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Vect$\big(\nabla f(x)\big)$ replaced by $\lbrace\lambda\nabla f(x)\rbrace_{\lambda\geq 0}$.
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Link between asymptotic cone and the boundary of a convex set

For $n\geq 1$, let $f\in\mathcal{C}^2(\mathbb{R}^n ; \mathbb{R})$ such that $\forall x\in\mathbb{R}^n$: $\nabla f(x)\neq 0$ and $\mathcal{Hess}f(x)$ is defined positive. Let $C=\lbrace x\in\mathbb{R}^n,f(x)\leq 0\rbrace$. $C$ is then a closed convex set, with boundary $\partial C=\lbrace x\in\mathbb{R}^n,f(x)= 0\rbrace$. Its asymptotic cone is denoted $C^{as}$, which has a polar cone denoted by $(C^{as})^\circ$.

Question: Do the following hold: $(C^{as})^\circ=\bigcup_{x\in\partial C} \lbrace\lambda\nabla f(x)\rbrace_{\lambda\geq 0}$ ?

The direct inclusion is difficult for me. Thanks for your interest.

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