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.