# Is strict convexity necessary and sufficient for non-degeneracy of the Hessian?

A function $f$ is called strictly convex if for $\lambda\in(0,1)$, $x\neq y,$

$$f(\lambda x + (1-\lambda)y) < \lambda f(x) + (1-\lambda)f(y)$$

If $f:\mathbb{R}^n\to\mathbb{R}$ is a twice continuously differentiable strictly convex function is the Hessian $H_f$ necessarily positive definite?

I know that in the case where $f:\mathbb{R}\to\mathbb{R}$ the inequality $f''(x)>0$ for all $x$ is only sufficient for strict convexity, not necessary (e.g. $f(x)=x^4$).

• I guess what I'm asking is, is there a possibility for $H_f$ to be singular even though $f(\lambda x + (1-\lambda)y) < \lambda f(x) + (1-\lambda)f(y)$ is satisfied for all $\lambda\in(0,1)$, $x\neq y$? – jonem Aug 16 '14 at 20:50
• Doesn't your example of $f(x) = x^4$ show that it's possible for a strictly convex function to have a singular Hessian? – littleO Aug 16 '14 at 21:06
• Right, obviously true, thanks. – jonem Aug 16 '14 at 21:11

As your own example shows non-degeneracy of the Hessian is not necessary. You can easily extend it to any $\mathbb{R}^n$ by taking $f(x)=x_1^4+\dots+x_n^4$. The problem is that non-degeneracy of the Hessian guarantees convexity to quadratic order near each point which is stronger than just strict convexity. In fact, strict convexity can be arbitrarily close to flatness, i.e. there can't be any necessary condition for it in terms of derivatives.