continuous functional calculus; spectrum of an self adjoint element in a c*algebra

Question

Let A be a C$^*$-Algebra, $a\in A$ selfadjoint and $\|a^2-a\|<\frac{1}{4}$. The claim is: $\sigma(a)\subseteq (-\frac{1}{2},\frac{1}{2}) \cup (\frac{1}{2},\frac{3}{2})$ and there is a projection $p\in A$ such that $\|a-p\|<\frac{1}{2}$.
I have found out: $a^2-a$ is selfadjoint too, $\sigma(a^2-a)\subseteq (-\frac{1}{4},\frac{1}{4})$, and i think i have to use the functional calculus to find out $\sigma(a)\subseteq (-\frac{1}{2},\frac{1}{2})\cup (\frac{1}{2},\frac{3}{2})$, but i dont know how. If $\Phi:C_0(\sigma(a))\to A$ is the continuous functional calculus, $\Phi(z\mapsto z^2-z,\;\sigma(a)\to \mathbb{C})=a^2-a=f(a)=\sigma(f(a))$. Can you help me what to do next? Regards
If $z^2-z=-1/4\ $ I get $z=1/2$ and if $z^2-z=1/4$, I get $z=1/2+\sqrt{2}$ and $z=1/2-\sqrt{2}$..

Answer 1

What you did is correct, $||a^2 - a || < \frac{1}{4}$ means $\sigma(a^2 - a) \subset (-1,1)$, that is the function $t^2 - t$ takes $\sigma(a)$ to the subset $(-1,1)$, and so, $\sigma(a) \subset (\frac{1-\sqrt{2}}{2},\frac{1}{2}) \cup (\frac{1}{2} , \frac{1+\sqrt{2}}{2}) \subset (-\frac{1}{2},\frac{1}{2})\cup (\frac{1}{2}, \frac{3}{2})$.

Define $f\colon (-\frac{1}{2},\frac{1}{2})\cup (\frac{1}{2}, \frac{3}{2}) \to \mathbb{R}$, $f(x) = x$ and $g\colon (-\frac{1}{2},\frac{1}{2})\cup (\frac{1}{2}, \frac{3}{2}) \to \{0,1\}$, $g(x) = $ closest integer to $x$ ( no ties on this set). We have $|f(x) - g(x)| < 1$ for all $x$. Hence on the compact set $\sigma(a)$ we have $\sup |f-g| < 1$. Define $p = g(a)$ ( functional calculus). We have $p^2 = p$ and $|a-p| < 1$.