On the augmented Lagrangian dual for integer programming

Abstract

We consider the augmented Lagrangian dual for integer programming, and provide a primal characterization of the resulting bound. As a corollary, we obtain proof that the augmented Lagrangian is a strong dual for integer programming. We are able to show that the penalty parameter applied to the augmented Lagrangian term may be placed at a fixed, large value and still obtain strong duality for pure integer programs.

Appendix

We will prove the cases of Lemma 1 separately. In all cases the orthogonal decomposition follows from Proposition 1, so we do not explicitly mention this again. We assume throughout that $z^{LP}<+\infty$. We precede these cases by showing the local uniform boundedness of $\kappa \mapsto {C} (\kappa )$ for all cases.

Proof

local uniform boundedness for all $0 \le \kappa <\overline{\kappa }$

We prove the local uniform boundeness of $\kappa \mapsto P_{V^{\circ }} K (\kappa )$ for the general case using an arbitrary norm $\Vert \cdot \Vert$ in the penalty. Denote the Euclidean norm by $\Vert \cdot \Vert _2$. As the constraint $Q(\kappa ) :=\left\{ y\mid \left\| b-Ay\right\| \le \kappa \right\}$ only allows $y$ to deviate from $\{x\mid Ax =b \}$ by a component in $\mathrm{Range} A^T$ of maximum norm $\Vert A^T (A A^T )^{-1} (Ay-b) \Vert _2 \le \Vert A^T (A A^T )^{-1} \Vert _2 \Vert Ay - b \Vert _2 \le \Vert A^T (A A^T )^{-1} \Vert _2 \frac{1}{\gamma } {\kappa }$ (we have invoked the equivalence of norms). Let $\eta := \frac{1}{\gamma }\Vert A^T (A A^T )^{-1} \Vert _2$. As $K(\kappa ) \subseteq X^{LP} \cap Q (\kappa )$ for all $\overline{\kappa } \ge \kappa \ge 0$ using [35, Corollary 8.3.3], $0^{+} K(\kappa ) \subseteq 0^{+} X^{LP} \cap \mathrm{ker} A:=V$ we have

$$\begin{aligned} P_{V^{\circ }} K(\kappa ) \subseteq P_{V^{\circ }} (X^{LP} \cap Q (\overline{\kappa })) \subseteq P_{V^{\circ }} (X^{LP} \cap Q (0)) +\overline{B}_{\eta \overline{\kappa }}(0)\,{:=}\,C . \end{aligned}$$

(12)

As $X_0:=X^{LP} \cap Q (0)$ is a polyhedral convex set with a recession cone $V$ and lineality space $L:=(-V)\cap V$ we may express $X_0 = \mathrm{conv}( \mathrm{ext} (X_0)) +V$ where $\mathrm{ext}$ denotes the finite set of extremal points of the convex set $X_0 \cap L^{\perp }$. Thus $P_{V^{\circ }} (X^{LP} \cap Q (0))\subseteq P_{V^{\circ }} ( \mathrm{conv}( \mathrm{ext} (X_0 ))$ which is a bounded set. Hence the right hand side set $C$ in (12) is closed and uniformly bound for all $\overline{\kappa } \ge \kappa \ge 0$).$\square$

Proof

[Lemma 1: Case 2 of Assumption 2]

In order to prove this case of Lemma 1, we follow the approach of Meyer [33]. Recall Case 2) of Assumption 2 is that $A$ and $D$ have rational entries and the penalty function uses the $\infty$-norm. Thus we investigate the continuity of $\kappa \mapsto K_{\infty } (\kappa )$ where

$$\begin{aligned} \left\| b- Ax \right\| _\infty \le \kappa \Leftrightarrow x\in Q_{\infty } (\kappa ) := \{ y \mid b - \kappa \mathbf{1} \le Ay \le b + \kappa \mathbf{1}\}, \end{aligned}$$

for $\mathbf 1$ the vector of all ones. In this case, $K_\infty ^{LP}\left( \kappa \right) :=X^{LP} \cap \, Q_{\infty } ( \kappa )$ is a polyhedron with rational constraint matrices. By Theorem 2.19 of [10] (see in particular the subsequent Remark 2.20; see also [33]), and since $K_\infty (\kappa ) \,{:=}\,\mathrm{conv} (X \cap Q_{\infty } ( \kappa ) )\supseteq K(0)$ is non-empty by our standing assumption that the IP is feasible, it must be that $K_\infty (\kappa )$ is a polyhedron with recession cone coinciding with that of $K_\infty ^{LP}\left( \kappa \right)$. Observe that the recession cone of $K_\infty ^{LP}\left( \kappa \right)$ is given by

