= Solution
In a two-player <symmetric finite game>, represent <mixed strategies> by <probability> <vectors> and let $e(x,y)=x^TAy$ be the <payoff> to a player using $x$ against one using $y$. A resident strategy $x^*$ is an <evolutionarily stable strategy> if every distinct mutant strategy $y$ has an invasion barrier $\varepsilon_0(y)>0$ such that, for $0<\varepsilon<\varepsilon_0(y)$,
$$
e\bigl(x^*,(1-\varepsilon)x^*+\varepsilon y\bigr)>e\bigl(y,(1-\varepsilon)x^*+\varepsilon y\bigr).
$$
Thus when mutants are sufficiently rare, the resident has strictly larger fitness against the population mixture. The barrier may depend on the mutant; a common barrier is not part of this definition.
For fixed $y\ne x^*$ put
$$
a_y=e(x^*,x^*)-e(y,x^*),\qquad b_y=e(x^*,y)-e(y,y).
$$
By <bilinearity> the resident-minus-mutant fitness difference is
$$
(1-\varepsilon)a_y+\varepsilon b_y.
$$
If this is positive for all sufficiently small positive $\varepsilon$, taking $\varepsilon\downarrow0$ gives $a_y\geq0$. If $a_y=0$, strict positivity requires $b_y>0$. Conversely, if $a_y>0$, choose $0<\varepsilon_0\leq a_y/[2(a_y+|b_y|)]$; then the difference is at least $a_y-\varepsilon(a_y+|b_y|)>a_y/2>0$. If $a_y=0$ and $b_y>0$, it is $\varepsilon b_y>0$ for every $0<\varepsilon<1$. This proves the <two-condition criterion for evolutionary stability>: \b[for every $y\ne x^*$, either]
$$
\boxed{e(x^*,x^*)>e(y,x^*)}
$$
\b[or]
$$
\boxed{e(x^*,x^*)=e(y,x^*)\quad\text{and}\quad e(x^*,y)>e(y,y).}
$$
The weak first comparison is the symmetric <Nash equilibrium> condition; the second excludes neutrally competitive mutants when that comparison is an equality.
For the <Hawk-Dove game>, let $x^*=(1/2,1/2)$ and $y=(p,1-p)$, $0\leq p\leq1$. Against the resident, both <pure strategies> give $1/2$, so
$$
e(x^*,x^*)=e(y,x^*)=\frac12.
$$
Direct multiplication of the <payoff matrix> gives
$$
e(x^*,y)=\frac32-2p,\qquad e(y,y)=1-2p^2.
$$
Consequently
$$
e(x^*,y)-e(y,y)=2\left(p-\frac12\right)^2>0\qquad(p\ne1/2).
$$
The equality case of the <two-condition criterion for evolutionary stability> therefore applies to every distinct mutant. More explicitly, the fitness difference in a mixed population is $2\varepsilon(p-1/2)^2$, positive for every $0<\varepsilon<1$. Hence \b[$\boxed{(1/2,1/2)\text{ is an evolutionarily stable strategy}}$], with the common invasion barrier one.
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