Let $u$ be an element of a Banach algebra $A$ such that $u^2-u \in \text{Rad}(A)$. I am trying to show that there exists a projection in $A$ which is equal to $u$ modulo the radical.
We have that $\sigma(u^2-u)=\{0\}$ and by the spectral mapping theorem we can show $\sigma(u)=\{0,1\}$.
Let $U=B(0,\frac{1}{4})$ and $V=B(1,\frac{1}{4})$ then $U$ and $V$ are two disjoint open sets which contains $0$ and $1$ respectively. Therefore $\sigma(u)\subset U\cup V$ with $U\cap V = \emptyset$. Now let $\Gamma_1$ and $\Gamma_2$ be two closed curves surrounding $0$ and $1$, contained within $U$ and $V$ respectively. Then $\Gamma=\Gamma_1\cup \Gamma_2$ encloses the spectrum and is contained within an open set surrounding the spectrum. Let $f:U\cup V \to \mathbb{C}$ be defined by $$ f(\lambda)=\begin{cases} 1, & \lambda \in V \\ 0, & \lambda \in U \end{cases}. $$ Then $f$ is analytic on $U\cup V$. By the holomorphic functional calculus, we have that $$f(u)=\frac{1}{2\pi i} \int_\Gamma f(\lambda)(u-\lambda \textbf{1})^{-1}d\lambda,$$ which is a projection in $A$.
Let $h(\lambda)= \lambda - f(\lambda)$, then using a similar argument, we can define $r=h(u)$.
Then we have that $$u=f(u) + (u-f(u))= f(u) + h(u) = f(u)+r.$$
Do we have that $r \in \text{Rad}(A)$?
If so, then the result will be proven, but I am having some difficulty seeing whether or not $r \in \text{Rad}(A)$ is true.
I do know that $r$ is quasinilpotent, since it has zero spectrum. But I am not sure how to show that $r \in \text{Rad}(A)$.
Define $h(\lambda )= \lambda ^2-\lambda $, and $$ g(\lambda ) = \left\{\matrix{ \displaystyle {1}/({1-\lambda }) & \text{, if } \lambda \in U. \cr \displaystyle {1}/{\lambda } & \text{, if } \lambda \in V, } \right. $$ and notice that $\lambda -f(\lambda ) = g(\lambda )h(\lambda )$. Therefore $$ r = u -f(u) = g(u)h(u) = g(u)(u^2-u), $$ which lies in the radical because $u^2-u$ does and the radical is an ideal.