Let $y = g(X,U)$, with $(X,U)$ two random variables. Suppose that for all values of $U$, $g(x;U)$ is convex as a function of $x$. It seems to me that the conditional expectation of $m(x) = E[Y|X=x] = E[g(X,U)|X=x]$ does not to be a convex function.
In fact, take $\alpha \in [0,1]$ and let $f(u|x) = f(U = u|X=x)$ be the conditional distribution of $U$ given $X$. Moreover, define $\tilde x = \alpha x + (1-\alpha)x'$. Then we have
\begin{aligned} m(\alpha x + (1-\alpha)x') = m(\tilde x) &= E[g(X,U)|X=\tilde x] \\ % &= \int g(\alpha x + (1-\alpha)x,u)f(u|\tilde x) \,d u \\ &\le \int [\alpha g(x,u) + (1-\alpha) g(x',u)]f(u|\tilde x) \,d u \\ &= \alpha \int g(x,u)f(u|\tilde x) \,d u + (1-\alpha)\int g(x',u)f(u|\tilde x)\,d u\\ &\ne \alpha m(x) + (1-\alpha)m(x'). \end{aligned}
Therefore, unless $X$ is independent from $U$, (i.e. $f(u|\tilde x) = f(u|x) = f(u)$) we cannot say that the conditional expectation function of a convex function is convex.
Is this right, or I am missing something?
You are right, take $g(x,u)=x^2-2u^2$ and suppose that $X=U$. Then $g(\,\cdot\,, u)$ is convex for all $u$ and $m(x)=\mathbb E[g(X,U)|X=x]=g(x,x)=-x^2$ which is not convex.