# Tag Info

1

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 ...

1

Of course, we are assuming an infinite-dimensional Hilbert space $H$ here. Since both $H$ and $H^n$ have the same dimension, there exists a unitary map $V:H\to H^n$. So we can define $\pi:B(H)\to B(H^n)$ by $$\pi(T)=VTV^*.$$ Since $V$ is a unitary, $\pi(T)$ is an injective $*$-homomorphism. So it remains to show that $\pi(\mathbb K)=M_n(\mathbb K)$. ...

2

They are equivalent to the identity, so they are also infinite. Explicitly, note that $S_jS_jS_j^*S_j^*$ is a subprojection of $S_jS_j^*$. It is proper, because if $S_jS_jS_j^*S_j^*=S_jS_j^*$, multiplying by $S_j^*$ on the left and $S_j$ on the right we get $S_jS_j^*=I$, a contradiction. And it is equivalent to $S_jS_j^*$, because if $V=S_jS_jS_j^*$, ...

0

As Paul remarked above- this is not true. Take $h_1$ and $h_2$ to be two orthogonal vectors in $\mathcal{H}$, and $T_1=T_2=I$. Then, your condition 1) holds, but condition 2) doesn't.

1

Sorry for a late reply but perhaps this will be helpful. Let $X$ be a locally compact metric space with metric $d$. Let $\rho \colon C_0(X) \to \mathcal{H}_X$ be a non-degenerate ample $*$-representation. Let $T$ be a bounded operator on $\mathcal{H}_X$. -The support of $T$ is the complement of the open subset in $X \times X$ $$\{ (x,y) \in X \times X ... 0 If you consider L=-\Delta +V on the linear space of complex functions, then the involution operator of complex conjugation on this space commutes with L. If you start with some domain \mathcal{D}(L) which is invariant under this involution on which L is symmetric, then L^{\star} commutes with conjugation. I cannot conceive of a reason that you ... 5 I might be missing something trivial- but this result seems to be false. Let's take the C^* algebra to be M_2(\mathbb{C}), a=\begin{pmatrix} 1 & 0\\ 0 & 2 \end{pmatrix}, b=\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}. Then b^*b=I, and b^*ab=\begin{pmatrix} 2 & 0\\ 0 & 1 \end{pmatrix}, and it's not true that b^*ab \leq ... 1 Yes |f|^2 is continuous, since compositions of continuous functions are continuous. Since it vanishes off a bounded set and is bounded, it's integrable. 3 The Spectral Theorem tells you that for every \varepsilon>0 there exists a partition \{\Delta_1,\ldots,\Delta_n\} of \sigma(N) and complex numbers \lambda_1,\ldots,\lambda_n such that$$ \left\|N-\sum_{j=1}^n\lambda_j\,E(\Delta_j)\right\|<\varepsilon. $$As A commutes with \sum_j\lambda_j\,E(\Delta_j), you get that ... 1 Robert already answered your question but let me give an extra illustration. Think of a normal operator with spectrum [0,1]. So you are in C[0,1]. Now you can compose functions from C[0,1] with Borel functions on C[0,1]. For example, take your favourite discontinuous Borel function f. Then  f = f\circ {\rm id}_{[0,1]}. This takes you out of ... 2 No, the range for Borel functional calculus is not C^*(N), in fact in general it takes you to the strong operator closure of C^*(N). 0 I'm assuming you are considering the lattice to be the closed invariant subspaces. Otherwise I don't think the result is true. Take a subspace M\in\text{Lat}(A). Then B_nM\subset M. So, for each x\in M, the sequence \{B_nx\} is in M and it converges to Bx. As M is closed, Bx\in M. The n-amplification version runs exactly the same. 0 You can get a trivial proof if you understand that the bilateral shift can be seen as the operator of multiplication by the identity on L^2(\mathbb T):$$ M_zf(z)=zf(z). $$It is an easy exercise that the spectrum of a multiplication operator by a function g is the closure of the range of g, which is \mathbb T in this case. 2 Since the unit ball of QH is compact, we can find a finite set \bar\chi\subset QH such that for every w\in QH with \|w\|\leq1, we have some v\in\bar\chi with \|w-v\|<\delta. Then (recall that we assume that \|\bar Pv-v\|<\delta for all v\in \bar\chi) for any w\in H with \|w\|\leq1 we have v\in\bar\chi with \|Qw-v\|\leq\delta, ... 0 What is being shown is that U-\lambda I cannot have a bounded inverse for \lambda\in\mathbb{T}, even though \mathcal{N}(U-\lambda I)=\{0\}. If U-\lambda I were to have a bounded inverse, then there would exist m > 0 such that \|(U-\lambda I)x\| \ge m\|x\| for all x. By showing that there is a sequence of unit vectors \{\varphi_{n}\} such ... 4 You can also do this explictly: for any h\in \Gamma,$$ \Phi(\lambda_s T\lambda_s^*)\delta_h=\sum_g e_{gg}\lambda_s T\lambda_{s^{-1}} e_{gg}\delta_h=e_{hh}\lambda_s T\lambda_{s^{-1}}\delta_h=e_{hh}\lambda_s T\delta_{s^{-1}h}\\=\sum_ge_{hh}\lambda_s\langle T\delta_{s^{-1}h},\delta_g\rangle\delta_g=\sum_g \langle T\delta_{s^{-1}h},\delta_g\rangle ...

