Let$\Omega \subset \mathbb{R}^n$ be open, bounded and strictly convex, i.e. for all$x, y \in \overline{\Omega}$and$\lambda \in (0, 1)$we have that
$$ \lambda x + (1 - \lambda)y \in \Omega.$$
Furhermore, let$g \colon \partial \Omega \rightarrow \mathbb{R}$be continuous and define$$G(x) :=\sup\{ h(x) \mid h \colon \overline{\Omega} \rightarrow \mathbb{R} \text{ is convex and } h|_{\partial \Omega} \leq g \}$$for all $x \in \overline{\Omega}$.I want to show that $G$ is continuous and satisfies $G|_{\partial \Omega} = g$.
I know that $G$ is convex and therefore (locally Lipschitz) continuous on $\Omega$. Hence I must fix an arbitrary point $x_0 \in \partial \Omega$ and show that $G$ is continuous at $x_0$ and satisfies $G(x_0) = g(x_0)$. How can do this?
I have some ideas of my own here. I may be able to show that $G$ is lower semicontinuous at $x_0$ and that $G(x_0) = g(x_0)$. To do so I want to show that for every $\varepsilon > 0$ there exists an affine map $P \colon \mathbb{R}^n \rightarrow \mathbb{R}$ such that
$$g(x_0) - \varepsilon < P(x_0) < g(x_0)\quad \text{and} \quadP|_{\partial\Omega} < g.$$
In general I'm thinking that perhaps the supporting hyperplane theorem might become neccecary in order to construct such an affine map $Q$. If additional smoothness assumption are imposed upon $\partial \Omega$ then perhaps a more straightforward argument will suffice.
Even if the above works, I must still prove that $G$ is upper semicontinuous at $x_0$. At the moment I have no idea how to do this. I will accept ideas that require $\partial \Omega$ to be smooth in some sense, but I prefer the more general case. Any help is appreciated.