$$\begin{aligned} 0^+ K_\infty ^{LP}\left( \kappa \right)&= 0^+\{x \in X^{LP} \mid b -\kappa \mathbf{1} \le Ax \le b + \kappa \mathbf{1} \} \\&= \{r \in 0^{+} X^{LP} \mid Ar = 0 \} = \{r \ge 0 \mid Dr = 0, \, Ar = 0 \}\\&= 0^{+} X^{LP} \cap \mathrm{ker} A =V \end{aligned}$$

independent of $\kappa$. We have thus shown that for all $\kappa \ge 0$, $K_\infty (\kappa )$ is a polyhedron with recession cone precisely $V$.

Suppose initially that we have a pure integer program, i.e. $|J|=n$. Choose a positive constant $M$ such that $\tilde{A} := M A$ is an integer matrix. Then we have (where $\lfloor \cdot \rfloor$ rounds down and $\lceil \cdot \rceil$ rounds up)

$$\begin{aligned} K_{\infty } (\kappa ) =\mathrm{conv} \{ x \in X \mid \lceil M(b - \kappa \mathbf{1}) \rceil \le \tilde{A} x \le \lfloor M(b+ \kappa \mathbf{1} )\rfloor \}. \end{aligned}$$

Considering $\kappa _q \downarrow \kappa$, we see that there must exist $\bar{q}$ such that for all $q \ge \bar{q}$ the lower and upper bounds on $\tilde{A}x$ in the definition of $K_\infty (\kappa _q)$ are precisely those in the definition of $K_\infty (\kappa )$, i.e. $\lceil M(b - \kappa _q \mathbf{1} )\rceil = \lceil M(b - \kappa \mathbf{1}) \rceil$ and $\lfloor M(b+ \kappa _q \mathbf{1}) \rfloor = \lfloor M(b+ \kappa \mathbf{1} )\rfloor$, and hence $K_\infty (\kappa _q ) = K_\infty (\kappa )$. Continuity follows.

For the mixed integer case we proceed as in [33] Lemma 3.8 amd Theorem 3.9. Without loss of generality suppose the integer variables are those with smallest indices, i.e. $J=\{1,\ldots ,n_1\}$, and write $A = [A_1 \ A_2]$ where $A_1=A_J$ is the matrix formed from columns associated with $J$, and similarly $D = [D_1 \ D_2]$ where $D_1=D_J$. Observe $K_\infty (\kappa )$ has the form

$$\begin{aligned} K_\infty (\kappa )&= \mathrm{conv}(\{ (x,y) \in \mathbf {Z}^{n_1}_+\times \mathbf {R}^{n-n_1} \mid b - \kappa \mathbf{1} \le A_1x + A_2 y \\&\qquad \quad \,\,\le b + \kappa \mathbf{1} d \le D_1 x + D_2 y \le d 0 \le y \}) \end{aligned}$$

Making use of the theory of linear programming, we call the pair $(B,r)$ a basis if $B = [B_1 B_2]$ is a subset of the rows of

$$\begin{aligned} \left[ \begin{array}{c@{\quad }c} A_1 &{} A_2 \\ D_1 &{} D_2 \\ 0 &{} I \end{array} \right] \end{aligned}$$

chosen so that $B_2$ is nonsingular and $r$ is the corresponding subvector of

$$\begin{aligned} \left[ \begin{array}{c} b \pm \kappa \mathbf{1} \\ d \\ 0 \end{array} \right] , \end{aligned}$$

i.e. if the $j$th row of $B$ corresponds to row $i$ in the original matrix with $i \le m$, then $r_j \in \{b_i - \kappa , b_i + \kappa \}$. Let $\mathcal {B}$ denote the set of all such bases. Define $S_{\infty } (\kappa ) := \{ (x,y) \in X \mid b - \kappa \mathbf{1} \le A_1 x + A_2 y \le b + \kappa \mathbf{1} \}$ and for basis $(B,r)$ define $S_{\infty }(\kappa )(B,r) := \{ (x,y) \mid (x,y) \in S_{\infty } (\kappa ), \, B_1x + B_2 y = r \}$. Then we have

$$\begin{aligned} S_{\infty }(\kappa )(B,r)&= \{ (x,y) \in \mathbf {Z}^{n_1}_+\times \mathbf {R}^{n-n_1} \mid B_1x + B_{2}y = r,\, y\ge 0,\, D_1x + D_2y = d \\&\text { and } b - \kappa \mathbf{1} \le A_1x + A_2y \le b + \kappa \mathbf{1} \}. \end{aligned}$$

