A constraint-reduced MPC algorithm for convex quadratic programming, with a modified active set identification scheme

Abstract

A constraint-reduced Mehrotra-predictor-corrector algorithm for convex quadratic programming is proposed. (At each iteration, such algorithms use only a subset of the inequality constraints in constructing the search direction, resulting in CPU savings.) The proposed algorithm makes use of a regularization scheme to cater to cases where the reduced constraint matrix is rank deficient. Global and local convergence properties are established under arbitrary working-set selection rules subject to satisfaction of a general condition. A modified active-set identification scheme that fulfills this condition is introduced. Numerical tests show great promise for the proposed algorithm, in particular for its active-set identification scheme. While the focus of the present paper is on dense systems, application of the main ideas to large sparse systems is briefly discussed.

Appendices

Appendix

The following results are used in the proofs in Appendices A and B. Here we assume that $Q\subseteq {\mathbf {m}}$ and W is symmetric, with $W\succeq H\succ {\mathbf {0}}$. First, from (10) and (13), the approximate MPC search direction $(\varDelta {\mathbf {x}}, \varDelta \varvec{\lambda }_Q, \varDelta {\mathbf {s}}_Q)$ defined in (16) solves

$$\begin{aligned} J(W,A_Q,{\mathbf {s}}_Q,\varvec{\lambda }_Q) \left[ \begin{array}{c} \varDelta {\mathbf {x}}\\ \varDelta \varvec{\lambda }_Q\\ \varDelta {\mathbf {s}}_Q \end{array}\right] = \left[ \begin{array}{c} -\nabla f({\mathbf {x}})+(A_Q)^T\varvec{\lambda }_Q \\ {\mathbf {0}}\\ -S_Q\varvec{\lambda }_Q + \gamma \sigma \mu _{(Q)}{\mathbf {1}} - \gamma \varDelta S_Q^{\text {a}}\varDelta \varvec{\lambda }_Q^{\text {a}} \end{array}\right] , \end{aligned}$$

(32)

and equivalently, when ${\mathbf {s}}_Q>{\mathbf {0}}$, from (20), (21) and (15),

$$\begin{aligned} \begin{aligned} M_{(Q)}\varDelta {\mathbf {x}}&= -\nabla f({\mathbf {x}}) + (A_Q)^TS_Q^{-1} \left( \gamma \sigma \mu _{(Q)}{\mathbf {1}}-\gamma \varDelta S_Q^{\text {a}}\varDelta \varvec{\lambda }_Q^{\text {a}}\right) ,\\ \varDelta {\mathbf {s}}_Q&=A_Q \varDelta {\mathbf {x}},\\ \varDelta \varvec{\lambda }_Q&=-\varvec{\lambda }_Q + S_Q^{-1}\left( -\varLambda _Q\varDelta {\mathbf {s}}_Q+\gamma \sigma \mu _{(Q)}{\mathbf {1}}-\gamma \varDelta S_Q^{\text {a}}\varDelta \varvec{\lambda }_Q^{\text {a}}\right) . \end{aligned} \end{aligned}$$

(33)

Next, with ${\tilde{\varvec{\lambda }}}^+$ and ${{\tilde{\varvec{\lambda }}}}^{\text {a},+}$ given by

$$\begin{aligned} {\tilde{\lambda }}^+_i := {\left\{ \begin{array}{ll} \lambda _i + \varDelta \lambda _i &{} \quad i\in Q,\\ 0 &{} \quad i\in Q^{\text {c}}, \end{array}\right. } \quad \text { and }\quad {\tilde{\lambda }}_i^{\text {a},+} := {\left\{ \begin{array}{ll} \lambda _i + \varDelta \lambda _i^\text {a}&{} \quad i\in Q,\\ 0 &{} \quad i\in Q^{\text {c}}, \end{array}\right. } \end{aligned}$$

(34)

from the last equation of (33) and from (21), we have

$$\begin{aligned}&{\tilde{\varvec{\lambda }}}_Q^+ = S_Q^{-1} \left( -\varLambda _Q \varDelta {\mathbf {s}}_Q + \gamma \sigma \mu _{(Q)}{\mathbf {1}} - \gamma \varDelta S^{\text {a}}_Q \varDelta \varvec{\lambda }_Q^{\text {a}}\right) , \end{aligned}$$

(35)

$$\begin{aligned}&{{\tilde{\varvec{\lambda }}}}^{\text {a},+}_Q = - S_Q^{-1} \varLambda _Q \varDelta {\mathbf {s}}^{\text {a}}_Q = - S_Q^{-1} \varLambda _Q A_Q \varDelta {\mathbf {x}}^{\text {a}}_Q , \end{aligned}$$

(36)

and hence

$$\begin{aligned} (\varDelta {\mathbf {x}}^\text {a})^T(A_Q)^T{{\tilde{\varvec{\lambda }}}}_Q^{\text {a},+} = -(\varDelta {\mathbf {x}}^\text {a})^T(A_Q)^T S_Q^{-1}\varLambda _Q A_Q \varDelta {\mathbf {x}}^\text {a}\le 0, \end{aligned}$$

(37)

so that, when in addition $\varvec{\lambda }>{\mathbf {0}}$, $(\varDelta {\mathbf {x}}^\text {a})^T(A_Q)^T{{\tilde{\varvec{\lambda }}}}^{\text {a},+} =0$ if and only if $A_Q\varDelta {\mathbf {x}}^\text {a}={\mathbf {0}}$. Also, (20) yields

$$\begin{aligned} \nabla f({\mathbf {x}})^T\varDelta {\mathbf {x}}^\text {a}{\,=\,} -{(\varDelta {\mathbf {x}}^\text {a})}^T M_{(Q)}\varDelta {\mathbf {x}}^\text {a}{\,=\,} - {(\varDelta {\mathbf {x}}^\text {a})}^T W\varDelta {\mathbf {x}}^\text {a}-{(\varDelta {\mathbf {x}}^\text {a})}^T (A_Q)^T S_Q^{-1}\varLambda _Q A_Q\varDelta {\mathbf {x}}^\text {a}. \end{aligned}$$

(38)

Since $W\succeq H$, it follows from (37) that,

$$\begin{aligned} \nabla f({\mathbf {x}})^T\varDelta {\mathbf {x}}^\text {a}+ {(\varDelta {\mathbf {x}}^\text {a})}^T H \varDelta {\mathbf {x}}^\text {a}\le 0. \end{aligned}$$

(39)

In addition, when $S_Q\succ {\mathbf {0}}$, $\varLambda _Q\succ {\mathbf {0}}$ and since $W\succeq {\mathbf {0}}$, the right-hand side of (38) is strictly negative as long as $W\varDelta {\mathbf {x}}^\text {a}$ and $A_Q\varDelta {\mathbf {x}}^\text {a}$ are not both zero. In particular, when $[W~(A_Q)^T]$ has full row rank,

$$\begin{aligned} \nabla f({\mathbf {x}})^T\varDelta {\mathbf {x}}^\text {a}<0\quad \mathrm{if}~\varDelta {\mathbf {x}}^\text {a}\not ={\mathbf {0}}. \end{aligned}$$

(40)

Finally, we state and prove two technical lemmas.

Lemma 3

Given an infinite index set K, $\{\varDelta {\mathbf {x}}^k\}\rightarrow {\mathbf {0}}$ as $k\rightarrow \infty$, $k\in K$ if and only if $\{\varDelta {\mathbf {x}}^{\text {a},k}\}\rightarrow {\mathbf {0}}$ as $k\rightarrow \infty$, $k\in K$.

Proof

We show that $\Vert \varDelta {\mathbf {x}}^k\Vert$ is sandwiched between constant multiples of $\Vert \varDelta {\mathbf {x}}^{\text {a},k}\Vert$. We have from the search direction given in (16) that, for all k, $\Vert \varDelta {\mathbf {x}}^k-\varDelta {\mathbf {x}}^{\text {a},k}\Vert = \Vert \gamma \varDelta {\mathbf {x}}^{\text {c},k}\Vert \le \tau \Vert \varDelta {\mathbf {x}}^{\text {a},k}\Vert$, where $\tau \in (0,1)$ and the inequality follows from (7). Apply triangle inequality leads to $(1-\tau )\Vert \varDelta {\mathbf {x}}^{\text {a},k}\Vert \le \Vert \varDelta {\mathbf {x}}^k\Vert \le (1+\tau )\Vert \varDelta {\mathbf {x}}^{\text {a},k}\Vert$ for all k, proving the claim. $\square$

Lemma 4

Suppose Assumption 1 holds. Let $Q\subset {\mathbf {m}}$, ${\mathcal {A}}\subseteq Q$, ${\mathbf {x}}\in {\mathcal {F}}_P^o$, ${\mathbf {s}}:=A{\mathbf {x}}-{\mathbf {b}}~(>{\mathbf {0}})$, and $\varvec{\lambda }>{\mathbf {0}}$ enjoy the following property: With $\varDelta \varvec{\lambda }_Q$, $\varDelta \varvec{\lambda }^{\text {a}}_Q$, $\varDelta {\mathbf {s}}$, and $\varDelta {\mathbf {s}}^{\text {a}}$ produced by Iteration CR-MPC, $\lambda _i+\varDelta \lambda _i>0$ for all $i\in {\mathcal {A}}$ and $s_i+\varDelta s_i>0$ for all $i\in Q\setminus {\mathcal {A}}$. Then

$$\begin{aligned} {\bar{\alpha }}_\text {d}\ge \min \left\{ 1 , \, \min _{i\in Q\setminus {\mathcal {A}}} \left\{ \frac{s_i}{|s_i+\varDelta s_i^{\text {a}}|}\right\} , \, \min _{i\in Q\setminus {\mathcal {A}}} \left\{ \frac{s_i-|\varDelta s_i^\text {a}|}{|s_i+\varDelta s_i|} \right\} \right\} \end{aligned}$$

(41)

and

$$\begin{aligned} {\bar{\alpha }}_\text {p}\ge \min \left\{ 1 , \, \min _{i\in {\mathcal {A}}} \left\{ \frac{\lambda _i}{|\lambda _i+\varDelta \lambda _i^\text {a}|}\right\} , \, \min _{i\in {\mathcal {A}}} \left\{ \frac{\lambda _i-|\varDelta \lambda _i^\text {a}|}{|\lambda _i+\varDelta \lambda _i|}\right\} \right\} . \end{aligned}$$

(42)

Proof

If ${\bar{\alpha }}_\text {d}\ge 1$, (41) holds trivially, hence suppose ${\bar{\alpha }}_\text {d}< 1$. Then, from the definition of ${\bar{\alpha }}_\text {d}$ in (24), we know that there exists some index $i_0\in Q$ such that

$$\begin{aligned} \varDelta \lambda _{i_0}<-\lambda _{i_0}<0 \quad \text { and }\quad {\bar{\alpha }}_\text {d}= \frac{\lambda _{i_0}}{|\varDelta \lambda _{i_0}|}. \end{aligned}$$

(43)

Since $\lambda _i+\varDelta \lambda _i>0$ for all $i\in {\mathcal {A}}$, we have $i_0\in Q\setminus {\mathcal {A}}$. Now we consider two cases: $|\varDelta \lambda _{i_0}^\text {a}|\ge |\varDelta \lambda _{i_0}|$ and $|\varDelta \lambda _{i_0}^\text {a}|<|\varDelta \lambda _{i_0}|$. If $|\varDelta \lambda _{i_0}^\text {a}|\ge |\varDelta \lambda _{i_0}|$, then, since the second equation in (21) is equivalently written as $\lambda _i s_i+s_i\varDelta \lambda _i^\text {a}+\lambda _i\varDelta s_i^\text {a}=0$ for all $i\in Q$ and since $\lambda _s s_i>0$ for all $i\in {\mathbf {m}}$, it follows from (43) that

$$\begin{aligned} {\bar{\alpha }}_\text {d}= \frac{\lambda _{i_0}}{|\varDelta \lambda _{i_0}|} \ge \frac{\lambda _{i_0}}{|\varDelta \lambda _{i_0}^\text {a}|} = \frac{s_{i_0}}{|s_{i_0}+\varDelta s_{i_0}^{\text {a}}|}, \end{aligned}$$

proving (41). To conclude, suppose now that $|\varDelta \lambda _{i_0}^\text {a}|<|\varDelta \lambda _{i_0}|$. Since (i) $s_i+\varDelta s_i>0$ for $i\in Q\setminus {\mathcal {A}}$; (ii) $\gamma$, $\sigma$, and $\mu _{(Q)}$ in (35) are non-negative; and (iii) $\varDelta \lambda _{i_0}<0$ (from (43)), (34), (35) yield

$$\begin{aligned} {\lambda }_{i_0} (s_{i_0}+\varDelta s_{i_0}) \ge s_{i_0}|\varDelta \lambda _{i_0}| - \gamma |\varDelta s_{i_0}^{\text {a}}| |\varDelta \lambda _{i_0}^{\text {a}}|. \end{aligned}$$

Applying this inequality to (43) leads to

$$\begin{aligned} {\bar{\alpha }}_\text {d}= \frac{\lambda _{i_0}}{|\varDelta \lambda _{i_0}|} \ge \frac{s_{i_0}}{|s_{i_0}+\varDelta s_{i_0}|} - \frac{\gamma |\varDelta \lambda _{i_0}^\text {a}| |\varDelta s_{i_0}^\text {a}|}{|s_{i_0}+\varDelta s_{i_0}||\varDelta \lambda _{i_0}|} \ge \frac{s_{i_0}-|\varDelta s_{i_0}^\text {a}|}{|s_{i_0}+\varDelta s_{i_0}|}, \end{aligned}$$

where the last inequality holds since $\gamma \le 1$ and $|\varDelta \lambda _{i_0}^\text {a}|<|\varDelta \lambda _{i_0}|$. Following a very similar argument that flips the roles of ${\mathbf {s}}$ and $\varvec{\lambda }$, one can prove that (42) also holds. $\square$

A Proof of Theorem 1 and Corollary 2

Parts of this proof are inspired from [16, 18, 29, 34, 35]. Throughout, we assume that the constraint-selection rule used by the algorithm is such that Condition CSR is satisfied and (except in the proof of Lemma 13) we let $\varepsilon =0$ and assume that the iteration never stops.

