I have concluded something about a class of functions but need mathematical assurance i.e. supporting theorem perhaps.
Suppose $(x_1, x_2, \cdots, x_{i-1},x_i,x_{i+1},\cdots,x_n) \in \mathcal{R^n_{++}}$.
$h_1(x_2,x_3,\cdots,x_n), h_2(x_1,x_3,\cdots,x_n), h_i(x_{-i}) \in \mathcal{R}$ where $x_{-i}= (x_1,x_2,\cdots,x_{i-1}, x_{i+1},\cdots,x_n) $ It is known that $x_1 \geq x_2 \geq \cdots \geq x_n$ and $h_{i+1}(x_{-(i+1)}) \geq h_{i}(x_{-i}) $ always. When $x_i = x_{i+1}, x_{-i}=(x_1,x_2,\cdots,x_{i-1}, x_{i+1},\cdots,x_n) =x_{-(i+1)}$. The vectors are identical in that case. This leads to $h_{i+1}(x_{-(i+1)}) = h_{i}(x_{-i})$.
Now suppose $h_i(x_{-i})$ is a linear polynomial like: $h_1(x_2,x_3,\cdots,x_n)= a_{12}x_2 + a_{13}x_3 + \cdots + a_{1n}x_n$.
Then: $h_{i+1}(x_{-(i+1)}) \geq h_{i}(x_{-i}) \implies a_{(i+1)i}x_i- a_{i(i+1)x_i+1} +\sum_{k \neq i,i+1} [a_{(i+1)k} - a_{ik}]x_k \geq 0$. and $\forall k \neq i,i+1$ and $\forall x_k \in \mathcal {R_{++}}$ whenever $x_i=x_{i+1}$, $ a_{(i+1)i}x_i- a_{i(i+1)x_i+1} +\sum_{k \neq i,i+1} [a_{(i+1)k} - a_{ik}]x_k = 0$
Conclusion: $a_{ik}=a_{(i+1)k} \forall k \neq i, i+1$ and $a_{i(i+1)}=a_{(i+1)i}$
The $a_{ik}$ do not depend upon x_i for sure. Is the conclusion valid?