Let, $\mathcal{E}$ be a Completely-Positive Trace-Preserving Map, i.e, for linear operators $\rho$ on a Hilbert-Space, $$ \mathcal{E}(\rho) = \sum_i E_i\rho E_i^{\dagger}\qquad {\rm s.t.}\qquad \sum_i E_i^{\dagger}E_i = \mathbb{I} $$ Show that, there exists an isometry $U$ (i.e, $U^{\dagger}U = \mathbb{I}$), such that, for all CPTP Maps $\mathcal{E}$ and all traceless Hermitian operators $\nu$, $$ ||\mathcal{E}(\nu)||_1 \leq ||\mathcal{E}\otimes \mathcal{I}(U\nu U^{\dagger})||_1 $$ where, $||A||_1 = \rm{Tr}(|A|) = \rm{Tr}(\sqrt{A^{\dagger}A})$ is the Trace-Norm of $A$.
Note: I am not sure if the traceless condition is necessary to prove the inequality. What conditions does $U$ need to satisfy in general so that this inequality holds for all CPTP maps $\mathcal{E}$ and (traceless) operators $\nu$ ?
The question—as worded currently—needs neither that $\nu$ is Hermitian and traceless, nor that $\mathcal E$ is completely positive or trace-preserving. The reason for this is that the isometry $U:\mathbb C^n\to\mathbb C^n\otimes\mathbb C^m$, $x\mapsto x\otimes\psi$ (where $\psi\in\mathbb C^m$ is an arbitrary unit vector) satisfies $$ (\mathcal E\otimes\mathcal I)(U\nu U^\dagger)=(\mathcal E\otimes\mathcal I)(\nu \otimes|\psi\rangle\langle\psi|)=\mathcal E(\nu)\otimes|\psi\rangle\langle\psi| $$ for all linear maps $\mathcal E:\mathbb C^{n\times n}\to\mathbb C^{k\times k}$ and all matrices $\nu\in\mathbb C^{n\times n}$ This implies the desired trace norm (in)equality via: \begin{align*} \|\mathcal E(\nu)\|_1=\|\mathcal E(\nu)\|_1\cdot\|\psi\|^2=\|\mathcal E(\nu)\|_1\cdot\|\,|\psi\rangle\langle\psi|\,\|_1&=\| \mathcal E(\nu)\otimes|\psi\rangle\langle\psi|\, \|_1\\&=\| (\mathcal E\otimes\mathcal I)(U\nu U^\dagger) \|_1 \end{align*} Note that I chose $\mathcal E$ to act on finite dimensional spaces for simplicity, but this argument works for arbitrary Hilbert spaces.