Let $\mathcal{P} = \{P_1, \cdots,P_n\}$ be a family of finite point sets in $\mathbb{R}^d$, each having at most $m$ points. Consider the following objective function
\begin{align} cost(\mathcal{P},c) = \sum_{i\in[n]}\max_{p\in P_i}\Vert p-c\Vert^2 \end{align}
Let $c^\star$ be the point which minimizes the above objective function and the optimal cost is $opt(\mathcal{P}) = cost(\mathcal{P},c^\star)$.
For a point $c\in\mathbb{R}^d$ and $i \in [n]$, let $p_i^{(c)}$ be the point of $P_i$ farthest from $c$, i.,e. \begin{align*} \Vert p_i^{(c)} -c\Vert^2 = \max_{p\in P_i}\Vert p-c\Vert^2 \end{align*} Define $c$-mean to be the mean of these farthest points \begin{align*} \mu^{(c)} = \frac{1}{n}\sum_{i=1}^n p_i^{(c)} \end{align*}
So we have \begin{align*} cost_2(\mathcal{P},c) = \sum_{i=1}^n \Vert p_i^{(c)} -c\Vert^2 \end{align*} For ease of notation, let $p_i^\star = p_i^{(c^\star)}$ and $\mu^\star = \mu^{(c^\star)}$. So \begin{align*} opt_2(\mathcal{P}) = \sum_{i=1}^n \Vert p_i^\star - c^\star\Vert^2 \end{align*}
Note that if every set in $\mathcal{P}$ is singleton, then this is the well-known $k$-means objective and $c^\star = \mu^\star$.
It is easy to show the following, for any $c\in \mathbb{R}^d$ \begin{align} cost(\mathcal{P},c) \leq opt(\mathcal{P}) + n\Vert c - \mu^{(c)}\Vert^2 \end{align}
It follows that if there is a point $c$ such that $c = \mu^{(c)}$, then $c$ is an optimum center. This already helps to find the optimal solution in some simple scenarios, such as $\mathcal{P} = \{P_1,P_2\}$, where $P_1=\{(1,1),(1,-1)\}$ and $P_2=\{(-1,1),(-1-1)\}$. However it seems to me that the reverse direction may not hold: meaning for any optimal center $c^\star$, it may not be the case that $c^\star = \mu^{(c^\star)}$. So far I have not been able to come up with a counter example.
Does anyone have some idea as to finding a counter example.
P.s. Is this objective function already known?
