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3

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 ...


2

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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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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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.


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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} ...


2

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 ...


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 ...


1

Approximate Unit Regard ball cone: $$\mathcal{B}_+:=\{A\in\mathcal{A}:\|A\|<1:A\geq0\}$$ Order elements: $$E,E'\in\mathcal{I}\cap\mathcal{B}_+:\quad E\leq E'$$ Then one has: $$I\in\mathcal{I}:\quad\|I-IE\|,\|I-EI\|\stackrel{E\to1}{\longrightarrow}0$$ (That is the hard part!) Quotient Norm Note that it holds: $$1-\sigma(E)\geq0\implies\|1-E\|\leq1$$ ...


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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 $P \in k[X]$ , ...


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Let $\lambda_a \colon b \mapsto a\cdot b$. Then it's easy to see that the operator norm of the matrix is at most $$\max_i \sum_j \lVert \lambda_{a_{ij}}\rVert_{\operatorname{op}}.$$ For a general Banach algebra, it is possible that $\lVert \lambda_a \rVert_{\operatorname{op}} < \lVert a\rVert$ for some $a$. Still, even if we take the operatornorm of the ...


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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 ...



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