Let $H$, $K$ be two self-adjoint operators of a unital $C^*$-algebra.
(1) Suppose that $HK = KH$ and $H \le K$, show that for any continuous and monotonically increasing function $f$ on $\mathbb{R}$, we have $f(H)\le f(K)$.
(2) Suppose that $0\le H\le K$, $H$ is invertible, show that $K$ is invertible and $K^{-1}\le H^{-1}$。
Proof: (1) We have $\sigma(H)\subset \mathbb{R}$，$\sigma(H)\subset \mathbb{R}$ since $H$ and $K$ are self-adjoint operators. $\sigma(K-H)\subset [0,+\infty)$ since $H\le K$. If we can prove that $\sigma(K-H)=\sigma(K)-\sigma(H)$, then we can write $\sigma(H)$ and $\sigma(K)$ as $$ \sigma(H)=\{\lambda_1,\lambda_2,\dots\}, \sigma(K)=\{\mu_1,\mu_2,\dots\} , \lambda_i\le \mu_i $$ since $\sigma(K)-\sigma(H)=\sigma(K-H)\subset [0,+\infty)$.
By spectral mapping theorem, we have $\sigma(f(H))=f(\sigma(H))$ and $\sigma(f(K))=f(\sigma(K))$. Hence if we can prove $\sigma(f(K)-f(H)) = \sigma(f(K))-\sigma(f(H))$, then $f(H)\le f(K)$ since $f$ is continuous and monotonically increasing.
(2)...
Could ypu please help me? Thanks in advance!
