I have the following problem: Let $A$ be a $C^*$-Algebra, let $x\in A$ be self-adjoint and let $p\in A$ be a projection, i.e. $p^*=p=p^2$. Moreover, assume that $||x-p||<\frac{1}{16}$.
Claim: $||x^2-x||<\frac{1}{4}$.
I know that the result is true, it comes up in a proof about the relation of UHF-Algebras and supernatural numbers. However, I don't know how to prove it, because $||x^2-x||\leq||x^2-p^2||+||p-x||$ does hold, but I don't succeed in bounding $||x^2-p^2||$.
Note that $\|p\|=1$ unless $p=0$. $$ \|x^2-x\| = \|(x-p)^2 - (x-p) + xp+px-2p\| \le \|x-p\|^2 + \|x-p\| + \|xp+px-2p\| $$ $$ =\|x-p\|^2 + \|x-p\| + \|(x-p)p + p(x-p)\| \le \|x-p\|^2 + \|x-p\| + 2\|x-p\| $$ $$ < (1/16)^2+3/16 < 1/4. $$