Geometric interpretation of optimality

Question

Consider the nonlinear problem $$\min_{x\in\Omega} f(x)$$ where $\Omega\subset R^n$ is convex and $f:\Omega\rightarrow R^n$. If $\bar{x}$ is a local min, why is it that $\nabla f(\bar{x})$ is in the dual cone of the set $\Omega - \bar{x}$?

Answer 1

Both dual cones coincide.

This can be seen as follows. Let us denote the dual cone by $(\cdot)^+$, the conic hull by $\mathrm{cone}$ and the closure by $\mathrm{cl}$. Then, one has \begin{align*} A^+ &= (\mathrm{cone} A )^+, \\ A^+ &= (\mathrm{cl} A )^+, \\ T_\Omega(\bar x) &= \mathrm{cl} \, \mathrm{cone} (\Omega - \bar x). \end{align*} Combining these facts, we get the desired $$ T_\Omega(\bar x)^+ = (\Omega - \bar x)^+. $$

Answer 2

As you mentioned in your comment $\nabla f(\bar{x})$ is in the tangent cone $T_\Omega(\bar{x})$. However, contrary to the statement in your comment it is left to proof that $T_\Omega(\bar{x})$ is a subset of the dual cone of $\Omega - \bar{x}$. This is, at least geometrically, clear.