# Differential of the gradient of a strictly convex function

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 $$n-1$$. 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 $$\mathcal{D}_x(\nabla f)(u)\,\not\perp \, u$$?

• 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? Jan 22, 2021 at 20:13
• (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.) Jan 22, 2021 at 20:19
• 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! Jan 22, 2021 at 20:58