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Is the problem

\begin{equation} \min_X \max_Y -\operatorname{tr}(X^TY)-\operatorname{tr}(Y^TYX) \end{equation}

Jointly convex in $X$ and $Y$? Can we solve it globally? Why or Why not? $X$ and $Y$ are real matrices

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1 Answer 1

up vote 1 down vote accepted

One can show that $\langle Y, X \rangle = \mathbb{tr}( Y^TX)$ defines an inner product on the set of matrices, from which we have the cost function $f(X,Y) = -\langle Y, X \rangle - \langle Y, Y X \rangle$. It is linear in $X$, but indeterminate in $Y$ (eg, $Y \mapsto F(I,Y)$ is concave, $Y \mapsto F(-I,Y)$ is convex).

Since $f(X,0) = 0$, we always have $\max_Y F(X,Y) \geq 0$, and since $f(0,Y) = 0$, we have $\min_X \max_Y f(X,Y) = 0$. Hence $(0,0)$ is a (global) solution.

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