Lagrangian duality for a quadratic threshold (${\min }_{x\ge 4}((x-1{)}^2+2)=11$)

Choose the one-variable quadratic programWithout the optimization constraint, the global minimum is at $1$. On the feasible set $[4,\infty )$, write $x=4+s$ with $s\ge 0$: then $f(x)=11+6s+s^2\ge 11$. Hence $x_{\ast }=4$ is the unique primal optimizer and $p_{\ast }=11$.

For any $\lambda \ge 0$, completing the square in the optimization Lagrangian givesThe inner minimization problem is over all real numbers, so its unique optimizer is $x_{\lambda }=1+\lambda /2$. ThereforeThe best lower bound is $d_{\ast }=11$, attained at ${\lambda }_{\ast }=6$. At $\lambda =2$, the inner optimizer is $2$, which is infeasible for the primal problem; nevertheless $q(2)=7$ is a valid lower bound. At $\lambda =9$, the inner optimizer is $5.5$, which is feasible, but $q(9)=8.75$ is weaker than $11$. Feasibility of an inner optimizer alone does not make the bound exact.

At $(x_{\ast },{\lambda }_{\ast })=(4,6)$, the Karush-Kuhn-Tucker conditions reduce toThe last equality is complementary slackness. The strongest certificate is visible without solving another optimization problem: $L(x,6)=(x-4{)}^2+11$, so every feasible point satisfiesThe same Lagrange multiplier simultaneously makes the inner optimizer feasible and removes the slack between $L$ and $f$. This is a convex KKT optimality certificate; $x=5$ also verifies Slater's condition.

The domain in the inner infimum is essential. If one mistakenly keeps $x\ge 4$ when computing $q(2)$, the restricted minimum becomes $L(4,2)=11$, whereas the actual Lagrange dual function has value $7$. Constraints deliberately retained in $X$ are allowed in a Lagrangian relaxation, but that is a different choice of which restrictions to dualize.