While reading Holevo's book on Quantum Information Theory, there was an exercise question.
Let $\Phi$ be a 2-positive linear operator map (between $\mathcal{L}(H)$ to $\mathcal{L}(H)$, $H$ being a Hilbert space). That is: $$\Phi\otimes I_2 \geq 0$$ where $I_2$ is the identity map for $2\times 2$ matrices. Then show that for any operator $X\in \mathcal{L}(H)$, $$\Phi(X^*)\Phi(X)\leq \|\Phi(I)\|\Phi(X^*X)$$
The hint was to exploit 2-positivity. I then checked some of the answers on the site, but I observed that either those were special cases of above or they gave some alternate proof. I'm not very comfortable manipulating tensors, but this is what I got so far.
Proof: Let $T$ be defined as $$\tag{a}T=\begin{bmatrix} I && X \\ 0 && 0 \end{bmatrix} $$ Then $T^*T\geq 0$ and $$T^*T=\begin{bmatrix} I && X \\ X^* && X^*X \end{bmatrix} $$ Now $$\tag{b}(\Phi \otimes I_2)(T^*T)=\begin{bmatrix} \Phi(I) && \Phi(X) \\ \Phi(X^*) && \Phi(X^*X) \end{bmatrix} \geq 0 $$
Now for any vector $U$ in the tensor space we have that $\langle U^*(\Phi \otimes I_2)( T^*T)U\rangle \ge 0$. Now if $\Phi(I) = \|\Phi(I)\|I = \alpha I$, then pick $$U=\begin{bmatrix}-\Phi(X) \\ \alpha I \end{bmatrix}$$ Substitute in the inner product above to get the result.
Now I have a few doubts about this proof:
$(a)$ Is $T$ as defined above an element in the tensor space? I have trouble visualizing it.
$(b)$ Is step (b) valid. All i've done is apply Phi to every element in the matrix.
$(c)$ Is $\Phi(I)=\alpha I$ as I've mentioned above? I only have a proof if this is true. The proofs I've seen earlier assume $\Phi$ is unital, so this is not an issue there.
In short, I have trouble visualizing $\Phi \otimes I_2$. I'd appreciate it if someone can clear the aforementioned doubts.
Update: I think I should view $T$ as $$T=I\otimes \begin{bmatrix}1 && 0\\ 0 && 0\end{bmatrix} + X\otimes \begin{bmatrix}0 && 1\\ 0 && 0\end{bmatrix}$$
Your update is correct. What you did in step $(b)$ is precisely to apply that $\Phi$ is $2$-positive.
To finish your proof, what you need to notice is that $$ \begin{bmatrix} \|\Phi(I)\|\,I && \Phi(X) \\ \Phi(X^*) && \Phi(X^*X) \end{bmatrix} \geq\begin{bmatrix} \Phi(I) && \Phi(X) \\ \Phi(X^*) && \Phi(X^*X) \end{bmatrix} \geq 0. $$ Then you can finish the proof with your $U$, or you use the result that if a matrix $\begin{bmatrix}I&A\\ A^*&B\end{bmatrix}$ is positive, then $A^*A\leq B$.