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For inequalities $c_j(x)\ge 0$, the active constraints at a feasible point are those holding with equality. Inactive constraints remain strictly satisfied in a sufficiently small neighborhood and do not restrict first-order feasible velocities there. The active gradients determine the linearized tangent cone and appear in the linear independence constraint qualification. At a KKT point, inactive constraints have zero Lagrange multipliers by complementary slackness.

Suppose strictly feasible stationary points of $f(x)-{\mu }_k{\sum }_j\log c_j(x)$ form a bounded sequence. Every cluster point satisfying the linear independence constraint qualification is a KKT point. Inactive multipliers tend to zero. The active gradients near the limit have uniformly bounded dual vectors, so pairing barrier stationarity with those vectors bounds the active multipliers. A convergent multiplier subsequence then gives stationarity, nonnegative multipliers and complementary slackness, since ${\lambda }_{k,j}{c}_j(x_k)={\mu }_k\to 0$. A stationarity residual tending to zero is also allowed.

This conclusion is stationarity, not global optimality. If the subproblems are globally minimized and strictly feasible points are dense in the feasible set, any cluster point is instead a global minimizer. Compare each barrier minimum with a fixed strictly feasible point; boundedness near the cluster point bounds $\log c_j(x_k)$ above. Taking the limit gives the objective comparison on strictly feasible points, then continuity and density extend it to all feasible points.

For smooth inequalities $c_j\ge 0$, the displayed intersection of half-spaces contains the contingent tangent cone. This follows by differentiating each active feasibility inequality along a feasible tangent sequence. The linear independence constraint qualification, and more generally the Mangasarian–Fromovitz constraint qualification, ensure equality in the inequality-only setting. Without a qualification the inclusion can be strict: for $c(x)=-x^2\ge 0$, the feasible set is $\{0\}$ and its tangent cone is $\{0\}$, while $c^{\prime }(0)=0$ makes the linearized cone the whole real line.

Write $x_1=t$, $x_2=t^2$. The two inactive constraints are strict exactly for $-2<t<2$. Complementary slackness gives ${\lambda }_1={\lambda }_2=0$. The KKT conditions reduce toIf $t=0$ these give ${\lambda }_3=0$. Otherwise ${\lambda }_3=2t$ and $t=-3/2$, which gives a negative multiplier and is inadmissible. Thereforeis the only KKT point with this active set. Its active gradient $(0,1)$ is nonzero, so the linear independence constraint qualification, and hence the Mangasarian–Fromovitz constraint qualification, holds.

It is not a local maximizer: $(0,\epsilon )$ is feasible for small $\epsilon >0$ and has objective ${\epsilon }^2>0=F(0,0)$. The feasible boundary path $(t,t^2)$ for small negative $t$ has objective $2t^3+t^4<0$, so it is not a local minimizer either.

All three active constraints force $x_1=2$ and $x_2=4$, which also satisfy $x_1^2+x_2^2=20$. For maximization use $\mathcal{L}=F+{\lambda }^{T}g$, $\lambda \ge 0$. At $x^{\ast }=(2,4)$,Stationarity gives the complete familyIn particular $({\lambda }_1,{\lambda }_2,{\lambda }_3)=(2,3/2,2)$ is strictly positive. Three gradients in two dimensions cannot be linearly independent, so the linear independence constraint qualification fails. But $d=(-1,-1)$ gives $\nabla g_i\cdot d=(1,12,3)>0$, so the Mangasarian–Fromovitz constraint qualification holds.

For a nonzero linearized feasible direction $d$, each $\nabla g_i\cdot d\ge 0$, andThe inequality is strict because equality would require all three inner products to vanish, and the first and third gradients already span ${\mathbb{R}}^2$. This proves a strict local maximum: otherwise a sequence of nearby feasible points with no smaller objective would, after normalizing their differences from $x^{\ast }$ and taking a convergent subsequence, give a unit linearized feasible direction with nonnegative objective derivative, contradicting the displayed inequality. Thus

With $L(x,y)=f(x)-y^{T}c(x)$, the Karush-Kuhn-Tucker conditions areUnder a suitable constraint qualification, such as the linear independence constraint qualification on active constraint gradients, they are necessary for a local minimum. Smoothness alone is insufficient. The sign convention here corresponds to constraints written as $c_i\ge 0$ and a minus sign in the optimization Lagrangian.

