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4

Since the spectrum does not depend on the (C$^*$) algebra, we can assume that $\mathcal A=C^*(x)$. Using the Gelfand transform, we can identify $C^*(x)$ with $C(\sigma(x))$, with $x$ mapped to the function $z\mapsto z$. The point evaluation states $f\mapsto f(t)$ are precisely the pure states The pure states are the extremal points of the set of states of ...

3

A bounded linear operator $A$ on a separable Hilbert space $X$ with orthonormal basis $\{ e_{j} \}_{j=1}^{\infty}$ is a Hilbert-Schmidt operator if $$\sum_{j=1}^{\infty}\|Ae_{j}\|^{2} < \infty.$$ This condition is true for one orthonormal basis iff it is true for every other orthonormal basis. If you're studying $X=L^{2}[a,b]$, then an ...

3

The classical operator $$L = -\frac{d^{2}}{dx^{2}}+V,\;\;\; a \le x \le b,$$ is different if $V$ is very singular. If $V \in L^{1}[a,b]$, then things are nice because there are 2 linearly-independet classical solutions of $Lf = \lambda f$ for every $\lambda$. That is, such solutions are continuous on $[a,b]$, their first derivatives are ...

2

This is a continuation of what I posted earlier. What follows are two examples of how the theory is applied to the trigonometric functions where $V=0$. The equation is in the limit point case on $[0,\infty)$ because $e^{i\sqrt{\lambda}x}\in L^{2}[0,\infty)$ while $e^{-i\sqrt{\lambda}x}$ is not, where $\sqrt{\lambda}$ is the branch whose branch cut is along ...

2

Consider $L^{2}(\mathbb{R})$. The Fourier transform and its inverse implement the Spectral Theorem for the selfadjoint operator $Af = \frac{1}{i}\frac{d}{dx}f$ on the domain $\mathcal{D}(A)$ consisting of absolutely continuous $f \in L^{2}(\mathbb{R})$ for which $f'\in L^{2}(\mathbb{R})$. The spectral measure $E$ is $$E[a,b]f = ... 2 For a matrix A \in \mathbb{R}^{n \times n} with strictly positive entries (actually you just need it to be irreducible), you can apply the Perron-Frobenius Theorem which asserts that the eigenvalue \lambda with largest magnitude is positive, simple and the associated eigenvector has strictly positive entries. In particular, with these special settings ... 2 This requires only the inverse of the Cayley transform. Start with$$ (U-I)=(A-iI)(A+iI)^{-1}-(A+iI)(A+iI)^{-1}=-2i(A+iI)^{-1}. $$It follows that \mathcal{N}(U-I)=\{0\} and \mathcal{R}(U-I)=\mathcal{D}(A). Similarly,$$ (U+I) = 2A(A+iI)^{-1} = iA(U-I). $$Let U=\int_{T}\lambda dF(\lambda), and, for each 0 < \delta < \pi, ... 1 Because nobody is answering, I thought I'd offer a start. Your equation for the resolvent of A is$$ (\lambda I -A)\left(\begin{array}{c}f \\ g\end{array}\right)= \left(\begin{array}{cc}\lambda & -I \\ I-\Delta & \lambda\end{array}\right) \left(\begin{array}{c}f \\ ...

1

Suppose $T^{n}=I$. Let $p_{k}$ be the Lagrange polynomials $$p_{k}=\prod_{j=0,j\ne k}^{n-1}(\lambda-e^{2\pi ji/n})\left/\prod_{j=0,j\ne k}^{n-1}(e^{2\pi ki/n}-e^{2\pi ji/n})\right.\;.$$ Notice that $p_{0}+\cdots+p_{n-1}=1$ because it is an (n-1)-st degree polynomial that equals $1$ at all n-th roots of unity. Define $P_{k}=p_{k}(T)$. ...

1

Let $N=\{ (Ax,x) : \|x\|=1\}$. Suppose $\lambda \notin N^{c}$ so that there exists $\delta$ such that the following holds whenever $\|x\|=1$: $$0 < \delta \le |(Ax,x)-\lambda|=|((A-\lambda I)x,x)|\le \|(A-\lambda I)x\|\|x\| = \|(A-\lambda I)x\|$$ Then $\|(A-\lambda I)x\| \ge \delta \|x\|$ for all $x$. (The same holds for ...

1

Using $P_i^2 = P_i$ we get $$e^{P_i\log\rho_i} = \sum_{k=0}^\infty \frac{(P_i\log\rho_i)^k}{k!} = \mathbb{1}-P_i +P_i\sum\frac{\log^k\rho_i}{k!} = \mathbb{1}-P_i+P_i\rho_i$$ For the whole expression we find (using that $P_iP_k=\delta_{i,k}P_k$, and $\sum_iP_i=\mathbb{1}$) $$\exp \sum_iP_i\log\rho_i = \prod_i\mathbb{1}-P_i+P_i\rho_i = \mathbb{1} -\sum_iP_i + ... 1 Classical Solutions: First, assume V\in L^{1}[a,b], and show the existence of classical solutions of$$ -f''+Vf = \lambda f,\;\;\; f(a)=A,\;f'(a)=B. $$This can be done by considering the equivalent integral equation$$ f(x)=A+B(x-a)+\int_{a}^{x}\int_{a}^{t}(V(u)-\lambda)f(u)\,du\,dv.  This is a fixed point problem for $C[a,b]$ ...

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