Is it true that for all trace preserving positive maps $\phi$ that $\phi^\dagger (\rho)\leq 1$, where $Tr(\rho)=1$ and $\rho$ is semi-positive?

Question

I would conclude it from the following reasioning:

Take $\rho$ and $Y$, both semi-positive. We have (with Cauchy-Schwarz inequality) $Tr(Y \phi^\dagger(\rho))= Tr(\phi(Y)\rho)\leq Tr(\phi(Y))Tr(\rho)=Tr(Y)$.

Accordingly $0\geq Tr(Y(1- \phi^\dagger(\rho))) $, for all semi-positive $Y$, with this $\phi^\dagger(\rho)\leq 1$.

Is the reasoning correct?

Answer 1

I don't see how you apply the Cauchy-Schwarz inequality there. You should get squares inside the trace.

However, the statement is still true. Note that since $\mathrm{Tr}(\rho)=1$ and $\rho\geq 0$, all eigenvalues of $\rho$ are in $[0,1]$. Thus $\rho\leq 1$ and you have $$ \mathrm{Tr}(Y\phi^\dagger(\rho))=\mathrm{tr}(\phi(Y)\rho)\leq \mathrm{Tr}(\phi(Y))=\mathrm{Tr}(Y) $$ for $Y\geq 0$. Hence $\phi^\dagger(\rho)\leq 1$.