A simple $C^*$ algebra whose all idempotents are contained in a given disc

Question

Assume that $A$ is a simple $C^*$ algebra and $K>0$ is a fixed number. Assume that all idempotents $e\in A$ satisfy $|e|<K$. Does this imply that $A$ has no non trivial idempotent?

Answer 1

Any C$^*$-algebra with trivial center that has a nontrivial idempotent has an unbounded sequence $\{e_n\}$ of idempotents.

Indeed, let $p$ be a non-trivial idempotent. This implies that there exists an idempotent $e$ that is not selfadjoint (i.e., not a projection). To see this, consider the orbit $\{apa^{-1}:\ a\ \text{ invertible}\}$. All elements are idempotents; if they were all projections, then in particular for positive invertible $a$ we would have $(apa^{-1})^*=apa^{-1}$, so $a^{-1}pa=apa^{-1}$; this gives $a^2p=pa^2$. So $p$ commutes with all squares of strictly positive elements, and so with all strictly positive elements, and then with all positive elements. As any elements is a linear combination of positives, we get that $ap=pa$ for all $a\in A$, so $p$ is central. As $A$ is assumed to be simple, its center is trivial, we get that $p=0$ or $p=1$, a contradiction. Thus a non-selfadjoint idempotent $e$ exists.

Given a non-selfadjoint idempotent $e$, it is not to hard to check that $$\tag1 p=ee^*(I+(e-e^*)(e^*-e))^{-1} $$ is a projection, and that $$\tag2ep=p,\ \ \ \ pe=e. $$ Now let, for some $n\in\mathbb N$ $$\tag3 f=p+n(e-p). $$ The identities $(2)$ guarantee that $f^2=f$. We have $$ n\,\|e-p\|=\|n(e-p)\|=\|f-p\|\leq\|p\|+\|f\|=1+\|f\|. $$ So $$ \|f\|\geq n\|e-p\|-1, $$ and so by choosing $n$ large enough we can make $\|f\|$ as large as we want.