If $0\leq a \leq b$ and $a$ is invertible, then $b$ is invertible

Question

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(?).

Can you prove that if $c>1$, then $c$ is invertible (and $c^{-1}<1$)? Perhaps using the Gelfand representation theorem. — Mankind, Dec 16 '15 at 21:07
Gelfand representation should not be needed. In the case with c, one can deduce that $\sigma(c)\subseteq [1,\infty)$. @HowDoIMath — Ingvar, Dec 16 '15 at 21:39
Then $1 = a^{-1/2}aa^{-1/2} \leq a^{-1/2}ba^{-1/2}$, and then $a^{-1/2}ba^{-1/2}$ is invertible. — Mankind, Dec 16 '15 at 21:58
Here is one of several proofs of this that can be found on this site. — Martin Argerami, Dec 24 '15 at 21:42

score 1 · Accepted Answer · edited Apr 13 '18 at 13:52

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, $0 \not \in \sigma(b)$.

cqfd · Answer 2 · 2020-06-17T11:48:56.440

Here's a detailed answer based on the comments above.

Suppose $c\geq 1$. Then $\sigma(c-1)=\sigma(c)-1\subseteq [0,\infty)$, so $\sigma(c)\subseteq [1,\infty)$ and hence $c$ is invertible. Notice that $\sigma(c^{-1})\subseteq (0,1]$. Thus $\sigma(c^{-1}-1)= \sigma(c^{-1})-1\subseteq (-1,0]$. So $\sigma(1-c^{-1})\subseteq [0,1)$ and hence $1-c^{-1}$ is positive, i.e., $1\geq c^{-1}$.

Notice that if $0\leq a\leq b$, then $ c^{*}ac\leq c^*bc$. Indeed, $b-a\geq 0$ implies that $b-a=d^*d$ for some $d\in\mathscr A$. Now the result follows immediately.

Next, suppose that $0\leq a\leq b$ and $a$ is invertible. Let $a^{1/2}$ denote the positive square root of $a$. Since $a$ is invertible, $a^{1/2}$ is invertible(hint: $a^{1/2}$ is the limit of a sequence of polynomials in $a$, so commutes with the inverse of $a$). Now $0\leq a\leq b$ implies that $$1=a^{-1/2}aa^{-1/2} \leq a^{-1/2}ba^{-1/2}=c\;\;\text{(because $a^{-1/2}$ is self-adjoint).}$$ Then $c$ is invertible and $1\geq c^{-1}$. So there exists $d\in\mathscr A$ such that $cd=dc=1$. Now it is easy to see that $a^{-1/2}da^{-1/2}$ is the inverse of $b$. Thus $b$ is invertible and $1\geq a^{1/2}b^{-1}a^{1/2}$. Furthermore, an application of the above lemma gives $a^{-1}\geq b^{-1}\geq0$.

If $0\leq a \leq b$ and $a$ is invertible, then $b$ is invertible

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