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For a convex function $f: S \to \mathbb{R}$, where $S \subset \mathbb{R}^{n}$, if $X^{*}$ is a local min, (hence global min), we can check along every line restriction to verify $X^{*}$ is indeed the local min. However, oftentimes in optimization, the function is not convex, or no longer convex after a change of variable. Therefore this does not always work

So I wonder in general, for $C^{2}$ function, $f:S \to \mathbb{R}, S \subset \mathbb{R}^{n}$, does there exist a necessary sufficient, or very weak sufficient condition to guarantee that if $X^{*}$ is the local min along every line restriction, it is actually a local min?

Of course, checking for gradient and hessian works in the interior, but this is not possible on the boundary. Also, what about the case if we lift the $C^{2}$ restriction?

Edit: I have changed this question from the original version so it can be more focused on my question.

wsz_fantasy
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    If there was a point of lower value, then the directional derivative in the direction of that point would be strictly negative. This is where convexity comes in. – copper.hat Jun 07 '23 at 03:39
  • @copper.hat Thank you. What about in general though? Does there exist some condition that can be imposed on the function such that checking every directional derivative would be sufficient? – wsz_fantasy Jun 07 '23 at 16:54
  • Slight hyperbole here, but to some extent, the beauty of convex programming is that $x^$ solves $\min_{x \in C} f(x)$ iff* $\operatorname{df}(x^,x-x^) \ge 0$ for all $x \in C$. $C$ is a closed convex set, $f$ convex, Gateaux differentiable. – copper.hat Jun 07 '23 at 17:14

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