5

Note that $\Phi$ has norm one and $\Phi\circ \Phi = \Phi$. Then use Tomiyama's Theorem (Theorem II.6.10.2 from Blackadar's book Operator algebras; thanks Martin!). Actually, the proof of this theorem will show you how to prove the module property.

1

Suppose that $A$ is quasi-diagonal, that is, $A$ embeds into $B=\prod M_k / \sum M_k$. It is enough to show that $M_n ( B )$ embeds into $B$. This however is clear. We can identify elements $X = ( [ [X^k_{ij}] ]_{i,j=1,\ldots n})_{k=1}^\infty$ of $M_n(B)$ with sequences in $B$ which are zero apart from terms whose index is divisible by $n$. More ...

2

Definition of projection operator is $P\circ P = P$, for (a) you can simply expand $(1-P)\circ (1-P)$ and find it holds true. You have to define the notation $U^+$ to get an answer for (b).

1

HINT: We have $$||I - T^* T || = || I - T T^*||$$ Indeed for any self adjoint operator $S$ we have $$||S|| = \sup \, \{ |\lambda| \ \mid \ \lambda\ \text{eigenvalue of } S \}= \rho(S)$$ Moreover, for any $U$, $V$ operators on a finite dimensional space we have $$\sigma(I - UV) = \sigma( I - VU)$$ Alternatively, use the singular decomposition of $T$ ...

3

Any finite rank operator is a compact operator, and it's a known result that the only points in the spectrum of a compact operator are the eigenvalues. See theorem 35.17 here for the general statement.

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

The conditions guarantee that the linear extension of $\Phi$ is an isometry in the $2$-norm. Indeed, $$\left\|\Phi\left(\sum_k\alpha_k\,g_k\right)\right\|_2^2=\left\|\sum_k\alpha_k\Phi(g_k)\right\|_2^2=\tau_{\mathcal U}\left(\sum_{k,j}\overline{\alpha_j}\alpha_kg_j^{-1}g_k\right) =\sum_{k,j}\overline{\alpha_j}\alpha_j\tau_{\mathcal U}(g_j^{-1}g_k)\\ ... 2 I don't understand why you cannot get an answer from the eigenvalues of A^*A? Evaluating directly the maximal eigenvalue of this matrix gives$$ \|A\|_2^2=|a|^2+\frac{|b|^2}{2}+\frac{|b|}{2}\sqrt{4|a|^2+|b|^2}. $$Now put the assumption |b|\leq 1-|a|^2 (with the one that |a|\leq 1 to make it possible) in there to verify that \|A\|_2\leq 1. The ... 1 Yes. It is tracial because it is tracial in the dense subalgebra \pi_\tau(A), and it extends to \pi_\tau(A)'' since it is weak operator continuous. B(H), for infinite-dimensional H, admits no tracial state: because when it is infinite-dimensional, we can find pairwise orthogonal projections p,q with p\sim q\sim I=p+q. So there are partial ... 2 Certainly not, if any of the \phi_i is not a homomorphism. If \phi_k(ab)\ne\phi_k(a)\phi_k(b) for certain a,b\in A, then$$\bigoplus\phi_n(ab)\ne\left(\bigoplus\phi_n(a)\right)\left(\bigoplus\phi_n(b)\right),  as they differ in the $k^{\rm th}$ coordinate.

1

From $\|\phi'_n(1_A)^2-\phi'_n(1_A)\,\|<\varepsilon_n$ we deduce that all spectral elements of $\phi'_n(1_A)$ satisfy the equation $|\lambda^2-\lambda|<\varepsilon_n$. As we are talking positive operators here, we get that $\sigma(\phi'_n(1_A))\subset [0,\varepsilon_n)\cup(1-\varepsilon,1]$. There is not part above $1$ because $\phi'_n$ is completely ...

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