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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 ...


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This thesis contains a lot of different proofs of Hahn-Banach theorem and much more.


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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 ...


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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 ...


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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 ...


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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 ...


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


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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, ...


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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} ...


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$(x,u^*\alpha y)=(ux,\alpha y)=\bar\alpha(x,u^*y)=(x,\alpha u^*y)$.


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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 ...



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