In a C*-algebra, does $a \leq b$ imply $a^2 \leq b^2$?

Question

While attempting to fill in the gaps in a proof of the Gelfand-Naimark-Segal representation theorem that I was given in a course in operator algebras, I found myself wondering whether, if $(u_\lambda)_\lambda$ is an approximate unit of a C*-algebra $\mathcal{A}$, it is true that so is $(u_\lambda^2)_\lambda$. I need this fact to complete my version of the proof of existence of a cyclic vector for the GNS representation of a non-unital C*-algebra.

Clearly, $(u_\lambda^2)_\lambda$ is a net of positive elements, bounded in norm by 1. But how to show that this net is increasing, knowing that $(u_\lambda)_\lambda$ is?. I wouldn't even know how to start proving something like this for an arbitrary approximate unit, but it's enough for my proof to show it for the canonical approximate unit of $\mathcal{A}$ with the usual order.

So this amounts to showing (I cannot find a counterexample) that if $u_\lambda \leq u_{\lambda'}$ then $u_\lambda^2 \leq u_{\lambda'}^2$. More generally, one can ask if $a \leq b$ implies $a^2 \leq b^2$, or even $a^n \leq b^n$.

As I said I'm a bit stuck with this. Thanks for your help!

Answer 1

The general statement ($0 \le a \le b$ implies $a^2 \le b^2$) is false, even for $2 \times 2$ matrices. Consider e.g. $$a = \pmatrix{1 & -1\cr -1 & 1\cr},\ b = \pmatrix{ 2 & -3\cr -3 & 5\cr}$$

Answer 2

It is true in a commutative unital C* algebra because you use the gel'fand transform to relate the abstract notion of positivity to one on functions mapping from the maximal ideal space into C, and this is a completely concrete and usual thing that is easy to deal with.

Even more in [1] it is proved that a $C^*$-algebra is commutative iff this implication is always true.

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[1] T. Ogasawara, A Theorem on Operator Algebras, Journal of Science of The Hiroshima University, Ser A, vol 18, No. 3, 1955, pp. 307-309.