# Tag Info

1

That's the easy part. The topology you need to consider on $\Phi_{C(K)}$ is the weak* topology; that is, pointwise convergence. So, if $x_j\to x$ in $K$, you want to show that $\delta_{x_j}\to\delta_x$. This means that $\delta_{x_j}(f)\to\delta_x(f )$ for all $f\in C(K)$. But this is $f(x_j)\to f(x)$, which is precisely the continuity of $f$. Conversely, ...

1

I think your argument is fine. I fail to see why Murphy feels the need to use approximate units in this argument.

1

Your argument is not correct. You say that $\|\phi (a^*a)\|=\|\phi (a)^*\phi (a)\|$ implies that $\phi (a^*a)=\phi (a)^*\phi (a)$, which makes no sense. Also, without the unital condition the statement is trivially false: take $\phi (x)=-x$. Now, here is an argument using all conditions. Note that $\phi$ maps selfadjoints to selfadjoints. For a ...

2

The space $C_0$ is a Banach space. This implies that absolutely convergent sequences are convergent, i.e. if $$\sum_n \Vert \frac{p_n}{3^n} \Vert_\infty$$ is finite, then $\sum_n \frac{p_n}{3^n} \in C_0$. But projections have norm at most one, which means $$\sum_n \Vert \frac{p_n}{3^n} \Vert \leq \sum_n 3^{-n} = \frac{1}{1-1/3} < \infty.$$ This ...

-1

Ok, I think I got it now... Both work perfectly fine as they are always nondegenerate: $$\mathcal{A}_\text{CAR}:\quad a(f\neq0)\neq0$$ $$\mathcal{W}:\quad W(f)\neq0$$ A counterexample is provided by the angular momentum algebra: $$\mathcal{J}:\quad [J_i,J_j]=\imath\varepsilon_{ijk}J_k$$ There one has one trivial representation: $J_x=J_y=J_z=0$

0

Aah, sometimes things are so simple. :) Suppose it vanishes: $a(f_0)=0$ Then one has by the CAR relations: $$0=\{a(f),a(f)^*\}=\|f\|^2\neq0$$ That is a contradiction!

1

It's not true at least if $p >\frac{\ln 3}{\ln 2}$. Take the following $2\times 2$ matrices: $$A = \left(\begin{array}{cc} 1 & 1 \\ 0 & 1\end{array}\right), B = \left(\begin{array}{cc} 1 & 0 \\ 1 & 1\end{array}\right).$$ Then for your first $p$-norm, $||A|| = ||B|| = 2^{\frac{1}{p}}$ and $$||AB|| = \left| \left(\begin{array}{cc} 2 ... 2 This is not true in general, e.g. let a_1=\begin{pmatrix}\sqrt{2}/2&-\sqrt{2}/2\\\sqrt{2}/2&\sqrt{2}/2\end{pmatrix} and a_2=a_1^* in M_2(\mathbb{C}). Since a_1 is unitary, then A=C^*(1,a_1,a_2)=C^*(a_1). But a_1 has two eigenvalues (so a_2 also has two eigenvalues), thus \Omega(A)=\sigma(a_1) has two elements, but ... 2 This thesis contains a lot of different proofs of Hahn-Banach theorem and much more. 1 I don't think you can get too far with your approach, because you want to deal with the set of all measures on X, and there is nothing explicit about it. So you want to use fewer states. 1) Following on what Phoenix87 said, here is an example of a faithful representation (denomination way more common in the literature than "injective"). It is based on ... 1 If I do understand your question correctly, you want a representation \pi of C_0(X) on some Hilbert space, and you want \pi to be injective. As injectivity in this sense is equivalent to "isometric", I'll give you three constructions of isometric representations of C^*-algebras that I know of. Construction 1 This is a universal construction, known ... 0 If \varphi(a)\geq0 for all states \varphi, you can do the following: Note that a state is selfadjoint, i.e. it maps selfadjoints to real numbers. This, because any selfadjoint is a difference of two positives. Write a=b+ic with b,c selfadjoint. Then \varphi(b) and \varphi(c) are real. So \varphi(a)=\varphi(b)+i\varphi(c) is positive, which ... 0 This is not true. For example you could take A unital, a=\frac12\,1. Then$$ (1-u_n)^2=\left(1-\frac 1{\frac12+\frac1n}\right)^2\to 1. $$So \varphi(b(1-u_n)^2b^*)\to\varphi(bb^*) for any b. 0 Hint: a^2 - a = 0 and the spectral mapping theorem strongly restrict the possible members of the spectrum of a. 0 The exact same idea as in the answer to your other question works. That is, now take a Hamel basis of B(H) that extends a Hamel basis of \mathbb RI, and you can still get a \mathbb Q-linear map (so additive) such that \mathbb RI\subsetneq \Phi(\mathbb RI). 0 Let A=B=\mathbb C. Fix a Hamel basis X of \mathbb R as a vector space over \mathbb Q. Then X\cup Xi\  is a Hamel basis of \mathbb C over \mathbb Q. Since X and X\cup\{i\} have the same cardinality, there exists a bijection \gamma:X\to X\cup\{i\} with \gamma(1)=i. Let \eta:iX\to iX\setminus\{i\} be a bijection. These bijections induce ... 0 Asyou remarked, one direction is trivial from definitions. For the other direction you can use the following well known facts: 1) An element is positive if and only if its spectrum is contained in [0,\infty ) 2) If \lambda belongs to spectrum of a then there exists a state f such that f(a)=\lambda. 0 If \phi(a)=0 for some state \phi, then \phi(u_n)=0 for all n, because 0\leq\phi(u_n)\leq2^n \phi(a). Now, for any b\in A_+,$$ \phi(b)=\lim_n \phi(u_nbu_n)=\lim_n\phi(u_n^{1/2}[u_n^{1/2}bu_n^{1/2}]u_n^{1/2})\leq\limsup_n\|u_n^{1/2}bu_n^{1/2}\|\,\phi(u_n) \\ \leq\|b\|\limsup_n\phi(u_n)=0. So \phi(b)=0 for all b\geq0, and thus \phi=0 as ... 1 In the indicated line, we can just expand the right hand side \begin{align} (x-\lambda)^{-1}[(x-\lambda_0) - (x-\lambda)](x-\lambda_0)^{-1} &= [(x-\lambda)^{-1}(x-\lambda_0) - (x-\lambda)^{-1}(x-\lambda)](x-\lambda_0)^{-1}\\ &=[(x-\lambda)^{-1}(x-\lambda_0) - I](x-\lambda_0)^{-1}\\ &= (x-\lambda)^{-1}(x-\lambda_0)(x-\lambda_0)^{-1} ... 1 I'll use X, U instead of x, u because I want to use vectors, and it just looks wrong otherwise. I'm assuming you have some sort of functional calculus in order to form |X|=(X^{\star}X)^{1/2}. And that means you can verify that |X|(|X|+\epsilon I)^{-1} is uniformly bounded in norm by 1 for \epsilon > 0. I'll show you that ...

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 ...

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