# Nash equilibrium conditions

In a 2 player game outlined here player $i$ strategy corresponds to some choice of effort $x_i$ (a continious variable) $x_i\geq0$. The Nash equilibrium for player $i$ is some strategy $x_1^*$ that maximizes profit $\pi_i(x_1,x_2)$:

$$\frac{\partial\pi_i(x_1,x_2)}{\partial{x_1}}=0\iff x_1=x_1^*$$

Now in order for this to be a sufficient condition for maximization, we must also satisfy the second order condition (to ensure profit is concave):

$$\frac{\partial^2\pi_i(x_1,x_2)}{\partial{x_1^2}}<0 \iff x_1^*\in[a,b]$$

where $a, b$ are some positive real numbers. However, in this game it may still be that

$$\pi_1(x^*_1\in[a,b],x_2)<0$$

And in this game, player $1$ receives a payoff of $0$ if $x_1=0$ so that she is better off 'dropping out' of the game. Hence $x_1^*\in[a,b]$ is not a Nash Equilibrium. How is it that $x_1^*\in[a,b]$ satisfies both the first and second order condition, but not constitute a nash equilibrium?

If $x_1 \in [a,b]$ then a deviation to $x_1=0$ would lead to such an improvement so it cannot be a part of a Nash equilibrium.
The reason why the first order (and second order) conditions are not relevant is because the profit function is discontinuous at $x_1=0$.