Consider the following optimization problem
$u(x)=\text{argmin}_{-b_i<u_i<b_i} R(u)+g(x)u $
where $u=[u_1,...,u_m]^T$, $R(u):\mathbb{R}^m \rightarrow \mathbb{R}$ is a strictly convex function, $R(\textbf{0})=\textbf{0}$ and $R(u)>0,\: \forall u \neq \textbf{0}$, $g(x)$ is a Lipschitz continuous function and the bound $b_i$ is constant for $i=1,...,m$. I would like to ask under which conditions on $R(u)$ and $g(x)$, the solution $u(x)$ is continuous w.r.t $x$?
Let $u_n=u(x_n)$. First see that the function is well defined that is there is a unique minimizer. Suppose that $u$ is not continuous at $x_0$. Therefore there is a sequence $x_n \to x_0$ s.t. $u_n=u(x_n)\to u\neq u_0$. But since the sequence moves over the minimum of the function over $u$ for each $x_n$, the continuity implies that it should converge to the minimum for $x_0$ which is $u=u_0$. So $u(x)$ is continuous and only continuity of $g$ and strict convexity of $R$ is needed.