Now define the projection onto the integer variables $S_{\infty }(\kappa )(B,r;J) := \{ x\in \mathbf {Z}^{n_1}_+ \mid (x,y) \in S_{\infty }(\kappa )(B,r)\}$ and observe that

$$\begin{aligned} \begin{aligned} S_{\infty }(\kappa )(B,r;J)&= \{ x\in \mathbf {Z}^{n_1}_+ \mid B_2^{-1} r - B_2^{-1} B_1x \ge 0, (D_1 - D_2B_2^{-1}B_1)x \\&= d - D_2B_2^{-1}r b - \kappa \mathbf{1} - A_2B_2^{-1}r\\&\le (A_1 - A_2B_2^{-1}B_1)x \le b + \kappa \mathbf{1} - A_2B_2^{-1}r \}. \end{aligned} \end{aligned}$$

This is a pure integer feasible set defined by constraints with a rational constraint matrix, $G(B)$, say, independent of $\kappa$ (and $r$), and a vector $g(\kappa ;B,r)$ such that $S_{\infty }(\kappa )(B,r;J) = \{ x\in \mathbf {Z}^{n_1}_+ \mid G(B)x \le g(\kappa ;B,r)\}$. By Corollary 3.5 of [33], $S_{\infty }(\kappa )(B,r;J)$ must have a finite number of extreme points. Observe that $g(\kappa ;B,r)$ has the form

$$\begin{aligned} g(\kappa ;B,r) = \left[ \begin{array}{c} B_2^{-1}r \\ d - D_2B_2^{-1}r \\ -d + D_2B_2^{-1}r \\ b+ \kappa \mathbf{1} - A_2B_2^{-1}r \\ -b + \kappa \mathbf{1} + A_2B_2^{-1}r\end{array}\right] . \end{aligned}$$

Furthermore there is a positive integer $M$ such that $S_{\infty }(\kappa )(B,r;J) = \{ x\in \mathbf {Z}^{n_1}_+ \mid MG(B)x \le \lfloor Mg(\kappa ;B,r) \rfloor \}$. Thus for small $\kappa < \kappa _{B,r}$, $S_{\infty }(\kappa )(B,r;J)$ is a constant set independent of $\kappa$; in fact for sufficiently small $\kappa$, $S_{\infty }(\kappa )(B,r;J) = S_{\infty }(0)(B,r;J)$. Hence $\mathrm{conv} (S_{\infty }(\kappa )(B,r;J))$ is a fixed polyhedral set for small $\kappa < \kappa _{B,r}$. Indeed for such small $\kappa$ it must be that $\mathrm{conv} (S_{\infty }(\kappa )(B,r;J)) = \mathcal {F}(B) := \{x \mid \ F(B)x \le f(B) \}$ for some rational matrix $F(B)$ and rational vector $f(B)$ (by well known results, e.g. in [34], since $MG(B)$ and $\lfloor Mg(0;B,r) \rfloor$ are integer).

Now we claim that

$$\begin{aligned} \mathrm{conv} (S_{\infty }(\kappa )(B,r)) \!=\! \{ (x,y) \mid x \in \mathrm{conv} (S_{\infty }(\kappa )(B,r;J)), y\!=\! B_2^{-1}r - B_2^{-1}B_1x \}.\nonumber \\ \end{aligned}$$

(13)

Containment of $\mathrm{conv} (S_{\infty }(\kappa )(B,r))$ in the right-hand side follows easily from first principles. Any extreme point $(x,y)$ of the right-hand side set must by Lemma 3.3 of [33] have $x$ an extreme point of $\mathrm{conv} (S_{\infty }(\kappa )(B,r;J))$, and hence $x \in S_{\infty }(\kappa )(B,r;J)$ (by Lemma 3.1 of [33]). Denote the finite set of extremal points of $\mathrm{conv} (S_{\infty }(\kappa )(B,r;J))$ by $\mathcal {F}(B)$. It follows (by the definition of $y$ in the right-hand side) that $(x,y)\in S_\infty (\kappa )(B,r) \subset \mathrm{conv} (S_{\infty }(\kappa )(B,r))$ as required (if the latter set contains all extreme points of the right-hand side set then it’s convex hull contains that set).

