In Convex Optimization by Boyd & Vandenberghe (page 230), they try to prove that in mixed-strategy matrix games, there is no advantage to knowing the opponent's strategy beforehand.
When the strategy is known beforehand, we have the following situation:
Let $u$ be the vector of Player 1's strategy and $v$ be the vector pf Player 2's strategy (probability vectors), and $\mathbf{P}$ be the payoff matrix. Expected payoff is $u^\top \mathbf{P}v$. Player 2 will try to maximize the expected Payoff, which results in the optimization problem
\begin{align} \sup \{u^\top\mathbf{P}v| v \succeq 0, \sum v_i = 1\} \end{align}
Up to this point, things are very clear.
But then the authors say that above equation is equal to $\max_{i}(\mathbf{P}^\top u)$
I do not understand the last part clearly. It will be equal to the supremum only if we assume that all $v_i$ are equal. Otherwise how can they drop the $v$ term from the equation ?.
Assuming that the supremum is taken over all $v$'s (i.e. over all strategies of Player 2), and that we know $u$ and $P$, what this means is that the strategy which would give player 2 (who wants to maximize the payoff) the highest payoff would be the entry that corresponds to the entry with the highest value in the vector $P^Tu$. To see this, observe that:
$$u^TPv = v^TP^Tu = v^T(P^Tu)$$
Where the first equality follows from the fact that a scalar is its own transpose. If I want to find the vector $v$ that gives me the highest value of $v^T(P^Tu)$, whilst ensuring that $\sum_i v_i = 1$, then I should put set $v_i = 1$ on the entry that corresponds to the maximum entry in $P^Tu$, and zero everywhere else.