The active constraint set is $\boxed{\mathcal{A}(x)=\{j:c_j(x)=0\}}$, and the linearized tangent cone isDifferentiating each active feasibility inequality along a tangent sequence gives $T_{\Omega }(x)\subseteq F(x)$. A sufficient condition for equality is the linear independence constraint qualification: the active constraint gradients are linearly independent. The weaker Mangasarian–Fromovitz constraint qualification also suffices in this inequality-only setting.

Under the stated independence condition, the necessary first-order conditions become the Karush-Kuhn-Tucker conditions: there are multipliers ${\lambda }_j$ withUse the allowed span assertion to write $\nabla f(x_{\ast })={\sum }_{j\in \mathcal{A}}{\lambda }_j\nabla c_j(x_{\ast })$. Independence gives vectors ${\phi }_i$ with $\nabla c_j(x_{\ast }{)}^T{\phi }_i={\delta }_{ji}$ for the active indices. Each ${\phi }_i$ lies in $F=T_{\Omega }$, so part (a) gives ${\lambda }_i=\nabla f(x_{\ast }{)}^T{\phi }_i\ge 0$. Assign zero multipliers to inactive constraints. This proves all displayed conditions, including complementarity. Conversely, their stationarity and sign conditions imply $\nabla f(x_{\ast }{)}^{T}d={\sum }_{j\in \mathcal{A}}{\lambda }_j\nabla c_j(x_{\ast }{)}^{T}d\ge 0$ for every $d\in F$. Thus they reformulate the first-order inequality; they do not alone prove a minimum for a nonconvex problem.

On the strictly feasible set ${\Omega }_c^{\circ }=\{x:c_j(x)>0\text{ for all }j\}$, use the logarithmic barrierChoose a strictly feasible starting point and parameters ${\mu }_n\downarrow 0$. For each parameter minimize the barrier functional by an unconstrained method confined to ${\Omega }_c^{\circ }$, warm-starting from the preceding point. In the exact version take a stationary point $x_n$ of the barrier subproblem. The stationarity equation isThese are the natural Lagrange multiplier estimates; the basic method need not perform a separate primal-dual correction.

The barrier method KKT convergence theorem is a global stationarity result: if the exact barrier stationary points form a bounded sequence, then every cluster point satisfying the linear independence constraint qualification is a KKT point. To prove it, take any subsequence $x_n\to x_{\ast }$. Continuity gives feasibility of $x_{\ast }$. For an inactive index, $c_j(x_{\ast })>0$ and therefore ${\lambda }_{n,j}={\mu }_n/c_j(x_n)\to 0$.

The active gradients at $x_{\ast }$ are independent. The supplied dual-basis result gives bounded vectors ${\phi }_{n,i}$, converging as $n\to \infty$, with $\nabla c_j(x_n{)}^T{\phi }_{n,i}={\delta }_{ji}$ for active $i,j$. Pair stationarity with these vectors:Thus all active multipliers are bounded as well. Extract a convergent multiplier subsequence. Its limit is nonnegative, satisfies stationarity by continuity, and satisfies complementarity because ${\lambda }_{n,j}{c}_j(x_n)={\mu }_n\to 0$. This proves the claimed KKT conclusion. The same proof permits approximate barrier stationary points whose stationarity residual tends to zero. Bounded iterates and the constraint qualification are substantive hypotheses, not automatic consequences of writing a barrier functional.

For a global-minimum conclusion one needs global minimization of the barrier subproblems and appropriate access to every feasible point. Specifically, suppose $x_n$ globally minimizes ${\Phi }_{{\mu }_n}$, has a cluster point $x_{\ast }$, and $\overline{{\Omega }_c^{\circ }}=\Omega$. For any fixed strictly feasible $z$,Along the convergent subsequence the $c_j(x_n)$ are bounded above, so taking the upper limit gives $f(x_{\ast })\le f(z)$. Density of strictly feasible points and continuity of $f$ extend the inequality to every feasible $z$. Hence $x_{\ast }$ is a global minimizer under these stronger hypotheses.

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.