# Tag Info

0

If $\varphi(a)\geq0$ for all states $\varphi$, you can do the following: Note that a state is selfadjoint, i.e. it maps selfadjoints to real numbers. This, because any selfadjoint is a difference of two positives. Write $a=b+ic$ with $b,c$ selfadjoint. Then $\varphi(b)$ and $\varphi(c)$ are real. So $\varphi(a)=\varphi(b)+i\varphi(c)$ is positive, which ...

0

This is not true. For example you could take $A$ unital, $a=\frac12\,1$. Then $$(1-u_n)^2=\left(1-\frac 1{\frac12+\frac1n}\right)^2\to 1.$$ So $\varphi(b(1-u_n)^2b^*)\to\varphi(bb^*)$ for any $b$.

1

It is not true that a linear functional $f\in A^*$ is positive if it is bounded and $f(I)\geq0$. For instance, let $A=M_2(\mathbb C)$ and define $$f\left(\begin{bmatrix}x&y\\ z&w\end{bmatrix}\right)=x-w/2.$$ The $f(I)=1/2\geq0$, but $f(E_{22})=-1/2$. What is true is that a linear functional is positive if and only if $f(I)=\|f\|$. To extend ...

0

Asyou remarked, one direction is trivial from definitions. For the other direction you can use the following well known facts: 1) An element is positive if and only if its spectrum is contained in $[0,\infty )$ 2) If $\lambda$ belongs to spectrum of $a$ then there exists a state $f$ such that $f(a)=\lambda$.

3

The basic idea is that $\varphi(f)$ behaves like evaluation of $f$ at $a$ for any continuous function. The immediate application is that we can "evaluate" well-known functions with elements of a $C^{*}$-algebra, e.g. $f(x) = e^x$, logarithms, square roots, etc. They have analogous uses that these functions have when dealing with numbers. For example, if we ...

0

The spectrum is invariant under unitary transformations, thus it is rotational invariant: $$\sigma(W(f\neq0))=\sigma(W(g)W(f\neq0)W(g)^*)=e^{-i\sigma(g,f\neq0)}\sigma(W(f\neq0))$$ Note that this excludes the special case: $$\sigma(W(0))=\sigma(1)=\{1\}$$

0

If $\phi(a)=0$ for some state $\phi$, then $\phi(u_n)=0$ for all $n$, because $0\leq\phi(u_n)\leq2^n \phi(a)$. Now, for any $b\in A_+$, $$\phi(b)=\lim_n \phi(u_nbu_n)=\lim_n\phi(u_n^{1/2}[u_n^{1/2}bu_n^{1/2}]u_n^{1/2})\leq\limsup_n\|u_n^{1/2}bu_n^{1/2}\|\,\phi(u_n) \\ \leq\|b\|\limsup_n\phi(u_n)=0.$$ So $\phi(b)=0$ for all $b\geq0$, and thus $\phi=0$ as ...

1

Take any $B$. Then $B+\mathbb C\,1_A$ is a C$^*$-subalgebra of $A$ that contains $1_A$, so $\sigma_{B+\mathbb C\,1_A}(b)=\sigma_A(b)$. Now by page 44, applied to the inclusion $B\subset B+\mathbb C\,1_A$, you have $\sigma_B(b)=\sigma_{B+\mathbb C\,1_A}(b)$, and you are done.

