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3

The sequence converges in the strong topology of a Hilbert space, but not in the uniform topology. You can start with continuous functions on $[0,1]$ and find a sequence of real continuous functions that converges up to $\chi_{(1/2,1]}$, while all of the functions remain uniformly bounded by $1$.


3

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 ...


3

Define $\phi:(B+I)/I\to B/B\cap I$ by $$\phi(b+j+I)=b+B\cap I,\ \ \ \ \ b\in B,\ j\in I.$$ Of course we need to check that this is well-defined. If $b_1+j_1=b_2+j_2$, then $$ b_1-b_2=j_2-j_1\in B\cap I, $$ so $b_1+B\cap I=b_2+B\cap I$. The map is obviously linear, multiplicative, $*$-preserving, and onto. As for injectivity, if $b_1+B\cap I=b_2+B\cap I$, ...


2

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 ...


2

The key is that, for any injective representations (i.e., $*$-homormophisms, so in particular $f$ and $g$), $$ \|\sum a_j\otimes b_j\|_\min=\|(f\otimes g)\left(\sum a_j\otimes b_j\right)\| $$ (technically, this might require using a further set of faithful representations to get embeddings $A\hookrightarrow B(H)$, $B\hookrightarrow B(K)$, but it doesn't ...


2

No, and by a long shot. Let $A=B=\mathbb C $, and $\varphi (z)=2z $. Or, let $A=C [0,1] $, $B=\mathbb C $, and $\varphi (f)=\int_0^1 f $. Or, $A=M_n (\mathbb C) $, $B=\mathbb C $, and $\varphi (x)=\text {Tr}\, (x) $. Or, $A=M_n (\mathbb C) $, $B=M_n (\mathbb C )$, and $\varphi (x)=cxc^*$ with $c $ not a unitary. There are countless examples. And a ...


2

You cannot use that technique to construct a C$^*$-algebra. The spectral radius does not give a norm in an arbitrary $*$-algebra. For an easy example, consider $\mathbb C[x]$ with complex conjugation as an involution. The only invertible polynomials are the constants, so for any polynomial $p$ of degree at least 1 $$ \sigma(p^*p)=\mathbb C. $$ The ...


1

I'm assuming that the statement in your question is that "for every $b\in B^+$ there exists a $\varphi\in Y$ such that $\|b\|=\varphi(b)$". The positivity condition is necessary, because otherwise it may be impossible to have $\varphi(b)$ positive. We can assume that $B$ is unital, since the state space of the unitization agrees with the state space of $B$. ...


1

Its not true, consider the algebra of bounded operators on $\mathscr l^2(\mathbb N)$. Let $P_i$ be the orthogonal projection onto the span of $e_i$. Let $A_n=\sum_{i=1}^n P_i$, then $A_{n+1}≥A_n$ and $\mathbb 1 ≥ A_n$ $\forall n$. So the conditions are satisfied. But $A_{n+1}-A_n=P_{n+1}$ has norm $1$, so this is not a Cauchy-Sequence and not convergent. ...


1

Your reasoning in point 3 works because you are using an isomorphism induced by that spatial isomorphism $\ell^2(\mathbb Z)\simeq L^2(\mathbb T)$, so masas go back and forth. In 4 and 5, the isomorphisms you are considering are not spatial (in terms of the original environment where you had $N$). For instance, let $$ ...


1

You have already done all the necessary work: $$ u=\begin{bmatrix}\begin{matrix}u_1&&&\\&\ddots&&\\&&u_j&\\ &&&I_d\end{matrix} \end{bmatrix}. $$


1

Let $A=\left\{\begin{pmatrix} a&0\\0&b \end{pmatrix}\mid a,b\in \mathbb{C}\right\}$ and $B=\left\{\begin{pmatrix}a&0&0\\0&b&0\\0&0&c\end{pmatrix}\mid a,b,c\in \mathbb{C}\right\}.$ Then $A$ and $B$ are two unital $*$-algebras in the usual way. Consider the map $$\phi:A\rightarrow B:\begin{pmatrix} a&0\\0&b ...


1

I think I have a solution to my answer: First, assume $ind(A)=ind(B)=0$. Then there exist $K_1,K_2 \in K(E)$ and $I_1,I_2$ invertible operators in $B(E)$ such that $A=K_1+I_1$ and $B=K_2+I_2$. Define $C=I_2^{-1}I_1$, then $C$ is invertible and $A-BC=K_1+I_1-(K_2+I_2)I_2^{-1}I_1=K_1-K_2I_2^{-1}I_1 \in K(E)$, as desired. Now, for the general case: Assume ...


1

The notation in the C$^*$ literature for tensor products is far from uniform. But yes, as $K(H)$ is nuclear, it doesn't matter which C$^*$-norm you put on $K(H)\otimes A$, they are all the same.


