Strong duality without dual attainment (${\sup }_{\lambda \ge 0}9\lambda /(1+\lambda )=9$)

Consider an objective function that is a quadratic polynomial, with a singular optimization constraint:Its only feasible point is $x_{\ast }=0$, giving $p_{\ast }=9$. For $\lambda \ge 0$, completing the square yieldsEvery finite Lagrange multiplier gives $q(\lambda )<9$, but $q(\lambda )\to 9$ as $\lambda \to \infty$. Thus $d_{\ast }=p_{\ast }=9$: strong duality holds, and the primal optimum is attained, yet dual attainment fails. The inner minimizers $3/(1+\lambda )$ approach $0$ without ever becoming feasible at finite $\lambda$.

Both the objective function and the constraint function are convex functions. However, Slater's condition fails because $x^2<0$ is impossible. At the optimum, stationarity would require $-6+\lambda \cdot 0=0$, which no finite Lagrange multiplier satisfies. Compare KKT regularity at a local minimum: a zero gradient of an active constraint can prevent a multiplier certificate. This example has zero duality gap without the Strong Lagrangian property.