The Nash equilibrium, an existence proof

Question

I do not follow here in the -4th line that the equality is achieved only if

$$p_i(s,\alpha)\leq 0$$

for every $s$.

Answer 1

If $p_i(s,\alpha)\leq 0$ for every $s$, then $\alpha'_{i} = \alpha_{i}$, thus we get equality (proving "if" but not "only if").

Suppose there exists an $\hat{s}$ where $p_i(\hat{s},\alpha)> 0$, then for any $s$ where $p_i(s,\alpha)\leq 0$ and $\alpha_{i}^{(s)} > 0$ we have $\alpha_{i}^{\prime(s)} < \alpha_{i}^{(s)}$, which implies: $$\sum_{s:p_{i}(s, \alpha)\leq0} \alpha_{i}^{\prime(s)} < \sum_{s:p_{i}(s, \alpha)\leq0} \alpha_{i}^{(s)}$$ (Note, there must exist such an $s$ because $\alpha_{i}$ cannot result in strictly lower payoff than all pure strategies in its support.) $$\Rightarrow \sum_{s:p_{i}(s, \alpha)>0} \alpha_{i}^{\prime(s)} > \sum_{s:p_{i}(s, \alpha)>0} \alpha_{i}^{(s)}$$ The last implication follows because $\sum\limits_{s}\alpha_{i}^{\prime(s)}=\sum\limits_{s}\alpha_{i}^{(s)}=1$.

Hence, we can say equality is achieved if and only if $p_i(s,\alpha)\leq 0$ for every $s$.