Despite Example 5.4.5, a number of papers have established the equivalence between maximal S-free sets and minimalCGF’s, for various forms ofS. This equivalence is indeed known to hold in a number of situations:
(a) whenS is a finite set of points inZq− b; see [106] and more recently [68];
(b) whenS is the intersection ofZnwith an affine space; this was considered in [38] and [17]; (c) whenS = P ∩ (Zq− b) for some rational polyhedron P ; this was considered in [18, 68].
Accordingly, we investigate in this section the question: when does minimality imply strong min- imality? So we consider anS-free set V , whose smallest representation µV = σV•is minimal, hence V is asymptoticaly maximal (Theorem 5.4.10); we want to exhibit conditions under which V is max- imal. We denote byL = (−V∞)∩ V∞the lineality space ofV (the largest subspace contained in the closed convex coneV∞) and our result is the following.
Theorem 5.5.1. Suppose0∈ S := conv (S). A minimal µV is strongly minimal whenever one of the following two properties(i) and (ii) holds:
(i) V∞∩ S∞={0} (in particular S bounded), (ii) (ii)1 S = U + S∞ withU bounded, and
(ii)2 V∞∩ S∞= L∩ S∞.
This theorem generalizes the above-mentioned results:(i) is a weakening of (a) and (ii) weakens (b) or (c). Note that(ii)2 generalizes(i) (to an unbounded V∞∩ S); the price to pay is assumption (ii)1, whose role is to exclude an asymptotic behaviour ofS similar to that of P in Figure 5.5 (see Proposition 5.3.11).
However, the interesting point does not lie in the above assumptions (a) –(ii). Recalling that the whole issue lies in unboundedness ofV , our proof of Theorem 5.5.1 uses Theorem 5.4.10 as follows. Starting from an S-free set V which is asymptotically maximal but not maximal, we construct a sequence of neighborhoodsVk satisfying V∞k ) V∞. Then Vk is not S-free: there is some rk ∈ S∩ int (Vk); see Figure 5.8.Conforti, Cornu´Mathematics of Operations Research xx(x), pp. xxx–xxx, cejols, Daniilidis, Lemar´echal, Malick: Cut-generating functions
⃝200x INFORMS 19 V∞ rk Vk ∞ S
Figure 8: Constructing in S an unbounded sequence “tending to” V
Obtaining rk and uk is a fairly complicate operation, which we divide into a series of lemmas. For a reason that will appear in (39) below, we may assume 0 /∈ V•. Then we enlarge V to Vkby chopping off a bit of V•as follows. Take an extreme rayR
+dV of V∞◦. By (28), its intersection with V•is a nonempty
segment [dV, tVdV], with 1! tV < +∞. Given a positive integer k, we introduce the open neighborhood of [dV, tVdV]: Nk:= [d V, tVdV] + B ! 0,1 k " = # 1!t!tV B!tdV, 1 k " , (32)
where B(d, δ) is the open ball of center d and radius δ. We remove Nk from V•, thus obtaining a set C, closed hence compact; its convex hull
Gk:= conv C , with C := V•\Nk=$d∈ V• : ∥d − td V∥ " 1 k for all t∈ [1, tV] % (33) is convex compact. Figure 9 illustrates our construction.
V•
R+Gk
Nk
tVdV
dV
Figure 9: Chopping off V•near an extreme ray
Note for future use that the distance from every d ∈ [dV, tVdV] to C does not exceed 1/k; and the same holds for Gk⊃ C. Formally:
∀ ¯d∈ [dV, tVdV] , ∃dk∈ Gk such that ∥dk− ¯d∥ ! 1
k. (34)
Remark 5.2 The above construction would become substantially simpler, and Nk would reduce to the open ball B&dV,k1', if V•∩ R+dV reduced to a singleton, i.e. if tV = 1; but this property need not hold when σV is not continuous.
