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In convex analysis, the second-order condition for strictly convex functions has a distinct loss of symmetry in that the implication is only one way.

Recall that: a twice differentiable function $f$, $\nabla^2 f(x) \succ 0$ (positive definite) for all $x$ in the domain of $f$ implies that $f$ is strictly convex.

This seems to be a "one-off" in the sense that the second-order condition for convexity and strong convexity are all if and only if. Similarly, the first-order condition for convexity, strict and strong convexity are all if and only if. Why the sudden loss of symmetry?

Is there a intuitive way of thinking about (or explaining) why possessing positive definite Hessian is not sufficient for strict convexity?

Your neighbor Todorovich
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  • The function $x \mapsto x^4$ is strictly convex but the Hessian is $0$ at $x=0$. It is not clear to me what symmetry you are referring to. Note that a function with a positive definite Hessian is necessarily strictly convex. – copper.hat Jul 19 '17 at 05:06
  • "But a function that has a positive definite Hessian is not necessarily strictly convex." appears to be backwards. –  Jul 19 '17 at 14:46

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Underlying reason: strict inequalities do not necessarily survive passing to the limit, while nonstrict ones do. (Consider $\lim_{x\to 0} x^2 = 0$ and so on.)

Compare $$ f\left(\frac{a+b}{2}\right) \le \frac{f(a)+f(b)}{2} \tag{convex} $$ $$ f\left(\frac{a+b}{2}\right) < \frac{f(a)+f(b)}{2}, \quad a\ne b \tag{strictly convex} $$ $$ f\left(\frac{a+b}{2}\right) \le \frac{f(a)+f(b)}{2} - c(a-b)^2 \tag{strongly convex} $$ and it becomes apparent which of these properties play well with the limiting process by which derivatives are obtained.

The situation is the same as in the relation between strictly increasing functions on the real line and functions with strictly positive derivative.

$f'>0$ everywhere $\implies$ $f$ is strictly increasing, but not conversely.

(Note that in one dimension, strict convexity is equivalent to the derivative being strictly increasing).

Because of the aforementioned limit properties, it's possible for a strictly increasing function to have a horizontal tangent, the standard example being $f(x)=x^3$. Whereas, if there is a tangent with negative slope, the function is definitely not increasing.

Move this one level up, to the second derivatives, and the same story plays out again.

  • https://math.stackexchange.com/questions/2908903/stable-strict-local-minimum-implies-local-convexity – Blind Sep 13 '18 at 11:50