Let $\{A_i,\Phi_{ij} \}_{i\in \mathcal{I}}$ a directed system of C* algebras and $A:=\varinjlim A_i$ its limit. I know that if $x\in A$ is self-adjoint, it can be approximated with another self-adjoint $x_i\in A_i$ for some $i$ such that $$\|x-\Phi_i(x_i)\|<\varepsilon$$ There's a proof in Wegge-Olsen's K-Theory and C*-Algebras that shows that if $x$ is positive (i.e. $\sigma_A(x)\subset \mathbb{R}^+$ and $x$ self-adjoint) then we can choose $x_i$ to be positive aswell. To show this he considers the real function $f$ $$t\mapsto \begin{cases}t &, t\geq 0 \\ 0 &, t<0\end{cases}$$ Since this function is continuous in $\mathbb{R}$, we can extend $f$ to $A_{sa}$ (the set of self-adjoint elements of $A$). Moreover, for any $y\in A_{sa}$ we have that $f(y)$ is positive and if $y$ is positive then $f(y)=y$. Now, taking $x$ positive we know that there's $x'_i\in A_i$ self-adjoint that approximates it. We set $x_i:=f(x' _i)$, which is positive and he states that: $$\|x-\Phi_i(x_i)\|=\|f(x)-f(\Phi_i(x_i'))\|\stackrel{?}{=}\|f(x-\Phi_i(x'_i))\|\leq\|x-\Phi_i(x'_i)\|<\varepsilon$$ I'm not sure why the middle inequality holds since $f$ is clearly non linear. If by any chance $x-\Phi_i(x_i)$ was positive then the equality would hold but I don't think that's the case.

I'm pretty sure the fact that we're dealing with a direct limit doesn't matter that much. Rather than if we have $x,y\in A$ such that $x$ is positive, $y$ self-adjoint and $\|x-y\|<\varepsilon$ then $\|x-f(y)\|<\varepsilon$. If we take $x=1/8$, $y=1/4$, $\varepsilon=1/2$ and $A=\mathbb{C}$ the $(?)$ equality doesn't hold but the overall inequality does.


There is something off with that argument as it is written. Nothing prevents, for instance, that $x-\Phi_i(x_i')\leq0$ and nonzero. In that case, $$f(x-\Phi_i(x_i'))=0,$$ and then $$ 0<\|x-\Phi_i(x_i)\|,\ \ \ \ \|f(x-\Phi_i(x_i'))\|=0. $$

This is how I think that argument can be saved. If $x\geq0$, $y=y^*$, and $\|x-y\|<\varepsilon$, then $\sigma(y)\subset(-\varepsilon,\|x\|+\varepsilon]$. In particular, $\|y^-\|\leq\varepsilon$. Thus $$ \|x-f(y)\|=\|x-y^+\|=\|x-y-y^-\|\leq\|x-y\|+\|y^-\|<2\varepsilon. $$

For the inclusion of the spectrum, if $\lambda<-\varepsilon$, then $x-\lambda\geq x+\varepsilon\geq\varepsilon I$ is invertible and $$ \|(x-\lambda)^{-1}\|\leq\|(\varepsilon I)^{-1}\|=\frac1\varepsilon. $$ Then $$ \|(x-\lambda)-(y-\lambda)\|=\|x-y\|<\varepsilon<-\lambda\leq\frac1{\|(x-\lambda)^{-1}\|}. $$ So $y-\lambda$ is invertible.

  • $\begingroup$ I suppose $y^+=f(y)$ and $y^-=y-y^+$. :) $\endgroup$ – Julio Cáceres May 3 '16 at 15:41
  • $\begingroup$ Yes. It is canonical notation. They are the preferred decomposition of a selfadjoint as a difference of two positives because their product is zero. $\endgroup$ – Martin Argerami May 3 '16 at 16:15
  • $\begingroup$ One last question, does it matter if $\Phi_i$ are not unitary morphisms to make the switch $\Phi_i(f(x'_i)) =f(\Phi_i(x'_i))$? $\endgroup$ – Julio Cáceres May 3 '16 at 16:18
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    $\begingroup$ No, because $f(0)=0$, so $f$ is a limit of polynomials with $p(0)=0$; that's all you need. $\endgroup$ – Martin Argerami May 3 '16 at 16:36

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