= Bounded barrier multipliers under Slater's condition
Let $c_i$ be <concave> and <continuously differentiable>, and let a strict <feasible point> $z$ satisfy $\delta=\min_i c_i(z)>0$. Suppose $x_k\to x_*$, $c_i(x_k)>0$, $\mu_k\to0$, and the <logarithmic barrier> residuals $r_k=\nabla f(x_k)-\sum_i\lambda_{k,i}\nabla c_i(x_k)$ tend to zero, where $\lambda_{k,i}=\mu_k/c_i(x_k)$. By <concavity>,
$$
\delta\|\lambda_k\|_1\leq\sum_i\lambda_{k,i}c_i(z)\leq p\mu_k+(\nabla f(x_k)-r_k)^T(z-x_k).
$$
The right-hand side is bounded when $\nabla f$ is <continuous> near $x_*$. Thus the nonnegative <Lagrange multiplier> estimates are bounded. A convergent <subsequence> gives $\lambda_*\geq0$; passing to the <limit> in stationarity and $\lambda_{k,i}c_i(x_k)=\mu_k$ proves the <Karush-Kuhn-Tucker conditions> at $x_*$. An inactive <optimization constraint> has $c_i(x_*)>0$ and hence $\lambda_{k,i}\to0$ even without selecting a <subsequence>. With a <convex> <objective function>, these conditions certify a <global minimum>. This result is conditional on the stated residual and primal convergence assumptions; it does not assert that every barrier subproblem has a <minimizer>.
Back to article page