A central feature of Algorithm CR-MPC, which plays a key role in the convergence proofs, is that it forces descent with respect of the primal objective function. The next proposition establishes some related facts.

Proposition 2

Suppose $\varvec{\lambda }>{\mathbf {0}}$ and ${\mathbf {s}}>{\mathbf {0}}$, and W satisfies $W\succ {\mathbf {0}}$ and $W\succeq H$. If $\varDelta {\mathbf {x}}^\text {a}\ne {\mathbf {0}}$, then the following inequalities hold:

$$\begin{aligned} f({\mathbf {x}}+\alpha \varDelta {\mathbf {x}}^\text {a})<f({\mathbf {x}}),\quad \forall \alpha \in (0,2), \end{aligned}$$

(44)

$$\begin{aligned} \frac{\partial }{\partial \alpha } f({\mathbf {x}}+\alpha \varDelta {\mathbf {x}}^\text {a})< 0,\quad \forall \alpha \in [0,1], \end{aligned}$$

(45)

$$\begin{aligned} f({\mathbf {x}})-f({\mathbf {x}}+\alpha \varDelta {\mathbf {x}})\ge \frac{\omega }{2}(f({\mathbf {x}})-f({\mathbf {x}}+\alpha \varDelta {\mathbf {x}}^\text {a})),\quad \forall \alpha \in [0,1], \end{aligned}$$

(46)

$$\begin{aligned} f({\mathbf {x}}+\alpha \varDelta {\mathbf {x}})<f({\mathbf {x}}),\quad \forall \alpha \in (0,1]. \end{aligned}$$

(47)

Proof

When $f({\mathbf {x}}+\alpha \varDelta {\mathbf {x}}^\text {a})$ is linear in $\alpha$, i.e., when $(\varDelta {\mathbf {x}}^\text {a})^T H\varDelta {\mathbf {x}}^\text {a}=0$, then in view of (40), (44), (45) hold trivially. When, on the other hand, $(\varDelta {\mathbf {x}}^\text {a})^T H\varDelta {\mathbf {x}}^\text {a}>0$, $f({\mathbf {x}}+\alpha \varDelta {\mathbf {x}}^\text {a})$ is quadratic and strictly convex in $\alpha$ and is minimized at

$$\begin{aligned} {\hat{\alpha }} = -\frac{\nabla f({\mathbf {x}})^T \varDelta {\mathbf {x}}^\text {a}}{{(\varDelta {\mathbf {x}}^\text {a})}^T H\varDelta {\mathbf {x}}^\text {a}} = 1+\frac{{(\varDelta {\mathbf {x}}^\text {a})}^T\left( W-H + (A_Q)^T S_Q^{-1}\varLambda _Q A_Q\right) \varDelta {\mathbf {x}}^\text {a}}{(\varDelta {\mathbf {x}}^\text {a})^T H\varDelta {\mathbf {x}}^\text {a}} \ge 1, \end{aligned}$$

where we have used (38), (37), and the fact that $W\succeq H$, and (44), (45) again follow. Next, note that, since $\omega >0$,

$$\begin{aligned} \psi (\theta ):= \omega (f({\mathbf {x}})-f({\mathbf {x}}+\varDelta {\mathbf {x}}^\text {a})) - (f({\mathbf {x}})-f({\mathbf {x}}+\varDelta {\mathbf {x}}^\text {a}+\theta \varDelta {\mathbf {x}}^\text {c}))\, \end{aligned}$$

is quadratic and convex. Now, since $\gamma _1$ satisfies the constraints in its definition (8), we see that $\psi (\gamma _1)\le 0$, and since $\omega \le 1$, it follows from (44) that $\psi (0)= (\omega -1)(f({\mathbf {x}})-f({\mathbf {x}}+\varDelta {\mathbf {x}}^\text {a}))\le 0$. Since $\gamma \in [0,\gamma _1]$ (see (7)), it follows that $\psi (\gamma )\le 0$, i.e., since from (16) $\varDelta {\mathbf {x}}=\varDelta {\mathbf {x}}^\text {a}+\gamma \varDelta {\mathbf {x}}^\text {c}$,

$$\begin{aligned} f({\mathbf {x}})-f({\mathbf {x}}+\varDelta {\mathbf {x}})\ge \omega (f({\mathbf {x}})-f({\mathbf {x}}+\varDelta {\mathbf {x}}^{\text {a}})), \end{aligned}$$

i.e.,

$$\begin{aligned} -\nabla f({\mathbf {x}})^T \varDelta {\mathbf {x}}-\frac{1}{2}{\varDelta {\mathbf {x}}}^T H\varDelta {\mathbf {x}}\ge \omega \left( -\nabla f({\mathbf {x}})^T \varDelta {\mathbf {x}}^\text {a}-\frac{1}{2}{(\varDelta {\mathbf {x}}^\text {a})}^T H\varDelta {\mathbf {x}}^\text {a}\right) . \end{aligned}$$

(48)

Now, for all $\alpha \in [0,1]$, invoking (48), (39), and the fact that $H\succeq {\mathbf {0}}$, we can write

$$\begin{aligned} f({\mathbf {x}})-f({\mathbf {x}}+\alpha \varDelta {\mathbf {x}})= & {} -\alpha \nabla f({\mathbf {x}})^T\varDelta {\mathbf {x}}- \frac{\alpha ^2}{2}{\varDelta {\mathbf {x}}}^T H \varDelta {\mathbf {x}}\ge \alpha \left( -\nabla f({\mathbf {x}})^T\varDelta {\mathbf {x}}- \frac{1}{2}{\varDelta {\mathbf {x}}}^T H \varDelta {\mathbf {x}}\right) \\\ge & {} \omega \alpha \left( -\nabla f({\mathbf {x}})^T\varDelta {\mathbf {x}}^\text {a}- \frac{1}{2}{(\varDelta {\mathbf {x}}^\text {a})}^T H \varDelta {\mathbf {x}}^\text {a}\right) \\= & {} \frac{\omega \alpha }{2}\left( -\nabla f({\mathbf {x}})^T\varDelta {\mathbf {x}}^\text {a}- \left( \nabla f({\mathbf {x}})^T\varDelta {\mathbf {x}}^\text {a}+ {(\varDelta {\mathbf {x}}^\text {a})}^T H \varDelta {\mathbf {x}}^\text {a}\right) \right) \\\ge & {} \frac{\alpha \omega }{2} \left( -\nabla f({\mathbf {x}})^T\varDelta {\mathbf {x}}^\text {a}\right) \\\ge & {} \frac{\alpha \omega }{2} \left( -\nabla f({\mathbf {x}})^T\varDelta {\mathbf {x}}^\text {a}-\frac{\alpha }{2}{(\varDelta {\mathbf {x}}^\text {a})}^T H\varDelta {\mathbf {x}}^\text {a}\right) = \frac{\omega }{2}(f({\mathbf {x}})-f({\mathbf {x}}+\alpha \varDelta {\mathbf {x}}^\text {a})), \end{aligned}$$

proving (46). Finally, since $\omega >0$, (47) is a direct consequence of (46) and (44). $\square$

In particular, in view of (23), (24), $\{f({\mathbf {x}}^{k})\}$ is monotonic decreasing. Since the iterates are primal–feasible, an immediate consequence of Proposition 2, stated next, is that, under Assumption 1, the primal sequence is bounded.

Lemma 5

Suppose Assumption 1 holds. Then $\{{\mathbf {x}}^k\}$ is bounded.

We are now ready to prove a key result, relating two successive iterates, that plays a central role in the remainder of the proof of Theorem 1.

Proposition 3

Suppose Assumptions 1 and 2 hold, and either $\{({\mathbf {x}}^k,\varvec{\lambda }^k)\}$ is bounded away from ${\mathcal {F}}^*$, or Assumption 3 also holds and $\{{\mathbf {x}}^k\}$ converges to the unique primal solution ${\mathbf {x}}^*$. Let K be an infinite index set such that

$$\begin{aligned} \left( \inf _{k\in K} \{\chi _{k-1}\}=\right) ~ \inf \left\{ \Vert \varDelta {\mathbf {x}}^{\text {a},k-1}\Vert ^\nu + \left\| \left[ {\tilde{\varvec{\lambda }}}^{\text {a},k}_{Q_{k-1}}\right] _{-} \right\| ^\nu :k\in K\right\} >0. \end{aligned}$$

(49)

Then $\{\varDelta {\mathbf {x}}^k\}\rightarrow {\mathbf {0}}$ as $k\rightarrow \infty$, $k\in K$.

Proof

From Lemma 5, $\{{\mathbf {x}}^k\}$ is bounded, and hence so is $\{{\mathbf {s}}^k\}$; by construction, ${\mathbf {s}}^k$ and $\varvec{\lambda }^k$ have positive components for all k, and $\{\varvec{\lambda }^k\}$ ((26), (27)) and $\{W_k\}$ are bounded. Further, for any infinite index set $K^\prime$ such that (49) holds, (26) and (27) imply that all components of $\{\varvec{\lambda }^k\}$ are bounded away from zero on $K^\prime$. Since, in addition, $Q_k$ can take no more than finitely many different (set) values, it follows that there exist ${\hat{{\mathbf {x}}}}$, ${\hat{\varvec{\lambda }}}>{\mathbf {0}}$, ${\hat{W}}\succeq {\mathbf {0}}$, an index set ${\hat{Q}}\subseteq {\mathbf {m}}$, and some infinite index set ${\hat{K}}\subseteq K^\prime$ such that

$$\begin{aligned} \{{\mathbf {x}}^k\}&\rightarrow {\hat{{\mathbf {x}}}} \text { as } k\rightarrow \infty ,\, k\in {\hat{K}}, \nonumber \\ \{{\mathbf {s}}^k\}&\rightarrow {\hat{{\mathbf {s}}}}:= \{A{\hat{{\mathbf {x}}}}-{\mathbf {b}}\}\ge {\mathbf {0}}\text { as } k\rightarrow \infty ,\, k\in {\hat{K}}. \end{aligned}$$

(50)

$$\begin{aligned} \{\varvec{\lambda }^k\}&\rightarrow {\hat{\varvec{\lambda }}}>{\mathbf {0}}\text { as } k\rightarrow \infty \,,\, k\in {\hat{K}}, \end{aligned}$$

(51)

$$\begin{aligned} \{W_k\}&\rightarrow {\hat{W}} \text { as } k\rightarrow \infty \,,\, k\in {\hat{K}},\nonumber \\ Q_k&= {\hat{Q}},\, \forall k\in {\hat{K}}. \end{aligned}$$

(52)

Next, under the stated assumptions, $J({\hat{W}},A_{{\hat{Q}}}, {\hat{{\mathbf {s}}}}_{{\hat{Q}}},{\hat{\varvec{\lambda }}}_{{\hat{Q}}})$ is non-singular. Indeed, if $\{({\mathbf {x}}^k,\varvec{\lambda }^k)\}$ is bounded away from ${\mathcal {F}}^*$, then $E({\mathbf {x}}^k,\varvec{\lambda }^k)$ is bounded away from zero and since $H+R\succ {\mathbf {0}}$, $W_k=H+\varrho _k R = H + \min \left\{ 1,\frac{E({\mathbf {x}}^k,\varvec{\lambda }^k)}{{\bar{E}}}\right\} R$ is bounded away from singularity and the claim follows from Assumption 2 and Lemma 1. On the other hand, if Assumption 3 also holds and $\{{\mathbf {x}}^k\}\rightarrow {\mathbf {x}}^*$, then the claim follows from Condition CSR(ii) and Lemma 1. As a consequence of this claim, and by continuity of J, it follows from Newton-KKT systems (10) and (32) that there exist $\varDelta {\hat{{\mathbf {x}}}}^\text {a}$, $\varDelta {\hat{{\mathbf {x}}}}$, ${\bar{\varvec{\lambda }}}^{\text {a}}_{{\hat{Q}}}$, ${{\bar{\varvec{\lambda }}}}_{{\hat{Q}}}$ such that

$$\begin{aligned} \{\varDelta {\mathbf {x}}^{\text {a},k}\}&\rightarrow \varDelta {\hat{{\mathbf {x}}}}^\text {a}\text { as } k\rightarrow \infty ,\, k\in {\hat{K}}, \end{aligned}$$

(53)

$$\begin{aligned} \{\varDelta {\mathbf {x}}^k\}&\rightarrow \varDelta {\hat{{\mathbf {x}}}} \text { as } k\rightarrow \infty \,,\, k\in {\hat{K}}\,, \nonumber \\ \{\varDelta {\mathbf {s}}^k\}&\rightarrow \varDelta {\hat{{\mathbf {s}}}}:=A\varDelta {\hat{{\mathbf {x}}}} \text { as } k\rightarrow \infty \,,\, k\in {\hat{K}}, \end{aligned}$$

