Jtdylktuy

In convex analysis optimization, I would like to show that the necessary conditions of the multiplier rule correspond to the nonexistence of a decrease direction. I would like to prove the following theorem:

Theorem: Let ${zeta_i : i = 1,2,...,k }$ finite subset in $X^{*}$ - dual space of the normed space $X$. The following are equivalent:

$(a)$ There is no $v in X$ such that $langle zeta_i , v rangle < 0 , forall i = 1,...,k$;

$(b)$ The set ${zeta_i : i = 1,2,...,k }$ is positively linearly dependent : there exists a nonzero nonnegative vector $gamma in mathbb{R}^k$ such that $sum_{1}^{k} gamma_i zeta_i = 0$.

$quad$ To prove $(a) implies (b)$, I will construct two subsets and I try to use the separation theorem. Before, I enunciate such theorem:

(Hahn-Banach separation theorem)

Let $K_1$ and $K_2$ be nonempty, disjoint
convex subsets of the normed space $X$. If $K_1$ is open, then the sets can be separated. That is, there exist $gamma$ ∈ $X^{*}$ and $eta$ ∈ R such that
$$langle gamma , x rangle < eta leq langle gamma , y rangle, quad forall x in K_1 , y in K_2.$$

To apply the separation theorem, let

$$K_1 = {y in mathbb{R}^k : y_i < 0, forall i}, quad K_2 = { (langle zeta_1 , v rangle , ..., zeta_k , v rangle) : v in X }$$

The sets $K_1 , K_2$ are both convex and nonempty. By $(a)$ the sets are disjoint. Then, by the separation theorem, exist $eta in mathbb{R}$ and $gamma in mathbb{R}^k$. It is straightforward show that $eta = 0$ and

$$ 0 leq sum gamma_i langle zeta_i , v rangle, quad forall v in X.$$

I can't show the nonnegative of $gamma$ and the other inequality

$$ 0 geq sum gamma_i langle zeta_i , v rangle, quad forall v in X.$$

Some help?

Hints:

1. Assume $gamma_m<0$ for some $m$. Then fix $x_0in K_1$ and note that $x_t=x_0-t e_min K_1$, $forall tge 0$ ($e_m$ is the $m$th vector of the canonical basis of $Bbb R^k$). What $langlegamma, x_trangle<0$ would mean for $tto+infty$?


2. Other inequality: $-vin X$.