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I have the following minmax problem and I am wondering if the order of the minimum and maximum can be interchanged and if yes, why?

Let $\|\cdot\|_1$ be the trace norm defined as $\|\rho\|_1 = \text{Tr}(\sqrt{\rho^\dagger \rho})$. This gives us a distance measure between quantum states i.e. $\|\rho - \sigma\|_1$. Consider Hilbert spaces $A,B,E$ and $R$.

The optimization problem I have is

$$\max\limits_{\phi_{AR}} \min_{V_{E\rightarrow E}}\|(I_A\otimes V_{E\rightarrow E})(U_{A\rightarrow BE}\otimes I_R)\phi_{AR} - (U'_{A\rightarrow BE}\otimes I_R)\phi_{AR} \|_1$$

where $U_{A\rightarrow BE}, U'_{A\rightarrow BE}, V_{E\rightarrow E}$ are all isometries and $\phi_{AR}$ is a pure state. So my optimization is over the set of isometries and pure states neither of which is a convex set. However, note that the optimizations involved are on different Hilbert spaces.

Is it true or false that \begin{align*} \max_{\phi_{AR}} &\min_{V_{E\rightarrow E}}\|(I_A\otimes V_{E\rightarrow E})(U_{A\rightarrow BE}\otimes I_R)\phi_{AR} - (U'_{A\rightarrow BE}\otimes I_R)\phi_{AR} \|_1 \\ &= \min_{V_{E\rightarrow E}}\max_{\phi_{AR}} \|(I_A\otimes V_{E\rightarrow E})(U_{A\rightarrow BE}\otimes I_R)\phi_{AR} - (U'_{A\rightarrow BE}\otimes I_R)\phi_{AR} \|_1 \end{align*} and if true, what is the justification? I am aware that I can always upper bound the LHS with the RHS using the maxmin inequality.

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  • $\begingroup$ This question needs clarification: $(U_{A\to BE}\otimes I_R)\phi_{AR}$ is a state on $BER$ so it cannot be an input to $I_A\otimes V_{E\to E}$. If anything the expression should read $(I_{\bf B}\otimes V_{E\to E}{\bf \otimes I_R})(U_{A\to BE}\otimes I_R)\phi_{AR}$ or maybe $V_{E\to E}$ is supposed to be an isometry/unitary $V_{ER\to ER}$? Kind of hard to tell without more detail. Also are you really optimizing in trace norm $\|_1$? If so you probably mean that things like $U_{A\to BE}$ are not isometric operators but rather the induced channel $X\mapsto U_{A\to BE}XU_{A\to BE}^\dagger$ $\endgroup$ Commented Jun 9 at 10:17
  • $\begingroup$ (2/2) Once the above problem is clarified the formula $\|\,|u\rangle\langle u|-|v\rangle\langle v|\,\|_1=2\sqrt{1-|\langle u|v\rangle|^2}$ (which holds for all unit vectors $u,v$, cf. Eq. (1.186) in "The Theory of Quantum Information" by Watrous) can be used to reduce the problem to a simpler, equivalent one. $\endgroup$ Commented Jun 9 at 13:05

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