Sion's MinMax theorem over matrices

Question

Suppose $A\in R^{n\times n}$ is positive definite, i.e., $A>0$. Can we apply Sion's theorem to conclude the following:

$\underset{Y\in R^{n\times n}}{\min}\underset{X\in \mathcal{S}}{\max} \text{Trace}(YAY^T+YXY^T)=\underset{X\in \mathcal{S}}{\max}\underset{Y\in R^{n\times n}}{\min}\text{Trace}(YAY^T+YXY^T)$

where $\mathcal{S}$ is the set of all positive definite matrices? How about if $\mathcal{S}$ is the set of all positive semi-definite matrices?

Answer 1

$f(Y,B):=\text{Tr}(YBY^T)\ge0$ for all positive semi-definite $B$. $f(0,B)=0$. So $\min_{Y\in R^{n\times n}}f(Y,B)=0$ for every $B$. Therefore both sides of your desired equation are $0$ and thus the equation holds. There is no need for Sion's theorem. Besides, the premise of Sion's theorem is not satisfied as $R^{n\times n}$ is not compact.

Answer 2

I am not sure if you would need Sion' minmax theorem to prove this.

Note that the max on the left-hand side when fixing $Y\neq 0$ is $+\infty$ as the objective is linear in $X$ and $Y^TY\succeq 0$. The left-hand side is only bounded for $Y=0$, so the left-hand side is 0.

Now let's look at the right-hand side: $Tr(AB) \geq 0$, when $A$ and $B$ are PSD. So the min on the right-hand side should be 0, which is achieved at $Y=0$, regardless of X. So the right-hand side is also 0.