# Schwarz inequality for unital completely positive maps

I came across the following form of Schwarz inequality for completely positive maps in Arveson's paper:

Let $\delta:\mathcal{A}\to\mathcal{B}$ be a unital completely positive linear map between two $C^*$-algebras, then $$\delta(A)^*\delta(A)\le \delta(A^*A),$$

which is a crucial part in proving proposition 3.

I do not know this inequality and thus searched on wiki, something related is

(Kadison-Schwarz) If $\phi$ is a unital positive map, then $$\phi(a^*a)\ge\phi(a^*)\phi(a)$$ for all normal elements $a$.

However, I cannot find a proof of Kadison-Schwarz. Also, since Kadison-Schwarz works only for normal elements, there seems to be a gap between Kadison-Schwarz and Schwarz.

I wonder where I can find a proof to the first inequality. Thanks!

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Note that Kadison-Schwarz is stated for positive (not necessarily cp) maps; that's why the restriction to normals is required (the C$^*$-algebra generated by a normal is abelian, and then any positive map is completely positive).

Also, for Schwarz inequality all that is required is 2-positivity (which of course is implied by complete positivity).

This is the proof of the inequality as in Paulsen's book:

We have $$\begin{bmatrix}1&a \\ a^*& a^*a\end{bmatrix}=\begin{bmatrix}1&a\\ 0&0\end{bmatrix}^*\begin{bmatrix}1&a\\ 0&0\end{bmatrix}\geq0.$$ Then $$0\leq\delta^{(2)}\left(\begin{bmatrix}1&a \\ a^*& a^*a\end{bmatrix}\right) =\begin{bmatrix}1&\delta(a) \\ \delta(a)^*& \delta(a^*a)\end{bmatrix}$$ Applying this in particular to the vector $\begin{bmatrix}-\delta(a)\eta\\ \eta\end{bmatrix}$, we get $$0\leq\left\langle\begin{bmatrix}1&\delta(a) \\ \delta(a)^*& \delta(a^*a)\end{bmatrix}\begin{bmatrix}-\delta(a)\eta\\ \eta\end{bmatrix},\begin{bmatrix}-\delta(a)\eta\\ \eta\end{bmatrix}\right\rangle=\langle(\delta(a^*a)-\delta(a)^*\delta(a))\eta,\eta\rangle.$$ As the vector $\eta$ can be chosen arbitrarily, we conclude that $\delta(a^*a)-\delta(a)^*\delta(a)\geq0$.

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Thanks! Quite elegant proof. But I guess in the second inequality it should be \le. – Hui Yu Jul 22 '12 at 21:14
It's corrected now. Thanks! – Martin Argerami Jul 22 '12 at 22:09