How to prove this optimization theorem?

Question

Let $X$ be an open set, $f:\mathbb R^n\to \mathbb R,g_i:\mathbb R^n\to\mathbb R,$ if $i\in I,$ then $g_i$ is differentiable at $\overline x$, if $i\notin I,$ then $g_i$ is continuous at $\overline x$. $F_0=\{d\in\mathbb R^n|\nabla f(\overline x)^td<0\}$ and $\in G_0=\{d\in\mathbb R^n|\nabla g_i(\overline x)^td<0\}$. $$min\ f(x) $$ $$s.t. g_i(x)\le 0$$ $$x\in X$$ $$\overline x\ feasible$$

1. If $\overline x$ is local minimum, then $F_0\cap G_0 =\emptyset$.

2. If $F_0\cap G_0=\emptyset$ and $f$ is strictly pseudo convex at $\overline x$ and there exists $\epsilon>0$ such that $g_i,i\in I$ is strictly pseudo convex in $B_\epsilon(\overline x),$ then $\overline x$ is local minimum.

Attempt

Let $d^*\in \mathbb R^n$ arbitrary. We have to show $d^*\in F_0=\{d\in\mathbb R^n|\nabla f(\overline x)^td<0\}$ and $d^*\in G_0=\{d\in\mathbb R^n|\nabla g_i(\overline x)^td<0\}$.

As $\overline x$ is local minimum then, there exists $\epsilon>0$ such that $f(\overline x)<f(x),\forall x\in B_\epsilon (\overline x)$.

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How do I use the local minimum definition to see that $F_0\cap G_0=\emptyset$?

Help me please

Answer 1

Okay, then we can "easily" solve the first problem by showing that $ F_0 \cap G_0 \neq \emptyset $ implies that $ \bar{x} $ is no local minimizer.

Let $ d \in F_0 \cap G_0 $ be given, then we have

(1) $ \nabla f(\bar{x})^\top d < 0 $

(2) $ \nabla g_i(\bar{x})^\top d < 0 $ (for all $ i \in I $)

Note that the scalar products represent the directional derivatives of the functions $ f,g_i $ at $ \bar{x} $ in direction $ d $. Hence, $ d $ is a recession direction at $ \bar{x} $ for all these functions. This means, that the values of $ f $ and $ g_i $ decrease if we move (a small positive stepsize $t$) in direction $ d $.

More precisely, we have:

(1) implies: $ 0>f(\bar{x})^\top d = \lim_{t \to 0, t>0} \frac{f(\bar{x}+td)-f(\bar{x})}{t} $

(2) implies: $ 0>g_i(\bar{x})^\top d = \lim_{t \to 0, t>0} \frac{g_i(\bar{x}+td)-g_i(\bar{x})}{t} $

Since the limits are negative , we can find a sufficiently small value of $ t^\star $, so that

(A) $ 0>f(\bar{x}+td)-f(\bar{x}) $

(B) $ 0>g_i(\bar{x}+td)-g_i(\bar{x}) $

(C) $ \bar{x}+td \in X $ (since $ X $ is open)

holds for all $ t \in [0,t^\star] $.

But this means that any of these points $ \bar{x}+td $ (for $ t \in [0,t^\star] $) are feasible (which follows from (B) and (C)) and possesses a lower objective value than $ \bar{x} $ (which follows from (A)). Obviously, then $ \bar{x} $ cannot be a local minimizer.