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Given $f$ and $\epsilon$, choose a polynomial $p$ with $\Vert f-p\Vert_{\infty,X}<\epsilon$ (where $\Vert\cdot\Vert_{\infty,X}$ is the supremum norm oin $X$). Now see the corresponding polynomial function in $\mathcal{A}$, $p:\mathcal{A}\to\mathcal{A}$. (Remember: the functional calculus respects this notation, i.e., $p(a)$, in the functional calculus, is ...

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You can write the similarity as $NS=SM$. As $N$ and $M$ are normal, the Fuglede-Putnam theorem guarantees that $N^*S=SM^*$. Taking adjoints, $S^*N=MS^*$. Then $$S^*SM=S^*NS=MS^*S.$$ Using this identity repeteadly, $p (S^*S)M=Mp (S^*S )$ for all polynomials; taking limits, $f (S^*S)M=Mf (S^*S)$ for all continuous functions $f$. In particular, if ...

1

They are not equivalent on an infinite-dimensional Hilbert space. The weak convergence for operators in this case is in the usually called "weak operator topology": $$A_n\xrightarrow{wot} A\ \ \iff\ \ \langle A_nx,y\rangle\to\langle Ax,y\rangle,\ \ \forall x,y\in H.$$ The weak operator topology is known to be coarser than the $\sigma$-weak operator ...

2

I don't know why you say that $f(\sigma(x))=F_1$. A point in $\sigma(x)$ is either in $F_1$ or in $F_2$, and so $f(x)$ is either $0$ or $1$; and then $f(\sigma(x))=\{0,1\}$.

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The inclusion $i$ is surjective, because both $B(\mathbb C^2)\otimes B(\mathbb C^2)$ and $B(\mathbb C^4)$ have the same dimension (concretely, 16). The issue, and what you showed, is that if you restrict $i$ to the subset of elementary tensors, then $i$ is not surjective. Your element $p$ is not a sot limit of elementary tensors. For starters, because we ...