Thus for $\kappa < \kappa _{B,r}$ it must be that $\mathrm{conv} (S_{\infty }(\kappa )(B,r))$ can be represented as a polyhedral set dependent on $\kappa$ only via the continuous variables (recall $r$ may depend on $\kappa$):

$$\begin{aligned} \mathrm{conv} (S_{\infty }(\kappa )(B,r)) = \{ (x, y) \mid x \in \mathcal {F}(B),\, B_1x + B_2y =r\}. \end{aligned}$$

(14)

By Theorem 3.9 of [33] we have $K_{\infty } (\kappa )= \mathrm{conv}\left( \bigcup _{(B,r)\in \mathcal {B}} \mathrm{conv} (S_{\infty }(\kappa )(B,r))\right)$+$V$ and as $\mathcal {B}$ has a finite cardinality we deduce that $K_{\infty } (\kappa )$ is a polyhedral set that only depends on $\kappa$ via the continuous variables for $0< \kappa < \min \{ \kappa _{B,r} \mid (B,r) \in \mathcal {B} \}:=\bar{\kappa }$. For each $x$ the system of equations in (14) yields a continuous convex, polyhedral valued multifunction of $\kappa$ (apply the Hoffman’s bound [41]). The upper semi-continuity of $K_{\infty } (\kappa )$ follows from the classical facts that finite intersections and finite unions preserves upper semi-continuity as does convex hulls and the addition of $V$.$\square$

For Case 1) of Assumption 2 we need another lemma. Denote the complement of a set $R$ by $R^c$. In a set that has a pointed recession cone the contribution of integral points to a convex combination must reduce as they become more distant from the point represented.

Lemma 2

Suppose $K$ is a convex set with $0^{+} K$ a pointed cone. Let $J$ denote a subset of indices $\{1,\ldots ,n\}$ to be constrained to integral values in the mixed variable set $I :=\{x \in \mathbf {R}^n \mid x_i \in \mathbf {Z}, \, i \in J \}$. Let $\tilde{c}\in \mathrm{int} (0^{+} K)^{\circ }$ so that $K \cap R(\alpha )$ is bounded for all $\alpha$, where $R(\alpha ) := \{ x \mid \tilde{c} x \ge \alpha \}$. Then for all $\varepsilon > 0$, $\alpha \in \mathbf {R}$ there exists fixed numbers $N_{\varepsilon } \in \mathbf {N}$, $\alpha _{\varepsilon }\in \mathbf {R}$, such that for all $x \in \overline{\mathrm{conv}} ( K \cap I ) \cap R(\alpha )$ there exists multipliers $\lambda _i \ge 0$, $\sum _{i=1}^{\infty } \lambda _i =1$ such that $x = \sum _{i=1}^{\infty } \lambda _i x^i$, for $x^i \in K \cap I$ and for which $\lambda _{\varepsilon }:= \sum _{i=1}^{N_{\varepsilon }} \lambda _i \ge 1- \varepsilon$, $x_1 := \sum _{i=1}^{N_{\varepsilon }} \frac{\lambda _i}{\lambda _{\varepsilon } }x^i \in \mathrm{conv} ( K \cap R(\alpha _{\varepsilon } ) \cap I )$.

