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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 ...


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.


1

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 ...


1

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 ...


1

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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