To make a counterexample, start from the parabola of Figure 5. We already know that σP(dk) can tend to any nonnegative value when dk → 0. However 0 ∈ P•; alternatively, the domain of σP is not closed. In fact, we need a discontinuous sublinear function which is locally bounded on its domain – and this requires three variables. Thus, we first bound σP by defining
f (d) := 1 + (
σP(d) if σP(d)! 1 , +∞ otherwise
(the “1+” appearing above is just aimed at getting 0 in the interior of V ). Although no longer positively
Figure 5.8: Constructing inS an unbounded sequence “tending to” V
Besides, our construction is organized in such a way thatVk“tends to”V and, by non-maximality ofV , rkis unbounded but “tends to”V . More precisely,
the cluster points of the normalized sequence{rk} lie in S∞∩ V∞.
Decomposingrk = `k + uk along L and L⊥, we also prove that uk is unbounded but “tends to” S∩ L⊥, more precisely
the cluster points of the normalized sequence{uk} lie in S
∞∩ L⊥.
We believe that these are key properties of non-maximalS-free sets. Having established them, the whole business is to find appropriate assumptions under which existence of our unbounded sequences is impossible; (a) – (ii) above are such ad hoc assumptions.
Obtaining rk anduk is a fairly complicate operation, which we divide into a series of lemmas. For a reason that will appear in (5.39) below, we may assume0 /∈ V•. Then we enlargeV to Vk by chopping off a bit ofV•as follows. Take an extreme rayR+dV ofV∞◦. By (5.28), its intersection with V•is a nonempty segment[dV, tVdV], with 16 tV < +∞. Given a positive integer k, we introduce the open neighborhood of[dV, tVdV]:
Nk := [dV, tVdV] + B 0,1 k = [ 16t6tV BtdV, 1 k , (5.32)
whereB(d, δ) is the open ball of center d and radius δ. We remove NkfromV•, thus obtaining a set C, closed hence compact; its convex hull
Gk := conv C , with C := V•\Nk =nd∈ V• : kd − tdVk > 1
k for allt∈ [1, tV] o
(5.33) is convex compact. Figure 5.9 illustrates our construction.
Conforti, Cornu´ejols, Daniilidis, Lemar´echal, Malick: Cut-generating functions
Mathematics of Operations Research xx(x), pp. xxx–xxx, c⃝200x INFORMS 19
V∞
rk
Vk ∞
S
Figure 8: Constructing in S an unbounded sequence “tending to” V
Obtaining rk and uk is a fairly complicate operation, which we divide into a series of lemmas. For a reason that will appear in (39) below, we may assume 0 /∈ V•. Then we enlarge V to Vkby chopping off a bit of V•as follows. Take an extreme rayR+dV of V∞◦. By (28), its intersection with V•is a nonempty
segment [dV, tVdV], with 1! tV < +∞. Given a positive integer k, we introduce the open neighborhood of [dV, tVdV]: Nk:= [dV, tVdV] + B!0,1 k " = # 1!t!tV B!tdV,1 k " , (32)
where B(d, δ) is the open ball of center d and radius δ. We remove Nkfrom V•, thus obtaining a set C, closed hence compact; its convex hull
Gk:= conv C , with C := V•\Nk=$d∈ V• : ∥d − tdV∥ " 1
k for all t∈ [1, tV] %
(33) is convex compact. Figure 9 illustrates our construction.
V•
R+Gk
Nk
tVdV
dV
Figure 9: Chopping off V•near an extreme ray
Note for future use that the distance from every d∈ [dV, tVdV] to C does not exceed 1/k; and the same holds for Gk⊃ C. Formally:
∀ ¯d∈ [dV, tVdV] , ∃dk∈ Gk such that ∥dk− ¯d∥ ! 1
k. (34)
Remark 5.2 The above construction would become substantially simpler, and Nk would reduce to the open ball B&dV,k1', if V•∩ R+dV reduced to a singleton, i.e. if tV = 1; but this property need not hold
when σV is not continuous.
