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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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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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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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No, it does not suffice! It can even severely fail to extend to the complex plane at all. (See the thread: Analyticity: Uniform Limit)

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Given the Fock space $\mathcal{F}(\mathcal{h})$. Consider the second quantization: $$\Gamma(e^{ith})=e^{it\mathrm{d}\Gamma(h)}$$ It acts continuous on the CAR-algebra: $$\quad\tau^t[a(\eta)]=a(e^{ith}\eta)\stackrel{t\to0}{\to}a(\eta)$$ (Remember that this won't work for the Weyl algebra.)

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The map $(a,b)\mapsto b\otimes a$ is bilinear and induces an isomorphism $\theta : A\otimes B \rightarrow B\otimes A$ satisfying $\theta(a\otimes b) = b\otimes a$ for all $a,b$. Any other such isomorphism takes sames values on simple tensors as $\theta$. As the simple tensors generate $A\times B$, this isomorphism is $\theta$. "Other proof" : $\theta$ ...

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After a while it's not as hard... In fact, local integrability is the key: $$\omega\circ\tau(A)\in\mathcal{C}(\mathbb{R}_+):\quad\omega_T(A)\stackrel{T\to\infty}{\to}\omega_\infty\implies\langle\omega\rangle_T\stackrel{T\to\infty}{\to}\omega_\infty$$ (See the thread: Asymptotic Convergence vs. Mean Convergence)

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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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$$(\lambda e-A-P+P)=(\lambda e-P)(e-(\lambda e-P)^{-1}(A-P)).$$ 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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I finally figured it out; both separately... Suppose it holds: $\omega(A^*A)\geq0$ Positivity can be proven by Kadison's inequality: ...

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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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This answer is community wiki. Feel free to edit, expand etc! Bratelli & Robinson exploit a difficult algorithm therefor: $$\sqrt{A}=\frac{1}{\pi}\int_0^\infty\frac{\mathrm{d}\lambda}{\sqrt{\lambda}}A(\lambda1+A)^{-1}$$ Kadison & Ringrose use a simple adjustment instead: ...

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Disclaimer My excuses; meanwhile I understood the proof. Proof Assume for contradiction that: $$A_n\geq0:\quad\omega(A_n)\geq n\quad(\|A_n\|\leq1)$$ Consider the absolutely convergent series: $$S_N:=\sum_{n=1}^N\frac{1}{n^2}A_n\to S\in\mathcal{A}$$ (Remember that the positive cone is closed!) By positivity one has an upper bound: $$S_N\leq ... 2 Something like this should work : Given \epsilon > 0, consider f(x) = x^{-1/2}. This is uniformly continuous on [1/2,2], so \exists 1/2>\delta > 0 such that for any positive invertible a \in A such that \sigma(a) \subset [1/2,2], we have$$ \|a - 1\| < \delta \Rightarrow \|\sqrt{a}^{-1} - 1\| < \frac{\epsilon}{\sqrt{2}} ...

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