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4

By the isoperimetric inequality on the sphere, a disk (spherical cap) has the smallest perimeter among all sets of given area on the sphere. It remains to minimize the ratio over circles of different radius. The isoperimetric inequality makes this easy by giving a convenient relation between the area and perimeter of a spherical cap: $p=\sqrt{a(4\pi -a)}$ ...


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

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 map $$A(\cdot)\colon \mathbb{C}\to \mathscr{B}(\mathcal{H}\oplus \mathcal{H})$$ such that $A(0)=A\oplus 0$ and $A(z)$ is unitary for $|z|=1$. Indeed, simply put $$A(z) = U(z)BU(z)$$ ...


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 ...


2

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


2

You probably do not mean the operator $TE_{\Delta}$ because this operator has a spectrum which now includes $0$ unless $E_{\Delta} = I$. I assume that you want to deal with the restriction $T_{\Delta}$ of $T$ to the range of $E_{\Delta}$ as an operator on the Hilbert space $\mathcal{H}_{\Delta}=E_{\Delta}\mathcal{H}$ under the induced norm; this makes sense ...


2

It is convinient to use the following notation $\mu_{x, y}(S) := \left<P(S)x, y\right>$ and shortly $\mu_{x}$ for $\mu_{x,x}$. Let $B(\mathbb{R})$ denote the abelian C*-algebra of bounded Borel functions on $\mathbb{R}$. Define a map $\Phi_P \colon B(\mathbb{R}) \rightarrow \mathcal{L}(\mathcal{H})$ by $$\left<\Phi_P(f)x, y\right> = ...


2

We can diagonalize $L$ (that is, make it a multiplication operator) with the help of the Fourier transform. More precisely, map $U:L^2(0,\infty)\to L^2(0,\infty)$, $Uf=\int f(x)\sin kx \, dx$; then $ULU^*$ is multiplication by $k^2$, and thus the cyclic vectors are exactly those for which $Uf\not=0$ almost everywhere. Clearly, $f(x)=e^{-x}$ has this ...


2

If you start with a unit vector $v \in \mathcal{H}$, then you can define a subspace $\mathcal{H}_{v}$ as the closure in $\mathcal{H}$ of all polynomials in $N$, $N^{\star}$ acting on $v$. $\mathcal{H}_{v}$ is invariant under $N$, $N^{\star}$. This defines a unitary map $$ \mathcal{F}_{v} : \mathcal{H}_{v}\rightarrow L^{2}_{\mu_{v}} $$ where ...


2

If you delete the first row and last column from an irreducible $n\times n$ tridiagonal matrix $T$, the resulting submatrix is triangular with non-zero diagonal entries. Hence it is invertible, and it follows that $\mathbb{rank}(T-\lambda I)$ is always at least $n-1$.


2

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.


2

The assertion applies not only to real tridiagonal matrices with positive sub and super diagonals, but also to every sign-symmetric real tridiagonal matrix whose sub or super diagonal entries are nonzero. What is missing in the paper you mentioned is the fact that every such tridiagonal matrix is diagonalisable (because $D^{-1}JD$ is real symmetric for some ...


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

I'm assuming a unit $1$. The resolvent $(x-\lambda 1)^{-1}$ is uniformly bounded near $\infty$ because $$ (x-\lambda 1)^{-1} = -\sum_{n=0}^{\infty}\frac{1}{\lambda^{n}}x^{n},\;\;\; |\lambda| > r_{\sigma}(x). $$ If $U$ is any open set containing $\sigma(x)$, then $M=\sup_{\lambda\in\mathbb{C}\setminus U}\|(x-\lambda 1)^{-1}\| < \infty$ because the ...


1

Yes, $C(X)_+$ is the set of continuous functions on $X$ such that $f(x)\in[0,\infty)$ for all $x\in X$. A positive map can be defined whenever you have a positive cone in the domain and a positive cone in the codomain. In this case, $\rho$ maps positive functions to positive operators, and so it is customary to call it "positive". In the case of a ...


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 ...


1

I'm going to assume the facts that I proved in a previous answer to a question of yours: Spectral Measure Integration: Unbounded Functions? For any Borel function $f$, I defined an operator $T_{f}$ on $\mathcal{D}(T_{f})$ consisting of all $x \in \mathcal{H}$ for which $\int |f(\lambda)|^{2}\,d\|E(\lambda)x\|^{2}<\infty$. I showed that $T_{f}$ is a ...



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