Let $a,b \in A^+_{\leq 1}$ then $\|a-b\|\leq 1$

Question

Let $A$ be a $C^*$-algebra and $a,b \in A^+$ such that $\|a\| \leq 1$ and $\|b\| \leq 1$. Then $\|a-b\| \leq 1$.

I have to prove the above, without assuming that $ab=ba$.

I don't know how to start with the proof. Any hints are welcome

Thanks

Answer 1

The element $a-b\in A$ is selfadjoint, and it satisfies $$ -b\leq a-b\leq a. $$ So $$ -1\leq-\|b\|\leq-b\leq a-b\leq a\leq\|a\|\leq 1. $$ This shows that $\sigma(a-b)\subset[-1,1]$ and so $\|a-b\|\leq1$.

Note that the inequalities above also prove the slightly sharper inequality $$ \|a-b\|\leq\max\{\|a\|,\|b\|\}. $$