= Solution
\b[The inequality printed in this part is false for general positive functionals.] In $A=M_2(\mathbb C)$ take the <state on a C-star algebra> $f(a)=a_{22}$ and $x=y=E_{12}$. It is positive since $f(z^*z)=\|ze_2\|^2\geq0$, and it has norm one. But
$$
x^*y=E_{22},\qquad xx^*=yy^*=E_{11},\qquad |f(x^*y)|^2=1,\quad f(xx^*)f(yy^*)=0.
$$
Thus the literal requested inequality would say $1\leq0$.
The correct <Cauchy–Schwarz inequality for positive C-star functionals> keeps the orders consistent:
$$
\boxed{|f(x^*y)|^2\leq f(x^*x)f(y^*y).}
$$
To prove it, set $a=f(x^*x)$, $d=f(y^*y)$ and $b=f(x^*y)$. By part (i) and positivity,
$$
0\leq f((x+ty)^*(x+ty))=a+2\operatorname{Re}(tb)+|t|^2d\qquad(t\in\mathbb C).
$$
If $d>0$, choose $t=-\overline b/d$ to obtain $|b|^2\leq ad$. If $d=0$, arbitrary choices of $t$ force $b=0$, and the inequality again follows. Replacing $x,y$ by $x^*,y^*$ also yields the valid alternative
$$
\boxed{|f(xy^*)|^2\leq f(xx^*)f(yy^*).}
$$
Either replacing the printed right-hand factors by $f(x^*x),f(y^*y)$, or replacing the left-hand product by $xy^*$, repairs the statement. No tracial hypothesis is supplied, so it cannot be silently used to exchange those products.
Back to article page