Proof

Let $x \in \overline{\mathrm{conv}} ( K \cap I ) \cap R(\alpha )$ be arbitrary. We first show that $\alpha _{\varepsilon }$ can be chosen so that we must have $\lambda _{\varepsilon } \ge 1- \varepsilon$ whenever $x = \lambda _{\varepsilon } x_1 + (1-\lambda _{\varepsilon }) x_2$ where $x_1 \in \mathrm{conv} ( K \cap R(\alpha _{\varepsilon } ) \cap I )$ and $x_2 \in \overline{\mathrm{conv}} ( K \cap R(\alpha _{\varepsilon } )^c \cap I ):=KI(\alpha _{\varepsilon })$. This follows easily from the fact that $x$ is constrained to a bounded region and that the recession cone of $K$ is pointed. [We argue by contradiction: if the proposition fails then there exists $\alpha$ and $\varepsilon >0$ and $x^k \in \overline{\mathrm{conv}} ( K \cap I ) \cap R(\alpha )$ with $x^k = \lambda ^k x_1^k + (1-\lambda ^k) x_2^k$ and $x^k_2 \in KI(\alpha ^k)$, $\Vert x_2^k \Vert \rightarrow + \infty$ as $|\alpha ^k| \rightarrow \infty$ with $\varepsilon < 1 - \lambda ^k$. Consequently we have $(1- \lambda ^k ) \frac{x_2^k}{\Vert x_2^k \Vert } = \frac{x}{\Vert x_2^k \Vert } - \lambda ^k \frac{x_1^k}{\Vert x_2^k \Vert }$ for all $k$. Taking convergent subsequences so that $\frac{x_2^k}{\Vert x_2^k \Vert } \rightarrow \hat{x}_2$, $\frac{x_1^k}{\Vert x_2^k \Vert } \rightarrow \hat{x}_1$, $\lambda ^k \rightarrow \lambda$ then $0\le \lambda <1$ and $(1-\lambda ) \hat{x}_2 = - \lambda \hat{x}_1$ with $\Vert \hat{x}_2 \Vert =1$ (implying $\lambda \ne 0$) and so $-\hat{x}_2, \hat{x}_2 \in 0^+ K$, a contradiction]. Having fixed $\alpha _{\varepsilon }$ we may consider whether there is a fixed number of components in the convex combination required to construct $x_1$. Note $K \cap R(\alpha _{\varepsilon } ) \cap I$ contains a finite number of integrally constrained values, i.e. the projection of $K \cap R(\alpha _{\varepsilon } ) \cap I$ onto the variables indexed by $J$ has a finite number of elements, say $M_{\varepsilon }$. It is not hard to show that since $K$ is convex, at most one component having a given integer subvector in this finite set is needed in the convex combination representing $x_1$. [To see why, take any convex combination of points in $K \cap R(\alpha _{\varepsilon } ) \cap I$ yielding $x_1$, and gather together the terms for each distinct integer subvector. Since $K \cap R(\alpha )$ is convex, the terms with the same integer subvector must be a convex combination of points in a convex set and hence can be represented by a single point in this set.] Thus we may take $N_{\varepsilon } := M_{\varepsilon }$.$\square$

Proof

(Lemma 1: Cases 1 and 3 of Assumption 2) Now assume that Case 1 of Assumption 2 holds, i.e. that the LP relaxation does not have a lineality space in its solution set. Note that $0^{+} K(\kappa ) \subseteq 0^{+} X^{LP} \cap \mathrm{ker} A = V$, a pointed cone. Let $\tilde{c} \in \mathrm{int}(V^{\circ })$ so that for all $\alpha$ the region $R(\alpha ) \cap X^{LP} \cap Q(\overline{\kappa } )$ is convex and bounded, where $\{x \mid \tilde{c} x \ge \alpha \}:=R(\alpha )$. Because $K(\kappa ^{\prime } ) \subseteq K(\kappa )$ for all $\kappa ^{\prime } \le \kappa$, to study graph closure of $\kappa \mapsto \overline{C(\kappa )}$ we need to establish for all $\alpha$ that $\limsup _q K (\kappa _q ) \cap R (\alpha ) \subseteq K (\kappa )$ for $\kappa _q \downarrow \kappa$. If this is so then $\{K(\kappa _q ) \}$ converges to $K ( \kappa )$ (see [37, Theorem 4.10]). It suffices to consider $x_q \in K (\kappa _q ) \cap R (\alpha )$ for a fixed $\alpha$ with $x_q \rightarrow x \in R(\alpha ) \cap X^{LP} \cap Q(\overline{\kappa } )$. Denoting $y_q = P_{V^{\circ }}(x_q)$ we then we wish to establish that $\lim _q y_q := y \in \overline{C(\kappa ) }$.

