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Let $\mathcal{A}$ be a unital C*-algebra (not necessarily commutative) and let $A^*=A\in \mathcal{A}$ be a self-adjoint element with $\vert\vert A \vert\vert \leq 2$?

I want to show that $\vert\vert \mathbb{1}-A \vert\vert \leq 1 \Leftrightarrow \sigma(A)\subset [0,\infty)$ (i.e. $A$ is positive), but don't really see the connection.

How can approach this problem?

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Let $\mathcal B\subseteq \mathcal A$ be the sub-$C^*$-algebra generated by $1$ and $A.$ Then $\mathcal B$ is commutative, so $\mathcal B\simeq C(X).$ Let $f_A\in C(X)$ be the function corresponding to $A.$ The following implications hold: \begin{gather*} ||1-A||\leq 1 \Longleftrightarrow ||1-f_A||_\infty\leq 1 \Longleftrightarrow |1-f_A|\leq 1\ \mbox{on}\ X\\ \Longleftrightarrow 0\leq f_A\leq 2\ \mbox{on}\ X\Longleftrightarrow 0\leq A\leq 2\ (\mbox{in } \mathcal B)\Longleftrightarrow 0\leq A\leq 2\ (\mbox{in } \mathcal A) \end{gather*} (Here $||\cdot||_\infty$ is the supremum norm on $X,$ i.e. norm in $C(X)$)

It solves your problem.

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  • $\begingroup$ I think you missed two norms in the last two steps. If you put an order relation on positive elements, you should define and clarify it, because it might confuse the OP. $\endgroup$ – Marc Palm Jul 2 '13 at 9:01
  • $\begingroup$ @Marc Palm: If madison54 knows about positive elements, I believe he knows about order relation on self-adjoint elements. $\endgroup$ – Yurii Savchuk Jul 2 '13 at 9:23
  • $\begingroup$ @YuriiSavchuk: Can I always take the spectrum $\sigma(A)$ for your compact Hausdorff space $X$? $\endgroup$ – madison54 Jul 2 '13 at 19:40
  • $\begingroup$ @YuriiSavchuk: I'm confused by your notation. Does $0\leq A \leq 2$ mean that $A$ and $(2\mathbb{1}-A)$ are positive elements in $\mathcal{A}$? $\endgroup$ – madison54 Jul 2 '13 at 19:49
  • $\begingroup$ @madison54: Yes, one can show that $X$ is homeomorphic to $\sigma(A).$ $\endgroup$ – Yurii Savchuk Jul 3 '13 at 7:07

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