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Algebras of operators on Hilbert space, $C^*-$algebras, von Neumann algebras, non-commutative geometry
6
votes
Accepted
Continuity of linear bijection that is continuous on ideal and quotient
No.
Counterexample. Let $C$ be an infinite-dimensional $C^*$-algebra and let $A = B = C \oplus C$. We set $I = J = C \oplus \{0\}$.
Let $\varphi: C \to C$ be a non-continuous linear mapping and define …
4
votes
Accepted
Status of finite-dimensional Ando's theorem
In their 2013 paper "Unitary $N$-dilations for tuples of commuting matrices" McCarthy and Shalit showed that there does indeed exist an $N$-dilation to a finite-dimensional space (see Theorem 1.2 in t …
3
votes
Are (completely) positive maps approximated by normal (completely) positive maps?
The answer to the first question is yes. This follows from the following more general result.
Terminology I: Ordered Banach spaces. By a pre-ordered Banach space I mean a pair $(X,X_+)$ where $X$ is a …
9
votes
Accepted
Removing the interior of spectrums
The answer is no, in general. Here is a counterexample:
Let $A$ be the algebra of bounded linear operators on $\ell^2(\mathbb{N})$, and let $a \in A$ be the left shift on $\ell^2(\mathbb{N})$. Then t …
4
votes
Accepted
Convergent bounded net of positive operators converges to a positive operator
Yes, $a$ is positive, too.
Proof.
For every $b \in A$ we have $b^* a_i b \to b^* a b$, so $b^*a b$ is positive.
Now, let $(e_j)$ be an approximate identity in $A$. Then it follows that $(e_j a e_j)$ i …
2
votes
Point spectrum of a positive invertible operator
Although the question was already answered by Robert Israel's post, here is a slightly more concrete counterexample:
Let $G$ be the additive group $\mathbb{Z}$ (endowed with the discrete topology). L …
5
votes
An example of non-invertible operator $F$ such that $P_nF$ is invertible on $\operatorname{I...
(In the following I assume that the word "invertible" in the question means "bijective".)
Your assumptions do not imply that $F$ is bijective (however, they imply that $F$ is injective and has closed …
4
votes
1
answer
468
views
Bicommutant theorem for commutative operator algebras
Let $\mathcal{B}(H)$ denote the space of bounded linear operators on a complex Hilbert space $H$. The von Neumann bicommutant theorem says:
Theorem. Suppose that $\mathcal{A}$ is a $C^*$-subalgebra o …
3
votes
Accepted
When is rank-1 perturbation to a positive operator still positive?
In "On Majorization, Factorization, and Range Inclusion of Operators on Hilbert Space (1966)", R. G. Douglas proved the following result (Theorem 1 in the paper):
Theorem. Let $C$ and $D$ be bounded …
1
vote
Accepted
Closure of the point spectrum of an unbounded diagonalizable operator
From your assumption you can easily see that $T$ is unitarily similar to a multiplication operator on $\ell^2$ (and thus, $T$ is normal, by the way). This shows that the answer is "yes" (as it is easy …
8
votes
3
answers
642
views
Commutant of the conjugations by unitary matrices
Let $\mathcal{L}(\mathbb{C}^{n \times n})$ denote the algebra of all linear mappings from $\mathbb{C}^{n \times n}$ to $\mathbb{C}^{n \times n}$ and let $\mathcal{C} \subseteq \mathcal{L}(\mathbb{C}^{ …
8
votes
1
answer
346
views
Order bounded version of monotone complete $C^*$-algebras
Let $A$ be a $C^*$-algebra with self-adjoint part $A_{\operatorname{sa}}$. Then $A$ is called monotone complete if every increasing norm bounded net in $A_{\operatorname{sa}}$ has a supremum (with res …
14
votes
1
answer
683
views
Criterion for a Banach algebra to be finite dimensional
Let $A$ be a Banach algebra (say, complex and unital) and suppose that every (closed) commutative subalgebra of $A$ is finite dimensional.
Question. Does it follow that $A$ is finite dimensional?
Re …