# Is the hessian negative semi-definite if we have an interior maximum?

Is it true that given a smooth scalar field f on a domain D , if f attains a maximum (minimum) on the interior of D then the hessian of f evaluated at this max (min) is negative (positive) semi-definite?

I have seen this quoted as a fact but usually when i try to find a proof all that comes up is the second derivative test (given hessian is negative definite then we have max)

is this actually just equivalent to the test?

EDIT:

if we consider an approximation via taylor's theorem we have (near $a$ ) :

$$f(x) = f(a) + Df(a)^{T} (x-a) + (x-a)^{T} D^2 f(a) (x-a) + o(|x-a|^3)$$

now if $a$ is max then$f(x) \le f(a)$ so $D^2 f(a)$ is negative semidef ?

Yes, at a local maximum the Hessian of a smooth (real) function will be negative semi-definite (and equivalently the Hessian will be positive semi-definite at a local minimum). If this is hard to find, being a weak converse of the second derivative test, it's likely because proving it requires some linear algebra material not yet covered at the time of a 3rd quarter in calculus where multivariable concepts are introduced.

Consider that the Hessian is a symmetric (real) matrix, and thus has a complete basis of eigenvectors. At a local maximum the function will have, on each line passing through the maximum point, a familiar one-dimensional local maximum. If the Hessian were not negative semi-definite, it would have a line (corresponding to an eigenvector of a positive eigenvalue) along which the restricted function would have a concave up appearance. But this would contradict the point being a local maximum.

• @hardmath Can we say the other way around, that is if $x$ is negative semidefinite then is it local max?
– OGC
Commented Oct 16, 2018 at 3:18
• @OGC: We need more than just the Hessian negative semidefinite; typically one needs the vanishing of the first derivatives and the Hessian negative definite to get sufficient conditions for a local maximum. If you are interested in the semi-definite cases, see this previous Question and this one. If you need further info, it's probably worth asking a Question. Commented Oct 16, 2018 at 4:03

Here is some explanation that may clarify what happens: For every $$u\in \mathbb R^2$$, if we define $$d^2_uf$$ to be the second derivative of $$f$$, then $$d^2_uf <0$$. By direct calculations, we obtain $$u^T Hess f u = d^2_uf<0.$$ So by definition $$Hess f$$ is negative definite at the point maximum happens.