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The constant function 1 is continuous no matter what $X$ is. This is extremely easy to prove from the basic definition of "continuous". The problem is in the phrase "vanishes at infinity". If you study the definition of this term carefully, you should be able to prove that the constant function 1 vanishes at infinity if, and only if, $X$ is compact. ...

3

You are almost done: \begin{align} ((\lambda m+\mu n)(a))^*b&=\overline{\lambda}m(a)^*b+\overline{\mu}n(a)^*b=\overline{\lambda}a^*m^*(b)+ \overline{\mu}a^*n^*(b)=a^*(\bar\lambda m^*(b)+\bar\mu n^*(b))\\ &=a^*(\bar\lambda m^* +\bar\mu n^*)(b). \end{align} But now you know that $((\lambda m+\mu n)(a))^*b=a^*(\lambda m+\mu n)^*(b)$. If you look at the ...

3

I'm assuming that by $\sigma(\Sigma P)\leq1$ you mean that $\|\Sigma P\|\leq1$. Suppose that $\|P+Q\|\leq1$. So $0\leq P+Q\leq 1$. Then $(P+Q)^2\leq P+Q$ (just conjugate with $(P+Q)^{1/2}$). That is, $$P+Q+QP+PQ\leq P+Q,$$ or $QP+PQ\leq0$. If we conjugate this inequality with $Q$, we get $QPQ+QPQ\leq0$. But $QPQ\geq0$, so $QPQ=0$. Then $$... 3 Indeed this is known as Ogasawara's theorem, as you pointed out. I like to think of this in terms of representation theory, though this might be total overkill. Basically, I can prove the result in three steps: Prove that B(H) contains self-adjoint operators x,y satisfying 0 \leq x \leq y and x^2 \not\leq y^2 whenever \dim(H) > 1. You already ... 3 Let \phi be a state. Suppose that \phi(a_1)=0. Then, for some fixed j,$$ 0\leq\phi(a_ja_j^*)\leq\phi(a_1)=0, $$so \phi(a_ja_j^*)=0. Now, since \phi is completely positive, it satisfies the Kadison-Schwarz inequality; thus$$\tag{1} 0\leq\phi(a_j)\phi(a_j)^*\leq\phi(a_ja_j^*)=0, $$so \phi(a_j)\phi(a_j)^*=0, which implies \phi(a_j)=0. Since ... 3 The assertion P\leq Q means Q-P\geq0. Then$$ 0\leq P(Q-P)P=PQP-P\leq P^2-P=P-P=0. $$So P=PQP. Now we can write this equality as$$0=P-PQP=P(I-Q)P=[(I-Q)P]^*[(I-Q)P],$$so (I-Q)P=0, i.e. P=QP. Taking adjoints, P=PQ. The converse also holds: if P=PQ=QP , then$$Q-P=Q^2-QPQ=Q (I-P)Q\geq0. $$Edit: for the equivalence Q-P\geq0 iff Q-P is a ... 2 I will write f_k=r_1+\cdots+r_k, because otherwise we get M_{k+1}(\mathbb C). The projections r_j are each equivalent to 1_A, so there exist partial isometries e_{1,j},\ldots,e_{1,k} with$$ e_{1,j}^*e_{1,j}^\vphantom{*}=r_j,\ \ \ \ e_{1,j}^\vphantom{*}e_{1,j}^*=r_1. $$Now define$$ e_{k,j}^\vphantom{*}=e_{1,k}^*e_{1,j}^\vphantom{*}. $$This is ... 2 There may well be a simpler solution. (But see a certain comment below.) Since a C^* algebra is semi-simple it's enough to show that \phi(x^*)=\phi(x) for every complex homomorphism \phi of A. Since we're given that \phi(x) is real, it's enough to show$$\phi(x^*)=\overline{\phi(x)}.$$I was surprised to find a few years ago when I was working this ... 2 Since every element x of the closed unit ball B of a C^* algebra can be written as (1/2) \sum_{j=1}^4 u_j where u_1, \ldots, u_4 are unitary, if \mathcal U(\mathcal A) is compact (in a topology where \mathcal A is a topological vector space and B is closed) then so is B. A C^*-algebra \mathcal A is a W^*-algebra iff it can be ... 2 Represent \mathcal U as a sub-algebra of \mathcal B(H), the bounded linear operators on some Hilbert space H. As s is non-unitary, s(H)^\perp\neq \{0\} in H. Choose an index 1\leq j\leq n such that \lambda_j\geq \frac 1 n and a unit vector \xi\in H such that u_j(\xi)\in s(H)^\perp (this is possible since u is surjective). Then we have ... 2 The fact that A is weak^*-dense in A^{**} is basic functional analysis. I will be surprised if there is a functional analysis book that doesn't contain this result. For your second question, it is not true as you stated it: \pi cannot be any faithful representation but it is rather a very special one, the universal representation. The way it ... 1 Since p is infinite, there exists r\leq p with r\ne p and r equivalent to p. That is, there exists v with v^*v=r, vv^*=p. Note that v=pvr, so in particular (q-p)v=0, and v(q-p)=0. Now let s=r+q-p. Then s is a proper subprojection of q. Let w=v+q-p. Then$$ w^*w=(v^*+q-p)(v+q-p)=v^*v+q-p=r+q-p=s,  ww^*=vv^*+q-p=p+q-p=q. ...

