I have the following functional:
\begin{equation}
g\left(F_{X,U}(x,u)\right) = -\int_{\mathcal{X}}\int_{\mathcal{U}}\left[\int_{\mathcal{Y}} p_{Y\vert X}(y\vert x)\log p_{Y\vert U}(y\vert u)\,dy - \mu\int_{\mathcal{Z}} p_{Z\vert X}(z\vert x)\log p_{Z\vert U}(z\vert u)\,dz \right]dF_{X,U}(x,u),
\end{equation}
where $F_{X,U}(x,u)$ is the joint cumulative distribution function (CDF) of random variables $X$ and $U$, i.e., it satisfies $\int_{\mathcal{X}}\int_{\mathcal{U}}dF_{X,U}(x,u)=1$. Notice that random variables $X$ and $U$ are not necessarily independent; $\mu\geq 0$ is some nonnegative coefficient; $p_{Y\vert X}(y\vert x)$, $p_{Z\vert X}(z\vert x)$, $p_{Y\vert U}(y\vert u)$, and $p_{Z\vert U}(z\vert u)$ are respectively given by:
\begin{align}
p_{Y\vert X}(y\vert x) &= \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{(y-x)^2}{2}\right),\\
p_{Z\vert X}(z\vert x) &= \frac{1}{\sqrt{4\pi}}\exp\left(-\frac{(z-x)^2}{4}\right),\\
p_{Y\vert U}(y\vert u) &= \int_{\mathcal{X}} p_{Y\vert X}(y\vert x) \frac{p_{X,U}(x,u)}{p_U(u)}\,dx,\\
p_{Z\vert U}(z\vert u) &= \int_{\mathcal{X}} p_{Z\vert X}(z\vert x) \frac{p_{X,U}(x,u)}{p_U(u)}\,dx,
\end{align}
where $p_{X,U}(x,u) = \frac{\partial^2 F_{X,U}(x,u)}{\partial x \partial u}$ is the joint pdf and $p_U(u) = \int_{\mathcal{X}} p_{X,U}(x,u) \,dx$ is the pdf of the random variable $U$.
I would like to establish that the functional $g(F_{X,U}(x,u))$ is concave in its argument by proving that: \begin{equation} g(\lambda F_{X,U}^1(x,u) + (1 - \lambda) F_{X,U}^2(x,u)) \geq \lambda\, g(F_{X,U}^1(x,u)) + (1 - \lambda)\, g(F_{X,U}^2(x,u)), \end{equation} where $\lambda\in[0,1]$, $F_{X,U}^1(x,u)$ and $F_{X,U}^2(x,u)$ are two different joint CDFs.
Any help is highly appreciated.