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3

Let $A$ be ${\mathbb C}^2$ with the pointvise multiplication and the involution $(x,y)^*=(\overline{x},\overline{y})$, i.e., continuous functions on two points. But the norm let be $\|(x,y)\|=|x|+|y|$.


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Note: This is community wiki. One has the equivalence: $$(W^*W)^2=W^*W\iff WW^*W=W$$ For shorthand denote: $$P:=W^*W$$ So on the one hand one has: $$P^2=W^*(WW^*W)=W^*W=P$$ By the C*-property it holds: $$X=0\iff X^*X=0$$ So on the other hand one has: $$(WW^*W-W)^*(WW^*W-W)=P^4-P^2-P^2+P=0$$


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It all works a bit more generally than that. Proposition: Let $A$ and $B$ be C*-algebras and suppose $\pi: A \to M(B)$ is a $*$-homomorphism that is "nondegenerate" in the sense you described. That is, $\mathrm{span}\{ \pi(a) b : a \in A, b \in B\}$ is dense in $B$. Then $\pi$ extends uniquely to a $*$-homomorphism $\overline \pi : M(A) \to M(B)$. How ...


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Something like this should work : Given $\epsilon > 0$, consider $f(x) = x^{-1/2}$. This is uniformly continuous on $[1/2,2]$, so $\exists 1/2>\delta > 0$ such that for any positive invertible $a \in A$ such that $\sigma(a) \subset [1/2,2]$, we have $$ \|a - 1\| < \delta \Rightarrow \|\sqrt{a}^{-1} - 1\| < \frac{\epsilon}{\sqrt{2}} ...


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I think your unitary goes from $\mathcal H_2\to\mathcal H_1$, or equivalently is an element in the set of intertwiners $(\mathcal B(\mathcal H_2),\mathcal B(\mathcal H_1))$. Given this, the unitary in question allows you to identify operators between those two Hilbert spaces, i.e. it induces a bijection through its adjoint action. Any Cauchy net in ...


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Given $Z=X+iY$ with $X$ and $Y$ self-adjoint, note that $X=\frac12(Z+Z^*)$, and $Y=\frac1{2i}(Z-Z^*)$. Take norms, apply the triangle inequality, and recall that $\|Z^*\|=\|Z\|$ to conclude that $\|X\|\leq \|Z\|$ and $\|Y\|\leq \|Z\|$.


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$$(\lambda e-A-P+P)=(\lambda e-P)(e-(\lambda e-P)^{-1}(A-P)).$$ Since $e-a$ is invertible for $\|a\|<1$ with $(e-a)^{-1}=\sum a^k$, it's enough to prove that $$\|(\lambda e-P)^{-1}(A-P)\|\le \delta\|(\lambda e-P)^{-1}\|<1.$$ By Gelfand-Neumark theorem, it's sufficient to consider $\mathcal{A}$ as a subalgebra of the space $L(H)$ of operators on some ...



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