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5

By using the polar decomposition, we can write $T=V|T|$. So $|T|=V^*T\in J$, and then $J$ contains a positive non-compact operator. On a side note, this argument also shows that $J$ contains all adjoints of its operators, since now $T^*=|T|V^*\in J$. So from now on we assume $T\geq0$, non-compact, $T\in J$. This means that there is $\lambda>0$ with ...

5

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 $$... 5 You can do this in any II_1-factor. Note that Q_j is a II_1-factor. In any II_1-factor M you can always get a sequence of pairwise orthogonal projections that add to the identity (those could be the q_j in your setup). So now you want to embed M_n(\mathbb C)\hookrightarrow q_jMq_j. Since you are in a II_1-factor, you can divide q_j as a ... 3 The basic idea is that \varphi(f) behaves like evaluation of f at a for any continuous function. The immediate application is that we can "evaluate" well-known functions with elements of a C^{*}-algebra, e.g. f(x) = e^x, logarithms, square roots, etc. They have analogous uses that these functions have when dealing with numbers. For example, if we ... 2 The situation is the following: you have U=\lambda I+T, with T compact. And \{P_M\} is a sequence of finite-rank projections such that P_M\nearrow I. So, given \varepsilon>0, by the compactness of T you can write T=P_MTP_M+T_0, with \|T_0\|<\varepsilon. Then$$ \|P_MU-UP_M\|=\|P_MT_0-T_0P_M\|<2\varepsilon. $$For your second ... 2 No. It wouldn't make sense if it were independent of the algebra. The properties you mention depend on the relation of the identity with the rest of the projections. For example, M_2(\mathbb C) is finite because I is not equivalent to any proper projection (because equivalence of projections is given by rank). On the other hand, on ... 2 The normal functionals of B(H) can be identified with the elements of the predual of B(H), which are the trace-class operators T(H), via the duality$$ T(H)\ni X\longmapsto \text{Tr}(X\ \cdot). $$So the question is why T(H) is weakly dense in B(H). There are several ways of proving this. The easiest is to notice that T(H) contains all ... 2 I'm writing this as an answer to have a little more space to write. What you want to prove is not true: for a *-homomorphism to be necessarily contractive, you need the domain to be a C^*-algebra. For instance, let \mathcal A=C[0,1], \mathcal B=\mathbb C, \mathcal D=\{\text{polynomials}\}, and \pi(p)=p(2). Then \pi is clearly a ... 2 It is not true that a linear functional f\in A^* is positive if it is bounded and f(I)\geq0. For instance, let A=M_2(\mathbb C) and define$$ f\left(\begin{bmatrix}x&y\\ z&w\end{bmatrix}\right)=x-w/2. $$The f(I)=1/2\geq0, but f(E_{22})=-1/2. What is true is that a linear functional is positive if and only if f(I)=\|f\|. To extend ... 2 Yes. The tensor product of separable algebras is separable. You can construct a dense subset of the tensor product by taking the algebraic tensor of two countable dense subsets of each algebra. 1 It is true for the min-norm, because you can construct the minimal tensor product explicitly by using two faithful representations. Namely, if you fix two faithful representations \pi:A\to B(H) and \mu:B\to B(K), then you have$$ A\otimes_\min B=\overline{\pi(A)\otimes\mu(B)}\subset B(H\otimes K).  Now you can use the restrictions $\pi|_{A_1}$ and ...

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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.

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If I do understand your question correctly, you want a representation $\pi$ of $C_0(X)$ on some Hilbert space, and you want $\pi$ to be injective. As injectivity in this sense is equivalent to "isometric", I'll give you three constructions of isometric representations of $C^*$-algebras that I know of. Construction 1 This is a universal construction, known ...

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I don't think you can get too far with your approach, because you want to deal with the set of all measures on $X$, and there is nothing explicit about it. So you want to use fewer states. 1) Following on what Phoenix87 said, here is an example of a faithful representation (denomination way more common in the literature than "injective"). It is based on ...

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Reposting my comment: in order for the map you describe to be defined it seems like you at least need $D(A)$ to contain $A$ (I don't see the point of the fraktur here; most of the time it just makes things harder to read). I don't see why this needs to hold if $A$ is noncommutative.

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