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I need help on proving the following. If $A$ is a C* algebra with unit, not necessarily commutative, then why is it that $(b^*)b\leq||b||^21$ wrt the partial order induced by the cone $A_+$? Also, why is it then true that for all $a$ we have $a^*b^*ba\leq||b||^2a^*a$? In your response, please help me with a solution that does not involve any sort of embedding theorem. I seem to remember that there is an embedding of all C* algebras into B(H) for some Hilbert space H, but I don't know the proof. If this is true, please let me know, but I do not want to have it used if at all possible. Thanks!

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I should clarify that my definition of positive element in a C* algebra is one that is $a^*a$ for some $a$. In the B(H) case or for any of its norm-closed C* subalgebras with unit, this aligns with the idea that $<f, Tf>\geq 0$ is equivalent by definition to $T$ being positive. It is the abstract definition in the general case that gives me problems, as I do not know how to wield it to solve this problem. – Jeff May 17 '12 at 0:58
The second quickly follows from the first: if $T\ge 0$ then $a^*Ta\ge 0$ for any $a$. (Precisely, if $T=b^*b$, then $a^*Ta=(ba)^*(ba)$. – user31373 May 17 '12 at 1:08
First question: open any book covering $C^\ast$-algebras in a serious way e.g. Arveson recommended here. Second question: order is stable under conjugation by $a$. Last question: which will also be covered in that book. – t.b. May 17 '12 at 1:11
I take it you do not have at your disposal the fact that $b$ is positive if and only if $b=b^*$ and $\sigma(b)\subseteq [0,\infty)$? – Jonas Meyer May 17 '12 at 3:25
Thank you Jonas. that was very helpful. – Jeff May 17 '12 at 4:40
up vote 1 down vote accepted

The easiest way to prove the first inequality is indeed by embedding in $B(H)$ (via the universal representation).

Anyway, it can be done in different ways depending on what you want to use. One more algebraic way would be to notice that the spectrum of $\|b\|^2-b^*b$ is $$\tag{1} \{\|b\|^2-\lambda:\ \lambda\in\sigma(b^*b)\}. $$ Then you can also notice that the spectral radius of $b^*b$ is $$ r(b^*b)=\lim_{n\to\infty}\|(b^*b)^n\|^{1/n}\leq\|b^*b\|=\|b\|^2. $$ This tells us that any $\lambda$ in $\sigma(b^*b)$ satisfies $-\|b\|^2\leq\lambda\leq\|b\|^2$ (I'm using here that $\sigma(b^*b)\subset\mathbb{R}$, which is easy to obtain from the fact that $b^*b$ is selfadjoint); so all the elements in the set (1) are non-negative. Now we can apply functional calculus to the selfadjoint operator $c=\|b\|^2-b^*b$ with the function $g(t)=t^{1/2}$ (possible since the spectrum of $c$ is nonnegative); the operator $g(c)$ is selfadjoint and $\|b\|^2-b^*b=g(c)^*g(c)\geq0$.

The second inequality follows from the first one and the fact that if $x\leq y$, then $a^*xa\leq a^*ya$ (because $y-x=z^*z$, so $a^*(y-x)a=a^*z^*za=(za)^*za\geq0$.

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I thought that GNS referred to the particular representation induced by a state, while the Gelfand-Neumark theorem is the one that says every C*-algebra can be embedding in $B(H)$. The Gelfand-Neumark theorem can be proved roughly by using Hahn-Banach to get a bunch of states, and then taking the direct sum of their GNS representations. – Jonas Meyer May 17 '12 at 3:53
A potential problem here is that it is nontrivial to show from the abstract definition of a C*-algebra that $b^*b$ has spectrum contained in $[0,\infty)$. This fact is also useful in proving the Gelfand-Neumark theorem. – Jonas Meyer May 17 '12 at 3:55
I haven't read about GNS for many years, so I don't claim that I'm using the names in the "proper" way. Regarding the spectrum thing, I think that my argument also works if we only use that $\lambda$ is real, and that one is easier to obtain; I'll edit accordingly. – Martin Argerami May 17 '12 at 4:01
Yes, good point. $\sigma(b^*b)\subseteq [-\|b\|^2,\|b\|^2]$ suffices to show that $\sigma(\|b\|^21-b^*b)\subseteq[0,\infty)$, and then to finish it suffices to know just enough functional calculus to get a positive square root of $\|b\|^21-b^*b$. – Jonas Meyer May 17 '12 at 4:06
Good point too. I'll add the functional calculus part. – Martin Argerami May 17 '12 at 4:34

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