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Let $f$ be a function of $n$ variables with continuous partial derivatives and cross partial derivatives on the convex open set $S$ and denote the Hessian of $f$ at the point $x$ by $H(x)$.

We know for concavity:

$f$ is concave if and only if $H(x)$ is negative-semidefinite for all $x ∈ S$

But, strict concavity is not IFF, in fact, if strictly concave, it does not necessarily mean the Hessian evaluated at the optimum is strictly negative.

Consider the following function:

$f(x)=-x^4$.

This function is strictly concave.

We cannot say, the Hessian evaluated at its optimum, which is $H(0)=0$ is strictly negative.

How would you explain that strict concavity is not IFF condition in relation to curvature of function to someone who is just taking Calc 3?

Notice the actual condition for strictly concave function is:

If $H(x)$ is negative definite for all $x ∈ S$ then f is strictly concave.

Basically I want to illustrate the point not only it fails in strict concavity case, but relate this to the curvature information of function what we cannot simply "do" IFF condition on strict concavity in comparison with concave one.

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  • $\begingroup$ Even your definition of concavity is incorrect. It is not necessary for a function to be differentiable to be concave; see, for instance, $f(x)=-|x|$. $\endgroup$ – Michael Grant Oct 19 '17 at 17:27
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    $\begingroup$ @MichaelGrant oh forgot to say assuming $f/in C^2$ on S. $\endgroup$ – Frank Swanton Oct 19 '17 at 21:04
  • $\begingroup$ Ah, of course, I see what you wrote in the first sentence now, sorry! $\endgroup$ – Michael Grant Oct 19 '17 at 22:40
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Note that concavity, like monotonicity, has nothing to do with derivatives per se. Your example $f(x)=-x^4$ shows that the negative definiteness of the Hessian is not an IFF-condition. This is a fact of life devoid of an "explanation". We have a similar fact with monotonicity: The function $g(x):=x^3$ is strictly monotone even though $g'(0)=0$.

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