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I know that if $\Lambda$ is a stochastic positive linear map, i.e., $\Lambda(I) = I$, it is true that

\[ \|\Lambda(B)\| \leq \| B \| \]

For any operator $B$, where $\|\cdot\|$ is the standard operator norm $\|B\| := \max_{|v|= 1} |Bv|$. Is it true for any other $p$-norm? Specifically, I want to prove it for the $2$-norm

\[ \|B\|^2_2 := \operatorname{tr} (B^*B)\]

also known as Hilbert-Schmidt norm, and I'm only interested in self-adjoint operators.

Naturaly, this question only makes sense if these operators have well-defined norms, so $\Lambda$ can be taken to act in this subalgebra.

It would be nice if the infinite-dimensional case could be done, but the main focus is on the finite-dimensional case.

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    $\begingroup$ I'm confused about your first claim. If stochastic just means that it maps the identity to the identity then this does not imply that $\Lambda$ has norm 1. On the 2x2 matrices consider, $\Lambda$= $2(x)-Tr(x)I$. Where Tr is the normalized trace. This sends I to I but has norm larger than 2. You're statement requires that $\Lambda$ be a positive operator. Under this situation (that $\Lambda$ is positive) then this should still work for all p-norms. $\endgroup$
    – Owen Sizemore
    Commented Sep 13, 2010 at 21:05
  • $\begingroup$ Yes! Sorry, I've been working only with positive maps, so that I forgot to specify. $\endgroup$
    – Mateus Araújo
    Commented Sep 13, 2010 at 21:33
  • $\begingroup$ Could you also clarify what $\Lambda$ is acting on. Is it all of B(H)? Or is it some subalgebra?. The reason I ask this is to clarify what you mean by the 2-norm. Do you mean the Hilbert-Schmidt norm? If so this means that we must restrict to those B with finite Hilbert Schmidt norm. $\endgroup$
    – Owen Sizemore
    Commented Sep 14, 2010 at 1:02

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Positivity is not enough (complete positivity is).

Indeed, in $M_2(\mathbb{C})$, let $\Lambda\left(\left[\begin{array}{cc}a&b \\\\ c&d\end{array}\right]\right)=\left[\begin{array}{cc}a&0 \\\\ 0&a\end{array}\right]$. Then $\Lambda$ is positive and $\Lambda(I)=I$; but if $B=\left[\begin{array}{cc}1&0 \\\\ 0&0\end{array}\right]$ we have, for any $1\leq p<\infty$, $\|B\|_p=1$, $\|\Lambda(B)\|_p=2^{1/p}$.

For a unital completely positive map, Stinespring leads to an easy proof that the inequality holds for $p=2$. For other $p$ it looks trickier, but I think it might work too.

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  • $\begingroup$ Great counterexample! Unfortunately, the positivity hypothesis is crucial to me. So although Stinespring does lead to an easy proof (actually more Choi's result), its useless to me. I'm trying to derive in a elegant way the bound [ \| I \otimes \Lambda(B) \| \le \| B \|_2, ] Which is valid for any positive stochastic Lambda. This would be a way. Guess I'll find another. $\endgroup$
    – Mateus Araújo
    Commented Sep 16, 2010 at 5:42

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