(54)

$$\begin{aligned} \{{{\tilde{\varvec{\lambda }}}}^{\text {a},k+1}_{{\hat{Q}}}\}&\rightarrow {\bar{\varvec{\lambda }}}^{\text {a}}_{{\hat{Q}}}\text { as } k\rightarrow \infty \,,\, k\in {\hat{K}}, \end{aligned}$$

(55)

$$\begin{aligned} \{{{\tilde{\varvec{\lambda }}}}^{k+1}_{{\hat{Q}}}\}&\rightarrow {\bar{\varvec{\lambda }}}_{{\hat{Q}}} \text { as } k\rightarrow \infty \,,\, k\in {\hat{K}}, \end{aligned}$$

(56)

The remainder of the proof proceeds by contradiction. Thus suppose that, for the infinite index set K in the statement of this lemma, $\{\varDelta {\mathbf {x}}^k\}\not \rightarrow {\mathbf {0}}$ as $k\rightarrow \infty$, $k\in K$, i.e., for some $K^{\prime \prime }\subseteq K$, $\Vert \varDelta {\mathbf {x}}^k\Vert$ is bounded away from zero on $K^{\prime \prime }$. Use $K^{\prime \prime }$ as our $K^\prime$ above, so that (since ${\hat{K}}\subseteq K^\prime$), $\Vert \varDelta {\mathbf {x}}^k\Vert$ is bounded away from zero on ${\hat{K}}$. Then, in view of Lemma 3 (w.l.o.g.),

$$\begin{aligned} \inf _{k\in {\hat{K}}}\Vert \varDelta {\mathbf {x}}^{\text {a},k}\Vert >0. \end{aligned}$$

(57)

In addition, we have ${\mathcal {A}}({\hat{{\mathbf {x}}}})\subseteq {\hat{Q}}$, an implication of Condition CSR(i) when $\{({\mathbf {x}}^k,\varvec{\lambda }^k)\}$ is bounded away from ${\mathcal {F}}^*$ and of Condition CSR(ii) when Assumption 3 holds and $\{{\mathbf {x}}^k\}$ converges to ${\mathbf {x}}^*$. With these facts in hand, we next show that the sequence of primal step sizes $\{\alpha _\text {p}^k\}$ is bounded away from zero for $k\in {\hat{K}}$. To this end, let us define

$$\begin{aligned} {{\tilde{\varvec{\lambda }}}}^{\prime ,k+1} :=-(S^{k})^{-1}\varLambda ^k\varDelta {\mathbf {s}}^k,\quad \forall k, \end{aligned}$$

(58)

so that, for all $i\in {\mathbf {m}}$ and all k, ${{\tilde{\lambda }}}_i^{\prime ,k+1}>0$ if and only if $\varDelta s_i^k<0$, and the primal portion of (24) can be written as

$$\begin{aligned} \begin{aligned} {\bar{\alpha }}_\text {p}^k&:= {\left\{ \begin{array}{ll} \infty &{} \quad \text {if } {{\tilde{\varvec{\lambda }}}}^{\prime ,k+1}\le {\mathbf {0}},\\ \min _i\left\{ \frac{\lambda _i^k}{{{\tilde{\lambda }}}_i^{\prime ,k+1}} : {\tilde{\lambda }}_i^{\prime ,k+1}>0\right\} &{} \quad \text {otherwise.} \end{array}\right. }\\ \alpha _\text {p}^k&:= \min \left\{ 1,\,\max \{\varkappa {\bar{\alpha }}_\text {p},\,{\bar{\alpha }}_\text {p}-\Vert \varDelta {\mathbf {x}}^k\Vert \}\right\} . \end{aligned} \end{aligned}$$

Clearly, it is now sufficient to show that, for all i, $\{{\tilde{\lambda }}_i^{\prime ,k+1}\}$ is bounded above on ${\hat{K}}$. On the one hand, this is clearly so for $i\not \in {\hat{Q}}$ (whence $i\not \in {\mathcal {A}}({\hat{{\mathbf {x}}}})$), in view of (58) and (54), since $\{\varvec{\lambda }^k\}$ is bounded and $\{s_i^k\}$ is bounded away from zero on ${\hat{K}}$ for $i\not \in {\mathcal {A}}({\hat{{\mathbf {x}}}})$ (from (50)). On the other hand, in view of (52), subtracting (58) from (35) yields, for all $k\in {\hat{K}}$,

$$\begin{aligned} {{\tilde{\varvec{\lambda }}}}'^{k+1}_{{\hat{Q}}} = {{\tilde{\varvec{\lambda }}}}^{k+1}_{{\hat{Q}}} - \gamma _k\sigma _k\mu _{({{\hat{Q}}})}^k\left( S_{{\hat{Q}}}^{k}\right) ^{-1}{\mathbf {1}} + \gamma _k \left( S_{{\hat{Q}}}^{k}\right) ^{-1}\varDelta S_{{\hat{Q}}}^{\text {a},k}\varDelta \varvec{\lambda }_{{\hat{Q}}}^{\text {a},k}. \end{aligned}$$

From (56), $\{{{\tilde{\varvec{\lambda }}}}^{k+1}_{{\hat{Q}}}\}$ is bounded on ${\hat{K}}$, and clearly the second term in the right-hand side of the above equation is non-positive component-wise. As for the third term, the second equation in (21) gives $(S_{Q_k}^{k})^{-1}\varDelta S_{Q_k}^{\text {a},k} = (\varLambda _{Q_k}^{k})^{-1} {\tilde{\varLambda }}_{Q_k}^{\text {a},k+1}$, so that we have

$$\begin{aligned} \gamma _k \left( S_{{\hat{Q}}}^{k}\right) ^{-1}\varDelta S_{{\hat{Q}}}^{\text {a},k}\varDelta \varvec{\lambda }_{{\hat{Q}}}^{\text {a},k} = \gamma _k \left( \varLambda _{{\hat{Q}}}^{k}\right) ^{-1} {\tilde{\varLambda }}_{{\hat{Q}}}^{\text {a},k+1}\varDelta \varvec{\lambda }_{{\hat{Q}}}^{\text {a},k}, \quad \forall k \in {\hat{K}}, \end{aligned}$$

which is bounded on ${\hat{K}}$ since, from (51), (55), and the definition (34) of $\{{\tilde{\lambda }}^{\text {a},+}\}$, both $\{{\tilde{\varLambda }}_{{\hat{Q}}}^{\text {a},k+1}\}$ and $\{\varDelta \varvec{\lambda }_{{\hat{Q}}}^{\text {a},k}\}$ are bounded, and from (51), $\{\varvec{\lambda }^k_{{\hat{Q}}}\}$ is bounded away from zero on ${\hat{K}}$. Therefore, $\{{\tilde{\lambda }}_i^{\prime ,k+1}\}$ is bounded above on ${\hat{K}}$ for $i\in {\hat{Q}}$ as well, proving that $\{{\alpha }_\text {p}^k\}$ is bounded away from zero on ${\hat{K}}$, i.e., that there exists ${\underline{\alpha }}>0$ such that $\alpha _\text {p}^k>{\underline{\alpha }}$, for all $k\in {\hat{K}}$, as claimed. Without loss of generality, choose ${\underline{\alpha }}$ in (0, 1).

Finally, we show that $\{f({\mathbf {x}}^k)\}\rightarrow -\infty$ as $k\rightarrow \infty$ on ${\hat{K}}$, which contradicts boundedness of $\{{\mathbf {x}}^k\}$ (Lemma 5). For all $k\in {\hat{K}}$, since $\varDelta {\mathbf {x}}^{\text {a},k}\ne {\mathbf {0}}$ (by (57)) and $\alpha _\text {p}^k\in ({\underline{\alpha }}, 1]$, Proposition 2 implies that $\{f({\mathbf {x}}^k)\}$ is monotonically decreasing and that, for all $k\in {\hat{K}}$,

$$\begin{aligned} f({\mathbf {x}}^k+\alpha _\text {p}^k\varDelta {\mathbf {x}}^{\text {a},k}) < f \left( {\mathbf {x}}^k+{\underline{\alpha }}\varDelta {\mathbf {x}}^{\text {a},k}\right) . \end{aligned}$$

Expanding the right-hand side yields

$$\begin{aligned} f({\mathbf {x}}^k+{\underline{\alpha }}\varDelta {\mathbf {x}}^{\text {a},k})&= f({\mathbf {x}}^k) + {\underline{\alpha }}\nabla f({\mathbf {x}}^k)^T \varDelta {\mathbf {x}}^{\text {a},k}+ \frac{{{\underline{\alpha }}}^2}{2} (\varDelta {\mathbf {x}}^{\text {a},k})^T H\varDelta {\mathbf {x}}^{\text {a},k}\\&= f({\mathbf {x}}^k) + {\underline{\alpha }} \left( \nabla f({\mathbf {x}}^k)^T \varDelta {\mathbf {x}}^{\text {a},k}+ (\varDelta {\mathbf {x}}^{\text {a},k})^T H\varDelta {\mathbf {x}}^{\text {a},k}\right) \\&\quad -\left( {{\underline{\alpha }}} - \frac{{{\underline{\alpha }}}^2}{2}\right) (\varDelta {\mathbf {x}}^{\text {a},k})^T H\varDelta {\mathbf {x}}^{\text {a},k}, \end{aligned}$$

where the sum of the last two terms tends to a strictly negative limit as $k\rightarrow \infty$, $k\in {\hat{K}}$. Indeed, in view of (39), the second term is non-positive and (i) if $(\varDelta {\hat{{\mathbf {x}}}}^{\text {a}})^T H\varDelta {\hat{{\mathbf {x}}}}^{\text {a}}>0$, since ${{\underline{\alpha }}}>{{\underline{\alpha }}}^2/2$, from (53) and (57), the third term tends to a negative limit, and (ii) if $(\varDelta {\hat{{\mathbf {x}}}}^{\text {a}})^T H\varDelta {\hat{{\mathbf {x}}}}^{\text {a}}=0$ then the sum of the last two terms tends to ${{\underline{\alpha }}}\nabla f(\hat{{\mathbf {x}}})^T\varDelta {\hat{{\mathbf {x}}}}^{\text {a}}$ which is also strictly negative in view of (40), since we either have ${\hat{W}} \succ {\mathbf {0}}$ (in the case that $\{({\mathbf {x}}^k,\varvec{\lambda }^k)\}$ bounded away from ${\mathcal {F}}^*$) or at least $[{\hat{W}} \,(A_{\hat{Q}})^T]$ full row rank (in the case that Assumption 3 holds and using the fact that ${\mathcal {A}}({\hat{{\mathbf {x}}}})\subseteq \hat{Q}$). It follows that, for some $\delta >0$, $f({\mathbf {x}}^k+\alpha _\text {p}^k\varDelta {\mathbf {x}}^{\text {a},k})<f({\mathbf {x}}^k)-\delta$ for all $k\in {\hat{K}}$ large enough. Proposition 2 (Eq. (46)) then gives that $f({\mathbf {x}}^{k+1}):=f({\mathbf {x}}^k+\alpha _\text {p}^k\varDelta {\mathbf {x}}^k)<f({\mathbf {x}}^k)-\frac{\omega }{2}\delta$ for all $k\in {\hat{K}}$ large enough, where $\omega >0$ is an algorithm parameter. Since $\{f({\mathbf {x}}^k)\}$ is monotonically decreasing, the proof is now complete. $\square$

We now conclude the proof of Theorem 1 via a string of eight lemmas, each of which builds on the previous one. First, on any subsequence, if $\{\varDelta {\mathbf {x}}^{\text {a},k}\}$ tends to zero, then $\{{\mathbf {x}}^k\}$ approaches stationary points. (Here both $\{{{\tilde{\varvec{\lambda }}}}^{\text {a},k+1}\}$ and $\{{{\tilde{\varvec{\lambda }}}}^{k+1}\}$ are as defined in (34).)

Lemma 6

Suppose that Assumption 1 holds and that $\{{\mathbf {x}}^k\}$ converges to some limit point ${\hat{{\mathbf {x}}}}$ on an infinite index set K. If $\{\varDelta {\mathbf {x}}^{\text {a},k}\}$ converges to zero on K, then (i) ${\hat{{\mathbf {x}}}}$ is stationary and

$$\begin{aligned} \nabla f({\mathbf {x}}^k) - \left( A_{{\mathcal {A}}({\hat{{\mathbf {x}}}})}\right) ^T{{\tilde{\varvec{\lambda }}}}^{\text {a},k+1}_{{\mathcal {A}}({\hat{{\mathbf {x}}}})} \rightarrow {\mathbf {0}}, \text { as } k\rightarrow \infty ,\quad k\in K. \end{aligned}$$

(59)

If, in addition, Assumption 2 holds, then (ii) $\{{{\tilde{\varvec{\lambda }}}}^{\text {a},k+1}\}$ and $\{{{\tilde{\varvec{\lambda }}}}^{k+1}\}$ converge on K to ${\hat{\varvec{\lambda }}}$, the unique multiplier associated with ${\hat{{\mathbf {x}}}}$.

