Let $G$ be a positive semidefinite matrix acting on the space $\mathcal{A}\otimes\mathcal{B}$ and let $A,B$ be rank-1, trace-1 matrices on $\mathcal{A}$ and $\mathcal{B}$. Which way (if any) should the inequality simbol go? $$\max_{A,B}\langle G,A\otimes B\rangle \lesseqqgtr \max_{A,B}\langle G^\mathcal{A}\otimes G^\mathcal{B},A\otimes B\rangle=\max_A\langle G^\mathcal{A},A\rangle\max_B\langle G^\mathcal{B},B\rangle=||G^\mathcal{A}||_\infty||G^\mathcal{B}||_\infty$$ where $G^\mathcal{A}$ and $G^\mathcal{B}$ are the restrictions of $G$ to $\mathcal{A}$ and $\mathcal{B}$ (i.e. $G^\mathcal{A}=\mathrm{tr}_\mathcal{B}(G)$ and vice versa). The last equality holds because (if I'm not mistaken) $G^\mathcal{A}$ and $G^\mathcal{B}$ are normal, since $G\geq0$)
Does an inequality hold here?
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matrices
functional-analysis
inequality
norm
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0Is $G^\mathcal{A}=\text{tr}_{\mathcal{B}}(G)$? – 2017-01-30
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0Yes, let me add it to avoid confusion. – 2017-01-30