Assume that $ u(\cdot) $ is concave, $ u'(x_1) > 0 $ and $ u''(x_1) < 0 $. Assume that $\theta_1, \theta_2$ are i.i.d. uniformly distributed on the square $\theta_1, \theta_2 \in [0,1]^2$.
Denote
\begin{align} U = & \int_0^{\frac{1}{u(x_1)-u(x_2)}} \int_0^1 \left( \theta_1 u(x_1)+ \theta_2 u(x_2) \right) \,d\theta_1 \,d\theta_2 \\[6pt] {}+{} & \int_{\frac{1}{u(x_1)-u(x_2)}}^1 \int_{\theta_2 - \frac{1}{u(x_1)-u(x_2)}}^1 \left(\theta_1 u(x_1)+\theta_2 u(x_2)\right) \, d\theta_1\,d\theta_2 \\[6pt] {}+{} & \int_{\frac{1}{u(x_1)-u(x_2)}}^1 \int_0^{\theta_2-\frac{1}{u(x_1)-u(x_2)}} \left(\theta_1 u(x_2)+\theta_2 u(x_1)\right) \,d\theta_1 \,d\theta_2 \end{align}
We restrict the case that $ x_2 = \overline{x} - x_1 $, and $\overline{x}$ is fixed. $u$ is increasing and concave in $x_1$, so is $u(\overline{x}-x_1)$ concave in $x_1$. And suppose $0 <\frac{1}{u(x_1) - u(x_2)} < 1$, so the line $\theta_1 = \theta_2 - \frac{1}{u(x_1) - u(x_2)}$ split the square $[0,1]^2$ into two regions.
Is $U$ still concave in $x_1$ locally? Maybe I can check the sign of the second order derivatives. But is there any properties to indicate the concavity of things like this?