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!

  • 1
    $\begingroup$ You need to work on your exposition. Please define: $\sigma$, Neumann series, Gelfand-Neimark. $\endgroup$ – Mario Carneiro Oct 19 '15 at 17:07
  • $\begingroup$ @MarioCarneiro: Definition of spectrum added. Defining the Neumann series and Gelfand-Naimark would definitely go way too far beyond this post. Those two were just meant as a hint for what proof I'm looking for. $\endgroup$ – C-Star-W-Star Oct 19 '15 at 17:12
  • $\begingroup$ Actually define is the wrong word for the last two. Rather, explain in what way they relate to your problem. (When you say "by Neumann series", do you mean that you have deduced the next line using the Neumann series, or you want to prove that line using Neumann series, or something else?) Also if you don't want to define certain terms, using links to wikipedia is a good way to make your post self-contained. We are not all experts in your field, so unraveling some (but not all) definitions can help make the problem understandable to a wider audience (some of which might be able to help you). $\endgroup$ – Mario Carneiro Oct 19 '15 at 17:19
  • $\begingroup$ @MarioCarneiro: Yes I know I were just phrasing your last comment... The referencing is a nice idea - thanks!!!! $\endgroup$ – C-Star-W-Star Oct 19 '15 at 17:20
  • $\begingroup$ If "functional calculus" includes the continuous functional calculus, part of that is that the Gelfand representation of a commutative $C^*$-algebra is an isometry. Is that meant to be excluded by your "not Gelfand-Naimark"? $\endgroup$ – Robert Israel Oct 19 '15 at 17:46

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, $\sigma (u+v)\subseteq\mathbb R^+$.

  • $\begingroup$ Thank you!!! :) ...I hope you don't mind that I posted an answer too?? In any case: (+1) $\endgroup$ – C-Star-W-Star Oct 20 '15 at 0:58
  • $\begingroup$ You're welcome! However, I don't understand your proof of transitivity. $\endgroup$ – Etienne Oct 20 '15 at 17:27
  • $\begingroup$ That is a nothing but: Flip the spectrum by $\sigma(-A)$ and shift it by $\sigma(-A+\Delta1)$ far enough $\Delta\geq\|A\|$ so that $\sigma(-A+\Delta1)\geq0$. Then $\sigma(-A+\Delta1)\leq R$ if and only if $r(-A+\Delta1)\leq R$. (Best visualize it!) $\endgroup$ – C-Star-W-Star Oct 20 '15 at 17:40
  • $\begingroup$ But what does this have to do with transitivity of the relation $\leq$? $\endgroup$ – Etienne Oct 20 '15 at 18:47
  • $\begingroup$ Transitivity is precisely related to the triangle inequality: For $A\geq0$ and $B\geq0$ if and only if $\|\|A\|1-A\|\leq\|A\|$ and $\|\|B\|1-B\|\leq\|B\|$. Thus $\|(\|A\|+\|B\|)1-(A+B)\|\leq\ldots\leq(\|A+\|B\|)$. Hence $A+B\geq0$. $\endgroup$ – C-Star-W-Star Oct 20 '15 at 19:54

Meanwhile I remember...


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.


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


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: $$A\geq0\iff\ldots\iff\exists r\geq0:\quad\|(\|A\|+r)1-A\|\leq\|A\|+r$$ Positive Cone Geometric characterization.


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


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