 Second-derivative test

Let $f$ have continuous second partial derivatives near an interior critical point $x_{\ast }$ of an open subset of ${\mathbb{R}}^n$, with $\nabla f(x_{\ast })=0$. The Hessian matrix gives the following sufficient conditions:

A positive semidefinite or negative semidefinite Hessian matrix with a zero eigenvalue does not settle the classification. The higher-order test for separated extrema gives two quartic examples with exactly the same zero Hessian. These conditions apply to an interior critical point; a boundary optimum or a constrained optimization problem requires attention to the allowed directions. ^[ref]

The proof outline is the Taylor expansionFor a positive-definite matrix, the smallest eigenvalue supplies a positive bound proportional to $\Vert h{\Vert }^2$, which dominates the remainder for small enough $h$. Negate the function for the local maximum case. For opposite-sign eigenvalues, choose the two corresponding eigenvectors: the leading terms have opposite signs along their lines.

There are also necessary conditions. At an interior local minimum, the Hessian matrix is positive semidefinite; at an interior local maximum, it is negative semidefinite. Restriction to each straight line reduces this to the one-variable second derivative condition. Since the Laplacian is the trace of the Hessian, it is nonnegative at such a local minimum and nonpositive at such a local maximum. The sign of the Laplacian alone is insufficient: $x^2-2y^2$ has negative Laplacian but a saddle point of a scalar function at the origin.