For $n\geq 2$, we consider $\mathbb{R}^n$ endowed with the usual scalar product. Let $f\in\mathcal{C}^2(\mathbb{R}^n,\mathbb{R})$ be a striclty convex function such that $\nabla f$ is nowhere vanishing and $C=f^{1}(\mathbb{R}_)$. Assuming that $C$ is not empty, it is a closed convex set of $\mathbb{R}^n$ whose boundary is $\partial C=f^{1}(\{0\})$ which is a manifold of dimension $n1$. At $x\in\partial C$, $T_x(\partial C)$ is the tangent space of $\partial C$ and $\mathcal{D}_x(\nabla f)$ is the differential of $\nabla f$. For any $u\in T_x(\partial C)\{0\}$, is it true that $\mathcal{D}_x(\nabla f)(u)\,\not\perp \, u$?
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$\begingroup$ I don't fully understand your notation. Is $\mathcal{D}_x(\nabla f)$ just simply the Hessian of $f$ (since $f$ is $C^2$ by assumption)? Or is it something else? Also, is the 0 vector considered to be $\perp u$ or not? $\endgroup$– Willie WongJan 22, 2021 at 20:13

1$\begingroup$ (To complete my thoughts: if the answer to my first question is "yes", then $\mathcal{D}_x(\nabla f)(u) \perp u$ implies $H_f(u,u) = 0$, and since $f$ is convex, the Hessian is pos. semidef and this requires $\mathcal{D}_x(\nabla f)(u) = 0$. And your answer to my second question answers your post. If the answer to my first question is "no", then please provide a definition.) $\endgroup$– Willie WongJan 22, 2021 at 20:19

$\begingroup$ Both answers are "yes". Then if I understand well, I have to assume that $H_f$ is defined positive to ensure that the result is correct. Thank you! $\endgroup$– G. PanelJan 22, 2021 at 20:58
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