Ordering: Definition

Question

This was a real question!

Given a unital C*-algebra $1\in\mathcal{A}$.

For $A\in\mathcal{A}$ denote its spectrum by $\sigma(A)$.

Consider the selfadjoints: $$\mathcal{A}_*:=\{A\in\mathcal{A}:A=A^*\}$$

Introduce an ordering: $$A,A'\in\mathcal{A}_*:\quad A\leq A':\iff\sigma(A'-A)\geq0$$

Then it is a partial order: $$A\leq A'\leq A\implies A=A'$$ $$A\leq A'\leq A''\implies A\leq A''$$ How can I prove this?*

*By results related to Functional Calculus and Neumann Series; but not Gelfand-Naimark!

Answer 1

Meanwhile I remember...

Reduction

By construction one has: $$A\leq B\implies A+C\leq B+C$$

So one may reduce to: $$0\leq A\leq0\implies A=0$$ $$0\leq A,B\implies0\leq A+B$$ That is a positive cone.

Reflexivity

It hold the implications: $$0\leq A\leq0\implies\sigma(A)=(0)\implies r(A)=0\implies\|A\|=0\implies A=0$$ Concluding transitivity.

Characterization

For spectrum one has: $$\sigma(A+1)=\sigma(A)+\sigma(1)\quad\sigma(\lambda A)=\lambda\sigma(A)\quad\sigma(1)=(1)$$

For normals one has: $$N^*N=NN^*:\quad\|N\|=\|\sigma(N)\|=:r(N)$$

For selfadjoints one has: $$S=S^*:\quad-r(S)\leq\sigma(S)\leq r(S)$$

All together one obtains: $$A\geq0\iff\ldots\iff\exists r\geq0:\quad\|(\|A\|+r)1-A\|\leq\|A\|+r$$ Geometric characterization.

Transitivity

It hold the inequalities: $$\|(\|A\|+\|B\|)1-(A+B)\|\leq\|\|A\|1-A\|+\|\|B\|1-B\|\leq\|A\|+\|B\|$$ Concluding reflexivity.

Answer 2

The only thing to be proved is transitivity. To do this, it is enough to show that if $u,v\in\mathcal A$ are self-adjoint and satisfy $\sigma (u), \sigma(v)\subseteq\mathbb R^+$, then $\sigma (u+v)\subseteq\mathbb R^+$. I reproduce the proof given in Arveson's book.

Without loss of generality, assume that $\Vert u\Vert, \Vert v\Vert\leq 1$. Then $\sigma (u), \sigma (v)\subseteq [0,1]$ because the spectral radius is not greater than the norm. It follows that $\sigma (\mathbf 1-u)$ and $\sigma (\mathbf 1-v)$ are also contained in $[0,1]$. But $\mathbf 1-u$ and $\mathbf 1-v$ are self-adjoint, and it is known (and provable by elementary means...) that the spectral radius of any self-adjoint $x\in\mathcal A$ is equal to its norm. So we have $\Vert\mathbf 1-u\Vert\leq 1$ and $\Vert\mathbf 1-v\Vert\leq 1$; and hence, by convexity of the norm, $\left\Vert \mathbf 1-\frac{u+v}2\right\Vert\leq 1$. Since $\mathbf 1-\frac{u+v}2$ is self-adjoint and the spectral radius does not exceed the norm (once again), it follows that the spectrum of $\mathbf 1-\frac{u+v}2$ is contained in $[-1,1]$, so that the spectrum of $\frac{u+v}2$ is contained in $[0,2]$. Hence, $\sigma (u+v)\subseteq\mathbb R^+$.