1

You don't need the square, which gives you the additional step of justifying that $(1-A)^2\leq 1-A$. You can simply do, since $0\leq A\leq 1$ (because $P\leq A\leq Q$), $$0\leq P(1-A)P=P-PAP=0,$$ so $$0=P(1-A)P=[(1-A)^{1/2}P](1-A)^{1/2}P,$$ from where $(1-A)^{1/2}P=0$, and then $(1-A)P=0$.

1

The category of $C^*$-algebras is complete like Martin Brandenburg explained. However, given an inverse system of $C^*$-algebras, it is often more convenient to consider its limit in the category of all topological $*$-algebras. The limit of a diagram $(A_i,\lVert - \rVert)_{i \in I}$ of $C^*$-algebras is then the topological $*$-subalgebra of the ...

1

It's not so much that the closed unit ball of $B(H)$ is never compact in the strong operator topology, but it is not compact in general. More precisely, it is compact if and only if $H$ is finite dimensional. For convenience, let $S$ denote the closed unit ball of $B(H)$. If $H$ is finite dimensional, then $B(H)$ is a finite dimensional normed space, so ...

1

Here is an argument for the last part. Assuming we already know that $I=\overline{\bigcup_{B}B\cap I}$: If $I,J$ are ideals in $A$ and $IJ=0$, then for any $B\in S$ we have $0=B\cap IJ=(B\cap I)(B\cap J)$. As $B$ is prime, $B\cap I=0$ or $B\cap J=0$. Suppose $B_1\cap I\ne0$, and $B_2\cap J\ne0$, with $B_1\subset B_2$; then $B_2\cap I\supseteq B_1\cap ... 1 Our main problem is that, in principle, we cannot just "map the question" to the elements of$S$by taking intersections. To solve this, we can use an argument similar to the one on the proof of Theorem 6.2.6 of the same book (and I will even use the same notation). Given an ideal$I$of$A$, let$J=\overline{\left(\bigcup_{B\in S}B\cap I\right)}$. Then$J$... 1 Given an element$a$of$A$, we denote its real and imaginary parts$a_r$and$a_i$, respectively. Given a self-adjoint$a$, we denote it's positive and negative parts by$a_+$and$a_-$, respectively. Suppose$\lim_{t\to 0}\frac{\Vert 1+itx\Vert-1}{t}=0$. We will first show that$(x_i)_-=0$. As you noted, the real part of$1+itx$is$1-tx_i$. Given ... 1 Perhaps an easier approach is as follows: Let$R$be any unital commutative ring, then there is a one-to-one correspondence between ideals in$R$and ideals in$M_2(R)$. In fact, if$J \subset M_2(R)$is an ideal, then $$I = \{a \in R : \begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix} \in J\}$$ is the corresponding ideal in$R$such that$J = M_2(I)$. ... 1 Note that all of the work is in Exercise 8, which says that if$0\leq x\leq y$in a C*-algebra and$x$is invertible, then$y^{-1}\leq x^{-1}$. You will apply this with$x=1+\alpha a$and$y=1+\alpha b$, and the rest follows easily. For a solution to Exercise 8, see Inversion in a unital C* algebra or positive invertible operators or Inverse of ... 1 The function$1_X: X \to \Bbb C$is in fact the multiplicative identity in$C(X)$. If$X$is compact, then clearly the function$1_X: x \to \Bbb C$which takes every$x \in X$to$1 \in \Bbb C$, i.e.$1_X(x) = 1$,$\forall x \in X$, has compact support, i.e.,$1_X \in C_0(X)$. Now if$1_X \in C_0(X)$, then by a careful reading of the definition of$C_0(X)$... 1 Let$A$be any$C^*$-algebra and let$A^{\text{sa}}$denote the set of self-adjoint elements. Note that$A^{\text{sa}}$is only a real vector space, not a complex one: for$x\in A^{\text{sa}} \setminus \{0\}$and$\lambda\in\mathbb{C}$we have$\lambda x \in A^{\text{sa}}$if and only if$\lambda\in\mathbb{R}$holds. As you pointed out,$A^{\text{sa}}\$ is ...

1

Unilateral shifts (including weighted shifts) are never normal.

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