algebra

As the comment of Andreas Thom indicated here, a separable $C^\star$ algebra $A$ can not contain a $Z^\star$ algebra.(A $Z^\star$ algebra is a $C^\star$ algebra which all elements are zero divis …
Ali Taghavi's user avatar
3 votes
1 answer
161 views

Simple $Z^{*}$ algebra

What is an example of a simple $C^{*}$ algebra which all elements are (two sided or equivalently one sided) zero divisor?
Ali Taghavi's user avatar
6 votes
3 answers
981 views

Invertible operator with countable spectrum

Let $H$ be a separable Hilbert space and $A$ is an invertible bounded operator on $H$. Can we approximate $A$ with an invertible operator $B$ such that $sp(B)$ is a countable set? Motivation: …
Ali Taghavi's user avatar
4 votes
2 answers
442 views

$C^{*}$ algebras which do not admit nontrivial idempotent morphism

In this question which I flag it as a community wiki, I search for a big list of $C^{*}$ algebras(and a big list of criterions) which do not admit a non trivial idempotent $C^{*}-$morphism. I kno …
6 votes
0 answers
242 views

For what kind of $C^*$ algebra $A$ every normal element $y\in A$ has a normal lift for every...

Is there a terminology for the following property of $C^*$ algebra $A$: For every $C^*$ algebra $B$ and surjective $C^*$ morphism $\phi: B\to A$, every normal element $y\in A$ admits a normal ele …
Ali Taghavi's user avatar
1 vote
2 answers
435 views

Fredholm $C^*$-algebras

Let $H$ be a Hilbert space. A vector subspace $W\subset B(H)$ is called a Fredholm subspace if there is an upper bound for the absolute value of Fredholm index of all Fredholm operators $T$ in $W$. …
Ali Taghavi's user avatar
1 vote
2 answers
397 views

A question on unbounded operators

Assume that $H$ is a separable Hilbert space. Is there a polynomial $p(z)\in \mathbb{C}[x]$ with $deg(p)>1$ with the following property?: Every densely defined operator $A:D(A)\to D(A),\;D(A)\subs …
Ali Taghavi's user avatar
4 votes
2 answers
394 views

The closure of all periodic homeomorphisms of circle

Let $X$ be the space of all continuous maps from $S^{1}$ to $S^{1}$. $X$ is equipped with $C^{0}$ topology. Let $Y\subset X$ be the union of all finite order homeomorphisms i.e: all homeomorphi …
Ali Taghavi's user avatar
4 votes
1 answer
300 views

Fredholm subvector spaces of $B(\mathcal{H})$

Let $\mathcal{H}$ be a separable Hilbert space with orthonormal base $\{e_i\}\;\;i\in \mathbb{N}$. Definition: We say a subvector space $W\subset B(\mathcal{H})$ is a Fredholm subspace if the …
Ali Taghavi's user avatar

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