Lemma 36.8.8.label Let $A$ be a unital $C^{*}$-algebra, $x, y \in G(A)$ be positive elements with $x \le y$, then $x^{-1}\ge y^{-1}$.
Proof. Since $y - x \ge 0$ and $x$ is invertible, $y^{-1/2}(y - x)y^{-1/2}\ge 0$ as well. As such,
\[y^{-1/2}xy^{-1/2}\le y^{-1/2}yy^{-1/2}= 1\]
Thus $\sigma_{A}(y^{-1/2}xy^{-1/2}) \subset \ol{B_\complex(0, 1)}$, and $\sigma_{A}(y^{1/2}x^{-1}y^{1/2}) \subset \complex \setminus B_{\complex}(0, 1)$. Hence $y^{1/2}x^{-1}y^{1/2}\ge 1$, and $x^{-1}\ge y^{-1}$.$\square$
Post a Comment