If $V:[0,1] \rightarrow \mathbb{R}$ is a convex and nondecreasing function, then for all $(p,q,r,t,\lambda,\mu) \in [0,1]^{6}$ such that
$p \geq q \geq r \geq t$
$\lambda p + (1-\lambda) t=\mu q + (1-\mu) r$
we have \begin{equation*} \lambda V(p) + (1-\lambda) V(t) \geq \mu V(q) + (1-\mu) V(r) \end{equation*}
I am trying to generalize this observation to the multidimensional case. Consider $n \geq 2$ and the simplex \begin{equation} \Delta=\{(p_1,\cdots,p_n) \in \mathbb{R}^{n} \mid \forall i, p_i > 0 \text{ and } \sum_{i=1}^{n}{p_i}=1\} \end{equation}
Suppose that $\Delta$ is endowed with a (partial) order $\succeq$ such that
\begin{equation*} p \succeq q \Rightarrow p \succeq \alpha p + (1-\alpha) q \succeq q \end{equation*} for all $\alpha \in [0,1]$. Consider a continuous function $V:\Delta \rightarrow \mathbb{R}$ that is convex and nondecreasing with respect to $\succeq$, i.e. $V(p) \geq V(q)$ whenever $p \succeq q$.
Finally, consider four vectors $p,q,r,t$ and two scalars $\lambda \in [0,1], \mu \in [0,1]$ such that:
- $p \succeq q \succeq r \succeq t$
- $\lambda p + (1-\lambda) t = \mu q + (1-\mu)r$
I am trying to prove that these conditions are sufficient to guarantee that \begin{equation*} \lambda V(p) + (1-\lambda) V(t) \geq \mu V(q) + (1-\mu) V(r) \end{equation*}
It is easy when $q=\alpha p + (1-\alpha) t$ and $r=\beta p + (1-\beta) t$ for some $(\alpha,\beta) \in [0,1]^2$ since this case essentially boils down to the unidimensional problem. But I haven't made any progress otherwise.
Any hint, help or reference would be greatly appreciated. Thank you!
