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).
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.
Last revised on January 20, 2012 at 05:10:01. See the history of this page for a list of all contributions to it.