Second derivative test in several variables

Statement

Let $U \subseteq \mathbb{R}^n$ be open, let $f \colon U \to \mathbb{R}$ be of class C^2 on $U$, and let $a \in U$ with $\nabla f(a) = 0$. Let $H = H_f(a)$ be the Hessian matrix of $f$ at $a$.

1. If $H$ is positive definite, then $f$ has a strict local minimum at $a$.
2. If $H$ is negative definite, then $f$ has a strict local maximum at $a$.
3. If $H$ has a positive and a negative eigenvalue, then $a$ is a saddle point of $f$.

In the other cases the test gives no conclusion.

Remarks

For $n = 2$, let $D = f_{xx}(a) f_{yy}(a) - f_{xy}(a)^2$ be the determinant of $H$. If $D > 0$ and $f_{xx}(a) > 0$, then $a$ is a local minimum. If $D > 0$ and $f_{xx}(a) < 0$, then $a$ is a local maximum. If $D < 0$, then $a$ is a saddle point. The proof uses the second-order Taylor expansion. If $H$ is positive definite, then $h^{\mathsf{T}} H h \ge m \|h\|^2$ for some $m > 0$, and this term dominates the remainder. The functions $x^4 + y^4$, $-(x^4 + y^4)$ and $x^4 - y^4$ all have zero Hessian at the origin. The origin is a minimum, a maximum and a saddle point, respectively, so the test cannot decide when $H$ is singular and semidefinite. By Sylvester's criterion, $H$ is positive definite if and only if all its leading principal minors are positive.

Explanations

L1 human

At a critical point the tangent plane is flat, so the second derivatives decide the shape of the graph. If the graph bends upward in every direction, the point is a minimum. If it bends down in every direction, the point is a maximum. If it bends up in some directions and down in others, the point is a saddle.