How to find Nash equilibrium in pure and mixed strategies? Problem.

Question

A two player game. Players choose a number from a segment $[0;1]$. A payoff function of the first player $f_1(x,y)=|x-y|,$ where $x$ and $y$ - the numbers chosen by the first and the second players respectively. A payoff function of the second player $f_2(x,y)=-|x-y|.$ How to find Nash equilibrium in pure(if it exists) and mixed strategies in the game?

Answer 1

Clearly no pure strategy equilibrium exists. If player one plays $x_0$ and player two plays $y_0$, and these values are unequal, then player two would benefit by playing $x_0$. If $x_0=y_0$, then player one would benefit by playing literally anything else.

Player two wants their choice to be as close to player one's ass possible. If player two plays $1/2$ with probability $1$, then they win at least $-1/2$ no matter what. No other pure strategy guarantees this payoff.

Player one wants to be as far away as possible; they need to be unpredictable, otherwise player two beats them by playing their strategy. A sensible idea is to choose one of the extremal points $0$ or $1$ with equal probability. If player two choose $y$, then the expected score for player $1$ is $\frac12(1-y)+\frac12(y-0)=\frac12$.

Since player one can guarantee their score is at least $1/2$, and player two can guarantee it is at most $1/2$, we have found a Nash equilibrium of the game, and the game's value. There are others; for example, both players could randomly choose between $0$ and $1$. I am not sure what the set of all Nash equilibria is.