To make a counterexample, start from the parabola of Figure 5. We already know that σP(dk) can tend to any nonnegative value when dk → 0. However 0 ∈ P•; alternatively, the domain of σP is not closed. In fact, we need a discontinuous sublinear function which is locally bounded on its domain – and this requires three variables. Thus, we first bound σP by defining
f (d) := 1 + (
σP(d) if σP(d)! 1 , +∞ otherwise
(the “1+” appearing above is just aimed at getting 0 in the interior of V ). Although no longer positively homogeneous, f is still convex, its domain is the compact convex set P•, on which 1! f ! 2; when
Figure 5.9: Chopping offV•near an extreme ray
Note for future use that the distance from everyd∈ [dV, tVdV] to C does not exceed 1/k; and the same holds forGk⊃ C. Formally:
∀ ¯d∈ [dV, tVdV] , ∃dk∈ Gk such that kdk− ¯dk 6 1
k. (5.34)
Remark 5.5.2. The above construction would become substantially simpler, andNkwould reduce to the open ballB dV,1k, ifV•∩ R+dV reduced to a singleton, i.e. iftV = 1; but this property need not hold whenσV is not continuous.
To make a counterexample, start from the parabola of Figure 5.5. We already know thatσP(dk) can tend to any nonnegative value whendk→ 0. However 0 ∈ P•; alternatively, the domain ofσP is not closed. In fact, we need a discontinuous sublinear function which is locally bounded on its domain – and this requires three variables. Thus, we first boundσP by defining
f (d) := 1 +
σP(d) ifσP(d)6 1 , +∞ otherwise
(the “1+” appearing above is just aimed at getting 0 in the interior of V ). Although no longer posi- tively homogeneous,f is still convex, its domain is the compact convex set P•, on which16 f 6 2;
whendk ∈ P•tends to 0,f (dk) can tend to any value in [1, 2]. To complete the construction, we take the so-called perspective off :
R2× R 3 (d, w) 7→ σ(d, w) := wfwd ifw > 0 , 0 if(d, w) = (0, 0) , +∞ otherwise
whose positive homogeneity is clear. Actually,σ is known to be convex and to support a closed convex setV ; see [103, § B.2.2] (in particular Remark 2.2.3), where our (d, w) is called (x, u). Besides, the propertyf > 1 implies that V is a neighborhood of the origin; remember (5.21).
Now take(d, w) ∈ bV◦ ⊂ dom σ, so that d0 := wd ∈ dom f and w > 0. Then use positive homogeneity:
1 = σ(d, w) =⇒ 1 w = σ(d
0, 1) = f (d0)∈ [1, 2] =⇒ w> 1 2.
Thus, bV◦ is separated from the origin (by the hyperplane w > 12) and this property is transmitted to its closed convex hullV•. On the other hand,σ inherits the discontinuities of f . In fact, choose α ∈ [1, 2] and construct a sequence {dk} in dom f tending to 0, such that f(dk) → α. Since σ(dk, 1) = f (dk) > 0, positive homogeneity gives
σ dk f (dk) , 1 f (dk) = 1 , hence dk f (dk) , 1 f (dk) ∈ bV◦. Pass to the limit:
dk f (dk) , 1 f (dk) →0,1 α ∈ cl bV◦⊂ V•.
Since α was arbitrary in [1, 2], the intersection of V• with the ray {0} × R+ contains the whole segment{0} × [1
2, 1].
ViewingGkof (5.33) as a prepolar, we set
Vk:= (Gk)◦.
Of course,V• ⊃ Gk+1 ⊃ GkandV ⊂ Vk+1 ⊂ Vk. The closed convex neighborhoodVkenjoys all of the properties listed in Section 5.3, in particular those coming from0 /∈ Gk.
Lemma 5.5.3 (EnlargingV∞). Assume0 /∈ V•; letR+dV be an extreme ray ofV∞◦ and assume that R+dV ( V∞◦ (R+dV is properly contained inV∞◦). Given an integerk > 0, construct Nk,Gk,Vkas above. ThenGk6= ∅ for k large enough (say k > k0) and
(i)V∞( Vk
∞fork> k0, (ii)∩k>k0V
k= V .
Proof. If Gk were empty for all k, we would have V• ⊂ Nk for allk, hence V• would reduce to [dV, tVdV]. In view of (5.28), this would implyR+dV = V∞◦, which our assumption rules out.
