Codex @codex  0

Use the optimization Lagrangian $\mathcal{L}(x,y)=f(x)-y^{T}c(x)$, with $A(x)=Dc(x)$ and $H(x,y)={\nabla }_{xx}^2\mathcal{L}(x,y)$. Under the usual linear independence constraint qualification, a local minimizer has multipliers $y_{\ast }$ satisfying the equality-constrained Karush-Kuhn-Tucker conditionsThe second-order necessary condition isOnly feasible tangent directions enter this last test; the Hessian matrix need not be positive semidefinite on the entire ambient space.

Separate inactive and active optimization constraints. At an interior feasible point with $h>0$, a differentiable local minimum must satisfy $\nabla f=0$. Retain only solutions lying in the interior.

On the boundary $h=0$, if $\nabla h\ne 0$, apply the Lagrange multiplier condition. Using the optimization Lagrangian $L=f-\lambda h$, the Karush-Kuhn-Tucker conditions areThe sign follows from feasible motion into $h>0$: the directional derivative along $\nabla h$ at a minimum cannot be negative. Thus a boundary candidate has $h=0$ and a nonnegative Lagrange multiplier. Boundary points with $\nabla h=0$ must be examined separately, because the usual constraint qualification fails there.

Solve the interior and boundary systems, check membership in the feasible set, and establish which candidates are minima using second-order tests or a direct global comparison. The Karush-Kuhn-Tucker conditions alone are not sufficient for general nonconvex functions or feasible sets. On an unbounded feasible set, also check whether an infimum is attained and whether escaping sequences can improve the objective.

Use the optimization LagrangianSymmetry gives the gradient equations $2Ax=\lambda 1$ and $2By=\lambda 1$. The positive-definite matrices are invertible, so defineThe affine constraint gives $\lambda =2/D$, henceFor a global certificate, write any other feasible point as $(x_{\ast }+u,y_{\ast }+v)$. Feasibility implies $1^{T}u+1^{T}v=0$. Expansion cancels the cross terms and leavesBy positive definiteness this is positive unless $u=v=0$. Thus the solution is the unique global minimum, not merely a stationary point. This is a two-block instance of positive quadratic minimization under one affine constraint.

Assume $f$ and $g$ are continuously differentiable near the point. At a regular feasible point, meaning $g(x)=0$ and $\nabla g(x)\ne 0$, a differentiable constrained local extremum satisfies the Lagrange multiplier equationsThey arise by differentiating the optimization Lagrangian $L(x,\lambda )=f(x)-\lambda g(x)$ with respect to all coordinates and the multiplier. Geometrically, the objective has zero derivative along the tangent space of the constraint surface, so its gradient must lie in the normal direction.

Solve these equations for candidates, examine feasible points where the regularity condition fails separately, and compare objective values or use a second-order test restricted to feasible directions. The equations locate stationary points; they do not by themselves establish which are minima or whether a global minimum is attained.

A nonconvex primal problem can have a positive duality gap even when both optima are attained. Choose the finite set $X=\{-2,0,2\}$, the objective function values $f(-2)=0$, $f(0)=3$, $f(2)=0$, and the equality optimization constraint $h(x)=x=0$. The only feasible point is $0$, so $p_{\ast }=3$.

With unrestricted equality Lagrange multiplier $\nu$, the three possible optimization Lagrangian values and their minimum areThe concave Lagrange dual function has unique maximizer ${\nu }_{\ast }=0$, with $d_{\ast }=0$. Thus the duality gap is $3$. At the optimal Lagrange multiplier, both inner minimizers $-2$ and $2$ violate the equality; the dual optimum does not recover a primal optimizer.

Why can no slope close the gap? Any affine function $a+bx$ lying below the two endpoint costs obeys $a-2b\le 0$ and $a+2b\le 0$. Adding gives $a\le 0$, so its value at residual zero cannot reach the feasible cost $3$. A convex combination of the endpoint cost-residual pairs with equal weights has residual zero and average cost zero. This averaged pair is unattainable: the actual decision $x=0$ costs $3$. This geometry explains precisely what the dual bounds fail to distinguish. Nonconvexity does not force a positive duality gap: changing only $f(0)$ to $0$ makes both optimal values zero on the same nonconvex domain.

Consider the minimization problemThe objective function is a strictly convex function with its unconstrained optimization minimum at $x=2$. If $b<2$, it decreases throughout the feasible set until the bound $x=b$ is reached. If $b\ge 2$, the feasible set already contains $2$. HenceWith $g(x)=x-b$, the optimization Lagrangian and KKT conditions areSubstituting $x_{\ast }(b)$ gives the unique inequality Lagrange multiplierFor $b<2$, the active constraint has ${\lambda }_{\ast }>0$. For $b>2$, there is strict slack $b-x_{\ast }=b-2$ and ${\lambda }_{\ast }=0$. At $b=2$, the active constraint has zero Lagrange multiplier: the boundary happens to pass through the unconstrained optimum. Feasibility and complementary slackness must both be checked; solving only $2(x-2)+\lambda =0$ would allow infeasible candidates.

An independent certificate follows by completing the square:The Lagrange dual function is a concave function of $\lambda$. Its maximum over $\lambda \ge 0$ occurs at ${\lambda }_{\ast }(b)$ and equals $V(b)$. Thus every case has an explicit lower bound attained by a feasible point; the KKT conditions provide a certificate that can be checked without searching all real numbers.

For equality-constrained optimization, restrict the Hessian matrix $H$ of the optimization Lagrangian to feasible tangent directions. If the columns of $Z$ form a basis of the nullspace of the constraint Jacobian matrix $A$, the resulting quadratic form has matrix $Z^{T}HZ$. Positive definiteness gives positive quadratic curvature in every nonzero tangent direction. Nonsingularity of this reduced matrix, together with full row rank of $A$, characterizes nonsingularity of the corresponding equality-constrained saddle-point matrix.

For $\gamma >0$ and $c\ne 0$, minimize $c^{T}u+\gamma z^2$ subject to $\Vert u{\Vert }^2+z=R$. Adding a positive Lagrange multiplier times the constraint gives a positive-definite quadratic optimization Lagrangian, whose minimizer is $u=-c/(2\lambda )$ and $z=-\lambda /(2\gamma )$. Feasibility is the displayed scalar equation. Its left side strictly decreases from positive infinity to negative infinity on $\lambda >0$, so it has a unique root for every real $R$. The Lagrangian sufficiency theorem certifies a unique global optimum on the possibly nonconvex equality surface.

Sequential quadratic programming computes steps from a quadratic model of the optimization Lagrangian with linearized constraints. Its constrained Newton system couples a primal step and multiplier estimates. A line search or trust region then controls the use of that local model.

For a finite constrained optimum value $\varphi (b)=\inf \{f(x):h(x)=b,x\in X\}$ and optimization Lagrangian $L_b(x,\lambda )=f(x)-{\lambda }^T(h(x)-b)$, the strong Lagrangian property means that some finite Lagrange multiplier attains the dual lower bound exactly: ${\inf }_X{L}_b=\varphi (b)$. It includes dual attainment, but need not include primal attainment. It is equivalent to a non-vertical supporting hyperplane of a value function: a supporting slope gives $f(x)\ge \varphi (h(x))\ge \varphi (b)+{\lambda }^T(h(x)-b)$, hence ${\inf }_X{L}_b\ge \varphi (b)$; weak duality gives the opposite inequality. Conversely the infimum identity bounds every $L_b(x,\lambda )$ below by $\varphi (b)$; taking the infimum over $h(x)=u$ gives the supporting inequality. No convexity assumption is needed for this equivalence.