2

Your two ideas would have been my first attempts; but I have no idea how to make them work (well, for the first one, I would try with $T$ the flip, but I still wouldn't know how to do it). Let $K\subset B(\ell^2)$ be the compact operators. On $\overline {B(\ell^2)\odot B(\ell^2)}$, consider the ideals $\overline{K\odot B(\ell^2)}$ and $\overline{K\odot ... 1 The inequality does not hold in general. Let $$T=\begin{bmatrix}1&1\\0&0\end{bmatrix},\ \ Q_1=\begin{bmatrix}0&0\\0&1\end{bmatrix}, \ \ Q_2=\begin{bmatrix}1&0\\0&0\end{bmatrix}.$$ Note that $$... 1 I think you make it unncessarily complicated. By the minimality of E, you have P_{\lambda_i}=P_{\lambda_j} for all i,j. But each \lambda_i is the particular projection corresponding to \lambda_i; so \lambda_i=\lambda_j for all i,j (projections corresponding to different eigenvalues are orthogonal to each other). Thus T=\lambda_1\,E. 2 This is not true. Let$$ T=\begin{bmatrix}1&2\\2&4\end{bmatrix},\ \ P=\begin{bmatrix}1&0\\0&0\end{bmatrix}. $$Then T is positive (selfadjoint, with eigenvalues 0 and 5), but$$ T-PTP=\begin{bmatrix}0&2\\2&4\end{bmatrix} $$is not positive (selfadjoint with eigenvalues 2\pm2\sqrt2). 0 Take H=\mathbb{C}^2, T=\begin{pmatrix}-1&-1\\6&4\\ \end{pmatrix}, P=\begin{pmatrix}0&0\\0&1\\ \end{pmatrix}. Then T is positive with eigenvalues 1,2, but T-PTP=\begin{pmatrix}-1&-1\\6&0\\ \end{pmatrix} is not positive. 0 I suggest you take a look at Arveson's "An Invitation to C^*-algebras", Chapter 2; in particular, section 2.2.3. 0 For the start make note:$$ \mathbb{R} \ni \phi((a+1)(a+1)^*) = \phi(aa^*)+\phi(a)+\phi(a^*)+ 1. $$Next on, since \phi(aa^*) \in {\mathbb R}, we must have \phi(a)+\phi(a^*) \in \mathbb{R}. Hence, \Im(\phi(a)) = - \Im(\phi(a^*)). Do the same trick for ia:$$ \mathbb{R} \ni \phi((ia+1)(ia+1)^*) = \phi(aa^*)+i\phi(a) - i\phi(a^*)+ 1 \\ ... 2 Note first that we may assume that all$T_k$are proper isometries; because if one of them, say$T_1$, is a unitary, we get$\sum_{k=2}^NT_kT_k^*=0$, which forces$T_k=0$for all$k\geq2$. Since$\sigma(T_k^*T_k)=\sigma(I)=\{1\}$, using that$\sigma(AB)\cup\{0\}=\sigma(BA)\cup\{0\}$we deduce that$\sigma(T_kT_k^*)=\{0,1\}$(the zero has to be there ... 3 No. For instance, you can take the representation$\pi\oplus\pi$on$H\oplus H$, and then$(\pi\oplus\pi)(x)$will have twice the index of$\pi(x)$. 3 Let$A \subseteq B(H)$be any AW*-algebra that is not a von Neumann algebra. (Actually, we don't need a full AW*-algebra, see below.) Let$t \in A$be any operator. By definition of AW*-algebra, every right-annihilator is generated by a projection. In particular, the right-annihilator of the singleton set$\{t\}$is generated by a projection$q \in A$. ... 2 Here's one possible take on the question:$K$-theory for$C^*$-algebras is motivated by topological$K$-theory. Topological$K_0$for a compact Hausdorff space$X$is the Grothendieck group of formal differences of isomorphism classes of locally trivial vector bundles over$X$, so we want the operator$K_0$-group of$C(X)$to match this. The Serre-Swan ... 1 I post here the answer I received for the same question in mathoverflow: Here I list some facts that may be useful for building your intuition: 1. Two commutative Morita equivalent$C^*$-algebra are in fact$*$-isomorphic. 2 If$A$is$C^*$-algebra and you take$B=M_n(A)$then$A$and$B$are Morita equivalent. 3 Many invariants for$C^*$-algebras such as ... 2 You can write$(\Bbb C^n, \lVert\cdot\rVert_p)=L^p(\Bbb Z/n\Bbb Z) = L^p(G)$, the inner group is compact, hence when$p\ge 1$you have a Banach space structure and the usual convolution on functions over a compact group give you an algebra structure, given as $$(f\star g)(x)=\int_G f(y)g(x-y)\,dy.$$ Here "-" means the subtraction in the group structure ... 3 If you have a linear map$F:V\to V((x))$, then the coefficient of$x^i$in$F(v)$depends linearly on$v$. Hence you can write$F(v)=\sum_{i\in\mathbb Z} f_i(v)x^i$where each$f_i:V\to V$is linear. (In addition, you know that for each$v\in V$, there is some$i_0$such that$f_i(v)=0$for$i<i_0$, but this does not play a role here.) Now you simply ... 2 Question (1) seems to be missing from your post. For the identification (2), the unitary$U : \ell^2(\Gamma \times \Gamma) \to \ell^2(\Gamma) \otimes \ell^2(\Gamma)$defined by$\delta_{(g,h)} \mapsto \delta_g \otimes \delta_h$intertwines$\lambda'$and$\lambda \otimes \lambda$. Let$F : H \otimes \ell^2(\Gamma) \to \ell^2(\Gamma) \otimes H$be the flip ... 1 If$\|\varphi(A)\|\leq c\,\|A\|$when$A$is selfadjoint, then for arbitrary$A$you have $$\|\varphi(A)\|=\|\varphi(\text{Re}\,A)+i\varphi(\text{Im}\,A)\| \leq\|\varphi(\text{Re}\,A)\|+\|\varphi(\text{Im}\,A)\|\\ \leq c\,(\|\text{Re}\,A\|+\|\text{Im}\,A)\|\leq 2c\|A\|,$$ since $$\|\text{Re}A\|=\frac12\,\|A+A^*\|\leq\frac12\,(\|A\|+\|A^*\|)=\|A\|.$$ 0 It was not me who posted in the comments (that got deleted), but let me answer your questions in the comments. Hatcher's vector bundles and$K$theory discusses clutching functions. Complex vector bundles over a torus minus a point correspond to complex vector bundles over the wedge of two circles. Such vector bundles are the same thing as two vector ... 3 There are many specific operations for matrices, usually given as a bilinear product from$M_n(K)\times M_n(K)\rightarrow M_n(K)$, like the matrix product or the Lie bracket$(X,Y)\mapsto [X,Y]=XY-YX$. Another operation is the Kronecker product from$M_{m,n}(K)\otimes M_{p,q}(K)\rightarrow M_{mp,nq}(K)\$.

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