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Let $\mathcal{E} \to CP(A \to B)$ (completely positive linear map) be a trace non-increasing $CP$ map. Show that any operator sum representation $\{M_x\}_{x=1}^{m}$ of $\mathcal{E}$ satisfes $\sum_{x=1}^{m}M^{*}_xM_x \leq I^A$. Show that the marginal of the Choi matrix $J^{AB}{\mathcal{E}}$ satisfies $J^A{\mathcal{E}}:= Tr_B[J^{AB}_{\mathcal{E}}] \leq I^A$.

The proof the second part is as follows, but I dont know how to prove first part. Let$\mathcal{E}:\mathcal{B}(\mathcal{H}^A)\to \mathcal{B}(\mathcal{H}^B)$ be a linear map. Clearly if $\mathcal{E}$ is completely positive then by definition $J^{AB}{\mathcal{E}}:=\text{id} \otimes \mathcal{E}(\phi^{AA^{\prime}}_{+})\geq0$.

Suppose now that $J^{AB}_{\mathcal{E}} \geq 0$. Let $k \in \mathbb{N}$, and $|\psi^{RA}\rangle \in \mathbb{C}^k \otimes \mathbb{C}^d$, where R is a k-dimensional (reference) system. $|\psi^{RA}\rangle = M_{\psi} \otimes I^A |\phi^{A A^{\prime}}_{+}\rangle$. Where $M_{\psi}:\mathcal{H}^{A^{\prime}} \to \mathcal{H}^R$is a linear operator. We therefore have $$ \text{id}_k \otimes \mathcal{E} |\psi^{RA}\rangle\langle\psi^{RA}| = (\text{id}_k \otimes \mathcal{E}) ( M_{\psi} \otimes I^A ) |\phi^{A A^{\prime}}_{+}\rangle \langle\phi^{A A^{\prime}}_{+}| ( M_{\psi}^{*} \otimes I^A ) $$ $$ = ( M_{\psi} \otimes I^A ) J^{AB}_{\mathcal{E}} ( M_{\psi}^{*} \otimes I^A ) $$ Finally any operator $\rho^{RA}\geq 0$ can be diagonalized as $\rho^{RA}= \sum_{x=1}^m |\psi_x^{RA}\rangle \langle\psi_x^{RA}|$, where $|\psi_x^{RA}\rangle \in \mathbb{C}^k \otimes \mathbb{C}^d$ are some (possibly unnormalized) pure states. Since $|\psi^{RA}\rangle$ above was arbitrary, we conclude that $(\text{id}_k \otimes \mathcal{E}) \rho^{RA} = \sum_{x=1}^m (\text{id}_k \otimes \mathcal{E}) (|\phi_x^{RA}\rangle \langle\phi_x^{RA}|) \geq 0$

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We are given $\mathcal{E}_{A\rightarrow B}$ to be some completely positive trace nonincreasing map with Kraus operators $\{M_i\}$ so that $\mathcal{E}(\rho) = \sum_i M_i\rho M_i^\dagger$.

For any $\rho$, it holds that \begin{align} \langle I_A, \rho\rangle &= \text{tr}(\rho)\\ &\geq \text{tr}(\mathcal{E}(\rho)) \\ &= \langle I_B, \mathcal{E}(\rho)\rangle \\ &= \langle I_B, \sum_i M_i \rho M_i^\dagger\rangle \\ &= \langle \sum_i M_i^\dagger M_i, \rho\rangle \end{align}

Thus, $\langle I_A - \sum_i M_i^\dagger M_i, \rho\rangle \geq 0$ for any $\rho$ and hence, $\sum_i M^\dagger_i M_i \leq I_A$.

For the second question, let us choose an orthonormal basis $\{\vert i\rangle\}$ for $\mathcal{H}_A$. Note that $J(\mathcal{E}) = \sum_{i,j} \vert i\rangle\langle j\vert_A\otimes\mathcal{E}(\vert i\rangle\langle j\vert)_B$. Since $\mathcal{E}$ is trace nonincreasing, it holds that $\text{tr}(\mathcal{E}(\vert i\rangle\langle j\vert))\leq \text{tr}(\vert i\rangle\langle j\vert) = \delta_{ij}$ and hence

$$\text{tr}_B\left(J(\mathcal{E})\right) = \sum_{i,j} \vert i\rangle\langle j\vert_A\otimes\text{tr}\left(\mathcal{E}(\vert i\rangle\langle j\vert)\right)\leq \sum_{i,j}\vert i\rangle\langle j\vert \delta_{ij} = I_A $$

This proof, and other useful identities, can be found in these lecture notes by John Watrous.

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