I have a set of linear combinations ;
$$L = \left\{ (x_1,x_2,x_3) \mid 2 x_1 + x_2 + 2 x_3 \geq 5, x_3 \geq x_1, x_1 \geq 0 \right\}$$
and wanted to show this set is convex or not. I followed this way;
Get two points and calculate their convex combination;
$P = \lambda(x_{11},x_{21},x_{31}) + (1-\lambda)(x_{12},x_{22},x_{32})$
for $0 \leq \lambda \leq 1$
and supposed that if this set convex, this inequality always should be satisfied;
$2(\lambda x_{11} +(1-\lambda)x_{12}) +(\lambda x_{21} + (1-\lambda)x_{22}) +2(\lambda x_{31} +(1-\lambda)x_{32}) \geq 5$
But here is where I am stuck. I didn't find a general way it is always satisfied or any counter example to prove the set is not convex.
Can I prove the convexity of this set by using convex combinations, how ?
Thanks in advance.
Each inequality is given by the dot product of some constant vector with a vector of variables. For example the first inequality is $$ c^Tx \geq 5$$ with $c = (2,1,2)$. With this notation let $x,y \in L$ and for $(1-\lambda)x + \lambda y \in L $ we need $$ (1-\lambda )c^Tx+ \lambda c^Ty \geq 5$$
but this is always the case since $(1-\lambda)$ and $\lambda$ are both positive and $c^Tx$ and $c^Ty $ are both greater or equal to $5$.
More generally your set $L$ can be rewritten in the form
$$ L = \{ x : Ax \geq b \}$$ for some matrix $A$ and vector $b$ and where the inequality is componentwise. In this way we have $$ (1- \lambda)Ax + \lambda Ay \geq b$$ by arguments similar to those given above.