Proof

Suppose $\{{\mathbf {x}}^k\}\rightarrow {\hat{{\mathbf {x}}}}$ on K and $\{\varDelta {\mathbf {x}}^{\text {a},k}\}\rightarrow {\mathbf {0}}$ on K. Let ${\mathbf {s}}^k:=A{\mathbf {x}}^k-{\mathbf {b}}(>{\mathbf {0}})$ for all $k\in K$ and $\hat{{\mathbf {s}}}:=A{\hat{{\mathbf {x}}}}-{\mathbf {b}}(\ge {\mathbf {0}})$, so that $\{{\mathbf {s}}^k\}\rightarrow \hat{{\mathbf {s}}}$ on K. As a first step toward proving Claim (i), we show that, for any $i\not \in {\mathcal {A}}({\hat{{\mathbf {x}}}})$, $\{{{\tilde{\lambda }}}_i^{\text {a},k+1}\}\rightarrow 0$ on K. For $i\not \in {\mathcal {A}}({\hat{{\mathbf {x}}}})$, since $\hat{s}_i>0$, $\{s_i^k\}$ is bounded away from zero on K. Since it follows from (34) and (36) that, for all k,

$$\begin{aligned} {\tilde{\lambda }}_i^{\text {a},k+1}=0,\, \forall i \not \in Q_k \quad \text {and}\quad {\tilde{\lambda }}_i^{\text {a},k+1}=-{\left( s_i^k \right) }^{-1}\lambda _i^k \varDelta s_i^{\text {a},k}, \quad \forall i\in Q_k, \end{aligned}$$

and since $\{\lambda _i^k\}$ is bounded (by construction) and $\varDelta {\mathbf {s}}^{\text {a},k}=A\varDelta {\mathbf {x}}^{\text {a},k}$ (by (21)), we have $\{{{\tilde{\lambda }}}_i^{\text {a},k+1}\}\rightarrow 0$ on K. To complete the proof of Claim (i), note that the first equation of (10) (with H replaced by W) yields

$$\begin{aligned} \nabla f({\mathbf {x}}^k) - (A_{Q_k})^T{{\tilde{\varvec{\lambda }}}}^{\text {a},k+1}_{Q_k} = - W_k \varDelta {\mathbf {x}}^{\text {a},k}. \end{aligned}$$

Since (i) $\{{\tilde{\lambda }}^{\text {a},k+1}_i\}\rightarrow 0$ on K for $i\not \in {\mathcal {A}}({\hat{{\mathbf {x}}}})$, (ii) $\{W_k\}$ is bounded (since $H\preceq W_k\preceq H+R$), (iii) $\{\varDelta {\mathbf {x}}^{\text {a},k}\}\rightarrow {\mathbf {0}}$ on K, and (iv) by definition (34), ${\tilde{\lambda }}_i^{\text {a},+}=0$ for $i\in Q^{\text {c}}$, we conclude that (59) holds, hence $\{\left( A_{{\mathcal {A}}({\hat{{\mathbf {x}}}})}\right) ^T{{\tilde{\varvec{\lambda }}}}^{\text {a},k+1}_{{\mathcal {A}}({\hat{{\mathbf {x}}}})}\}$ converges (since $\nabla f({\mathbf {x}}^k)$ does) as $k\rightarrow \infty$, $k\in K$, to a point in the range of $\left( A_{{\mathcal {A}}({\hat{{\mathbf {x}}}})}\right) ^T$, say $\left( A_{{\mathcal {A}}({\hat{{\mathbf {x}}}})}\right) ^T {\hat{\varvec{\lambda }}}_{{\mathcal {A}}({\hat{{\mathbf {x}}}})}$. We get $\nabla f({\hat{{\mathbf {x}}}}) - \left( A_{{\mathcal {A}}({\hat{{\mathbf {x}}}})}\right) ^T {\hat{\varvec{\lambda }}}_{{\mathcal {A}}({\hat{{\mathbf {x}}}})}={\mathbf {0}}$, proving Claim (i). Finally, Claim (ii) follows from (59), Assumption 2, and the fact that for $i\not \in {\mathcal {A}}({\hat{{\mathbf {x}}}})$, $\{{{\tilde{\lambda }}}_i^{\text {a},k+1}\}\rightarrow 0$ as $k\rightarrow \infty$, $k\in K$, noting that the same argument applies to $\{{\tilde{\varvec{\lambda }}}^{k+1}\}$, using a modified version of (59), with ${\tilde{\varvec{\lambda }}}$ replacing ${\tilde{\varvec{\lambda }}}^\text {a}$, obtained by starting from the first equation of (32) instead of that of (10) and using the fact, proved next, that $\{{\tilde{\lambda }}_i^{k+1}\}\rightarrow 0$ on K for all $i\not \in {\mathcal {A}}(\hat{{\mathbf {x}}})$. From its definition in (34) and the last equation in (33), we have that, for all k,

$$\begin{aligned} {\tilde{\lambda }}_i^{k+1}=0,\quad&\forall i \not \in Q_k, \\ {\tilde{\lambda }}_i^{k+1} = {\left( s_i^k \right) }^{-1}\left( -\lambda _i^k \varDelta s_i^k + \gamma _k\sigma _k\mu _{(Q_k)}^{(k)} - \gamma _k\varDelta s_i^{\text {a},k} \varDelta \lambda _i^{\text {a},k}\right) , \quad&\forall i\in Q_k. \end{aligned}$$

Since $\{{{\tilde{\lambda }}}_i^{\text {a},k+1}\}$ converges (to zero) on K, $\{\varDelta \lambda _i^{\text {a},k}\}$ is bounded on K. Furthermore, from its definition (7), (8) (see also (16)), $\{\gamma _k\}$ is bounded and $|\gamma _k\sigma _k\mu _{(Q_k)}^{(k)}|\le \tau \Vert \varDelta {\mathbf {x}}^{\text {a},k}\Vert$ for all k. Since $\varDelta {\mathbf {s}}^{\text {a},k}=A\varDelta {\mathbf {x}}^{\text {a},k}$ and $\varDelta {\mathbf {s}}^k=A\varDelta {\mathbf {x}}^k$, in view of Lemma 3, it follows that, for $i\not \in {\mathcal {A}}({\hat{{\mathbf {x}}}})$, $\{{{\tilde{\lambda }}}_i^{k+1}\}\rightarrow 0$ on K. $\square$

Lemma 6, combined with Proposition 3 via a contradiction argument, then implies that (on a subsequence), if $\{{\mathbf {x}}^k\}$ does not approach ${\mathcal {F}}_P^*$, then $\{\varDelta {\mathbf {x}}^k\}$ approaches zero.

Lemma 7

Suppose that Assumptions 1 and 2 hold and that $\{{\mathbf {x}}^k\}$ is bounded away from ${\mathcal {F}}_P^*$ on some infinite index set K. Then $\{\varDelta {\mathbf {x}}^k\}\rightarrow {\mathbf {0}}$ as $k\rightarrow \infty$, $k\in K$.

Proof

Proceeding by contradiction, let K be an infinite index set such that $\{{\mathbf {x}}^k\}$ is bounded away from ${\mathcal {F}}_P^*$ on K and $\{\varDelta {\mathbf {x}}^k\}\not \rightarrow {\mathbf {0}}$ as $k\rightarrow \infty$, $k\in K$. Then, in view of Proposition 3 and boundedness of $\{{\mathbf {x}}^k\}$ (Lemma 5), there exist ${\hat{Q}}\subseteq {\mathbf {m}}$, ${\hat{{\mathbf {x}}}}\not \in {\mathcal {F}}_P^*$, and an infinite index set ${\hat{K}}\subseteq K$ such that $Q_k = {\hat{Q}}$ for all $k\in {\hat{K}}$ and

$$\begin{aligned} \begin{aligned}&\{{\mathbf {x}}^{k}\} \rightarrow {\hat{{\mathbf {x}}}}, \text { as } k\rightarrow \infty ,\, k\in {\hat{K}},\\&\{\varDelta {\mathbf {x}}^{\text {a},k-1}\} \rightarrow {\mathbf {0}}, \text { as } k\rightarrow \infty ,\, k\in {\hat{K}},\\&\{[{\tilde{\varvec{\lambda }}}^{\text {a},k}_{{\hat{Q}}}]_{-}\} \rightarrow {\mathbf {0}}\,, \text { as } k\rightarrow \infty ,\, k\in {\hat{K}}. \end{aligned} \end{aligned}$$

On the other hand, from (25), (16) and (7), (8),

$$\begin{aligned} \Vert {\mathbf {x}}^k-{\mathbf {x}}^{k-1}\Vert = \left\| \alpha _\text {p}^{k-1}\varDelta {\mathbf {x}}^{k-1} \right\| \le \Vert \varDelta {\mathbf {x}}^{k-1}\Vert \le (1+\tau ) \Vert \varDelta {\mathbf {x}}^{\text {a},k-1}\Vert , \end{aligned}$$

which implies that $\{{\mathbf {x}}^{k-1}\}\rightarrow {\hat{{\mathbf {x}}}}$ as $k\rightarrow \infty$, $k\in {\hat{K}}$. It then follows from Lemma 6 that ${\hat{{\mathbf {x}}}}$ is stationary and that $[{{\tilde{\varvec{\lambda }}}}^{\text {a},k}_{{\mathcal {A}}({\hat{{\mathbf {x}}}})}]_+$ converges to the associated multiplier vector. Hence the multipliers are non-negative, contradicting the fact that ${\hat{{\mathbf {x}}}}\not \in {\mathcal {F}}_P^*$. $\square$

A contradiction argument based on Lemmas 6 and 7 then shows that $\{{\mathbf {x}}^k\}$ approaches the set of stationary points of (P).

Lemma 8

Suppose Assumptions 1 and 2 hold. Then the sequence $\{{\mathbf {x}}^k\}$ approaches the set of stationary points of (P), i.e., there exists a sequence $\{\hat{{\mathbf {x}}}^k\}$ of stationary points such that $\Vert {\mathbf {x}}^k-\hat{{\mathbf {x}}}^k\Vert$ goes to zero as $k\rightarrow \infty$.

Proof

Proceeding by contradiction, suppose the claim does not hold, i.e., (invoking Lemma 5) suppose $\{{\mathbf {x}}^k\}$ converges to some non-stationary point ${\hat{{\mathbf {x}}}}$ on some infinite index set K. Then $\{\varDelta {\mathbf {x}}^{\text {a},k}\}$ does not converge to zero on K (Lemma 6(i)) and nor does $\{\varDelta {\mathbf {x}}^k\}$ (Lemma 3). Since ${\hat{{\mathbf {x}}}}$ is non-stationary, this is in contradiction with Lemma 7. $\square$

The next technical result, proved in [29, Lemma 3.6], invokes analogues of Lemmas 5, 7 and 8.

Lemma 9

Suppose Assumptions 1 and 2 hold. Suppose $\{{\mathbf {x}}^k\}$ is bounded away from ${\mathcal {F}}_P^*$. Let ${\hat{{\mathbf {x}}}}$ and ${\hat{{\mathbf {x}}}}^{\prime }$ be limit points of $\{{\mathbf {x}}^k\}$ and let ${\hat{\varvec{\lambda }}}$ and ${\hat{\varvec{\lambda }}}^\prime$ be the associated KKT multipliers. Then ${\hat{\varvec{\lambda }}} = {\hat{\varvec{\lambda }}}^\prime$.

Convergence of $\{{\mathbf {x}}^k\}$ to ${\mathcal {F}}_P^*$ ensues, proving Claim (i) of Theorem 1.

Lemma 10

Suppose Assumptions 1 and 2 hold. Then $\{{\mathbf {x}}^k\}$ converges to ${\mathcal {F}}_P^*$.

Proof

We proceed by contradiction. Thus suppose $\{{\mathbf {x}}^k\}$ does not converge to ${\mathcal {F}}_P^*$. Then, since $\{{\mathbf {x}}^k\}$ is bounded (Lemma 5), it has at least one limit point $\hat{{\mathbf {x}}}$ that is not in ${\mathcal {F}}_P^*$, and since (by Proposition 2) $\{f({\mathbf {x}}^k)\}$ is a bounded, monotonically decreasing sequence, $f(\hat{{\mathbf {x}}})=\inf _{k}f({\mathbf {x}}^k)$. Then, by Lemmas 7 and 3, $\{\varDelta {\mathbf {x}}^k\}$ and $\{\varDelta {\mathbf {x}}^{\text {a},k}\}$ converge to zero as $k\rightarrow \infty$. It follows from Lemmas 6 and 9 that all limit points of $\{{\mathbf {x}}^k\}$ are stationary, and that both $\{{\tilde{\varvec{\lambda }}}^{\text {a},k}\}$ and $\{{\tilde{\varvec{\lambda }}}^k\}$ converge to ${\hat{\varvec{\lambda }}}$, the common KKT multiplier vector associated to all limit points of $\{{\mathbf {x}}^k\}$. Since ${\hat{{\mathbf {x}}}}\not \in {\mathcal {F}}_P^*$, there exists $i_0$ such that ${\hat{\lambda }}_{i_0}<0$, so that, for some $\hat{k}>0$,

$$\begin{aligned} {\tilde{\lambda }}_{i_0}^{\text {a},k+1}<0\,\text { and } {\tilde{\lambda }}_{i_0}^{k+1}<0, \quad \forall k>\hat{k}, \end{aligned}$$

(60)

which, in view of Step 8 of the algorithm, implies that $i_0\in Q_k$ for all $k>{\hat{k}}$. Then (36) gives

$$\begin{aligned} \varDelta s_{i_0}^{\text {a},k} = - \left( \lambda _{i_0}^k \right) ^{-1} s_{i_0}^k{\tilde{\lambda }}_{i_0}^{\text {a},k+1} ,\quad \forall k>{\hat{k}}, \end{aligned}$$

where $s_{i_0}^k>0$, $\lambda _{i_0}^k>0$ by construction. Thus, in view of (60), $\varDelta s_{i_0}^{\text {a},k}>0$ for all $k>{\hat{k}}$. On the other hand, the last equation of (33) gives

$$\begin{aligned} \varDelta s_{i_0}^k = \left( \lambda _{i_0}^k \right) ^{-1} \left( -s_{i_0}^k{\tilde{\lambda }}_{i_0}^{k+1} + \gamma _k \sigma _k\mu _{(Q_k)}^k - \gamma _k \varDelta s_{i_0}^{\text {a},k} \varDelta \lambda _{i_0}^{\text {a},k}\right) , \quad \forall k > {\hat{k}} , \end{aligned}$$

(61)

where $\gamma _k\ge 0$, $\sigma _k\ge 0$, and $\mu _{(Q_k)}^k\ge 0$ by construction. Further, for $k>{\hat{k}}$, $\varDelta \lambda _{i_0}^{\text {a},k}<0$ since $\lambda _{i_0}^k>0$ and ${\tilde{\lambda }}_{i_0}^{\text {a},k+1}(=\lambda _{i_0}^k+\varDelta \lambda _{i_0}^{\text {a},k})<0$. It follows that all terms in (61) are non-negative and the first term is positive, so that $\varDelta s_{i_0}^k>0$ for all $k>k^\prime$. Moreover, for all $k>{\hat{k}}$, we have $s_{i_0}^{k+1} = s_{i_0}^{k} + \alpha _\text {p}^k \varDelta s_{i_0}^k> s_{i_0}^{k} > 0$, where $\alpha _\text {p}^k>0$ since ${\mathbf {s}}^k > {\mathbf {0}}$. Since $\{{\mathbf {s}}^k\}$ is bounded (Lemma 5), we then conclude that $\{s_{i_0}^k\}\rightarrow \hat{s}_{i_0}>0$ so that $\hat{s}_{i_0}\hat{\lambda }_{i_0}<0$, in contradiction with the stationarity of limit points. $\square$

Under strict complementarity, the next lemma then establishes appropriate convergence of the multipliers, setting the stage for the proof of part (ii) of Theorem 1 in the following lemma.

