Inequality for inner product with constraints.

Question

Define $$\triangle := \ \{a\in \mathbb{R}^{n} \ \ : \ \ \sum_{i=1}^n a_i=1, \ \ a_i\ge0 \ \ \forall_{1\le i \le n} \}$$ Moreover, let $ \ L \ $ be any linear subspace of $ \ \mathbb{R}^{n}, \ $ and put
$$Q:= \ \ L \ \cap \ \triangle.$$ Prove that for any $x \in \mathbb{R}^n$ \begin{equation}\sup_{p\in Q} \ x^{T}p \ \ \ge \ \ \inf\Big\{y^Tp \ \ : \ \ y^Tp_1=y^Tp_2 \ \ \forall_{p_1,p_2 \in Q}, \ \ y \ge x, \ p \in Q \Big\}. \tag{$\star$}\end{equation}

Note that $ \ \inf \ $ is over $ \ y \ $ such that $ \ y\ge x \ $ and $ \ f(p)=<y,p> \ $ is constant on $ \ Q \ $. Then $ \ y^Tp \ $ does not depend on choice of $ \ p\in Q \ $.

This problem comes from a problem in financial mathematics. This is the reformulation of this problem with probability and financial parts deleted. I believe that this should be an easy problem in linear programing, which I do not have experience in. I have been trying to solve it with basic linear algebra methods, but failed so far. I would be highly interested in such elementary solution. I would be very thankful for any help, insight or hints. This seems as an easy problem...

As a matter of fact, I even suspect that $(\star)$ should turn out to be an equality.

Answer 1

I think that I have found my answer. This is a relatively easy problem, if we know Farkass' lemma:

Let $ \ A\in \mathbb{R}^{m\times n} \ $ and $ \ b\in \mathbb{R}^{m} \ $. Then exactly one of the following statements is true:

1. There exists $ \ y\in \mathbb{R}^{n} \ $ such that $ \ Ay=b \ $ and $ \ y\ge 0, \ $
2. There exists $ \ z\in \mathbb{R}^{m} \ $ such that $ \ A^Tz \ge 0 \ $ and $ \ b^Tz <0 \ $ and $\ z \ \ge 0, \ $

where $\ x \ \ge \ 0$ means that all components of the vector $ \ x \ $ are nonnegative.

Farkas' lemma turns out to be quite important in proving basic duality results in linear programming. I think, that applying this lemma is the most elementary way of solving my problem.