Proving two stubborn inequalities for completely positive maps in C*-algebras

Question

I recently came across this in my studies of functional analysis in C* algebras which got me stuck:

For a completely positive map between C* algebras $ \phi : A \to B $ we are to prove these two inequalities where $ a,b \in A $:

$ \phi(a) ^{*} \phi(a) \leq ||\phi(1)||\phi(a^{*}a) $

$ ||\phi(a^{*}b)||^2 \leq ||\phi(a^{*}a)|| \space || \phi(b^{*}b) || $

where * denotes of course the dual. I am truly stuck as I tried anything so I cannot really get past these, and I am truly in need of help. I thank all helpers.

Answer 1

1. Your first inequality is usually known as the Kadison-Schwarz inequality. It only requires $2$-positivity.

Claim. $A=\begin{bmatrix}I &a\\ a^*&b\end{bmatrix}\geq0$ if and only if $a^*a\leq b$.

Proof. If $A\geq0$, then for any $\xi\in H$, $$ \langle (b-a^*a)\xi,\xi\rangle=\left\langle \begin{bmatrix}I &a\\ a^*&b\end{bmatrix}\,\begin{bmatrix}-a\xi\\ \xi\end{bmatrix},\begin{bmatrix}-a\xi\\ \xi\end{bmatrix}\right\rangle \geq0, $$ so $a^*a\leq b$. Conversely, if $a^*a\leq b$, then for any $\xi,\eta\in H$, \begin{align*} \left\langle \begin{bmatrix}I &a\\ a^*&b\end{bmatrix}\,\begin{bmatrix}\xi\\ \eta\end{bmatrix},\begin{bmatrix}\xi\\ \eta\end{bmatrix}\right\rangle &=\|\xi\|^2+\langle b\eta,\eta\rangle+2\text{Re}\,\langle a\eta,\xi\rangle \geq\|\xi\|^2+\langle b\eta,\eta\rangle-2\langle a^*a\eta,\eta\rangle^{1/2}\,\|\xi\| \\ &\geq\|\xi\|^2+\langle b\eta,\eta\rangle-2\langle b\eta,\eta\rangle^{1/2}\,\|\xi\| =(\|\xi\|-\langle b\eta,\eta\rangle)^2\geq0.&\Box \end{align*} $$ \ $$

Now, since $a^*a\leq a^*a$, we have that $\begin{bmatrix}I&a\\a^*&a^*a\end{bmatrix}\geq0$. As $\phi$ is $2$-positive, $$ \begin{bmatrix}\|\phi(I)\|\,I &\phi(a)\\ \phi(a)^*&\phi(a^*a)\end{bmatrix}\geq\begin{bmatrix}\phi(I) &\phi(a)\\ \phi(a)^*&\phi(a^*a)\end{bmatrix}=\phi^{(2)}\left(\begin{bmatrix}I & a\\ a^*& a^*a\end{bmatrix}\right)\geq0. $$ Applying the Claim again, we get $$ \frac{\phi(a)^*}{\|\phi(I)\|}\,\frac{\phi(a)}{\|\phi(I)\|}\leq\frac{\phi(a^*a)}{\|\phi(I)\|}, $$ so $$ \phi(a)^*\phi(a)\leq\|\phi(I)\|\,\phi(a^*a). $$

On the other hand, a very simple proof can be obtained using Stinespring's Dilation: if you write $\phi(x)=V^*\pi(x)V$, then \begin{align} \phi(a)^*\phi(a)&=V^*\pi(a)^*VV^*\pi(a)V\leq\|VV^*\|\,V^*\pi(a)^*\pi(a)V\\ \ \\ &=\|V\|^2\,V^*\pi(a^*a)V=\|V^*V\|\,\phi(a^*a)\\ \ \\ &=\|\phi(I)\|\,\phi(a^*a). \end{align}

2. For the second inequality, only positivity is needed. Let $f$ be a state on $A$. Then $a\longmapsto f(\phi(a))$ is a positive linear functional. As such, it satisfies Cauchy-Schwarz. So $$ f(\phi(a^*b))^2\leq f(\phi(a^*a))\,f(\phi(b^*b))\leq\|\phi(a^*a)\|\,\|\phi(b^*b)\|. $$ As the norm of any operator can be realized via states, $$ \|\phi(a^*b)\|^2\leq \|\phi(a^*a)\|\,\|\phi(b^*b)\|. $$