I hope your day is going well.
This is a problem, I don't know how to solve it since 1 week. It's going to be a relief and a pleasure to get your help.
Problem : Let $X = \{x_{1},\ldots,x_{N+M} \}$ such as $\{x_{1},\ldots,x_{N} \} \subset \Omega := \text{int}( \text{conv} ( \{x_{N+1},\ldots,x_{N+M} \} ) )$ a convex polygon with $ \text{conv} $ the convex hull.
Let $u \in R^{X}$ such as $u(x_{N+i}) = 0$ ($u=0$ on the edge) and $\tilde{u}^{**}(x_{i}) = u(x_{i})$ ($u$ is its convex conjugate on $X$).
Here's some explications and definitions : Let $\tilde{u}$ which is $u(x_{i})$ when $x_{i} \in X$ and $+\infty$ otherwise. Its convex conjugate is $\tilde{u}^{*}(x) = \sup_{y \in \mathbb{R}^{n}} \{ \langle x,y \rangle - \tilde{u}(x)\} = \max_{x \in X} \{ \langle x,y \rangle - \tilde{u}(x) \}$. Then I take the convex conjugate once again, I called it $v$.
Question : Show $v=0$ on the edge $\partial{\Omega}$
I think we can use that $v$ is the supremum of affine function $\varphi$ such as $\varphi \le \tilde{u}$. Which can be written : If $\Sigma = \{(p,r) ; \forall y \in \mathbb{R}^{n}, \langle p,y \rangle + r \le \tilde{u}(y) \}$ we have : \begin{align*} \sup\{ \langle p,y \rangle + r ; (p,r) \in \Sigma \} = v(y) \end{align*}
I wish you a very good day.
I'm going to develop this idea : Let $x \in \partial{\Omega}$ let say $ x \in [x_{a} ; x_{b}]$ with $x_{a}$ and $x_{b}$ two vertexes. We want to show that $\tilde{v}(x)=0$.
It is enough to show that it exists a line $D$ such that $ D \le u$ and $D(x) = 0$. From the following result : $$ p \in \partial{u}(x) \Leftrightarrow x \in \partial{u^{*}}(p) \text{ et } u^{**}(x) = u(x) $$ It is enough to find a vector $p$ orthogonal to $x_{a}-x_{b}$ AND such that $$ p \in \partial{u^{*}}(x_{a}) \cap \partial{u^{*}}(x_{b}) = \text{co}( \{x_{j} ; j \in J(x_{a}) \}) \cap \text{co}( \{x_{j} ; j \in J(x_{b}) \}) $$ With $J(x_{a}) = \{ i ; \tilde{u}^{*}(x_{a}) = max_{z \in X} \{ \langle x_{a},z \rangle - u(z) = \langle x_{a},x_{i} \rangle - u(x_{i}) \} $.
But I'm not sur that a such $p$ may exist. Maybe someone will be inspired by theses ideas.