Given $x_i, x_j \in [0,1]$, and a payoff function $u_i(x_i, x_j) = (\theta_i + 3x_j - 4x_i)x_i$ if $x_j < 2/3$, and $= (3x_j-2)x_i$ if $x_j \geq 2/3$. My question is I don't understand why we have the best-response function $b_i(x_j) = {1}$ when $x_j > 2/3$?
My argument. For $x_j < 2/3$, $\partial(u_i)/\partial(x_i) = \theta_i + 3x_j - 8x_i = 0 \iff x_i = (3x_j + \theta_i)/8$. For $x_j \geq 2/3$, $\partial(u_i)/\partial(x_i) = 3x_j - 2 = 0 \iff x_j = 2/3$ for every $x_i \in [0,1]$.
Shouldn't this imply player i's pure best response function is: $b_i(x_j) = [0, 1]$ if $ x_j \geq 2/3$, $=(\theta_i + 3x_j)/8$ if $x_j < 2/3$?
We have
$$u_i(x_i, x_j) = \begin{cases}(\theta_i + 3x_j - 4x_i)x_i &\text{if } \quad x_j < \frac{2}{3}\\ (3x_j-2)x_i &\text{if } \quad x_j \ge \frac{2}{3}\end{cases} $$
for a given $x_j$ the best response correspondence (please notice that there is no reason that the best response be a function, for a given strategy, there is a set of best actions, thus they call it correspondence instead) for player $i$ can be obtained
$$\frac{\partial u_i(x_i, x_j)}{\partial x_i} =0= \begin{cases}(\theta_i + 3x_j - 8x_i) &\text{if } \quad x_j < \frac{2}{3}\\ (3x_j-2) &\text{if } \quad x_j \geq \frac{2}{3}\end{cases}$$
If $x_j < \frac{2}{3}$ the maximum of $u_i$ happens at either $x_i=0,1,\frac{\theta_i+3x_j}{8}$ (from the Fundamental theorem of calculus).
$$u_i(x_i, x_j) = \begin{cases}\theta_i + 3x_j - 4 &\text{at} \quad x_i=1 \\0 &\text{at} \quad x_i=0 \\ \left(\frac{\theta_i + 3x_j}{4}\right)^2 &\text{at} \quad x_i=\frac{\theta_i+3x_j}{8} \end{cases}$$
Here best response is $x_i=1$ if $\left(\frac{\theta_i + 3x_j}{4}\right)^2<\theta_i + 3x_j - 4$ and $x_i=\frac{\theta_i+3x_j}{8}$ otherwise. For the second case there is a subtle difference, our objective is linear. The player $i$ does not have the ability to enforce the choice of the player $j$ (so $x_j=\frac{2}{3}$ that we obtain from critical point of utility function, is not in the authority of player $i$) best response is either $x_i=0,1$, please notice that we have a linear objective here, the slope of the line decides the fate of $u_i$ and thus the strategy of the player $i$.
$$u_i(x_i, x_j) = \begin{cases}3x_j-2 &\text{at} \quad x_i=1 \\0 &\text{at} \quad x_i=0 \end{cases}$$
Therefore the best strategy here is $x_i=1$, because the slope of the line $(3x_j-2)$ is a positive value for $x_j>\frac{2}{3}$ thus the player $i$ gains profit by increasing $x_i$ in $(3x_j-2)x_i$. But the player $i$ can not exceed $x_i=1$ and thus $x_i=1$ is the best response here.
P.S. in the first part of solution, the best response $x_i=\frac{\theta+3x_j}{8}$ must satisfy the constraint $x_i\in[0,1]$, therefore $\theta+3x_j\le 8, \forall x_j \in [0,\frac{2}{3}]$ which results in $\theta\le6$ and by similar argument $\theta+3x_j\ge 0, \forall x_j \in [0,\frac{2}{3}]$ which results in $\theta \ge 0$. Also it is noteworthy that the inequality $\left(\frac{\theta_i+3x_j}{4}\right)^2>(\theta_i+3x_j-4)$ always holds true here, I live it to reader to prove it (simply plot the graphs, or check algebraically).