C* algebra inequalities

Question

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!

Answer 1

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 $a^*(y-x)a=a^*z^*za=(za)^*za\geq0$.