Lemma 11

Suppose Assumptions 1–3 hold and let $({\mathbf {x}}^*,\varvec{\lambda }^*)$ be the unique primal-dual solution. Then, given any infinite index set K such that $\{\varDelta {\mathbf {x}}^{\text {a},k}\}_{k\in K}\rightarrow {\mathbf {0}}$, it holds that $\{\varvec{\lambda }^{k+1}\}_{k\in K}\rightarrow \varvec{\xi }^*$, where $\xi _i^*:=\min \{\lambda ^*_i,\lambda ^{\max }\}$, for all $i\in {\mathbf {m}}$.

Proof

Lemma 10 guarantees that $\{{\mathbf {x}}^k\}\rightarrow {\mathbf {x}}^*$. Let K be an infinite index set such that $\{\varDelta {\mathbf {x}}^{\text {a},k}\}_{k\in K}\rightarrow {\mathbf {0}}$. Then, in view of Lemma 6(ii), $\{{\tilde{\varvec{\lambda }}}^{\text {a},k+1}\}_{k\in K}\rightarrow \varvec{\lambda }^*\ge {\mathbf {0}}$. Accordingly, $\{\chi _k\}=\{\Vert \varDelta {\mathbf {x}}^{\text {a},k}\Vert ^\nu +\Vert [{\tilde{\varvec{\lambda }}}^{\text {a},k+1}_{Q_k}]_-\Vert ^\nu \}\rightarrow 0$ as $k\rightarrow \infty$, $k\in K$. Hence, in view of (26) and (27), the proof will be complete if we show that $\{{\breve{\varvec{\lambda }}}^{k+1}\}_{k\in K}\rightarrow \varvec{\lambda }^*$, where

$$\begin{aligned} {\breve{\lambda }}^{k+1}_i := \lambda ^k_i + \alpha _\text {d}^{k}\varDelta \lambda ^k_i,\quad i\in Q_k, \quad \text{ and } \quad {\breve{\lambda }}^{k+1}_i := \mu ^{k+1}_{(Q_k)}/s_i^{k+1},\quad i\in Q^{\text {c}}_k, \end{aligned}$$

or equivalently, $\{{\tilde{\varvec{\lambda }}}^{k+1}-{\breve{\varvec{\lambda }}}^{k+1}\}_{k\in K}\rightarrow {\mathbf {0}}$, which we do now.

For every $Q\subseteq {\mathbf {m}}$, define the index set $K(Q):=\{k\in K:Q_k=Q\}$, and let ${\mathcal {Q}}:=\{Q\subseteq {\mathbf {m}}:|K(Q)|=\infty \}$. We first show that for all $Q\in {\mathcal {Q}}$, $\{{\tilde{\varvec{\lambda }}}^{k+1}_{Q}-{\breve{\varvec{\lambda }}}^{k+1}_Q\}_{k\in K(Q)}\rightarrow {\mathbf {0}}$. For $Q\in {\mathcal {Q}}$, the definition (34) of ${{\tilde{\varvec{\lambda }}}}^+$ yields

$$\begin{aligned} \left\| {\tilde{\varvec{\lambda }}}^{k+1}_{Q}-{\breve{\varvec{\lambda }}}^{k+1}_Q \right\| = \left( 1-\alpha _\text {d}^k\right) \left\| \varDelta \varvec{\lambda }^k_{Q}\right\| , \quad k\in K(Q). \end{aligned}$$

Since boundedness of $\{\varvec{\lambda }^k_Q\}$ (by construction) and of $\{{\tilde{\varvec{\lambda }}}^{k+1}_Q\}_{k\in K}$ (=$\{\varvec{\lambda }^k_Q+\varDelta \varvec{\lambda }^k_{Q}\}_{k\in K}$) (by Lemma 6(ii)) implies boundedness of $\{\varDelta \varvec{\lambda }^k_{Q}\}_{k\in K}$, we only need $\{\alpha _\text {d}^k\}_{k\in K(Q)}\rightarrow 1$ in order to guarantee that $\Vert {\tilde{\varvec{\lambda }}}^{k+1}_{Q}-{\breve{\varvec{\lambda }}}^{k+1}_Q\Vert \rightarrow 0$ on K(Q). Now, $\{\varDelta {\mathbf {x}}^{\text {a},k}\}_{k\in K}\rightarrow {\mathbf {0}}$ implies that $\{\varDelta {\mathbf {s}}^{\text {a},k}\}_{k\in K}\rightarrow {\mathbf {0}}$, and from Lemma 3 that $\{\varDelta {\mathbf {x}}^k\}_{k\in K}\rightarrow {\mathbf {0}}$, implying that $\{\varDelta {\mathbf {s}}^k\}_{k\in K}\rightarrow {\mathbf {0}}$; and $\{{\mathbf {x}}^k\}\rightarrow {\mathbf {x}}^*$ yields $\{{\mathbf {s}}^k\}\rightarrow {\mathbf {s}}^*:=A{\mathbf {x}}^*-{\mathbf {b}}$, so $s_i^k+\varDelta s_i^k>0$ for all $i\in {\mathcal {A}}({\mathbf {x}}^*)^\mathrm{c}$, $k\in K$ large enough. Moreover, Assumption 3 gives $\lambda _i^*>0$ for all $i\in {\mathcal {A}}({\mathbf {x}}^*)$ so that, for sufficiently large $k\in K$, ${\tilde{\lambda }}_i^{k+1}>0$ for all $i\in {\mathcal {A}}({\mathbf {x}}^*)$, and Condition CSR(ii) implies that ${\mathcal {A}}({\mathbf {x}}^*)\subseteq Q$, so Lemma 4 applies, with ${\mathcal {A}}:={\mathcal {A}}({\mathbf {x}}^*)$. It follows that $\{{\bar{\alpha }}_\text {d}^k\}_{k\in K}\rightarrow 1$, since all terms on the right-hand side of (41) converge to one on K. Thus, from the definition of $\alpha _\text {d}^k$ in (24) and the fact that $\{\varDelta {\mathbf {x}}^k\}_{k\in K}\rightarrow {\mathbf {0}}$, we have $\{\alpha _\text {d}^k\}_{k\in K}\rightarrow 1$ indeed, establishing that $\{{\tilde{\varvec{\lambda }}}^{k+1}_{Q}-{\breve{\varvec{\lambda }}}^{k+1}_Q\}_{k\in K(Q)}\rightarrow {\mathbf {0}}$.

It remains to show that, for all $Q\in {\mathcal {Q}}$, $\{{\tilde{\varvec{\lambda }}}^{k+1}_{Q^{\text {c}}}-{\breve{\varvec{\lambda }}}^{k+1}_{Q^{\text {c}}}\}_{k\in K(Q)}\rightarrow {\mathbf {0}}$. To show this, we first note that, since $\{\chi _k\}_{k\in K}\rightarrow 0$, it follows from (26) and (27) and the fact established above that $\{{\breve{\varvec{\lambda }}}^{k+1}_Q\}_{K(Q)}\rightarrow \varvec{\lambda }_Q^*$ that, for all $Q\in {\mathcal {Q}}$,

$$\begin{aligned} \{\varvec{\lambda }^{k+1}_{Q}\}\rightarrow \varvec{\xi }^*_{Q}, \quad k\rightarrow \infty , \quad k\in K(Q). \end{aligned}$$

(62)

Next, from (26), (27), and the definition (34) of ${{\tilde{\varvec{\lambda }}}}^+$, we have, for $Q\in {\mathcal {Q}}$ and sufficiently large $k\in K(Q)$,

$$\begin{aligned} \left| {{\tilde{\lambda }}}_i^{k+1} - {\breve{\lambda }}_i^{k+1} \right| = {\breve{\lambda }}_i^{k+1} = \frac{\mu _{(Q)}^{k+1}}{s_i^{k+1}}, \quad i\in Q^{\text {c}}. \end{aligned}$$

(63)

Clearly, since ${\mathcal {A}}({\mathbf {x}}^*)\subseteq Q$, we have $s_i^*>0$ for $i\in Q^{\text {c}}$. Hence, since $\{\varDelta {\mathbf {s}}^k\}_{k\in K}\rightarrow {\mathbf {0}}$, $\{s_i^{k+1}\}$ is bounded away from zero on K for $i\in Q^{\text {c}}$. When Q is empty, the right-hand side of (63) is set to zero [see definition (14) of $\mu _{(Q)}$]. When Q is not empty, since $\xi _i^*=0$ whenever $\lambda _i^*=0$, (62) and complementary slackness gives

$$\begin{aligned} \left\{ \mu _{(Q)}^{k+1}\right\} =\left\{ \frac{({\mathbf {s}}^{k+1}_{Q})^T\varvec{\lambda }^{k+1}_{Q}}{|Q|}\right\} \rightarrow \left\{ \frac{({\mathbf {s}}^*_Q)^T \varvec{\xi }^*_Q}{|Q|}\right\} =0, \quad k\in K(Q) , \end{aligned}$$

and it follows from (63) that $\{{\tilde{\varvec{\lambda }}}^{k+1}_{Q^{\text {c}}}-{\breve{\varvec{\lambda }}}^{k+1}_{Q^{\text {c}}}\}_{k\in K(Q)}\rightarrow {\mathbf {0}}$, completing the proof. $\square$

Claim (ii) of Theorem 1 can now be proved.

Lemma 12

Suppose Assumptions 1–3 hold and let $({\mathbf {x}}^*,\varvec{\lambda }^*)$ be the unique primal-dual solution. Then $\{{\tilde{\varvec{\lambda }}}^k\}\rightarrow \varvec{\lambda }^*$ and $\{\varvec{\lambda }^{k}\}\rightarrow \varvec{\xi }^*$, with $\xi _i^*:=\min \{\lambda ^*_i,\lambda ^{\max }\}$ for all $i\in {\mathbf {m}}$.

Proof

Again, Lemma 10 guarantees that $\{{\mathbf {x}}^k\}\rightarrow {\mathbf {x}}^*$ and $\{{\mathbf {s}}^k\}\rightarrow {\mathbf {s}}^*:=A{\mathbf {x}}^*-{\mathbf {b}}$. Note that if $\{\varDelta {\mathbf {x}}^{\text {a},k}\}\rightarrow {\mathbf {0}}$, the claims are immediate consequences of Lemmas 6 and 11. We now prove by contradiction that $\{\varDelta {\mathbf {x}}^{\text {a},k}\}\rightarrow {\mathbf {0}}$. Thus, suppose that for some infinite index set K, $\inf _{k\in K}\Vert \varDelta {\mathbf {x}}^{\text {a},k}\Vert >0$. Then, Lemma 3 gives $\inf _{k\in K}\Vert \varDelta {\mathbf {x}}^k\Vert >0$. It follows from Proposition 3 that, on some infinite index set $K^\prime \subseteq K$, $\{\varDelta {\mathbf {x}}^{\text {a},k-1}\}\rightarrow {\mathbf {0}}$ and $\{[{\tilde{\varvec{\lambda }}}^{\text {a},k}]_-\}\rightarrow {\mathbf {0}}$. Since $Q_k$ is selected from a finite set and $\{W_k\}$ is bounded, we can assume without loss of generality that $Q_k=Q$ on $K^\prime$ for some $Q \subseteq {\mathbf {m}}$, and that $\{W_k\}\rightarrow W^*\succeq H$ on $K^\prime$. Further, from Lemma 11, $\{\varvec{\lambda }^k\}_{k\in K^\prime }\rightarrow \varvec{\xi }^*$. Therefore, $\{J(W_k,A_{Q_k},{\mathbf {s}}_{Q_k},\varvec{\lambda }_{Q_k})\}_{k\in K^\prime } \rightarrow J(W^*,A_{Q},{\mathbf {s}}_Q^*,\varvec{\xi }_Q^*)$, and in view of Assumptions 2 and 3 and Lemma 1, $J(W^*,A_{Q},{\mathbf {s}}_Q^*,\varvec{\xi }_Q^*)$ is non-singular (since $({\mathbf {x}}^*,\varvec{\lambda }^*)$ is optimal). It follows from (10), with W substituted for H, that $\{\varDelta {\mathbf {x}}^{\text {a},k}\}\rightarrow {\mathbf {0}}$ on $K^{\prime }$, a contradiction, proving that $\{\varDelta {\mathbf {x}}^{\text {a},k}\}\rightarrow {\mathbf {0}}$. $\square$

Claim (iv) of Theorem 1 follows as well.

