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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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