4

For your first question, I think you are misunderstanding what the theorems say. When you restrict your $\sigma$-weakly continuous functional to the unit ball, you don't get a wot functional on the whole space: so 4.6.4 does not apply. For your second question, here is an example: fix an orthonormal basis $\{e_n\}$ and let $$... 2 I'm writing this as an answer to have a little more space to write. What you want to prove is not true: for a *-homomorphism to be necessarily contractive, you need the domain to be a C^*-algebra. For instance, let \mathcal A=C[0,1], \mathcal B=\mathbb C, \mathcal D=\{\text{polynomials}\}, and \pi(p)=p(2). Then \pi is clearly a ... 0 Remark As shown in the answer above the domain must be all. Thanks alot, Martin Argerami!! :) Proof Suppose the domain is all \mathcal{D}=\mathcal{A}. Choose a C*-subalgebra with identity:$$\mathrm{im}\pi\subseteq\mathcal{C}\subseteq\mathcal{B}:\quad1_\mathcal{C}=\pi[1_\mathcal{A}]$$A minimal candidate is the closure of the image: ... 0 Here is an example of a net that is weakly convergent, but not ultraweakly convergent. I will use the fact that X_j\to0 ultraweakly precisely when \text{Tr}(AX_j)\to0 for all trace-class operators A. Fix A to be an injective trace-class operator, so that AP\ne0 for all projections P. For example, you could take A=\sum_k\frac1{k^2}\,\langle\, ... 1 For 1) \implies 2), note that for an u\in C_0(X) and any \varepsilon > 0, the set \{x\in X : u(x) \geqslant \varepsilon \} is compact. Let K_n = \left\{x \in X : u_n(x) \geqslant \tfrac{1}{n+2}\right\}. Since u_n\in C_0(X), K_n is compact. Furthermore, we have K_n \subset \overset{\Large\circ}{K}_{n+1} since (u_n) is monotonically ... 0 I will assume that X is locally compact, because otherwise it is hard to determine what "vanishing at infinity" could mean. There is also a problem with the meaning of C_0(U). Because "vanishing at infinity" is defined in terms of compact sets, and compact subsets of U is the relative As you say, define$$ g(x)=\begin{cases}f(x),&\ x\in U\\ ...

2

Let $\{E_{kj}\}$ be the canonical matrix units and $P_n=\sum_1^nE_{kk}$. Take $S$ the unilateral shift, $S=\sum_kE_{k+1,k}$, and $T_1=S^*$, $T_2=S$. Then $$P_nT_1T_2P_n=P_nS^*SP_n=P_n,$$ while $$P_nS^*P_nSP_n=P_{n-1}.$$ Thus $$\|P_nT_1T_2P_n-P_nT_1P_nT_2P_n\|=\|P_n-P_{n-1}\|=1$$ for all $n$.