Lemma 13

Suppose Assumptions 1 and 2 hold and $\varepsilon >0$. Then Algorithm CR-MPC terminates (in Step 1) after finitely many iterations.

Proof

If $\{({\mathbf {x}}^k,\varvec{\lambda }^k)\}$ has a limit point in ${\mathcal {F}}^*$, then $\inf _k\{E_k\}=0$, proving the claim. Thus, suppose that $\{({\mathbf {x}}^k,\varvec{\lambda }^k)\}$ is bounded away from ${\mathcal {F}}^*$. In view of Lemmas 5 and 10, $\{{\mathbf {x}}^k\}$ has a limit point ${\mathbf {x}}^*\in {\mathcal {F}}_P^*$. Assumption 2 then implies that there exists a unique KKT multiplier vector $\varvec{\lambda }^*\ge {\mathbf {0}}$ associated to ${\mathbf {x}}^*$. If $({\mathbf {x}}^*,\varvec{\lambda }^*)\in {\mathcal {F}}^*$ is a limit point of $\{({\mathbf {x}}^k, {\tilde{\varvec{\lambda }}}^k)\}$, which also implies that $\inf _k\{E({\mathbf {x}}^k, {\tilde{\varvec{\lambda }}}^k)\}=0$, then in view of the stopping criterion, the claim again follows. Thus, further suppose that there is an infinite index set K such that $\{{\mathbf {x}}^k\}_{k\in K}\rightarrow {\mathbf {x}}^*$, but $\inf _{k\in K}\Vert {\tilde{\varvec{\lambda }}}^k-\varvec{\lambda }^*\Vert >0$. It then follows from Lemma 6(ii) that $\{\varDelta {\mathbf {x}}^{\text {a},k-1}\}_{k\in K}\not \rightarrow {\mathbf {0}}$, and from Lemma 3 that $\{\varDelta {\mathbf {x}}^{k-1}\}_{k\in K}\not \rightarrow {\mathbf {0}}$. Proposition 3 and Lemma 3 then imply that $\{\varDelta {\mathbf {x}}^{\text {a},k-2}\}_{k\in K^\prime }\rightarrow {\mathbf {0}}$ and $\{\varDelta {\mathbf {x}}^{k-2}\}_{k\in K^\prime }\rightarrow {\mathbf {0}}$ for some infinite index set $K^\prime \subseteq K$. Next, from Lemmas 5 and 10, we have $\{{\mathbf {x}}^{k-2}\}_{k\in K^{\prime \prime }}\rightarrow {\mathbf {x}}^{**}\in {\mathcal {F}}_P^*$ for some infinite index set $K^{\prime \prime }\subseteq K^\prime$, and in view of Lemma 6(ii) $\{{{\tilde{\varvec{\lambda }}}}^{k-1}\}_{k\in K^{\prime \prime }}\rightarrow \varvec{\lambda }^{**}$, where $\varvec{\lambda }^{**}$ is the KKT multiplier associated to ${\mathbf {x}}^{**}$. Since $\alpha _\text {p}^{k}\in [0,1]$ for all k, we also have $\{{\mathbf {x}}^{k-1}\}_{k\in K^{\prime \prime }} =\{{\mathbf {x}}^{k-2}+\alpha _\text {p}^{k-2}\varDelta {\mathbf {x}}^{k-2}\}_{k\in K^{\prime \prime }}\rightarrow {\mathbf {x}}^{**}$, i.e., $\{({\mathbf {x}}^{k-1},{{\tilde{\varvec{\lambda }}}}^{k-1})\}_{k\in K^{\prime \prime }}\rightarrow ({\mathbf {x}}^{**},\varvec{\lambda }^{**})\in {\mathcal {F}}^*$, completing the proof. $\square$

Proof of Theorem 1

Claim (i) was proved in Lemma 10 and Claim (ii) in Lemma 12, Claim (iii) is a direct consequence of Condition CSR(ii), and Claim (iv) was proved in Lemma 13. $\square$

Proof of Corollary 2

 From Theorem 1, $\{({\mathbf {x}}^k,\varvec{\lambda }^k)\}\rightarrow ({\mathbf {x}}^*,\varvec{\lambda }^*)$, i.e., $\{E_k\}\rightarrow 0$. It follows that (i) in view of Proposition 1 and Condition CSR(ii) $Q_k\supseteq {\mathcal {A}}(x^*)$ for all k large enough, and (ii) in view of Rule R, $\{\delta _k\}\rightarrow 0$, so that $Q_k$ eventually excludes all indexes that are not in ${\mathcal {A}}({\mathbf {x}}^*)$. $\square$

B Proof of Theorem 2

Parts of this proof are adapted from [16, 29, 34]. Throughout, we assume that Assumption 3 holds (so that Assumption 1 also holds), that $\varepsilon =0$ and that the iteration never stops, and that $\lambda ^*_i<\lambda ^{\max }$ for all i.

Newton’s method plays the central role in the local analysis. The following lemma is standard or readily proved; see, e.g., [29, Proposition 3.10].

Lemma 14

Let $\varPhi :{\mathbb {R}}^{n}\rightarrow {\mathbb {R}}^{n}$ be twice continuously differentiable and let ${\mathbf {t}}^*\in {\mathbb {R}}^n$ such that $\varPhi ({\mathbf {t}}^*)={\mathbf {0}}$. Suppose there exists $\rho >0$ such that $\frac{\partial \varPhi }{\partial {\mathbf {t}}}({\mathbf {t}})$ is non-singular for all ${\mathbf {t}}\in B({\mathbf {t}}^*,\rho )$. Define $\varDelta ^{\mathrm{N}}{\mathbf {t}}$ to be the Newton increment at ${\mathbf {t}}$, i.e., $\varDelta ^\mathrm{N}{\mathbf {t}}=-\left( \frac{\partial \varPhi }{\partial {\mathbf {t}}}({\mathbf {t}})\right) ^{-1}\varPhi ({\mathbf {t}})$. Then, given any $c>0$, there exists $c^*>0$ such that, for all ${\mathbf {t}}\in B({\mathbf {t}}^*,\rho )$, if ${\mathbf {t}}^+\in {\mathbb {R}}^n$ satisfies

$$\begin{aligned} \min \left\{ \left| t_i^+-t_i^* \right| , \left| t_i^+-\left( t_i+ \left( \varDelta ^{\mathrm{N}} t\right) _i \right) \right| \right\} \le c \, \max \left\{ \left\| \varDelta ^{\mathrm{N}}{\mathbf {t}}\right\| ^2, \Vert {\mathbf {t}}-{\mathbf {t}}^*\Vert ^2 \right\} ,\quad i=1,\ldots ,n, \end{aligned}$$

(64)

then

$$\begin{aligned} \Vert {\mathbf {t}}^+-{\mathbf {t}}^*\Vert \le c^* \Vert {\mathbf {t}}-{\mathbf {t}}^*\Vert ^2. \end{aligned}$$

For convenience, define ${\mathbf {z}}:=({\mathbf {x}}, \varvec{\lambda })$ (as well as ${\mathbf {z}}^*:=({\mathbf {x}}^*,\varvec{\lambda }^*)$, etc.). For ${\mathbf {z}}\in {\mathcal {F}}^o:=\{{\mathbf {z}}:{\mathbf {x}}\in {\mathcal {F}}_P^o,\,\varvec{\lambda }>{\mathbf {0}}\}$, define

$$\begin{aligned} \varrho ({\mathbf {z}}):=\min \left\{ 1,\frac{E({\mathbf {x}},\varvec{\lambda })}{{\bar{E}}}\right\} \quad \text{ and } \quad W({\mathbf {z}}):=H+\varrho ({\mathbf {z}})R. \end{aligned}$$

The gist of the remainder of this appendix is to apply Lemma 14 to

$$\begin{aligned} \varPhi _Q({\mathbf {z}}):=\left[ \begin{array}{c} H{\mathbf {x}}-(A_Q)^T\varvec{\lambda }_Q+{\mathbf {c}}\\ \varLambda _Q(A_Q{\mathbf {x}}-{\mathbf {b}}_Q) \end{array}\right] , \quad Q\subseteq {\mathbf {m}}. \end{aligned}$$

(Note that $\varPhi _Q({\mathbf {z}}^*)={\mathbf {0}}$.) Let ${\mathbf {z}}_Q:=({\mathbf {x}}, \varvec{\lambda }_Q)$, then the step taken on the Q components along the search direction generated by the Algorithm CR-MPC is analogously given by ${\breve{{\mathbf {z}}}}^+_Q : = ({\mathbf {x}}^+,{\breve{\varvec{\lambda }}}_Q^+)$ with ${\breve{\varvec{\lambda }}}_Q^+:=\varvec{\lambda }_Q + \alpha _\text {d}\varDelta \varvec{\lambda }_Q$. The first major step of the proof is achieved by Proposition 4 below, where the focus is on ${\breve{{\mathbf {z}}}}^+_Q$ rather than on ${\mathbf {z}}^+$. Thus we compare ${\breve{{\mathbf {z}}}}^+_Q$, with $Q\in {\mathcal {Q}}^*$ to the Q components of the (unregularized) Newton step, i.e., ${\mathbf {z}}_Q+(\varDelta ^\mathrm{N}{\mathbf {z}})_Q$. Define

$$\begin{aligned} {\mathscr {A}}:=\left[ \begin{array}{cc} \alpha _\text {p}I_n &{} \quad {\mathbf {0}}\\ {\mathbf {0}}&{} \quad \alpha _\text {d}I_{|Q|} \end{array} \right] ,\quad \text{ and } \quad \alpha :=\min \{\alpha _\text {p},\alpha _\text {d}\}. \end{aligned}$$

The difference between the CR-MPC iteration and the Newton iteration can be written as

$$\begin{aligned} \begin{aligned}&\left\| {\breve{{\mathbf {z}}}}^+_Q-\left( {\mathbf {z}}_Q+\left( \varDelta ^{\mathrm{N}}{\mathbf {z}}\right) _Q\right) \right\| \\&\quad \le \left\| {\breve{{\mathbf {z}}}}^+_Q-({\mathbf {z}}_Q+\varDelta {\mathbf {z}}_Q) \right\| + \left\| \varDelta {\mathbf {z}}_Q-\varDelta {\mathbf {z}}^{\text {a}}_Q \right\| + \left\| \varDelta {\mathbf {z}}^{\text {a}}_Q - \varDelta {\mathbf {z}}^0_Q\right\| +\left\| \varDelta {\mathbf {z}}^0_Q-\left( \varDelta ^{\mathrm{N}}{\mathbf {z}}\right) _Q \right\| \\&\quad = \left\| (I-{\mathscr {A}})\varDelta {\mathbf {z}}_Q \right\| + \gamma \left\| \varDelta {\mathbf {z}}^{\text {c}}_Q \right\| + \left\| \varDelta {\mathbf {z}}^{\text {a}}_Q - \varDelta {\mathbf {z}}^0_Q \right\| + \left\| \varDelta {\mathbf {z}}^0_Q- \left( \varDelta ^{\mathrm{N}}{\mathbf {z}}\right) _Q \right\| \\&\quad \le (1-\alpha ) \left\| \varDelta {\mathbf {z}}_Q \right\| + \Vert \varDelta {\mathbf {z}}^{\text {c}}_Q\Vert +\left\| \varDelta {\mathbf {z}}^{\text {a}}_Q - \varDelta {\mathbf {z}}^0_Q \right\| + \left\| \varDelta {\mathbf {z}}^0_Q- \left( \varDelta ^{\mathrm{N}}{\mathbf {z}}\right) _Q \right\| , \end{aligned} \end{aligned}$$

(65)

where $\varDelta {\mathbf {z}}_Q:=(\varDelta {\mathbf {x}},\varDelta \varvec{\lambda }_Q)$, $\varDelta {\mathbf {z}}^{\text {a}}_Q:=(\varDelta {\mathbf {x}}^{\text {a}},\varDelta \varvec{\lambda }^\text {a}_Q)$, $\varDelta {\mathbf {z}}^\text {c}_Q:=(\varDelta {\mathbf {x}}^{\text {c}},\varDelta \varvec{\lambda }^\text {c}_Q)$, and $\varDelta {\mathbf {z}}^0_Q$ is the (constraint-reduced) affine-scaling direction for the original (unregularized) system (so $\varDelta ^{\mathrm{N}}{\mathbf {z}}=\varDelta {\mathbf {z}}^0_{\mathbf {m}}$).

