# Why is this statemenet clear for the C*-algebra of continuous complex functions on a compact space?

Firstly, for a unital C*-algebra $A$, and an hermitian element $x \in A$, the following are equivalent: 1) The spectrum $Sp_A x \geq 0$; 2) $x$ is of the form $yy^*$ for some $y \in A$; 3) $x$ is of the form $h^2$ for some hermitian $h \in A$.

Let $P$ be the set of $x \in A$ satisfying any of these conditions.

On page 16 of Dixmier's C*-algebra, there is a lemma:

Let $A$ be a unital C*-algebra. If $x \in A$ is hermitian and $||1-x|| \leq 1$, then $x \in P$. If $x \in P$ and $||x|| \leq 1$ then $||1-x|| \leq 1$.

The author reduced the proof to the case of $A$ being the C*-algebra of continuous complex-valued functions on a compact space. Then he said that this was clear. But I don't know why is this clear in this case.

I'll be very grateful for the explanation.

-
Did you determine the functions in $P$ in the case you ask about? The claim comes down to the observation that $0 \leq r \leq 1$ is equivalent to $0 \leq 1-r \leq 1$ for real numbers $r$. Take $r = f(x)$ and then take the supremum over $x$ in the compact space. –  commenter Oct 7 '12 at 12:49

## 1 Answer

Since $x$ is hermitian, the unital C$^*$-subalgebra of $A$ generated by $x$ is abelian, and then it is isomorphic to $C(\sigma(x))$ via Gelfand-Neumark. So it is enough to show that the lemma holds in $C(\sigma(x))$. Here hermitian means real-valued; as @commenter said in his comment, you have a real function $f$ such that $|1-f(t)|\leq 1$ for all $t\in\sigma(x)$. This is the same as $$-1\leq f(t)-1\leq 1.$$ Adding $1$, we get $0\leq f(t)\leq 2$, so $f(t)$ is a non-negative number. This shows that $\sigma(f)\subset[0,\infty)$ (the spectrum of $f$ is the closure of its range), so $f$ is positive.

-