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The problem I am working on is:

Given an $n$ dimensional vector $r \in \mathcal{R}^n$, and a convex set $G=\{\mu \in \mathcal{R}^n | \mu_i \ge 0, ~ \mu^T \mathbf{1}=1, ~ A\mu =0 \}$ where $\mathbf{1}= [1, 1, ..., 1]^T$. ($A \in \mathcal{R}^{m \times n}$ and $m < n$) We assume that the set $G$ is not an empty set.

Suppose $\bar{\mu} = \arg \max_{\mu \in G} r^T\mu$ given $r$. Then can we say $\tilde{\mu} = \arg \max_{\mu \in G} \bar{\mu}^T\mu$ is identical to $\bar{\mu}$?

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In other words, you are asking if the scalar product with every vertex is maximized by the vertex itself (we can certainly create $r$ to make any vertex we want the maximizer of the scalar product with $r$). The answer is "No". We can define the segment between $(0,1,2,3)/6$ and $(1,0,0,5)/6$ in $\mathbb R^4$ this way, take $r=(0,1,1,0)$, say, and notice that still $15/36>14/36$.

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