Stationary point and classification of a nonsingular quadratic response surface

For any $x,h\in {\mathbb{R}}^k$, symmetry of $B$ gives $m(x+h)-m(x)=h^{\mathsf{T}}(\beta +2Bx)+h^{\mathsf{T}}Bh$. The second term is bounded in absolute value by a constant times $\mathrm{\parallel }h{\mathrm{\parallel }}^2$, so the differential at $x$ is $h\mapsto h^{\mathsf{T}}(\beta +2Bx)$; hence $x$ is stationary exactly when $\beta +2Bx=0$.

Since $B$ is nonsingular, this equation has the unique solution $x_0=-\frac{1}{2}{B}^{-1}\beta$.

At this point, for every $h$, $m(x_0+h)-m(x_0)=h^{\mathsf{T}}Bh$. Regard $B$ as an operator on ${\mathbb{R}}^k$ with the real coordinate inner product [F2], for which the standard basis is orthonormal. By [F3] it is self-adjoint, so [F4] supplies an orthonormal eigenbasis $v_1,\dots ,v_k$ with real eigenvalues ${\lambda }_i$; none is zero because $B$ is nonsingular. Writing $h={\sum }_i{z}_i{v}_i$ yields $h^{\mathsf{T}}Bh={\sum }_i{\lambda }_i{z}_i^2$.

If all ${\lambda }_i>0$, the increment in step 3.1 is positive for every nonzero $h$, and if all ${\lambda }_i<0$ it is negative for every nonzero $h$. If both signs occur, choosing $h$ along a positive-eigenvalue vector or a negative-eigenvalue vector gives increments of opposite signs, so $x_0$ is a saddle. The exact quadratic identity makes the first two classifications global for the polynomial on ${\mathbb{R}}^k$.

If $x_0$ lies in the interior of $\mathcal{R}$, sufficiently small perturbations in every direction remain in $\mathcal{R}$, so step 4.1 also gives the local classification there. Whenever $x_0\in \mathcal{R}$, the strict global inequalities in the definite cases of step 4.1 hold for every $x\in \mathcal{R}\setminus \{x_0\}$, proving strict constrained optimality even at the boundary. A boundary saddle need not persist: $m(x_1,x_2)=x_1^2-x_2^2$ has an ambient saddle at $0$, but its restriction to $\mathcal{R}=\{(t,0):t\in \mathbb{R}\}$ has a strict minimum there. If $x_0\mathrm{\notin }\mathcal{R}$, it is infeasible.