Now consider $\kappa _q \downarrow \kappa$. As $x_q \in K (\kappa _q )$ applying Lemma 2, for all $\varepsilon >0$ there exists $\alpha _{\varepsilon }$ and $N_{\varepsilon }$ with the following properties. We have $x_q = \sum _i \lambda ^{q}_i x_q^i$ for $x_q^i \in X$ with $\sum _i \lambda _i^q =1$, $\lambda _q^i \ge 0$ and $\Vert b - Ax_q^i \Vert \le \kappa _q$. Also $\lambda ^q := \sum _{i=1}^{N_{\varepsilon }} \lambda _i^q \ge 1- \varepsilon$, $x_1^q:= \sum _{i=1}^{N_{\varepsilon }} \frac{\lambda _i^q}{\lambda ^q} x_q^i$, $x_2^q := \sum _{i>N_{\varepsilon }}\frac{\lambda _i^q}{1-\lambda ^q} x_q^i$ then $x_q = \lambda ^q x_1^q+ (1-\lambda ^q ) x_2^q \rightarrow x$ and

$$\begin{aligned} x_1^q \in \mathrm{conv} ( X \cap R( \alpha _{\varepsilon }) \cap Q(\kappa _q ). \end{aligned}$$

Taking successive subsequences we may also (after a renumbering of these subsequences) assume $x_q^i \rightarrow x^i$ for all $i=1,\ldots ,N_{\varepsilon }$. As $y_q$, $P_{V^{\circ }} x_2^q$, $P_{V^{\circ }} x_1^q\in C$ are uniformly bounded we may take further subsequences (and after renumbering) assume $y_q \rightarrow y$, $\lim _q x_1^q =x_1$ and as $\lambda ^q \ge 1-\varepsilon$ we have $y_q = \lambda ^q P_{V^{\circ }} x_1^q + (1-\lambda ^q ) P_{V^{\circ }} x_2^q \rightarrow y \in P_{V^{\circ }} x_1 + B_{\varepsilon 2 M} (0).$ By the finiteness of $N_{\varepsilon }$ there exists $\varepsilon > 0$ such that $K( \kappa + \varepsilon ) \cap R(\alpha _{\varepsilon })$ contains no new integral points in the sense that if the $j$th component $(x_q^i)_j \in \mathbf {Z}$ is part of the integrality constraint we have $(x_q^i)_j =(x^i)_j$ for $\kappa _q < \kappa + \varepsilon$. On fixing these components to their limiting values and denoting by $A_{n-J}$ the matrix with derived from $A$ by deleting all columns in $A_J$ associated with the integrally constrained components (indexed by $J$) and $b^i := b-(A_J (x^i)_J)$, where $(x_q^i )_J$ are the fixed integrally constrained components $(x^i)_J$. We then have all the continuous variable constrained to satisfy $\Vert b^i - A_{n-J} (x_q^i)_{n-J} \Vert \le \kappa _q$ for all $i=1,\ldots ,N_{\varepsilon }$, where $(x_q^i)_{n-J}$ denotes the vector obtained from $x_q^i$ by deleting all the components indexed in $J$. Denote by $g^i(y) : =\Vert b^i - A_{n-J} y \Vert$ and $S^i (\kappa ) := \{ y \mid g^i(y) \le \kappa \text{ and } ((x^i)_J, y) \in X \cap R(\alpha _{\varepsilon } ) \}$. Then the Hoffman bound [41] supplies a constant $r^i>0$ such that $d( (x_q^i)_{n-J},S^i (\kappa )) \le r^i [ g^i( (x_q^i)_{n-J})-\kappa ]_+ \le r^i [\kappa _q- \kappa ]$. Thus in the limit we have $\kappa _q \downarrow \kappa$ and $(x^i)_{n-J}\in S^i (\kappa )$. Hence for all $i=1,\ldots ,N_{\varepsilon }$ we have $\Vert b - Ax^i \Vert \le \kappa$ and $x_1 \in \mathrm{conv} [ {K( \kappa ) \cap R(\alpha _{\varepsilon } )} \cap Q(\kappa )]$ and so $y \in {P_{V^{\circ }} K(\kappa )} + B_{\varepsilon \eta } (0)$. As $\varepsilon >0$ was arbitrary we have $y \in \overline{P_{V^{\circ }} K(\kappa )}=\overline{C(\kappa )}$. The uniform boundedness and the closed graph property of $\kappa \mapsto \overline{C(\kappa )}$ allows the application of [1, Proposition 6.3.2] to obtain the upper semi-continuity result, thus completing our proof under Case 1) of Assumption 2.

The above arguments follow through in a simplified form when we assume $\mathrm{conv}(X)$ is bounded and hence contains a finite number of integrally constrained values. This establishes the result under Case 3) of Assumption 2.$\square$