How to compute the best replies of this zero-sum game? I'm trying to solve this exercise in the last final exam of game theory:

Consider a zero-sum game $G=(X, Y, g)$, where $X=Y=[0,1]$, and $$\forall (x,y) \in X\times Y: g(x, y)=(x-y)^{2}$$
  
  
*
  
*Suppose that Player 1 plays the mixed strategy $\sigma=\frac{1}{2} \delta_{0}+\frac{1}{2} \delta_{1}$ (i.e., play $x=0$ with probability
  $1 / 2$ and play $x=1$ with probability $1 / 2$). Compute the best replies of player $2$.
  
*Compute the value of the game and give an optimal mixed strategy for each player.

I've successfully shown that $G$ doesn't have value in pure strategies, but it does in mixed strategies. Then I failed to finish above questions:
My attempt:
Let $\tau$ be a best reply of Player 2 against $\sigma$ of Player 1. Then $$\forall y \in [0,1]: g(\sigma,\tau) \le g(\sigma,y) = \frac{1}{2} g(0,y) + \frac{1}{2} g(1,y) = y^2-y+\frac{1}{2}$$
This is equivalent to $g(\sigma,\tau) \le 1/4$. Then I'm stuck to derive $\tau$ from the last inequality.

Could you please help me finish this exercise? Thank you so much!
 A: If player 2 plays the pure strategy $y$, then the payoff is, as you wrote, $y^2-y+0.5$. The minimum of this function is at $y=0.5$ so this is the best reply. Any other reply would yield a higher number (what player 2 pays to 1). Note that we can limit ourselves to pure actions, as there is always a pure best reply.
A: I will try to answer, even if the problem is not very clear (at least to me). 
1- X and Y are the sets of mixed actions, right? 
2- g(x,y) is the payoff of which player ? Since the game is zero sum, I would expect that payoff of one player plus the payoff of the opponent to be equal to zero, is that right? In this case we would have $g_1(x,y)=(x-y)^2$ and $g_2(x,y)=-(x-y)^2$. 
If the latter is correct, player 1 will choose $x=1$ whenever $y<1/2$, and player 2 will choose $y=1$ whenever $x>1/2$. Thus, the game has only one equilibrium in mixed strategies, that is $x=y=1/2$. The value of the game is then 1/2 for player1 and -1/2 for player 2. But probabily I m missing something..
