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I can't bring myself to write $fg$ for the convolution of $f$ and $g$. So I'm going to write $f\mapsto f'$ for the involution, so I can write $f*g$ for the convolution. Are you certain you got the definition of $f'$ straight? What would make much more sense to me would be $$f'(t)=\overline{f(-t).}$$ That seems to me is the "standard" involution on $L^1$. ...

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The part of theorem 3.31 that is used here is the equality $(3)$; for closed paths $\Gamma_1,\,\Gamma_2$ that are homologous in $\Omega$ (that is, they have the same winding number around all $w \in \mathbb{C}\setminus \Omega$), and any (weakly) holomorphic function $f$ on $\Omega$ we have $$\int_{\Gamma_1}f(\zeta)\,d\zeta = \int_{\Gamma_2} ... 1 It is easier to see without the clutter notation. What you want to show is that if in some C^*-algebra ab=0 with a\geq0, then a^{1/2}b=0; there is no positivity requirement for b. Here are two proofs: From ab=0, you get$$ (a^{1/2}b)^*a^{1/2}b=b^*ab=0,$$so a^{1/2}b=0. From ab=0, you get a^nb=0 for all n\geq 1, so p(a)b=0 for all ... 0 I take it A is a Banach algebra with identity and \Delta(A) is the maximal ideal space? The point is that$$\Delta(A)\subset K=\prod_{x\in A}\overline {D(0,||x||)}.$$The Gelfand topology is the relative topology inherited from K. So it's Hausdorff, just because each of those disks is Hausdorff. And K is compact, so yes to show \Delta(A) is ... 0 The Hilbert space for the GNS construction for that particular state is certainly not L^2[0,1]. What you are expected to do is to go through the proof of the GNS theorem and calculate the Hilbert space and the inner product for that concrete C^*-algebra and that particular state. It is not hard. 1 You should study Morita equivalence of C*-algebras (in the sense of Rieffel): the fact is that C(X,M_2) is Morita equivalent to C(X). And it is a general fact that Morita equivalent C*-algebras have same closed (two-sided) ideals. It follows that the closed ideals of C(X,M_2) are all of the form C_0(U,M_2) (functions vanishing outside U) for ... 1 Assume k field and A a finite dimensional k algebra. Then the spectrum of any element is finite. Step 1. Every element a of A satisfies a polynomial equation of degree at most n= \dim_{k} A. Indeed, the elements 1, a, \ldots, a^n are linearly dependent. Step 2. Let a satisfying a polynomial equation P(a) = 0, where P \in k[X] , ... 2 Consider abelian C*-algebra C^*(a) which is infinite dimensional ( because \sigma(a) is infinite). Also C^*(a) \subset A which implies that A is infinite dimensional. 2 If you represent C  in B (H)  faithfully, then it is well-known that$$ M (C)\simeq\{x\in B (H):\ xc\in C,\ cx\in C,\ \forall c\in C\}. 

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Based on Proposition 3.10, pp. 67 of the following Book "Functional Analysis and infinite-dimensional geometry By Vaclav Zizler and ...Springer 2001", if $X$ is an infinite-dimensional normed space, then weak and norm topologies do not coincide. Because $L^{1}(\mathbb{Z})^{*}$ is an infinite-dimensional normed space by using the mentioned theorem the ...

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You can repeat the proof with $\phi/\|\phi\|$, which does satisfy the inequality. If the authors didn't mention that their were sloppy, but the argument still works. What they mean by "hereditary" is that if $a,b$ are orthogonal and positive and $a'\leq a$, $b'\leq b$, then $a'$ and $b'$ are orthogonal. One way to see this last assertion is by ...

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If we assume that $\varphi(a^2)=\varphi(a)^2$ for all $a$, then for $a,b$, we have, using linearity and the square-preserving property, $\varphi((a+b))^2=(\varphi(a)+\varphi(b))^2=\varphi(a)^2+\varphi(a)\varphi(b)+\varphi(b)\varphi(a)+\varphi(b)^2.$ Now, observe that $\varphi((a+b)^2)=\varphi(a^2+ab+ba+b^2)=\varphi(a^2)+\varphi(ab+ba)+\varphi(b^2)$. By ...

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