Let $\mathscr A$ be a unital C*-algebra and let $a,b\in \mathscr A$ such that $0\leq a \leq b$ and $a$ is invertible. How to show that $b$ is invertible?
($0\leq a \leq b$ means that $a,b$ is positive and that $b-a$ is positive. Moreover, a positive element is a hermitian element with a spectrum which is a subset of $[0,\infty)$.)
Since $a$ is invertible, we have that $0 \not \in \sigma(a)$. I guess my question is why this implies that $0 \not \in \sigma(b)$. Maybe one should use functional calculus in some way(?).
Since $\sigma(a) \in [0,\infty)$ is compact, it contain all its limit Points. Put $$\varepsilon=\min\{z:z\in \sigma(a)\}.$$ Define $f:\sigma(a) \rightarrow \mathbb C$ by $f(z)=z-\varepsilon$. Then, by the Spectral Mapping Theorem, $$\sigma(f(a))=\sigma(a-\varepsilon)=\{z-\varepsilon:z\in\sigma(a)\}\subseteq [0,\infty),$$ where the inclusion follows since $\varepsilon>0$. So, $a-\varepsilon \geq 0$. Moreover, $$(b-\varepsilon)-(a-\varepsilon) =b-a\geq 0 \\ \implies b-\varepsilon \geq a-\varepsilon \\ \implies b\geq a. $$ Define $g:\sigma(b) \rightarrow \mathbb C$ by $g(z)=z-\varepsilon$. Then, by Spectral Mapping Theorem, $$\sigma(g(b))=\sigma(b-\varepsilon)=\{z-\varepsilon:z\in \sigma(b)\}\subseteq [0,\infty) \\ \implies z\geq \varepsilon > 0 \ \forall z \in \sigma(b)$$ Hence, $0 \not \in \sigma(b)$.