I know that given a strictly convex function $f$
There exists a one-way condition that relates the strict convexity of $f$ with the positive-definiteness of the Hessian $\nabla^2 f(x)$.
However I just cannot remember which way it was.
Is it,
- Hessian PD $\implies$ strict convexity
or
- Strict convexity $\implies$ Hessian PD
$f(x) = x^2$ is strictly convex, its Hessian is $1$, hence PD. In this case it seems both way works!
The Hessian being strictly positive definite is sufficient for strict convexity. It isn't necessary (even if you assume sufficient regularity that the Hessian makes sense) as you can see from, for example, the one dimensional $f(x)=x^4$.