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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.

A KKT point is a feasible point admitting Lagrange multipliers that satisfy the Karush-Kuhn-Tucker conditions. Some authors use the name for the full primal and multiplier pair; specify the multipliers when this distinction matters. For a convex optimization problem with a differentiable convex function as its objective function, convex inequality functions and affine equality constraints, the KKT conditions certify a global minimizer. Their necessity requires an appropriate constraint qualification.

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.

With inequalities written as $c_i\ge 0$, use the optimization Lagrangian $L(x,\lambda )=f(x)-{\sum }_i{\lambda }_i{c}_i(x)$. The Karush-Kuhn-Tucker conditions at $(x_{\ast },{\lambda }_{\ast })$ areFor any feasible point $z$, the first-order inequalities for a convex function and a concave function giveThe last equality uses complementary slackness. Thus $x_{\ast }$ is a global minimizer. Slater's condition is useful for necessity and existence of Lagrange multipliers, but is not required for this sufficiency argument once a KKT point is supplied.

A precise conditional convergence statement is: with the smoothness and convexity assumptions of the question, a strict feasible point, ${\mu }_k\downarrow 0$, and barrier stationarity residualsevery convergent sequence of strictly feasible points $x^k\to x_{\ast }$ has a global minimizer as its limit. The Lagrange multiplier estimates are bounded and have a subsequence converging to multipliers satisfying the KKT conditions. The theorem does not assert convergence of the primal sequence without further assumptions.

First, continuity gives $c_i(x_{\ast })\ge 0$. For an inactive optimization constraint, $c_i(x_{\ast })>0$, soFor the active constraints, the hint permits assuming ${\lambda }_i^k\to {\lambda }_i^{\ast }$. Combining this with the inactive limits gives ${\lambda }_{\ast }\ge 0$, and passage to the limit in the residual equation proves stationarity. Since ${\lambda }_i^k{c}_i(x^k)={\mu }_k\to 0$, complementary slackness also holds. All KKT conditions follow, and part (b) proves global optimality.

The hint's multiplier assumption can in fact be weakened to selecting a subsequence. To see why the estimates are bounded, choose a strict feasible point $z$ and put $\delta ={\min }_i{c}_i(z)>0$. Concavity givesThe right side is bounded because $x^k\to x_{\ast }$ and $r_k\to 0$. This is the bounded barrier multipliers under Slater's condition argument. A convergent multiplier subsequence then gives the same KKT point, which suffices to prove that the assumed primal limit is globally optimal.

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.