Let

$$\begin{aligned} J_a(W,A,{\mathbf {s}},\varvec{\lambda }) := \left[ \begin{array}{cc} W &{} \quad -A^T \\ \varLambda A &{} \quad S \end{array}\right] . \end{aligned}$$

The following readily proved lemma will be of help. (For details, see Lemmas B.15 and B.16 in [16]; also Lemmas 13 and 1 in [28])

Lemma 15

Let ${\mathbf {s}},\varvec{\lambda }\in {\mathbb {R}}^m$ and $Q\subseteq {\mathbf {m}}$ be arbitrary and let W be symmetric, with $W\succeq H$. Then (i) $J_a(W,A_Q,{\mathbf {s}}_Q,\varvec{\lambda }_Q)$ is non-singular if and only if $J(W,A_Q,{\mathbf {s}}_Q,\varvec{\lambda }_Q)$ is, and (ii) if ${\mathcal {A}}({\mathbf {x}}^*)\subseteq Q$, then $J(W,A_Q,{\mathbf {s}}_Q^*,\varvec{\lambda }_Q^*)$ is non-singular (and so is $J_a(W,A_Q,{\mathbf {s}}_Q^*,\varvec{\lambda }_Q^*)$).

With ${\mathbf {s}}:=A{\mathbf {x}}-{\mathbf {b}}$, $J_a(H,A_Q,{\mathbf {s}}_Q,\varvec{\lambda }_Q)$, the system matrix for the (constraint-reduced) original (unregularized) “augmented” system, is the Jacobian of $\varPhi _Q({\mathbf {z}})$, i.e.,

$$\begin{aligned} J_a(H,A_Q,{\mathbf {s}}_Q,\varvec{\lambda }_Q) \varDelta {\mathbf {z}}^0_Q = -\varPhi _Q({\mathbf {z}}), \end{aligned}$$

and its regularized version $J_a(W({\mathbf {z}}),A_Q,{\mathbf {s}}_Q,\varvec{\lambda }_Q)$ satisfies (among other systems solved by Algorithm CR-MPC)

$$\begin{aligned} J_a(W({\mathbf {z}}),A_Q,{\mathbf {s}}_Q,\varvec{\lambda }_Q) \varDelta {\mathbf {z}}^{\text {a}}_Q = -\varPhi _Q({\mathbf {z}}). \end{aligned}$$

Next, we verify that $J_a(W({\mathbf {z}}),A_Q,{\mathbf {s}}_Q,\varvec{\lambda }_Q)$ is non-singular near ${\mathbf {z}}^*$ (so that $\varDelta {\mathbf {z}}^0_Q$ and $\varDelta {\mathbf {z}}^{\text {a}}_Q$ in (65) are well defined) and establish other useful local properties. For convenience, we define

$$\begin{aligned} {\mathcal {Q}}^*:=\{Q\subseteq {\mathbf {m}}:{\mathcal {A}}({\mathbf {x}}^*)\subseteq Q\}. \end{aligned}$$

and

$$\begin{aligned} {\tilde{{\mathbf {s}}}}^+ := {\mathbf {s}}+ \varDelta {\mathbf {s}}, \qquad {\tilde{{\mathbf {s}}}}^{\text {a},+} := {\mathbf {s}}+ \varDelta {\mathbf {s}}^{\text {a}}. \end{aligned}$$

Lemma 16

Let $\epsilon ^*:=\min \{1,\min _{i\in {\mathbf {m}}}(\lambda _i^*+s_i^*)\}$. There exist $\rho ^*>0$ and $r>0$, such that, for all ${\mathbf {z}}\in {\mathcal {F}}^o\cap B({\mathbf {z}}^*,\rho ^*)$ and all $Q\in {\mathcal {Q}}^*$, the following hold:

• (i) $\Vert J_a(W({\mathbf {z}}),A_Q,\varvec{\lambda }_Q,{\mathbf {s}}_Q)^{-1}\Vert \le r$,

• (ii) $\max \{\Vert \varDelta {\mathbf {z}}^{\text {a}}_Q\Vert , \Vert \varDelta {\mathbf {z}}_Q\Vert , \Vert \varDelta {\mathbf {s}}^{\text {a}}_Q\Vert , \Vert \varDelta {\mathbf {s}}_Q\Vert \}<\epsilon ^*/4$,

• (iii) $\min \{\lambda _i, {\tilde{\lambda }}_i^{\text {a},+} , {\tilde{\lambda }}_i^+ \}>\epsilon ^*/2,\,\forall i\in {\mathcal {A}}({\mathbf {x}}^*)$,

  $\max \{ \lambda _i , {\tilde{\lambda }}_i^{\text {a},+} , {\tilde{\lambda }}_i^+ \}<\epsilon ^*/2,\,\forall i\in {\mathbf {m}}\setminus {\mathcal {A}}({\mathbf {x}}^*)$,

  $\max \{ s_i , \tilde{s}_i^{\text {a},+} , \tilde{s}_i^+ \}<\epsilon ^*/2,\,\forall i\in {\mathcal {A}}({\mathbf {x}}^*)$,

  $\min \{ s_i , \tilde{s}_i^{\text {a},+} , \tilde{s}_i^+ \}>\epsilon ^*/2,\,\forall i\in {\mathbf {m}}\setminus {\mathcal {A}}({\mathbf {x}}^*)$.

• (iv) ${{\tilde{\lambda }}}_i^+ < \lambda ^{\max },\,\forall i\in {\mathbf {m}}$.

Proof

Claim (i) follows from Lemma 15, continuity of $J_a(W({\mathbf {z}}),A_Q,\varvec{\lambda }_Q,{\mathbf {s}}_Q)$ (and the fact that $W({\mathbf {z}}^*)=H$). Claims (ii) and (iv) follow from Claim (i), Lemma 15, continuity of the right-hand sides of (10) and (32), which are zero at the solution, definition (34) of ${{\tilde{\varvec{\lambda }}}}^+$, and our assumption that $\lambda ^*_i < \lambda ^{\max }$ for all $i\in {\mathbf {m}}$. Claim (iii) is true due to strict complementary slackness, the definition of $\epsilon ^*$, and Claim (ii). $\square$

In preparation for Proposition 4, Lemmas 17–20 provide bounds on the four terms in the last line of (65). The $\rho ^*$ used in these lemmas comes from Lemma 16. The proofs of Lemmas 17, 18, and 20 are omitted, as they are very similar to those of Lemmas A.9 and A.10 in the supplementary materials of [34] (where an MPC algorithm for linear optimization problems is considered) and of Lemma B.19 in [16] (also Lemma 16 in [28]).

Lemma 17

There exists a constant $c_1>0$ such that, for all ${\mathbf {z}}\in {\mathcal {F}}^o\cap B({\mathbf {z}}^*,\rho ^*)$, and for all $Q\in {\mathcal {Q}}^*$,

$$\begin{aligned} \left\| \varDelta {\mathbf {z}}^{\text {c}}_Q \right\| \le c_1 \left\| \varDelta {\mathbf {z}}^{\text {a}}_Q \right\| ^2. \end{aligned}$$

Note that an upper bound on the magnitude of the MPC search direction $\varDelta {\mathbf {z}}_Q$ can be obtained by using Lemmas 17 and 16(ii), viz.

$$\begin{aligned} \left\| \varDelta {\mathbf {z}}_Q \right\| \le \left\| \varDelta {\mathbf {z}}^{\text {a}}_Q \right\| + \left\| \varDelta {\mathbf {z}}^{\text {c}}_Q \right\| \le \left\| \varDelta {\mathbf {z}}^{\text {a}}_Q \right\| +c_1\left\| \varDelta {\mathbf {z}}^{\text {a}}_Q \right\| ^2 \le \left( 1+c_1\frac{\epsilon ^*}{4}\right) \left\| \varDelta {\mathbf {z}}^{\text {a}}_Q \right\| . \qquad \end{aligned}$$

(66)

This bound is used in the proofs of Lemma 18 and Proposition 4.

Lemma 18

There exists a constant $c_2>0$ such that, for all ${\mathbf {z}}\in {\mathcal {F}}^o\cap B({\mathbf {z}}^*,\rho ^*)$, and for all $Q\in {\mathcal {Q}}^*$,

$$\begin{aligned} |1-\alpha | \le c_2 \left\| \varDelta {\mathbf {z}}^{\text {a}}_Q \right\| . \end{aligned}$$

Lemma 19

There exists a constant $c_3>0$ such that, for all ${\mathbf {z}}\in {\mathcal {F}}^o\cap B({\mathbf {z}}^*,\rho ^*)$ and all $Q\in {\mathcal {Q}}^*$,

$$\begin{aligned} \left\| \varDelta {\mathbf {z}}^{\text {a}}_Q - \varDelta {\mathbf {z}}^0_Q \right\| \le c_3 \Vert {\mathbf {z}}-{\mathbf {z}}^*\Vert ^2. \end{aligned}$$

Proof

We have

$$\begin{aligned} \varDelta {\mathbf {z}}^{\text {a}}_Q - \varDelta {\mathbf {z}}^0_Q = -\left( J_a(W({\mathbf {z}}),A_Q,{\mathbf {s}}_Q,\varvec{\lambda }_Q)^{-1} - J_a(H,A_Q,{\mathbf {s}}_Q,\varvec{\lambda }_Q)^{-1}\right) \varPhi _Q({\mathbf {z}}) \end{aligned}$$

so that there exist $c_{31}>0$ such that, for all ${\mathbf {z}}\in {\mathcal {F}}^o\cap B({\mathbf {z}}^*,\rho ^*)$ and all $Q\in {\mathcal {Q}}^*$,

$$\begin{aligned} \left\| \varDelta {\mathbf {z}}^{\text {a}}_Q - \varDelta {\mathbf {z}}^0_Q \right\| \le c_{31} \Vert W({\mathbf {z}})-H\Vert \Vert {\mathbf {z}}-{\mathbf {z}}^*\Vert , \end{aligned}$$

where the second inequality follows from Lemma 16(i). Since $W({\mathbf {z}})-H=\varrho ({\mathbf {z}})R$, $|\varrho ({\mathbf {z}})|\le c_{32} |E({\mathbf {z}})|$, and $|E({\mathbf {z}})|\le c_{33}\Vert {\mathbf {z}}-{\mathbf {z}}^*\Vert$, for some $c_{32}>0$ and $c_{33}>0$, the proof is complete. $\square$

Lemma 20

There exists a constant $c_4>0$ such that, for all ${\mathbf {z}}\in {\mathcal {F}}^o\cap B({\mathbf {z}}^*,\rho ^*)$, and for all $Q\in {\mathcal {Q}}^*$,

$$\begin{aligned} \left\| \varDelta {\mathbf {z}}^0_Q-\left( \varDelta ^{\mathrm{N}}{\mathbf {z}}\right) _Q \right\| \le c_4 \Vert {\mathbf {z}}-{\mathbf {z}}^*\Vert \left\| \left( \varDelta ^{\mathrm{N}}{\mathbf {z}}\right) _Q \right\| . \end{aligned}$$

With Lemmas 17–20 in hand, we return to inequality (65).

Proposition 4

There exists a constant $c_5>0$ such that, for all ${\mathbf {z}}\in {\mathcal {F}}^o\cap B({\mathbf {z}}^*,\rho ^*)$, and for all $Q\in {\mathcal {Q}}^*$,

$$\begin{aligned} \left\| {\breve{{\mathbf {z}}}}^+_Q-\left( {\mathbf {z}}_Q+\left( \varDelta ^{\mathrm{N}}{\mathbf {z}}\right) _Q \right) \right\| \le c_5 \max \left\{ \left\| \varDelta ^{\mathrm{N}}{\mathbf {z}}\right\| ^2, \left\| {\mathbf {z}}-{\mathbf {z}}^* \right\| ^2 \right\} . \end{aligned}$$

(67)

Proof

Let ${\mathbf {z}}\in {\mathcal {F}}^o\cap B({\mathbf {z}}^*,\rho ^*)$ and $Q\in {\mathcal {Q}}^*$. It follows from (65), Lemmas 17–20, and (66) that

$$\begin{aligned} \begin{aligned} \left\| {\breve{{\mathbf {z}}}}^+_Q-\left( {\mathbf {z}}_Q+\left( \varDelta ^{\mathrm{N}}{\mathbf {z}}\right) _Q\right) \right\|&\le (1-\alpha ) \left\| \varDelta {\mathbf {z}}_Q\right\| +\left\| \varDelta {\mathbf {z}}^{\text {c}}_Q\right\| \\&\quad + \left\| \varDelta {\mathbf {z}}^{\text {a}}_Q - \varDelta {\mathbf {z}}^0_Q\right\| + \left\| \varDelta {\mathbf {z}}^0_Q-(\varDelta ^{\mathrm{N}}{\mathbf {z}})_Q\right\| \\&\le c_2\left\| \varDelta {\mathbf {z}}^{\text {a}}_Q\right\| \left\| \varDelta {\mathbf {z}}_Q\right\| + c_1 \left\| \varDelta {\mathbf {z}}^{\text {a}}_Q\right\| ^2 + c_3 \left\| {\mathbf {z}}-{\mathbf {z}}^*\right\| ^2 \\&\quad + c_4 \left\| {\mathbf {z}}-{\mathbf {z}}^*\right\| \left\| (\varDelta ^{\mathrm{N}}{\mathbf {z}})_Q\right\| \\&\le \left( c_2\left( 1+c_1\frac{\epsilon ^*}{4}\right) + c_1\right) \left\| \varDelta {\mathbf {z}}^{\text {a}}_Q \right\| ^2 + c_3 \left\| {\mathbf {z}}-{\mathbf {z}}^* \right\| ^2 \\&\quad + c_4 \left\| {\mathbf {z}}-{\mathbf {z}}^*\right\| \left\| \left( \varDelta ^{\mathrm{N}}{\mathbf {z}}\right) _Q\right\| . \end{aligned} \end{aligned}$$

Also, by Lemmas 19 and 20, we have

$$\begin{aligned} \begin{aligned} \left\| \varDelta {\mathbf {z}}^{\text {a}}_Q \right\|&\le \left\| \varDelta {\mathbf {z}}^{\text {a}}_Q-\varDelta {\mathbf {z}}^0_Q \right\| + \left\| \varDelta {\mathbf {z}}^0_Q-(\varDelta ^\mathrm{N}{\mathbf {z}})_Q \right\| + \left\| (\varDelta ^{\mathrm{N}}{\mathbf {z}})_Q \right\| \\&\le c_3 \left\| {\mathbf {z}}-{\mathbf {z}}^* \right\| ^2 + c_4\left\| {\mathbf {z}}-{\mathbf {z}}^* \right\| \left\| (\varDelta ^\mathrm{N}{\mathbf {z}})_Q \right\| + \left\| (\varDelta ^{\mathrm{N}}{\mathbf {z}})_Q \right\| . \end{aligned} \end{aligned}$$

(68)

The claim follows (in view of boundedness of ${\mathcal {F}}^o\cap B({\mathbf {z}}^*,\rho ^*)$). $\square$

With Proposition 4 established, we proceed to the second major step of the proof of Theorem 2: to show that (67) still holds when ${\mathbf {z}}^+$ is substituted for ${\breve{{\mathbf {z}}}}^+_Q$.

