Active-set quadratic with an exact KKT certificate

The constraint is feasible exactly when $x_1\ge 0$. At $x_{\ast }=(0,2)$ it is active, and $\mathrm{\nabla }f(x_{\ast })=(1,0)$ while $\mathrm{\nabla }g(x_{\ast })=(-1,0)$. Therefore $\mathrm{\nabla }f(x_{\ast })+{\lambda }_{\ast }\mathrm{\nabla }g(x_{\ast })=0$ for ${\lambda }_{\ast }=1\ge 0$, and ${\lambda }_{\ast }g(x_{\ast })=0$. The active gradient is nonzero, so the one-row LICQ holds.

The quadratic is convex and differentiable. Primal feasibility and the multiplier checks in step 1.1 satisfy the hypotheses of [F2], so $x_{\ast }$ is a global minimizer. Directly, for every feasible $x$, $f(x)-f(x_{\ast })=x_1+\frac{1}{2}{x}_1^2+\frac{1}{2}(x_2-2{)}^2\ge 0,$ with equality only when $x_1=0$ and $x_2=2$. Thus the minimizer is unique, and [F1] makes ${\lambda }_{\ast }=1$ the unique multiplier.

The active-set KKT certificate is exact: it proves the unique global minimizer $x_{\ast }=(0,2)$ and its unique nonnegative multiplier ${\lambda }_{\ast }=1$.