1

I try to answer my own question. And I think this question is not hard to answer if we notice the fact that $\pi=\pi_{A}\times\pi_{B}$, where $\pi_{A}: A\rightarrow M_{n}(\mathbb{C})$ and $\pi_{B}: B\rightarrow M_{n}(\mathbb{C})$ are two $*$-representations with commuting ranges. Now, from universality, we have $$A\otimes_{max} B \xrightarrow{} ... 3 Consider the operator T:\mathcal H\oplus \mathcal H\to\mathcal H defined by$$T(h_1\oplus h_2):= T_1(h_1)+T_2(h_2)\, . $$Then (1) says exactly that \Vert T\Vert\leq 1. Since \Vert T\Vert=\Vert T^*\Vert, this is equivalent to the condition \Vert T^*\Vert\leq 1, which is again equivalent to$$TT^*\leq I\, .$$Now, you will easily verify that ... 3 These conditions are not equivalent. Take a non-zero operator x. Then certainly x\neq -x. However,$$|(-xh, k)| = |-(xh,h)| = |(xh, k)|$$for any h,k\in H. Separation in the context of locally convex spaces means the following: Let \mathscr{P} be a family of seminorms. Then \mathscr{P} is (by definition) separating if for each x\neq 0 ... 4 The first one is commutative, so... 0 You can bound \|T - T_n\| using the uniform boundedness principle. Fix x \in H and let f_n(y) = \langle T_n x, y \rangle. Clearly f_n is a bounded linear functional on H. Moreover, for each y, since f_n(y) \to \langle Tx, y\rangle by WOT convergence, in particular |f_n(y)| is bounded. Then the uniform boundedness principle asserts that ... 0 Key Case For unbounded normal operators one has:$$\langle\sigma(N)\rangle=\overline{\mathcal{W}(N)}$$(For rigorous treatments see numerical range for Normal Operators and Spectral Measures.) So the notions of positivity agree! Other Cases For bounded operators one has at least:$$\langle\sigma(T)\rangle\subseteq\overline{\mathcal{W}(T)}$$but also one ... 3 Below is one avenue. Since a=b^*b, we have a^*=a. Then C^*(a), the C^*-subalgebra generated by a, is commutative. In a commutative Banach algebra, we have$$ \sigma(a)=\{\tau(a):\ \tau \text{ is a character }\}. $$So, when a=b^*b, \tau(a)=\tau(b^*)\tau(b)=\overline{\tau(b)}\tau(b)=|\tau(b)|^2\geq0 for any character \tau. Thus, ... 2 Only anti-symmetry seems to be non-trivial here. Let us apply the spectral theorem. Suppose that b\leqslant a and a\leqslant b. Since a-b is positive, C^*(a-b) is commutative and of course b-a\in C^*(a-b). But now you can think of a-b and b-a as continuous functions on \sigma(a-b) which are both positive (non-negative, strictly speaking). ... 3 Assume (b-\lambda 1_B)c=1_B. When we use 1_A=(0,1),$$ [(b, 0) - \lambda (0,1)](x,t)=(b,-\lambda)(x,t)=\left(bx-\lambda x+tb, -\lambda t\right). $$Looking at the second coordinate, t=-1/\lambda. Then the equality in the first coordinate becomes$$ 0=bx-\lambda x -\frac1\lambda b=(b-\lambda 1_B)x-\frac1\lambda b,  So $x=\frac1\lambda ... 4 I’m assuming that$\varphi_x$is evaluation at$x$. Instead of just giving the argument or a hint, I’ll show how I actually worked my way to the argument. Let$U$be a non-empty open subset of$X$, and let$\Phi=\{\varphi_x:x\in U\}$; we want to show that$\Phi$is open. You already know that the map is bijective, so any point of$\Phi$is$\varphi_x$for ... 0 What you did is correct,$||a^2 - a || < \frac{1}{4}$means$\sigma(a^2 - a) \subset (-1,1)$, that is the function$t^2 - t$takes$\sigma(a)$to the subset$(-1,1)$, and so,$\sigma(a) \subset (\frac{1-\sqrt{2}}{2},\frac{1}{2}) \cup (\frac{1}{2} , \frac{1+\sqrt{2}}{2}) \subset (-\frac{1}{2},\frac{1}{2})\cup (\frac{1}{2}, \frac{3}{2})$. Define$f\colon ...

1

Certainly not in that generality. For instance, $C$ could have no finite-rank projections at all; it could be even projectionless. Even in cases where $C$ has finite-rank projections, what you want can fail, as the finite-rank projections may exists only in a direct summand.

0

If $p$ is normal and $\sigma(p)\subseteq \{ 0, 1\}$, then the spectral mapping theory gives $\sigma(p(p-1))=\{0\}$. Because $p$ is normal, then the spectral radius and the norm are the same, which gives $p(p-1)=0$ or $p^{2}=p$. A normal operator with spectrum $\sigma(p)\subset\mathbb{R}$ must be selfadjoint. Okay, the opposite direction. If ...

1

It is not true even for $T = I$. For example if $H = \ell^2$, and $P_n$ are the projections to the first $n$ entries, then $P_n$ converges to $I$ in strong operator topology but not in norm topology. Actually, your assertion is true only if $H$ is finite dimensional. From $P_n TP_n \to T$ in norm topology and plug in $T=I$, then $P_n$ converges to $I$ in ...

1

You have to look at Theorem 1.2.1, which is just a re-statement of I.9.16 in Davidson's book. Davidson's proof is a page long, not counting auxiliary results. Davidson's proof doesn't state it explicitly, but the condition $e_{j+1}e_j=e_j$ can be achieved by going far enough along the index set when defining $\mathcal F$ in Davidson's proof.

2

Since the rank of each $\phi_n$ is finite-dimensional, finite direct sums like $\bigoplus_1^m\phi_n$ are compact (actually, they are finite-rank. So, in the quotient, $\bigoplus_1^\infty\phi_n(a)$ and $\bigoplus_m^\infty\phi_n$ are equal, and so \begin{align} ...

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