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

convex-analysishessian-matrixpositive-semidefinite

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?

Best Answer

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).

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