1

The easy way is to notice that maps $a\longmapsto xax^*$ are always cp when they make sense (even if they are matrices and $x$ is rectangular). Then, with $x_j=\begin{bmatrix}0&\cdots0&&1&0\cdots&0\end{bmatrix}$ (the $1$ in the $j^{\rm th}$ position), $$ \text{Tr}(a)=\sum_{j=1}^nx_jax_j^* $$ is cp. The hard way is the following. ...


1

The proof of the converse is simple: if $\sigma(T)=\sigma(S)$, then $C(\sigma(T))=C(\sigma(S))$. Then $G_2\circ G_1^{-1}$ is the $*$-isomorphism you are looking for. Now, it is not true that the statement you gave implies unitary equivalence; you cannot ignore multiplicity. Consider $$ ...


1

It seems as if you confused what to assume and what to prove in the second part. If you want to show the existence of $\pi$, you cannot include it in the definition of $H$. But if you get your arguments sorted out, it's quite obvious. Let $\pi=G_2^{-1}\circ G_1^{-1}$ and $f\colon \sigma(S)=\sigma(T)\to \mathbb{C},\,z\mapsto z$. By definition, $G_1(f)=S$ and ...


1

In norm, it is clear that no. All II$_1$-factors contain a copy of $L^\infty[0,1]$, where the trace corresponds with integration against Lebesgue measure. You can take $x_n=\sqrt2\cdot 1_{[1/n,1/n+1/2]}$. Then $\|x_n\|=\sqrt2$, $\text{Tr}\,(x_n^2)=1$ for all $n$, and $\|x_n-x_m\|=1$ for all $n\ne m$.


1

Definitely no. There is no infinite-dimensional context where you could make that kind of deduction. If that property were true, the unit ball of $C_r^*(S_\infty)$ would be SOT-closed, which would make $C_r^*(S_\infty)$ a von Neumann algebra; and this is impossible, since $C_r^*(S_\infty)$ is separable, while any infinite-dimensional von Neumann algebra is ...


1

Tensor products of c.p. faithful maps with respect to the minimal norm are faithful: D. Avitzour, Free products of C*-algebras, Trans. Amer. Math. Soc. 271 (1982), 423–435. Certainly, *-homomorphisms are c.p.


1

There is something off with that argument as it is written. Nothing prevents, for instance, that $x-\Phi_i(x_i')\leq0$ and nonzero. In that case, $$f(x-\Phi_i(x_i'))=0,$$ and then $$ 0<\|x-\Phi_i(x_i)\|,\ \ \ \ \|f(x-\Phi_i(x_i'))\|=0. $$ This is how I think that argument can be saved. If $x\geq0$, $y=y^*$, and $\|x-y\|<\varepsilon$, then ...


1

Consider $M_2(\mathbb C)$. As vectorspace it is isomorphic to $\mathbb C ^4$. But $K_0(M_2(\mathbb C)) = K_0(\mathbb C) = \mathbb Z \neq \mathbb Z^4 = K_0(\mathbb C)^4.$


1

You are right: When taking a direct limit like $$A_1\overset{T_1}{\longrightarrow}A_2\overset{T_2}{\longrightarrow}\cdots,$$ the first elements of your direct sequence do not matter. In fact, you can take any subsequence $(n_k)$ (and compose the maps appropriatelly to get a directed sequence $A_{n_1}\to A_{n_2}\to\cdots$) and you still get the same direct ...


1

It's true that any finite number of maps at the beginning don't affect the direct limit, so you may ignore the first two. Then the direct limit is the stabilization of the image of your $3\times 3$-matrix. Explicitly, if you call $A$ this matrix, then the image of $A^n$ stabilizes for $n$ big enough, and the limit is this image. Now just looking at its ...


1

Your isometries are not such. Consider $$ \rho_1=\begin{bmatrix}1/2&0\\0&1/2\end{bmatrix},\ \ \rho_2=\begin{bmatrix}1&0\\0&0\end{bmatrix}. $$ Then $$\|\rho_1\|=1/2,\ \ \|\rho_2\|=1,$$ while $$ \|\rho_1\|_1=\|\rho_2\|_1=1.\ \ $$ For the proof, if $\rho=\sum_k\lambda_kP_k$ with $\sum_k\lambda_k=1$ and $0\leq\lambda_k$ for all $k$ and at ...


1

Let $\mathbb T$ be the circle and $\mathbb D$ the disc. The canonical unitary in $C(\mathbb T)$ does not lift to a partial isometry in $C(\mathbb D)$: such a lift would automatically be unitary, violating Brouwer's fixed-point theorem. The following is true, however (see Lemma 9.2.1 in Rordam-Larsen-Laustsen): Let $A$ be unital and $u\in A/I$ a unitary. ...



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