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In relation to this question of mine: C* algebra inequalities

I am wondering if it is true that if $0\leq a \leq b$ in a C* algebra, does one have $||a||\leq||b||$? If you need the C* algebra to be unital you can assume that since I could use the C* algebra unitization. If it is not true, please state any reasonable conditions under which it is true because I see things like this used all the time, and it is possible that it merely applies in those special cases.

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Yes it is true in general. As you mention, the question isn't affected by going to the unitization, so assume that the algebra is unital.

Note that $b\leq \|b\|1$, because $\sigma(b)\subseteq[0,\|b\|]$, so that $\sigma(\|b\|1-b)\subseteq[0,\|b\|]$. Thus $a\leq \|b\|1$, and since $\|a\|$ is in the spectrum of $a$, this implies that $\|a\|\leq \|b\|$.


You could use Gelfand theory here, because $a$ and $b$ both commute with $\|b\|1$, so you could consider $C^*(b,1)$ then $C^*(a,1)$. However, all that is used above are the facts that $x\geq 0$ if and only if $x=x^*$ and $\sigma(x)\subseteq [0,\infty)$, and if $x\geq 0$, then the spectral radius of $x$ is $\|x\|$ (true for all normal elements, and quickly derived from the spectral radius formula and the C*-identity). Regardless, the "trick" is to use $\|b\|1$ as an intermediary between the possibly noncommuting $a$ and $b$.

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    http://math.stackexchange.com/questions/141094/in-a-c-algebra-does-a-leq-b-imply-a2-leq-b2 is an example of when $0\leq a \leq b$ in a C* algebra of operators, even though one does not have $a^2\leq b^2$. The latter statement is equivalent in this case to having $||a||\leq ||b||$. Isn't this right? Edit: actually this remark I made is not right. I also made the same remark above, and I now retract it.2012-06-03
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    @Jeff: No [Edit: as you noticed]. If you want a sanity check, you could compute the norms in the example you linked to. Incidentally, note that you will also have $\|a^2\|\leq\|b^2\|$ in such cases (simply because $\|x^2\|=\|x\|^2$ when $x$ is self-adjoint, or even normal).2012-06-03
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    Thank you for this and your other comment.2012-06-03
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When $a$ and $b$ commute, it is true. You only need to look at the commutative C*-subalgebra generated by $a, b, 1$. But for the general case, I don't remember any thing right now.

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    Actually it just occurred to me that I once read a thread here where the OP asked if $0\leq a \leq b$ implies $a^2 \leq b^2$ and there a counterexample was given in the answer that was from 2x2 matrices. The connection I just made is that this question is equivalent to mine in the operator case, meaning that even if we have a C* algebra of operators, one can fail to have what I stated. I see the proof to which you are alluding in the commutative case. Perhaps that is what people use.2012-06-02
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    Ah actually I misunderstood the proof, but I think now I have it. Just for the sake of having a record, I'll include it here. Let $0\leq a \leq b$. WLOG suppose our C* algebra is unital. Then if $ab=ba$ then by definition we have $b-a \geq 0$. Since the positive elements form a cone, we have that $b+a\geq 0$. Therefore, we have that $b^2-a^2=(b+a)(b-a)=(b+a)^{1/2}(b-a)(b+a)^{1/2}\geq 0$. hm... actually this seems only to work for operators. Can you shed some light on this Vahid or anybody else?2012-06-02
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    Oh. I should be familiar with this trick by now... One always uses the Gel'fand transform in the commutative case to relate abstract positivity to a concrete positivity that obeys the common laws. Pity I was trying to use the functional calculus instead.2012-06-02
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    @Jeff: You may be interested in "operator monotone" functions. These are real-valued functions on $\mathbb R$ (or subsets of $\mathbb R$) that preserve inequalities when applied to self-adjoint elements of C*-algebras via functional calculus. For example, $0\leq a\leq b$ does imply that $\sqrt a\leq \sqrt b$, and if $a$ is invertible, then so is $b$, and $b^{-1}\leq a^{-1}$. This means that $t\mapsto \sqrt t$ is operator monotone on $[0,\infty)$, and $t\mapsto-\frac{1}{t}$ is operator monotone on $(0,\infty)$. Furthermore, $t\mapsto t^p$ is operator monotone if $0, never if $p>1$.2012-06-03
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    @Jeff: In the commutative case, $C^*(a,b,1)\cong C(X)$ for some compact Hausdorff space $X$, and the isomorphism preserves order, so it reduces to the statement that $0\leq f(x)\leq g(x)$ for all $x\in X$ implies $\sup f(X)\leq \sup g(X)$.2012-06-03
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    @Jeff: Sorry Jeff, for brevity. Jonas' answer completes the proof.2012-06-03