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Suppose we have a function $f(\boldsymbol{x})$ and its hessian, i.e $\nabla_{\boldsymbol{x}}^2f(\boldsymbol{x})$, equals $\mathbf{0}$.

We know that for convexity $\nabla_{\boldsymbol{x}}^2f(\boldsymbol{x})\succeq\mathbf{0}$.

The first question is whether $f(\boldsymbol{x})$ is strictly convex or just convex? Secondly, if the hessian is equal to $\mathbf{0}$, how does it change the first order condition of convexity? Does the inequality in the first order condition $f(\boldsymbol{y})\geq f(\boldsymbol{x}) + \nabla^\top f(\boldsymbol{x}) (\boldsymbol{y}-\boldsymbol{z})$ also reduces to equality?

Maybe if someone can tell how the second order condition is derived from the first order condition, it could answer my question.

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    $\begingroup$ maybe a better question for the pure math stack, like mathoverflow. $\endgroup$ – steveo'america Feb 13 at 17:27
  • $\begingroup$ Please be more precise: "its hessian... equals 0" - in a single point $x$ or everywhere? If the hessian vanishes everywhere, then the function is linear - and thus convex and concave at the same time. If it vanishes at a single point, not much can be said. For example, $x^4$ is strictly convex, and the hessian vanishes at $x=0$. Also, why should the condition be changed? The first order condition will remain necessary. Are you looking for the second order condition in case the first order condition is satisfied as equality? $\endgroup$ – jarauh Feb 13 at 21:09
  • $\begingroup$ The hessian equals zero at all x. Ok, I understand now. I just wanted to see if hessian equals zero implies equality in the first order condition as well. $\endgroup$ – Osama Feb 13 at 21:15
  • $\begingroup$ @Osama did my answer help you at all? $\endgroup$ – Lucas Farias Feb 13 at 23:38
  • $\begingroup$ @LucasFarias thanks for your answer. I could not comprehend it entirely $\endgroup$ – Osama Feb 14 at 16:03
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A real valued function is convex, using the first-order condition, if the inequality below holds

$$ f(\alpha x +(1-\alpha)y) \leq \alpha f(x) +(1-\alpha) f(y) $$

It is strictly convex if such inequality holds for $<$.

Now, the second-order condition can only be used for twice-differentiable functions (after all you'll need to be able to compute it's second derivatives), and strict convexity is evaluted like above; convex if $\nabla_{\boldsymbol{x}}^2f(\boldsymbol{x})\succeq\mathbf{0}$ and strictly convex if $\nabla_{\boldsymbol{x}}^2f(\boldsymbol{x})>\mathbf{0}$.

Finally, the second-order condition does not overlap the first-order one, as in the case of linear functions.

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  • $\begingroup$ According to your definition, it is impossible for any function to be strictly convex, because when you plug in $\alpha=0$ or $\alpha=1$ you do not get a strict inequality. You need to be clear about the quantifiers in your statements. $\endgroup$ – whuber Feb 25 at 14:01

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