An inequality for completely positive maps.

Question

Let $f\colon A\to B$ be a contractive completely positive, ${}^*$-preserving map between C*-algebras and take $a\in A$. How one can prove that $$0\leqslant f(a)f(a^*)\leqslant f(aa^*)?$$ Some authors take it for granted without any explanation.

Answer 1

One of the most important results about completely positive maps is Stinespring's Dilation Theorem.

Suppose that $f:A \to B$ is a completely positive map, where $A$ and $B$ are $C^*$-algebras. Then we can find a Hilbert space $H$ such that $B \subseteq B(H)$.

Stinespring's Theorem then states that there exists a Hilbert space $K$, $*$-homomorphism $\pi: A \to B(K)$, and a bounded operator $V: H \to K$ such that $$ f(a)= V \pi(a) V^*, $$ with $\Vert f \Vert = \Vert V \Vert^2$.

The desired inequality now follows easily: \begin{array}{ccc} f(aa^*)& =& V \pi(aa^*) V^* \\ &= &V \pi(a) \pi(a^*) V^* \\ & \geq &V \pi(a) V^* V \pi(a^*) V^* \\ &= &f(a) f(a^*). \end{array} Here we use that $ V^*V \leq \Vert V \Vert^2 1 \leq \Vert f \Vert \leq 1$ because $f$ is contractive.

In fact, one can prove this result only using the weaker hypothesis that $f$ is $2$-positive. This is known as Kadison's Inequality.

Answer 2

Just for fun, the matrix trick mentioned in the answer by @Tom Cooney:

Assume $A$ is unital. If not, extend $f$ to unitization of $A$.

For any $a \in A$,

$$ \left[ \begin{matrix} 1 & a^*\\ a & aa^* \end{matrix} \right] $$

is positive in $M_2(A)$. The $2$-positivity of $f$ then tells you

$$ f(a) f(a^*) \leq f(aa^*). $$