Proof of Theorem 2

Again, let $\rho ^*$ be as given in Lemma 16. Let ${\mathbf {z}}\in {\mathcal {F}}^o\cap B({\mathbf {z}}^*,\rho ^*)$ and $Q\in {\mathcal {Q}}^*$. Let $\rho :=\rho ^*$, ${\mathbf {t}}:={\mathbf {z}}$, and ${\mathbf {t}}^*:={\mathbf {z}}^*$. Then the desired q-quadratic convergence is a direct consequence of Lemma 14, provided that the condition (64) is satisfied. Hence, we now show that there exists some constant $c>0$ such that, for each $i\in {\mathbf {m}}$,

$$\begin{aligned} \min \left\{ \left| z_i^+-z_i^*\right| ,\left| z_i^+-(z_i+(\varDelta ^{\mathrm{N}} z)_i)\right| \right\} \le c \,\max \left\{ \left\| \varDelta ^{\mathrm{N}}{\mathbf {z}}\right\| ^2,\left\| {\mathbf {z}}-{\mathbf {z}}^*\right\| ^2 \right\} . \end{aligned}$$

(69)

As per Proposition 4, (69) holds for $i\in Q$ with $z_i^+$ replaced with ${\breve{\text {z}}}_i^+$. In particular, (69) holds for the ${\mathbf {x}}^+$ components of ${\mathbf {z}}^+$. It remains to show that (69) holds for the $\varvec{\lambda }^+$ components of ${\mathbf {z}}^+$. Firstly, for all $i\in {\mathcal {A}}({\mathbf {x}}^*)$, we show that $\lambda _i^+={\breve{\lambda }}_i^+$, thus (69) holds for all $\lambda ^+_i$ such that $i\in {\mathcal {A}}({\mathbf {x}}^*)$ by Proposition 4. From the fact that $\varvec{\lambda }>{\mathbf {0}}$ (${\mathbf {z}}\in {\mathcal {F}}^o$) and Lemma 16(ii), and since $\nu \ge 2$, it follows that

$$\begin{aligned} \chi :=\Vert \varDelta {\mathbf {x}}^\text {a}\Vert ^\nu + \Vert [{\tilde{\varvec{\lambda }}}^{\text {a},+}_Q]_-\Vert ^\nu \le \Vert \varDelta {\mathbf {x}}^\text {a}\Vert ^\nu + \Vert \varDelta \varvec{\lambda }^{\text {a}}_Q\Vert ^\nu \le 2\left( \frac{\epsilon ^*}{4}\right) ^\nu \le \frac{\epsilon ^*}{2}, \end{aligned}$$

(70)

so that $\min \{\chi ,{{\underline{\lambda }}}\}\le \epsilon ^*/2$. Also, from Lemma 16(iii) and the fact that ${\breve{\varvec{\lambda }}}_Q^+$ is a convex combination of $\varvec{\lambda }_Q$ and ${\tilde{\varvec{\lambda }}}_Q^+$, we have, for all $i\in {\mathcal {A}}({\mathbf {x}}^*)$,

$$\begin{aligned} \frac{\epsilon ^*}{2}< \min \{\lambda _i,{\tilde{\lambda }}_i^+\}\le {\breve{\lambda }}_i^+. \end{aligned}$$

(71)

Hence, from (70), (71), Lemma 16(iv), and (26), we conclude that $\lambda _i^+={\breve{\lambda }}_i^+$ for all $i\in {\mathcal {A}}({\mathbf {x}}^*)$. Secondly, we prove that there exists $d_1>0$ such that

$$\begin{aligned} \left\| \varvec{\lambda }_{Q\setminus {\mathcal {A}}({\mathbf {x}}^*)}^+ \right\|= & {} \left\| \varvec{\lambda }_{Q\setminus {\mathcal {A}}({\mathbf {x}}^*)}^+ - \varvec{\lambda }_{Q\setminus {\mathcal {A}}({\mathbf {x}}^*)}^* \right\| \nonumber \\\le & {} d_1\max \left\{ \Vert \varDelta ^\mathrm{N}{\mathbf {z}}\Vert ^2,\Vert {\mathbf {z}}-{\mathbf {z}}^*\Vert ^2 \right\} \quad \forall i\in Q\setminus {\mathcal {A}}({\mathbf {x}}^*), \end{aligned}$$

(72)

thus establishing (69) for $\lambda _i^+$ with $i\in Q\setminus {\mathcal {A}}({\mathbf {x}}^*)$. For $i\in Q\setminus {\mathcal {A}}({\mathbf {x}}^*)$, we know from (26) that, either $\lambda _i^+=\min \{\lambda ^{\max }, {\breve{\lambda }}_i^+\}$, or $\lambda _i^+ = \min \{{\underline{\lambda }},\Vert \varDelta {\mathbf {x}}^\text {a}\Vert ^\nu +\Vert [{\tilde{\varvec{\lambda }}}^{\text {a},+}_Q]_-\Vert ^\nu \}$. In the former case, we have

$$\begin{aligned} \begin{aligned} \left| \lambda _i^+ \right| \le \left| {\breve{\lambda }}_i^+\right| = \left| {\breve{\lambda }}_i^+-\lambda _i^*\right|&\le \left| {\breve{\lambda }}_i^+-(\lambda _i+\left( \varDelta ^{\mathrm{N}}\lambda \right) _i)\right| + \left| (\lambda _i+\left( \varDelta ^{\mathrm{N}}\lambda \right) _i)-\lambda _i^*\right| \\&\le d_2\max \left\{ \Vert \varDelta ^{\mathrm{N}}{\mathbf {z}}\Vert ^2,\Vert {\mathbf {z}}-{\mathbf {z}}^*\Vert ^2\right\} + d_3\Vert {\mathbf {z}}-{\mathbf {z}}^*\Vert ^2, \end{aligned} \end{aligned}$$

for some $d_2>0$, $d_3>0$. Here the last inequality follows from Proposition 4 and the quadratic rate of the Newton step given in Lemma 14. In the latter case, since $\varvec{\lambda }>{\mathbf {0}}$, we obtain

$$\begin{aligned} |\lambda _i^+|\le & {} \Vert \varDelta {\mathbf {x}}^\text {a}\Vert ^\nu + \Vert [[{\tilde{\varvec{\lambda }}}^{\text {a},+}_Q]_-\Vert ^\nu \le \Vert \varDelta {\mathbf {x}}^\text {a}\Vert ^\nu + \Vert \varDelta \varvec{\lambda }^{\text {a}}_Q\Vert ^\nu \nonumber \\= & {} \Vert \varDelta {\mathbf {z}}^{\text {a}}_Q\Vert ^\nu \le d_4\max \left\{ \Vert \varDelta ^\mathrm{N}{\mathbf {z}}\Vert ^2,\Vert {\mathbf {z}}-{\mathbf {z}}^*\Vert ^2\right\} , \end{aligned}$$

(73)

for some $d_4>0$. Here the equality is from the definition of $\varDelta {\mathbf {z}}^{\text {a}}$ and the last inequality follows from $\nu \ge 2$, (68), and boundedness of ${\mathcal {F}}^o\cap B({\mathbf {z}}^*,\rho ^*)$. Hence, we have established (72). Thirdly and finally, consider the case that $i\in Q^{\text {c}}$. Since ${\mathcal {A}}({\mathbf {x}}^*)\subseteq Q$, $\varvec{\lambda }_{Q^{\text {c}}}^*={\mathbf {0}}$ and it follows from (27) that, either $\lambda _i^+=\min \{\lambda ^{\max }, {\mu _{(Q)}^+}/{s_i^+}\}$, or $\lambda _i^+ = \min \{{\underline{\lambda }},\Vert \varDelta {\mathbf {x}}^\text {a}\Vert ^\nu +\Vert [{\tilde{\varvec{\lambda }}}^\text {a}_Q]_-\Vert ^\nu \}$. In the latter case, the bound in (73) follows. In the former case, we have

$$\begin{aligned} |\lambda _i^+ - \lambda _i^*| = |\lambda _i^+| \le {\mu _{(Q)}^+}/{s_i^+} . \end{aligned}$$

By definition, $s_i^+:=s_i+\alpha _\text {p}\varDelta s_i$ is a convex combination of $s_i$ and $\tilde{s}_i^+$. Thus, Lemma 16(iii) gives that $s_i^+\ge \min \{s_i,\tilde{s}_i^+\}>\epsilon ^*/2$. Then using the definition of $\mu _{(Q)}^+$ (see Step 10 of Algorithm CR-MPC) leads to

$$\begin{aligned} |\lambda _i^+ - \lambda _i^*| \le {\left\{ \begin{array}{ll} \frac{2}{\epsilon ^*|Q|} \left( \left( {\mathbf {s}}_{{\mathcal {A}}({\mathbf {x}}^*)}^+\right) ^T \left( \varvec{\lambda }_{{\mathcal {A}}({\mathbf {x}}^*)}^+\right) + \left( {\mathbf {s}}_{Q\setminus {\mathcal {A}}({\mathbf {x}}^*)}^+\right) ^T \left( \varvec{\lambda }_{Q\setminus {\mathcal {A}}({\mathbf {x}}^*)}^+\right) \right) , &{} \quad \text {if } |Q|\ne 0\\ 0, &{} \quad \text {otherwise} \end{array}\right. }. \end{aligned}$$

Since ${\mathbf {z}}\in B({\mathbf {z}}^*,\rho ^*)$, $\varvec{\lambda }_{{\mathcal {A}}({\mathbf {x}}^*)}^+$ and ${\mathbf {s}}_{Q\setminus {\mathcal {A}}({\mathbf {x}}^*)}^+$ are bounded by Lemma 16(ii). Also, by definition, ${\mathbf {s}}_{{\mathcal {A}}({\mathbf {x}}^*)}^*={\mathbf {0}}$. Thus there exist $d_5>0$ and $d_6>0$ such that

$$\begin{aligned} \left| \lambda _i^+ - \lambda _i^*\right| \le d_5\left\| {\mathbf {s}}_{{\mathcal {A}}({\mathbf {x}}^*)}^+-{\mathbf {s}}_{{\mathcal {A}}({\mathbf {x}}^*)}^*\right\| + d_6 \left\| \varvec{\lambda }_{Q\setminus {\mathcal {A}}({\mathbf {x}}^*)}^+\right\| . \end{aligned}$$

Having already established that the second term is bounded by the right-hand side of (72), and we are left to prove that the first term also is. By definition,

$$\begin{aligned} \left\| {\mathbf {s}}_{{\mathcal {A}}({\mathbf {x}}^*)}^+ - {\mathbf {s}}_{{\mathcal {A}}({\mathbf {x}}^*)}^*\right\| = \left\| A_{{\mathcal {A}}({\mathbf {x}}^*)}{\mathbf {x}}^+ - A_{{\mathcal {A}}({\mathbf {x}}^*)}{\mathbf {x}}^* \right\| \le \left\| A{\mathbf {x}}^+ - A{\mathbf {x}}^* \right\| \le \Vert A\Vert \left\| {\breve{{\mathbf {z}}}}_Q^+ - {\mathbf {z}}_Q^* \right\| . \end{aligned}$$

Applying Proposition 4 and Lemma 14, we get

$$\begin{aligned} \begin{aligned} \left\| {\mathbf {s}}_{{\mathcal {A}}({\mathbf {x}}^*)}^+ - {\mathbf {s}}_{{\mathcal {A}}({\mathbf {x}}^*)}^*\right\|&\le \Vert A\Vert \left\| {\breve{{\mathbf {z}}}}_Q^+ - \left( {\mathbf {z}}_Q+\left( \varDelta ^{\mathrm{N}}{\mathbf {z}}\right) _Q\right) \right\| + \Vert A\Vert \left\| \left( {\mathbf {z}}_Q+\left( \varDelta ^{\mathrm{N}}{\mathbf {z}}\right) _Q\right) - {\mathbf {z}}_Q^*\right\| \\&\le d_7 \max \left\{ \Vert \varDelta ^{\mathrm{N}}{\mathbf {z}}\Vert ^2, \left\| {\mathbf {z}}-{\mathbf {z}}^*\right\| ^2\right\} + d_8 \left\| {\mathbf {z}}- {\mathbf {z}}^* \right\| ^2, \end{aligned} \end{aligned}$$

for some $d_7>0$, $d_8>0$. Hence, we established (69) for all $i\in {\mathbf {m}}$, thus proving the q-quadratic convergence rate. $\square$