Do we have $u^2 \leq u$ when $\|u \| \leq 1$ in a $C^*$-algebra?

Question

Let $u$ be a positive element in a $C^*$-algebra $A$ with $\|u \|\le 1$. Then is it true that $u^2 \leq u$?

Attempt: This is true in the commutative $C^*$-subalgebra $C^*(u)$, since by the Gelfand-Naimark theorem, $C^*(u) \cong C_0(X)$ where $X$ is some locally compact Hausdorff space. Thus, this also holds in $A$ since positivity does not depend on the ambenient $C^*$-subalgebra.

Is this correct?

Answer 1

From $u$ positive and $\|u\|\leq1$, you get $0\leq u\leq 1$. Also, being positive, $u$ has a (unique) positive square root.

Then $$ u-u^2=u^{1/2}(1-u)u^{1/2}\geq0. $$

Answer 2

Without loss of generality, assume $A$ is unital. Then $C^*(u,1)\cong C(\sigma(u)$, the continuous functions on the spectrum of $u$, and this isomorphism takes $u$ to the inclusion function, $\sigma(u)\ni\lambda\mapsto\lambda\in\mathbb C$. The conditions that $u$ is positive and $\|u\|\leq1$ then imply that $0\leq \lambda\leq 1$ for all $\lambda\in\sigma$, and for such $\lambda$, $\lambda^2\leq\lambda$. Using the above isomorphism, this yields that $u^2\leq u$.