Everyd ∈ Gkis a convex combinationP
iαidi with eachdi inV•\Nk ⊂ V∞◦. None of these di’s can lie in[dV, tVdV]⊂ Nk, and none of their convex combinations either because of extremality ofR+dV. We conclude that
Gk∩ [dV, tVdV] =∅ . (5.35)
Now, we see from Theorem 5.3.8 that
but from Proposition 5.3.7, this is actually a chain of equalities:
R+(Vk)•=R+Gk. (5.36)
Besides,(Vk)• ⊂ Gk⊂ V•, hence0 /∈ (Vk)•and we can apply (5.28) toVk. Then we write V∞k◦ = R+(Vk)• = R+Gk ( R+V• = V∞◦ . [(5.28)] [(5.36)] [consequence of (5.35)] [(5.28) again] Thus,(Vk ∞ ◦ ( V◦
∞, which implies (i) since polarity is an involution between closed convex cones. To prove (ii), taker in¯ ∩kVk; we have to prove thatr¯∈ V (the other inclusion being obvious). If ¯
r /∈ V there is a separating hyperplane ¯d: σV( ¯d) < ¯d>r. Normalizing ¯¯ d via (5.28), we have altogether ¯
r∈\ k
Vk, d¯∈ bV◦, d¯>r > 1 ;¯ (5.37)
butσGk representsVk, so (5.37) gives
σGk(¯r)6 1 < ¯d>r ,¯ hence ¯d /∈ Gk.
Then ¯d∈ V•∩ Nkfor allk (large enough), i.e. ¯d∈ [dV, tVdV]. Introduce dk∈ Gkfrom (5.34): kdk− ¯dk 6
1
k and d
>
k¯r6 σGk(¯r)6 1 .
Passing to the limit, ¯d>¯r6 1; a contradiction to (5.37). Therefore ¯r ∈ V . Now we assume the existence of anS-free set W containing V ; it satisfies in particular
W• ⊂ W◦⊂ V◦ = [0, 1]V•. (5.38)
IfW• ⊂ V•, thisW is of no use to disprove maximality of V (Proposition 5.4.8). We are therefore in the situation
W• 6⊂ V•, which implies from (5.38): 0 /∈ V•. (5.39) Thus,W• contains some points out of V•. The key argument for our analysis is that one of these points lies on an extreme ray ofV∞◦ – which will be thedV of Lemma 5.5.3, crucial to construct the unbounded sequence{rk} of Figure 5.8.
Lemma 5.5.4 (Constructing an appropriate extreme ray). Let W ⊃ V satisfy (5.39). There is an extreme rayR+dV ofV∞◦ such that the setNkdefined by(5.32) satisfiesW◦∩ Nk = ∅ for k large enough.
Proof. From (5.39), we are in the framework of Corollary 5.3.10; Figure 5.10 is helpful to follow the proof. If cW◦ ⊂ V• thenW• = conv cW◦ ⊂ V•, contradiction. So there ise ∈ cW◦ (hence σW(e) = 1) which does not lie in V•; becauseV ⊂ W , i.e. σV 6 σW, thise satisfies σV(e) < 1 (otherwiseσV(e) = 1, hence e∈ bV◦⊂ V•).
Then constructde := σV1(e)e ∈ bV◦ (remember (5.21): σV(e) > 0). For every e0 ∈ [0, e], the segment [e0, de] contains e. Being a convex set, V• cannot contain such ane0 (otherwise it would
Conforti, Cornu´ejols, Daniilidis, Lemar´echal, Malick: Cut-generating functions
Mathematics of Operations Research xx(x), pp. xxx–xxx, c⃝200x INFORMS 21 Thus, (Vk
∞
!◦! V◦
∞, which implies (i) since polarity is an involution between closed convex cones. To prove (ii), take ¯r in ∩kVk; we have to prove that ¯r ∈ V (the other inclusion being obvious). If ¯
r /∈ V there is a separating hyperplane ¯d: σV( ¯d) < ¯d⊤¯r. Normalizing ¯d via (28), we have altogether ¯
r∈" k
Vk, d¯∈ #V◦, d¯⊤r > 1 ;¯ (37) but σGk represents Vk, so (37) gives
σGk(¯r)! 1 < ¯d⊤¯r , hence ¯d /∈ Gk.
Then ¯d∈ V•∩ Nk for all k (large enough), i.e. ¯d∈ [d
V, tVdV]. Introduce dk∈ Gk from (34): ∥dk− ¯d∥ ! 1 k and d ