Consider the following complete-information, auction game. There are two players $i=1,2$. Each bids simultaneously a value $b_i\in[0,\infty)$. The payoff function is symmetric:
$$ \pi_i =\begin{cases} a - b_i & \text{if } b_i > \max\{b_j,\bar b\} \\ (a - b_i)/2 & \text{if } b_i = b_j \geq \bar b \\ 0 & \text{otherwise.} \end{cases} $$ where $\bar b$ is a minimum bid required to win a prize $a>0$.
What is the mixed strategy equilibrium, if any, of this game?
Any pure strategy of bidding less than $\bar b$ or more than $a$ is dominated by respectively a pure strategy of bidding $\bar b$ or bidding $a$. Therefore we can assume that $\bar b \le b_i \le a$.
Consider the payoff function of the game if we offset both $\bar b$ and $a$ by $\Delta$. If we also offset $b_i$ and $b_j$ by $\Delta$ then the differences and inequalities are invariant. So wlog we can assume that $\bar b = 0$ and that $a$ is offset appropriately. Thus we simplify the payoff function to $$\frac{\pi_i} a =\begin{cases} 1 - \frac{b_i}a & \text{if } b_i > b_j \\ (1 - \frac{b_i}a)/2 & \text{if } b_i = b_j \geq 0 \\ 0 & \text{otherwise.} \end{cases}$$
and since the Nash equilibria are unaffected by a linear scaling of the payoff function we can simplify further by assuming wlog that $a = 1$, giving $$\pi_i =\begin{cases} 1 - b_i & \text{if } b_i > b_j \\ (1 - b_i)/2 & \text{if } b_i = b_j \geq 0 \\ 0 & \text{otherwise.} \end{cases}$$
Working through some small finite cases where the bids are drawn from $\{0, \frac 1 d, \frac 2 d, \ldots, 1\}$ I find that lrs gives three extreme equilibria, all of which are pure strategies:
In the limiting case $d \to \infty$ the first is clearly still an equilbrium (if I play $a$ then no matter what you play we both get a payoff of $0$); and the other two tend towards the first case.
So it is clearly the case that both players bidding $a$ is a Nash equilibrium; and I hypothesise that it is the only one.