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 ?
Best Answer
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.