If $1=\frac{1}{2}(x+y)$ for $x,y\in A_{sa}$ then $x$ and $y$ commute by spectral theory

Question

I am trying to understand example 1.1.13 given in C*-algebras and their automorphism groups by Pedersen:

The part that I do not understand is the following: For $1=\frac{1}{2}(x+y)$ where $x,y$ are self-adjoint elements of the $C^*$-algebra A of norm $\leq 1$ and 1 is the unit of $A$. They claim that it then follows by spectral theory that $x$ and $y$ commute and that $x=y=1$.

My question: how does commutativity follow?

Answer 1

You have $$x+y=2.$$ So $y=2-x$, and then it is obvious that $x$ and $y$ commute. you don't even need them to be selfadjoint for this to hold.

For the other part, from $y=2-x$ we get that $\|x\|\leq1$ and $\|2-x\|\leq1$. The first inequality gives us that $\sigma(x)\subset[-1,1]$, while the second one gives us $\sigma(x)\subset [1,3]$. Thus $\sigma(x)=\{1\}$.