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Let $f: D \to \mathbb{R}\ $ be a function on non-singular, convex domain $D \subseteq \mathbb{R}^d$ and let us assume the second-order derivatives of $f$ exist. It is well known that $f$ is convex if and only if its Hessian $\nabla^2 f(x)$ is positive semi-definite for all $x \in D$. It is also known that if $\nabla^2 f(x)$ is positive definite for all $x \in D$, we may conclude that $f$ is strictly convex (for a reference, see Boyd and Vandenberghe, 2004).

On the other hand, if $f$ is strictly convex, we still merely know that $\nabla^2 f(x)$ is positive semi-definite for all $x \in D$. That is, there may be $x \in D$ such that $y^T \nabla^2 f(x) y = 0\ $ for some $y\not=0$.

As an example, consider $f(x)=x^4$. In this case, $f$ is strictly convex, but $f''(x)=12x^2$ and, hence, $yf''(x)y=0$ for $x=0$ and $yf''(x)y>0$ for all $x\not=0$.

Yet, these points that ruin the complete positive definiteness seem to be very sparsely distributed within $D$. So, my question is as follows:

If $f$ is strictly convex, how can we characterize the set of points $X$ for which $\nabla^2f(x)$ is not positive definite for $x \in X$, and $\nabla^2f(x)$ is positive definite for $x \in D \setminus X$?

That is, has such a set $X$ been investigated before, what properties are known, and where can I learn more about it? Any reference is welcome.

In particular, my guess is that on can state the following:

Conjecture 1: The set $X$ is merely a discrete subset of $D$.

EDIT: Since Conjecture 1 has obviously been disproved by George Lowther below, allow me to restate my guess as the following (less bold) statement:

Conjecture 2: The set $X$ is does not contain a non-empty open ball.

or, even more cautious:

Conjecture 3: The set $D \setminus X$ does contain a non-empty open ball.

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I'm no expert, but this reminds me of Sard's theorem. The right notion of smallness for $X$ is probably that it has measure $0$. –  Greg Graviton Aug 26 '10 at 10:36
Thanks for the reference, Sard's theorem indeed seems kind of related. However, it seems that in a certain sense the set $X$ is even "smaller" than just a null set. For instance, $X$ may not contain any non-singular line segment (otherwise, $f$ is not strictly convex on this line segment). I edited my question to make this a little bit clearer. –  MRA Aug 26 '10 at 12:52
You could have $\partial^2 f/\partial y^2=0$, but $\partial^2 f/\partial x^2>0$, on the line $y=0$. –  George Lowther Aug 26 '10 at 23:40
A proof of Conjecture 2 in dimension d=1: Assume such an open interval exists, that is, there is a (closed, non-empty) interval $[a,b]$ such that $f''(t)=0$ for all $t \in [a,b]$. By Taylor expansion, we find $f(b) = f(a) + f'(a)\cdot(b-a)$. Let $c=(a+b)/2$. Again using Taylor expansion, we find $f(c) = f(a) + f'(a)\cdot(c-a) = f(a) + f'(a)\cdot(b-a)/2$. Hence, $(f(a)+f(b))/2 = f(c)$, which contradicts the strict convexity of $f$. –  MRA Aug 27 '10 at 12:45

1 Answer 1

I can't fully characterize such sets, but can say one thing - the conjecture is false. For a counterexample on one dimension, choose a closed and nowhere dense set X (such as the Cantor middle thirds set, which is not discrete) and let $g(x)=\min\{\vert x-y\vert\colon y\in X\}$ be the distance of a point x from X. Integrate it twice, $f(x)=\int_0^x\int_0^yg(z)\,dz\,dy$ to get a strictly convex function whose second derivative $f^{\prime\prime}=g$ vanishes precisely on X.

In fact, you could take X to be the Smith-Volterra-Cantor set, which has positive Lebesgue measure.

This can be used to define a strictly convex function on $\mathbb{R}^2$, $\tilde f(x,y)=f(x)+f(y)$, whose Hessian vanishes completely on a set $X\times X$ of positive Lebesgue measure.

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Thanks, this is a very convincing counter example. (I wasn't even aware that there are nowhere dense sets of positive Lebesgue measure.) I took the liberty of updating my question. –  MRA Aug 27 '10 at 11:19

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