# Extrema

## Idea

In general, a maximum is a top element, a minimum is a bottom element, and an extremum is either. However, these terms are typically used in analysis, where the order theory is secondary. One also usually speaks of extrema of a function, meaning a top or bottom element of the range of the function under its induced order as a subset of an ordered codomain. In this context, one also considers local extrema of functions; a local extremum of $f$ is an extremum of a restriction of $f$ to an open subspace of its original domain. In any case, the extremum is strict if the function takes the extreme value only once (in the relevant domain).

## Local extrema of differentiable functions

We list some sufficient and necessary conditions for a (nice) function $f$ on a smooth manifold to have a local extremum at a point $x$. As these are local conditions, we may assume $f$ is a function $U \to \mathbb{R}$ where $U$ is an open subset in a Cartesian space $\mathbb{R}^n$. These conditions fall under the rubric of “second derivative test”.

Assume $f$ to be a twice-differentiable function, and let $x$ in its domain be a critical point: a point where its derivative / Jacobian vanishes. Let $H_x(f)$ be the Hessian matrix of the function. Recall that $x$ is a nondegenerate critical point if the symmetric matrix $H_x$ is nondegenerate; equivalently, if $0$ is not an eigenvalue of $H_x$.

Let $x$ be a nondegenerate critical point. Then

• For $x$ to be a strict local minimum within some neighborhood, it is necessary and sufficient that $H_x(f)$ be a positive definite form?.

• For $x$ to be a strict local maximum within some neighborhood, it is necessary and sufficient that $H_x(f)$ be a negative definite form?.

(The only other possibility left for a nondegenerate critical point is that $H_x(f)$ be an indefinite form?, having a mix of positive and negative eigenvalues. In this case, $x$ is a saddle point. For more on this, see Morse theory.)

If $x$ is a degenerate critical point (so $0$ is an eigenvalue of $H_x$), we have:

• For $x$ to be a local minimum, it is necessary that $H_x(f)$ be a positive semidefinite form?, i.e., that all eigenvalues are nonnegative.

• For $x$ to be a local maximum, it is necessary that $H_x(f)$ be a negative semidefinite form?, i.e., that all eigenvalues are nonpositive.

These conditions are not sufficient. For a simple example, the origin in $\mathbb{R}^2$ is a critical point of $f(x, y) = x^3 - y^3$, where the Hessian is the zero matrix (hence positive semidefinite and negative semidefinite), but clearly the origin is neither a local maximum nor a local minimum.

Revised on January 20, 2012 05:10:01 by Todd Trimble (74.88.146.52)