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Questions tagged [functional-calculus]

Functional calculus allows the evaluation of a function applied to a linear operator or a matrix. The function could be a polynomial, a holomorphic function, a continuous function or a measurable function defined on the spectrum of an operator or a Banach algebra. Functional calculus is a basic and powerful tool in the spectral theory of operators and operator algebras and is part of functional analysis.

3 votes
2 answers
200 views

I am trying to fill in the details of the following (already discussed in these posts for example and also in Rudin, Reed and Simon, etc) fact: If $\lambda \in \sigma(A)$ is isolated for $A$ self-...
JGJ7's user avatar
  • 447
3 votes
1 answer
156 views

I'm currently trying to understand the action of a unitary operator on $L^2(\mathbb{R})$ (as a complex Hilbert space). For instance, let's assume we have an operator $P\colon D(P) \to L^2(\mathbb{R}), ...
Marcus's user avatar
  • 31
6 votes
1 answer
144 views

Let $A$ be an unitial $C^*$-algebra, $X \subseteq \mathbb{C}$ a compact set and $f : X \to \mathbb{C}$ a continuous function. Prove that for any $\epsilon > 0$ there is a $\delta > 0$ such that ...
Student 00's user avatar
7 votes
2 answers
273 views

Let $H$ be a Hilbert space. Let $A:D(A)\to H$ be a self-adjoint operator. When can we say $A^2:D(A^2) \to H$ is self-adjoint? Is it enough to define $D(A^2)=\{u\in D(A) | Au\in D(A)\}$? And can you ...
Alucard-o Ming's user avatar
6 votes
1 answer
259 views

The following is from Hsing/Eubank's Theoretical Foundations of Functional Data Analysis, in the proof of Theorem 11.1.1. I have a compact operator $\mathcal{T}$ from the Sobolev space $\mathbb{W}_q[0,...
cmperez024's user avatar
0 votes
0 answers
98 views

Let $T \in B(H)$ be positive and injective, so that $\log(T) \in B(H)$ makes sense. One can define $T^{iz} := \exp(iz\log(T))$. I am trying to verify that $z \mapsto T^{iz}$ is analytic for $z \in \{...
Isochron's user avatar
  • 2,004
14 votes
4 answers
1k views

The question is, as the title already says: Why do we introduce the continuous functional calculus for self-adjoint operators? To be more specific, why self-adjoint operators? It is introduced (in ...
Denis's user avatar
  • 1,203
3 votes
1 answer
94 views

Recently I learned about the Borel measurable functional calculus. Now I am learning about spectral measures. The motivation behind spectral measure was that for characteristic functions, $\Phi(\chi_A)...
NTc5's user avatar
  • 1,257
7 votes
0 answers
164 views

Context: For some function $f\in\mathcal{M}(\mathbb C)$ (meromorphic function on $\mathbb C$), I am interested in linear operators $T_f$ that act on functions of the form $g_a:x\mapsto \exp(ax)$ in ...
Kolakoski54's user avatar
  • 2,015
3 votes
1 answer
132 views

I am trying to understand the proof of the Borel measurable functional calculus, especially the construction of the map $\tilde{\Phi}:B_b(\sigma(T)) \rightarrow B(H)$. For that reason I'll try to ...
NTc5's user avatar
  • 1,257
7 votes
1 answer
242 views

I am learning about measurable functional calculus at the moment. I have learned that given a operator $T \in B(H)$ that is self adjoint, there exists a measurable functional calculus $\overline{\Phi}:...
NTc5's user avatar
  • 1,257
0 votes
1 answer
108 views

Title: When does $\{1, a, a^*\}$ being linearly dependent imply that $a$ is normal and $\sigma(a)$ lies on a line? I'm working through a problem in spectral theory involving elements of a unital $C^*$...
Matko Petričić's user avatar
5 votes
1 answer
103 views

I want to make sense to functions of diagonal operators $\mathrm{diag}[a_1, \dots, a_n, \dots]$ in $\ell_2$. Such operators are normal, not self-adjoint. So, generally, functional calculi may not ...
Leonid Aznaurian's user avatar
1 vote
1 answer
91 views

I have to proof that if $X$ is a C$^*$-algebra and $x\in X$ is invertible, there exist an hermitean $y\in X$ and an $u\in X^+$ such that $x=yu$. If $x$ is normal, I found this decompositian using ...
Vincent Batens's user avatar
0 votes
1 answer
82 views

Our teaching assistant claimed the following. If you are in a let's say commutative $C^*$-algebras $X,Y$ and a linear $*$-morphism $\varphi:X\to Y$, let $x\in X$ and let $f\in C(\text{sp} (x),\mathbb{...
Vincent Batens's user avatar

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