First, let us prove that every quantum channel is trace-norm contractive using the unitary Stinespring representation. For this we need the following lemma.
Lemma. For all $m,n\in\mathbb N$ and all $Z\in\mathbb C^{n\times n}\otimes\mathbb C^{m\times m}$ it holds that $\|{\rm tr}_2(Z)\|_1\leq\|Z\|_1$.
Proof. The easiest way to show this is to go via the adjoint channel/dual channel, that is, the fact that ${\rm tr}({\rm tr}_2(Z)Y)={\rm tr}(Z(Y\otimes{\bf1}))$ for all $Z\in\mathbb C^{n\times n}\otimes\mathbb C^{m\times m}$, $Y\in\mathbb C^{n\times n}$. This simplifies the norm computation by the duality of the trace norm and the $\infty$-norm:
\begin{align*}
\|{\rm tr}_2(Z)\|_1&={\rm sup}_{Y\in\mathbb C^{m\times m},\|Y\|_\infty=1}|{\rm tr}({\rm tr}_2(Z)Y) )|\\
&={\rm sup}_{Y\in\mathbb C^{m\times m},\|Y\|_\infty=1}|{\rm tr}(Z(Y\otimes{\bf1}))|\\
&\leq{\rm sup}_{Y\in\mathbb C^{m\times m},\|Y\|_\infty=1}\|Z\|_1\|Y\otimes{\bf1}\|_\infty\\
&={\rm sup}_{Y\in\mathbb C^{m\times m},\|Y\|_\infty=1}\|Z\|_1\|Y\|_\infty\|{\bf1}\|_\infty=\|Z\|_1\,.
\end{align*}
In the second-to-last line we used this inequality and in the last line we used that the $\infty$-norm factorizes under tensor products. $\square$
With this the statement we want to prove follows at once; for all $X\in\mathbb C^{n\times n}$, all states $\omega\in\mathbb C^{m\times m}$ and all unitaries $U\in\mathbb C^{n\times n}\otimes\mathbb C^{m\times m}$ we compute
\begin{align*}
\|{\rm tr}_2(U( X\otimes\omega )U^*)\|_1&\leq \|U( X\otimes\omega )U^*\|_1\\
&=\|X\otimes\omega\|_1\\
&=\|X\|_1\|\omega\|_1\\
&=\|X\|_1{\rm tr}(\omega)=\|X\|_1\,.
\end{align*}
In the second step we used that the trace norm is invariant under unitary channels (because the singular values do not change) and in the last line we used that because $\omega\geq 0$ its trace norm is equal to its trace (obvious from the definition of $\|\cdot\|_1$).
One can re-write this result in terms of superoperator-norms as $\|T\|_{1\to 1}\leq 1$ for all channels $T$ where $\|T\|_{1\to 1}:=\sup_{\|X\|_1=1}\|T(X)\|_1$; in fact because every channel has a fixed point one even gets $\|T\|_{1\to 1}=1$ for all channels.
The follow-up question now is whether this result continues to hold if the underlying trace norm is replaced by another norm. For this one defines the general Schatten $p$-norm as $$\|X\|_p:=\big({\rm tr}((X^\dagger X)^{p/2})\big)^{1/p}$$---or, equivalently, $\|X\|_p:=(\sum_j\sigma_j(X)^p)^{1/p}$ where $\sigma_j(X)$ are the singular values of $X$---as well as the corresponding operator norm $\|\Phi\|_{p\to p}:=\sup_{\|X\|_p=1}\|\Phi(X)\|_p$. Note that this Schatten-$p$ norm is the natural generalization of $\|\cdot\|_2$ and $\|\cdot\|_\infty$ to other indices.
In this language, asking whether $\|T(X)\|_p\leq\|X\|_p$ for all $X$ is equivalent to $\|T\|_{p\to p}\leq 1$ ("$T$ is $p$-norm contractive"). Indeed, one can characterize when this is the case: as shown in this paper (arXiv) the following statements are equivalent:
- $\Phi$ is unital, i.e. $\Phi({\bf1})={\bf1}$
- $\|\Phi\|_{p\to p}=1$ for all $p\in[1,\infty]$
- $\|\Phi\|_{p\to p}=1$ for some $p\in(1,\infty]$
In other words:
For all channels that are not unital and all $p>1$ there exists $X$ such that $\|T(X)\|_p>\|X\|_p$.
