Does the positive semidefiniteness of a function not necessarily imply the function is convex?

Question

It is well known that the Hessian matrix of a convex function is positive semidefinite (PSD) and positive definite (PD) for strict convexity. I have rarely thought about the reverse direction, but it seems it is not true.

If a Hessian matrix of an arbitrary function is PSD, can any conclusions be made about the convexity? For this, I am considering 2 cases:

1. PSD at a critical point.

2. PSD everywhere.

In the single variable case, e.g., $f(x) = x^2$, the Hessian reduces to just a single second derivative, and the second derivative test can be applied for convexity, but for the general multivariable case, it seems no conclusions can be made. Is that correct?

Answer 1

A twice continuously differentiable function over an open convex set is convex if and only if its Hessian is a PSD matrix. If indeed it is PSD over the feasible open convex set, then it must be PSD at a critical (stationary) point as well. And the main advantage of convex functions is that a critical point is also a local and global minimum of the function (although it need not be unique).