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Let $\mathcal{S}\subset\mathbb{R}^n$ be a nonempty open convex set and let $f:\mathcal{S}\to\mathbb{R}$ be quasiconvex and differentiable on $\mathcal{S}$. Consider the following sets: $$\tilde{\mathcal{L}}(\pmb{p}):=\{\pmb{x}\in\mathcal{S}\,|\,f(\pmb{x})< f(\pmb{p})\},$$ $$\mathcal{L}(\pmb{p}):=\{\pmb{x}\in\mathcal{S}\,|\,f(\pmb{x})\le f(\pmb{p})\},$$ $$\tilde{\mathcal{N}}(\pmb{p}):=\{\pmb{d}\in\mathbb{R}^n\,|\,\forall\pmb{x}\in\tilde{\mathcal{L}}(\pmb{p}):\pmb{d}\cdot(\pmb{x}-\pmb{p})\le0\},$$ $$\mathcal{N}(\pmb{p}):=\{\pmb{d}\in\mathbb{R}^n\,|\,\forall\pmb{x}\in\mathcal{L}(\pmb{p}):\pmb{d}\cdot(\pmb{x}-\pmb{p})\le0\}.$$ If $\pmb{p}\in\mathcal{S}$ and $\nabla f(\pmb{p})\neq\pmb{0}$, then $\mathcal{N}(\pmb{p})=\tilde{\mathcal{N}}(\pmb{p})=\{\lambda\nabla f(\pmb{p}) \,|\, \lambda\ge0\}$.

The statement $\tilde{\mathcal{N}}(\pmb{p})=\{\lambda\nabla f(\pmb{p}) \,|\, \lambda\ge0\}$ is given in [1, end of chapter 8 - The normal cone] but without proof. If we assume its truth, the rest is trivial: Since $\tilde{\mathcal{L}}(\pmb{p})\subset\mathcal{L}(\pmb{p})$, we have $\mathcal{N}(\pmb{p})\subset\tilde{\mathcal{N}}(\pmb{p})$. As further $\lambda\nabla f(\pmb{p})\in\mathcal{N}(\pmb{p})$ for all $\lambda\ge0$ (see the first note below), there holds $\tilde{\mathcal{N}}(\pmb{p})\subset\mathcal{N}(\pmb{p})$ and thus $\mathcal{N}(\pmb{p})=\tilde{\mathcal{N}}(\pmb{p})$.

Overall, how can one proof this statement?

The following might be helpful and can be found, for example, in [2]:

[1] Crouzeix, Jean-Pierre. "Continuity and differentiability of quasiconvex functions.", https://www.genconv.org/files/Crouzeix_continuity.pdf.

[2] Bazaraa, Mokhtar S., ... Nonlinear programming: theory and algorithms.

Beton
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