Positive Elements: Unique Decomposition

Question

Problem

Given a C*-algebra $A$.

Define positive resp. negative part: $$(\cdot)_\pm:\mathbb{R}\to\mathbb{C}:\quad2(x)_\pm:=|x|\pm x$$

Then for commuting positive elements: $$a_\pm\geq0:\quad a_\mp a_\pm=0\implies(a_+-a_-)_\pm=a_\pm$$ That is the decomposition is unique.

How can I prove this carefully?

Attempt

The issue I'm having here:

I can't just pull through the calculus: $$f:\mathbb{R}\to\mathbb{C}:\quad f(a+b)\neq f(a)+f(b)$$

Contrary Gelfand duality gives: $$a_\pm\in\mathcal{C}^*(a_+,a_-)\cong\mathcal{C}(\Omega)$$ But the Gelfand spectrum is abstract.

So how can I check the above?

Answer 1

Suppose that $a,b\geq0$ and $ab=0$. We have $$ (a-b)^2=a^2+b^2=(a+b)^2. $$ By the uniqueness of the positive square root, it follows that $$ |a-b|=(a^2+b^2)^{1/2}=a+b. $$ So $$ 2(a-b)_+=|a-b|+a-b=2a, $$ $$ 2(a-b)_-=|a-b|-(a-b)=2b. $$