Proof of the equivalence between two ways of defining ESS
We need to show that if $\mathbf{p}$ is a strict Nash equilibrium, $E(\mathbf{p},\mathbf{p})&gt; E(\mathbf{q},\mathbf{p})$, then $W(\mathbf{p}) &gt; W (\mathbf{q})$ for some sufficiently small $\epsilon$. When $E(\mathbf{p},\mathbf{q}) \geq E(\mathbf{q},\mathbf{q})$, $\mathbf{p}$ clearly dominates $\mathbf{q}$.The less obvious case is when $E(\mathbf{p},\mathbf{q}) &lt; E(\mathbf{q},\mathbf{q})$. 

Since $E(\mathbf{p},\mathbf{p}) - E(\mathbf{q},\mathbf{p})&gt;0$, there should be a strictly positive number $k$ such that
$$k[E(\mathbf{p},\mathbf{p}) - E(\mathbf{q},\mathbf{p})]+ E(\mathbf{p},\mathbf{q}) - E(\mathbf{q},\mathbf{q})&gt;0$$
is true. Whatever this number is, we can write $k$ as $(1-\epsilon)/\epsilon$ for $0&lt;\epsilon&lt;1$. Substituting this into the equation above gives us the desired result.