# Tag Info

5

The claim you are trying to prove is the following statement. Proposition. Let $H$ be a separable Hilbert space and let $(T_n)_{n=1}^\infty$ be a sequence of operators in $B(H)$ which converges to some $T$ in the weak operator topology, i.e., $$\langle T_nx, f\rangle \to \langle Tx, f\rangle$$ for all $x,f\in H$. Then there is a sequence ...

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

If you want your subalgebra to be unital, it is not true. For example, let $A=\mathbb C\oplus M_3(\mathbb C)$. Then the only proper unital subalgebras are $\mathbb C\oplus \mathbb C$ and---up to unitary equivalence---$\mathbb C\oplus M_2(\mathbb C)\oplus\mathbb C$ (note that $0\oplus M_2(\mathbb C)\oplus 0$ is a subalgebra, but not a unital one). If you ...

3

When $H$ is infinite-dimensional, the set of unitaries is weakly dense in the unit ball. So there exists a net of unitaries such that $u_n\to \frac12\,I$. As $|u_n|=I$ for all $n$, the net $|u_n|$ does not converge to $|\frac12\,I|=\frac12\,I$. For a proof of the density of the unitaries, let $x\in B(H)$ be a contraction. Given $\varepsilon>0$ and ...

3

Let $H=\ell^2(\mathbb{N})$ and let $\mathcal{A}$ be the set of all self-adjoint elements in the unit ball of $B(H)$. Define $S_n\colon H\to H, S_n \xi(k)=\xi(k+n)$. Then $\|S_n\|=1$, hence $A_n:=\frac 1 2(S_n+S_n^\ast)\in \mathcal{A}$. The adjoint of $S_n$ is given by $$S_n^\ast\xi(k)=\begin{cases}\xi(k-n)&\colon k\geq n\\0&\colon k<n\end{cases} ... 3 Define \phi:(B+I)/I\to B/B\cap I by$$\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, $$so b_1+B\cap I=b_2+B\cap I. The map is obviously linear, multiplicative, *-preserving, and onto. As for injectivity, if b_1+B\cap I=b_2+B\cap I, ... 3 Given f and \epsilon, choose a polynomial p with \Vert f-p\Vert_{\infty,X}<\epsilon (where \Vert\cdot\Vert_{\infty,X} is the supremum norm oin X). Now see the corresponding polynomial function in \mathcal{A}, p:\mathcal{A}\to\mathcal{A}. (Remember: the functional calculus respects this notation, i.e., p(a), in the functional calculus, is ... 2 The key is that, for any injective representations (i.e., *-homormophisms, so in particular f and g),$$ \|\sum a_j\otimes b_j\|_\min=\|(f\otimes g)\left(\sum a_j\otimes b_j\right)\| $$(technically, this might require using a further set of faithful representations to get embeddings A\hookrightarrow B(H), B\hookrightarrow B(K), but it doesn't ... 2 The arguments I know require that \Omega is both cyclic and separating (i.e., also cyclic for the commutant). When you only require that \Omega is also separating, I don't think that S_0 even makes sense. For instance, let \mathcal M=B(\mathcal H), \Omega=e_1. Consider the (constant) sequence E_{12}e_1=0. Then S_0E_{12}e_1=E_{21}e_1=e_2\ne0, ... 1 Outlining answer of TAE: (Thank you for the nice hint!) Bounded Functions:$$\mathcal{B}(\mathbb{R}):=\{f:\mathbb{R}\to\mathbb{C}:\|f\|_\infty<\infty\}$$Unital Subalgebra:$$\mathcal{A}_0:=\{f\in\mathcal{B}(\mathbb{R}):f(0)=\lim_{\varepsilon\to0}f(\varepsilon)\}$$Ordered Family: ... 1 Its not true, consider the algebra of bounded operators on \mathscr l^2(\mathbb N). Let P_i be the orthogonal projection onto the span of e_i. Let A_n=\sum_{i=1}^n P_i, then A_{n+1}≥A_n and \mathbb 1 ≥ A_n \forall n. So the conditions are satisfied. But A_{n+1}-A_n=P_{n+1} has norm 1, so this is not a Cauchy-Sequence and not convergent. ... 1 Your reasoning in point 3 works because you are using an isomorphism induced by that spatial isomorphism \ell^2(\mathbb Z)\simeq L^2(\mathbb T), so masas go back and forth. In 4 and 5, the isomorphisms you are considering are not spatial (in terms of the original environment where you had N). For instance, let$$ ...

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

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 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

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

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

Let $A_n=\left(1+\frac1n\right)\,I$. Then $A_n\to I$ in norm (and so, also strongly). We have $$E^{|A_n|}(1,\infty)=I,\ \ \ \ \ \ \ \ E^{|A|}(1,\infty)=0.$$

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

Martin gave a beautiful answer to your question but let me take this opportunity to advertise a solution to a more general question of which C*-algebras may be written as tensor products of two infinite-dimensional C*-algebras. In particular, it is proved that $B(\ell_2)$ as well as the Calkin algebra cannot be decomposed like that no matter which tensor ...

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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