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3

The Spectral Theorem tells you that for every $\varepsilon>0$ there exists a partition $\{\Delta_1,\ldots,\Delta_n\}$ of $\sigma(N)$ and complex numbers $\lambda_1,\ldots,\lambda_n$ such that $$ \left\|N-\sum_{j=1}^n\lambda_j\,E(\Delta_j)\right\|<\varepsilon. $$ As $A$ commutes with $\sum_j\lambda_j\,E(\Delta_j)$, you get that ...


3

The answer to your question is no, the spectrum can be infinite, and pairs of self-adjoint involutions are actually a good class of counterexamples because they can be fully described using the spectral theorem. The irreducible pairs of involutions occur in dimensions $1$ and $2$, and the rest are direct integrals of irreducibles. So every such pair is ...


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No, the range for Borel functional calculus is not $C^*(N)$, in fact in general it takes you to the strong operator closure of $C^*(N)$.


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Note that if $X$ is bounded, $X$ is contained in some bounded convex closed ball $B$. Then, since $f_{\mid B}$ sends $B$ to $B$, it must have a fixed point.


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What involution do you consider? If just complex conjugation, then $x\mapsto \overline{x}$ is not even differentiable. If you consider $f\mapsto f^*$ where $f^*(z) = \overline{f(\overline{z})}$ then it does not satisfy the C*-identity.


2

The Calculus distance scale factors in spherical coordinates are $s_{r}=1$, $s_{\theta}=r$, $s_{\phi}=r\sin\theta$. So the Laplacian is $$ \Delta=\frac{1}{s_{r}s_{\theta}s_{\phi}}\left[ \frac{\partial}{\partial r}\frac{s_{\theta}s_{\phi}}{s_{r}}\frac{\partial}{\partial r} ...


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The simplest example would be a nonzero algebra with any involution and zero multiplication.


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Any normal operator $T$ gives rise to some spectral measure $E:Bor(\sigma(T))\to\mathcal{P}(H)$ which maps Borel subsets of the spectrum of $T$ into orthogonal projections in $H$. If you take Borel subset $A\subset\sigma(T)$, then $E(A)$ is called a spectral projection. Search spectral theorem on this site.


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It is often difficult to prove that an operator is self-adjoint, which seems reasonable since you know a lot about an operator when it is self-adjoint. For instance it is unitarily equivalent to a multiplication operator by a real function. There are many possible approaches, one which is sometimes useful is the following Show that $D$ is symmetric, i.e. ...


2

Unfortunately, the multipliers of the Dirichlet space are not as easy to describe as of Hardy and Bergman spaces. The characterization was obtained by Stegenga in Multipliers of the Dirichlet space (free access). In addition to boundedness, it requires a Carleson-type condition in terms of capacity: $$ \iint_{\bigcup S(I_j)} |f'|^2 \le A ...


1

Assuming that $\mu$ is a regular Borel measure implies that $\mu F < \infty$ for any compact subset $F$. Suppose $K$ is the support of $\mu$ and that $\lambda \in K$. Then, for every $r \in (0,\infty)$, the closed disk $D_{r}[\lambda]$ centered at $\lambda$ of radius $r$ is compact and, therefore, has finite $\mu$ measure. Let ...


1

If you start with any topological space $\Omega $ then $ C_b (\Omega) $, the set of bounded continuous functions on $\Omega $, is a C $^*$-algebra. But the Gelfand transform allows you to show that there exists a locally compact $\Omega'$ with $ C_b (\Omega)\simeq C_b (\Omega') $. So when talking in abstract, you gain nothing by considering ...


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As I already stated in the comment, the cyclic vectors are exactly those $f \in L^2$ for which $f(x) \neq 0$ holds for almost every $x \in X$. It is easy to see that if $M := \{x \mid f(x) = 0\}$ has positive measure, then by $\sigma$-finiteness, there is some set $M' \subset M$ of finite positive measure. But then $$ \overline{\{\phi \cdot f \mid \phi \in ...


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Case 1: $p<1$ Consider the function \begin{equation*} f(t) = \begin{cases} L^{1/p}, & t<L^{-1} \\ 0, &t \geq L^{-1} \end{cases} \end{equation*} with $L_p$ quasinorm $1$. Its Fourier coefficients are \begin{equation*} c_N = \left< f, e^{2\pi i N \cdot}\right> = \frac{-i}{2\pi} \frac{L^{1/p}}{N} \left(1 - e^{-2\pi i ...


1

I now looked up the definition of a core of an operator and it could be that the answer actually depends on the exact definition. In Wikipedia, a core (of a closed operator) is defined as a subset $D$ of the domain of $A$ such that $A$ is the closure of $A|D$. If we replace this by requiring that $D$ be a subspace, I can prove your claim. By symmetry, it ...


1

Robert already answered your question but let me give an extra illustration. Think of a normal operator with spectrum $[0,1]$. So you are in $C[0,1]$. Now you can compose functions from $C[0,1]$ with Borel functions on $C[0,1]$. For example, take your favourite discontinuous Borel function $f$. Then $ f = f\circ {\rm id}_{[0,1]}$. This takes you out of ...


1

Define $E(S)f = \chi_{S}f$ for any $f \in L^{2}_{\mu}$ and for any Borel subset $S$ of $\mathbb{C}$. For any Borel subsets $S$, $T$, $$ E(S)=E(S)^{2}=E(S)^{\star},\\ E(S)E(T)=E(T)E(S)=E(T\cap S),\\ E(\emptyset)=0,\;\;\; E(\mathbb{C})=I. $$ So, $E$ is a spectral measure. And $E(S)M=ME(S)$ so that ...


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Since \begin{align} \left\|Af \right\|_{\infty} := \sup_{x\in [0,1]} \left| \int_{0}^{x} f(t) \, \mathrm{d}t \right| \le \sup_{x\in [0,1]} \int_{0}^{x} |f(t)| \, \mathrm{d}t \le \int_{0}^{1}|f(t)|\, \mathrm{d}t =: \left\|f\right\|_{L_1} \le \left\|f\right\|_{\infty}, \end{align} then \begin{align} \left\| A \right\| &:= \sup_{f \in C^{1}[0,1]} ...


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More generally, you can work with functions $p$ which are analytic in some open neighbourhood of the unit disc. The proof is based on the fact that you can find a norm-analytic maps $A(\cdot)\colon \mathbb{C}\to \mathscr{B}(\mathcal{H}\oplus \mathcal{H})$ such that $A(0)=A\oplus 0$ and $A(z)$ is unitary for $|z|=1$. Indeed, simply put $$A(z) = U(z)BU(z)$$ ...



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