Let $v_i \in \mathbb{R}^{n}, \ i=1, \ldots, m, \ \ $ $\mathcal{S}$ a convex polyhedron and $x \in \mathbb{R}^{n}$ be given. Consider the following solution $(s^{*},i^{*})$ to the problem

\begin{equation} \underset{s \in \mathcal{S}}{\max} \underset{j=1, \ldots, m}{\min} \langle v_j, s-x\rangle \end{equation}

I would like to conclude the following **Claim**:

Let $i \in \{1, \ldots, m\}$ be arbitrary, then it holds that

\begin{equation} \langle v_i, s^{*}-x \rangle \geq \langle v_{i^{*}}, s^{*}-x \rangle \ . \end{equation}

**Question 1**: Is the claim true?

**Motivation**: Why I think it might not be true: Consider first maximizing with repsect to $s \in S$, giving us an optimal $s$ for each $j=1, \ldots, m$, which we denote by $s^{(j)}$. Then minimzing with respect to $j$ the expressions $\langle v_j, s^{(j)}-x \rangle$, and denoting with $j^{(*)}$ the index where the minimum is attained. It is clear, that then

\begin{equation} \langle v_i, s^{(j^{*})}-x \rangle \geq \langle v_{j^{*}}, s^{(j^{*})}-x \rangle \end{equation}

Does **not** hold for arbitrary $i$, indeed it would only hold if we replace $s^{(j^{*})}$ with $s^{(i)}$ on the left hand side. So it is clear that if it is possible to exchange $\max$ and $\min$ in our expression, then the claim is false. This motivates my second question

**Question 2** Can we exchange $\max$ and $\min$ ans still get the same solution to the optimization problem?