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Define the cone

$G(\textbf{x} = \{\textbf{p} \in \mathbb{R}^n | \nabla g_i(\textbf{x}^T \textbf{p} \leq 0, i \in I(\textbf{x})\}$

So this is a cone associated with the inequality constraints that is fulfilled with equality at the point $\textbf{x}$. As I interpret it, it basically contains all the directions that are either tangent to the boundary, or pointing inward from it.

There's a theorem that states that $T_s(\textbf{x}) \subseteq G(\textbf{x})$, i.e. the tangent cone (containing all the directions that at $\textbf{x}$ are tangent to all sequences in $S$ that converges to $\textbf{x}$) is a subset of the above cone, or in other words, $G$ is an outer approximation for $T_s$.

I'm wondering if it is ever the case that $G$ contains directions that $T_s$ doesn't? I can't picture a scenario where the two cones differ.

Benjamin Lindqvist
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1 Answers1

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The constraint gradients are not guaranteed to point in every unfeasible direction - they just kind of point in the 'worst' direction in a way that makes sense to a gradient. So it may be that the direction $[-1,0]$ is unfeasible at the origin, but the constraint gradients point in other directions anyway, as is the case with this constraint set

$-x_1^3+x_2 \leq 0$

$x_1^5-x_2 \leq 0$

$-x_2 \leq 0$

Here, the gradients all point to $\pm[0,1]$, but $[-1,0]$ is still unfeasible. The tangent cone only contains the positive $x_1$ axis, while $G$ contains the entire $x_1$ axis and thus containts the tangent cone as well as other directions.

Benjamin Lindqvist
  • 3,112