Unconstrained Extrema Conditions¶

Theorem¶

Let $A\subseteq\mathbb{R}^n$ be open and let $f\colon A\to\mathbb{R}$ $R$ .

If $f$ is $C^1$ and has a local extremum at an interior point $\boldsymbol{x}^\ast$ , then

\boldsymbol{D}f(\boldsymbol{x}^\ast)=(0,\ldots,0).

If $f$ is $C^2$ , then a local maximum has $\boldsymbol{H}_f(\boldsymbol{x}^\ast)$ negative semidefinite and a local minimum has $\boldsymbol{H}_f(\boldsymbol{x}^\ast)$ positive semidefinite. Conversely, if $\boldsymbol{x}^\ast$ is a critical point, a negative definite Hessian gives a strict local maximum, a positive definite Hessian gives a strict local minimum, and an indefinite Hessian gives a saddle point.

Interpretation¶

At an interior unconstrained optimum, no direction may give a first-order gain, so the derivative must vanish. The Hessian then describes the local curvature and decides the type of critical point when it is definite or indefinite.

Level sets near three stationary points: closed curves around a local maximum and around a local minimum, and crossing hyperbolas around a saddle point.

Around a local maximum (Hessian negative definite) and a local minimum (positive definite) the level curves are closed; around a saddle point (indefinite) they bend away, and the level set through the point is the pair of dashed crossing lines. The level curves alone do not tell a maximum from a minimum; the definiteness of the Hessian does.

Drag $\boldsymbol{x}_0$ to a point where $\boldsymbol{\nabla} f=0$ $0$ ; the Hessian there separates minima, maxima and saddles.

At a critical point $\boldsymbol{\nabla} f(\boldsymbol{x}_0)=0$ and the tangent plane is flat; the Hessian $\boldsymbol{H}_f(\boldsymbol{x}_0)$ decides whether the surface bends up, down or both ways.

If $\boldsymbol{D}f(\boldsymbol{x}^\ast)\neq(0,\ldots,0)$ $(0, \dots, 0)$ , a small step along the gradient raises $f$ and a step against it lowers $f$ (steepest ascent direction), so $\boldsymbol{x}^\ast$ cannot be an extremum. This needs room in every direction, which is why $A$ is open: on $[0,1]$ , $f(x)=x$ has its maximum at the endpoint $1$ , where its one-sided derivative is $1$ . At a critical point, the second-order Taylor approximation reduces to $f(\boldsymbol{x}^\ast+\boldsymbol{h})\approx f(\boldsymbol{x}^\ast)+\tfrac{1}{2}\boldsymbol{h}^{\top}\boldsymbol{H}_f(\boldsymbol{x}^\ast)\boldsymbol{h}$ $\frac{1}{2} h^{⊤} H_{f} (x^{*}) h$ , so the sign of this quadratic form decides. Between the necessary (semidefinite) and sufficient (definite) conditions there is a gap: at the origin, $x^2+y^4$ has a minimum and $x^2+y^3$ has neither, though both have Hessian $\operatorname{diag}(2,0)$ .

This is the $n$ -variable form of the first-order condition $f'(x_0)=0$ and the second-order condition on $f''(x_0)$ ; definiteness is checked with Sylvester's criterion. The conclusions are local. For a convex function, however, every critical point is a global minimizer.

Key Results¶

First-order stationarity is necessary but not sufficient.
Semidefinite Hessians can make the second-order test inconclusive.
An indefinite Hessian at a critical point means the point is neither a local maximum nor a local minimum.

Examples¶

For $f(x,y)=x^2+y^2$ $y^{2}$ , the Hessian is positive definite and the origin is a strict local minimum.
For $f(x,y)=x^2-y^2$ $y^{2}$ , the Hessian is indefinite and the origin is a saddle point.

Prerequisites¶

Sources¶

Math IV: Multivariate Calculus · Week 4, Lecture 1

Keywords: FONC, SONC, SOSC, Hessian test for extrema Symbols: nabla gradient, real numbers, subset

Last updated: 20 October 2026 03:02