Consider $M$ finite sets of integer points $P_m$, $m=1,\ldots,M$. Let $$A = \left\{x_m\in\operatorname{conv}P_m, m=1,\dots,M, \sum_{m=1}^MN_mx_m=0\right\}$$ and $$B =\operatorname{conv}\left\{x_m\in P_m, m=1,\dots,M, \sum_{m=1}^MN_mx_m=0\right\}$$ where $N_m$ is a matrix of dimension compatible with $x_m$ and $\operatorname{conv}P_m$ is the convex hull of the points in $P_m$.

I was wondering if it is possible to prove that $$B\subseteq A$$ and how.


I think it should be possible.

Firstly, let us see if we can establish that $A$ is convex. Take

\begin{align}X &= (x_1,\ldots,x_M)\in A\\Y&=(y_1,\ldots,y_M)\in A.\end{align}

Let $0\leq\lambda\leq 1$. Then

$$\lambda X + (1-\lambda) Y = (\lambda x_1 + (1-\lambda)y_1,\ldots,\lambda x_M + (1-\lambda)y_M).$$

Since \begin{align}N_1x_1 + \ldots + N_Mx_M &= 0\\N_1y_1 + \ldots + N_My_M &= 0,\end{align} we have $$N_1(\lambda x_1 + (1-\lambda) y_1) +\ldots+ N_M(\lambda x_M + (1-\lambda) y_M) = 0.$$

Also, $\forall i=1,\ldots,M, \lambda x_i + (1-\lambda) y_i\in\operatorname{conv}(P_i)$.

Thus, convexity of $A$ is established.

$B$ is the convex hull of a subset of points of $A$. Let this subset be $K\subseteq A$.

So, $B = \operatorname{conv}(K)\subseteq\operatorname{conv}(A)=A$, since $A$ is convex.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.