I am reading an old paper of Fell's and I am having some problems sorting out one of his continuous functional calculus arguments. The essential problem is this.
Let $A$ be a C$^{*}$-algebra and let $b$ and $p$ be elements in $A$ such that $b$ is positive and $p$ is a projection with the additional assumptions that:
1) $pbp=b$
2) $\|p-b\|<1/8$.
Let $\psi\colon\mathbb{R}\to[0,\infty)$ be continuous and bounded function that satisfies $$ \begin{aligned} \psi(r)= \begin{cases} 0&\text{if } r=0,\\ 1&\text{if } |r-1|\leq 1/8. \end{cases} \end{aligned} $$
The claim is that $\psi(b)=p$. I am having trouble showing this. I'm not sure how to relate fact 2) with the fact that $\psi$ is identically $1$ on $[7/8,9/8]$ and, moreover, how to use this (or another method) to yield the result.
The $C^*$-subalgebra generated by $b$ and $p$ is commutative. So, we may assume without loss of generality that our $C^*$-algebra is $C(X)$ for some compact Hausdorff space $X$. We then have a function $p\in C(X)$ which only takes the values $0$ and $1$ and a nonnegative function $b\in C(X)$ such that $bp=b$ and $|p(x)-b(x)|<1/8$ for all $x\in X$. So, at points $x\in X$ such that $p(x)=0$, we have $b(x)=0$. Moreover, at points $x\in X$ such that $p(x)=1$, we have $b(x)\in [7/8,9/8]$.
On an algebra of the form $C(X)$, the continuous functional calculus is just composition: $\psi(b)$ is the composition $\psi\circ b$. By our description of $b$ above, we see that $\psi(b(x))=0$ when $p(x)=0$ and $\psi(b(x))=1$ when $p(x)=1$. Thus, $\psi\circ b=p$.