4
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.