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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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