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?



  1. Assume $\gamma_m<0$ for some $m$. Then fix $x_0\in K_1$ and note that $x_t=x_0-t e_m\in K_1$, $\forall t\ge 0$ ($e_m$ is the $m$th vector of the canonical basis of $\Bbb R^k$). What $\langle\gamma, x_t\rangle<0$ would mean for $t\to+\infty$?
  2. Other inequality: $-v\in X$.
  • $\begingroup$ Yes, this is true. I think I focused a lot on the $K_2$ set and not on the $K_1$ set. Thank you! $\endgroup$ – orrillo Nov 20 '18 at 2:21
  • $\begingroup$ Only one doubt. I tried to imagine the set $K_2 $ in small dimensions. For example. I think that in the case $k = 2$ we have lines $x = 0$ or $y = 0$ and in the case $k = 3$, maybe planes of the form $x = 0$, $y = 0$ or $z = 0$. Or maybe lines in the tridemsional space. $\endgroup$ – orrillo Nov 20 '18 at 2:26
  • 1
    $\begingroup$ @orrillo For $k=2$: if $\zeta_1$, $\zeta_2$ are linearly independent then $K_2={\Bbb R}^2$. If they are linearly dependent then $K_2$ is a line through the origin. It can be any line, not just the axes $x=0$ or $y=0$. If we denote $x=\langle\zeta_1,v\rangle$ and $y=\langle\zeta_2,v\rangle$, for dependent $\zeta_i$ we can find $\gamma_1$, $\gamma_2$ such that $\gamma_1x+\gamma_2y=\langle\gamma_1\zeta_1+\gamma_2\zeta_2,v\rangle=0$. However, if (a) is true, the line cannot go in the third quadrant, which is $K_1$. $\endgroup$ – A.Γ. Nov 20 '18 at 7:14

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