Let $A \in \mathbb{R}^n$ describe a symmetric game with $n$ strategies.
For the sake of clarity, I call symmetric game a two-player game where payoff matrices are the same for both players.
Suppose that $x^* = e_i$ is a non-strict pure Nash equilibria ($e_i$ is the $i$-th versor of the space $\mathbb{R}^n$).
In order to prove that $x^*$ is $ESS$ I have to check that:
$${x^*}^T A y > y^T A y ~\forall y \neq x^*$$
where $y$ is a generic strategy such that
$$\sum_{k=1}^n y_k = 1 \wedge y_k \geq 0 ~\forall k \in \{1, \ldots, n\}$$
I suppose the following fact: since $x^*$ is pure, then I have to check only that
$${x^*}^T A e_j > e_j^T A e_j ~\forall e_j \neq x^*$$
in order to say that $x^*$ is $ESS$.
Is this true? Is there some proof for this or for the contrary?
The inequality, $${x^*}^T A y > y^T A y ~\forall y \neq x^*$$ is not linear in $y$ so the answer to the question is no.
Counter-example:
$$ \pmatrix{& A & B & C \\ A & 5,5 & 0,0 & 0,0\\B& 0,0& -1,-1& 90,90\\C& 0,0 & 90,90 & -1,-1}$$
Notice that: