A feasible strictly convex linear-constrained QP has a unique KKT minimizer with independent active support

If $n=0$, then ${\mathbb{R}}^0=\{0\}$ and nonempty $C$ is this singleton, so $p_{\ast }=0$ is its unique minimizer. Every equality row is the zero vector, so independence forces $A$ to have zero rows; every inequality row is also zero. Taking the empty equality multiplier and ${\lambda }_i=0$ for all $i$ gives stationarity and complementarity, with empty positive support, so the required row-independence and KKT certificate hold. Now assume $n\ge 1$. The unit sphere is compact by [F4]. The continuous function $z\mapsto z^{T}Bz$ is positive on it, so it attains a positive minimum $\alpha$. Thus $q(p)\ge \frac{\alpha }{2}\mathrm{\parallel }p{\mathrm{\parallel }}_2^2-\mathrm{\parallel }c{\mathrm{\parallel }}_2\mathrm{\parallel }p{\mathrm{\parallel }}_2$ by [F6], which tends to $+\mathrm{\infty }$ as $\mathrm{\parallel }p{\mathrm{\parallel }}_2\to \mathrm{\infty }$. By [F1], the constraint rows are affine, so their closed half-spaces and affine equality sets intersect in a closed set; choosing one feasible $p_0$, the nonempty sublevel $\{p\in C:q(p)\le q(p_0)\}$ is bounded and closed, hence compact by [F4]. The minimum is attained there and therefore on $C$.

If two distinct feasible points $p_1,p_2$ minimized $q$, their midpoint is feasible and $q((p_1+p_2)\mathrm{/}2)=\frac{1}{2}q(p_1)+\frac{1}{2}q(p_2)-\frac{1}{8}(p_1-p_2{)}^{T}B(p_1-p_2)<q(p_1),$ contradicting minimality. Hence the minimizer $p_{\ast }$ is unique.

Let $I=\{i:G_i{p}_{\ast }+d_i=0\}$. For any $z$ with $Az=0$ and $G_{i}z\le 0$ for $i\in I$, inactive constraints have positive slack, so for all sufficiently small $t>0$, $p_{\ast }+tz\in C$. Minimality and the one-sided derivative at $t=0$ give $(B{p}_{\ast }+c{)}^{T}z\ge 0$. Therefore $v:=-(B{p}_{\ast }+c)$ satisfies $v^{T}z\le 0$ for every such $z$.

If $v$ were not representable as $A^T\eta +{\sum }_{i\in I}{\mu }_i{G}_i^T$ with ${\mu }_i\ge 0$, apply [F2] to the matrix whose columns are $G_i^T$ for $i\in I$, the columns of $A^T$, and their negatives, with right-hand side $v$. The alternative gives $y$ with $G_{i}y\ge 0$, $Ay=0$, and $v^{T}y<0$. Then $z=-y$ satisfies the direction conditions of step 2.2 but $v^{T}z>0$, a contradiction. Thus $v=A^T\eta +{\sum }_{i\in I}{\mu }_i{G}_i^T$ for some ${\mu }_i\ge 0$. Since $v=-(B{p}_{\ast }+c)$, this is KKT stationarity with equality multiplier $\eta$; setting inactive multipliers to zero gives complementarity.

If $A$ has positive row count, full row rank makes $A{A}^T$ positive definite, so set $P=I-A^T(A{A}^T{)}^{-1}A$; if it has zero rows, set $P=I$. Direct multiplication gives $AP=0$, $P^T=P$, and $P^2=P$, and $Pz=z$ for every $z\in \ker A$. Thus $P$ is the orthogonal projection onto $\ker A$ and $\ker P=\mathrm{range}(A^T)$. Project the multiplier representation from step 3.1. If the projected active normals with positive coefficients are dependent, choose a nonzero dependence, change its sign to give a positive coefficient, and subtract $t$ times that relation, where $t$ is the minimum ratio of a current multiplier to a positive dependence coefficient. All coefficients remain nonnegative and at least one becomes zero. Repeating finitely many times leaves independent projected normals. Their unprojected remainder lies in $\ker P=\mathrm{range}(A^T)$, so it is absorbed into the equality multiplier. If a linear combination of the equality rows and supported active rows is zero, projecting it by $P$ first forces all active coefficients to be zero by projected independence; full row rank then forces the equality coefficients to be zero.

The feasible tuple from steps 3.1–3.1 satisfies the hypotheses of [F3], because $q$ is differentiable and convex and every constraint is affine. Thus it certifies a global minimizer; step 2.1 makes that global minimizer unique.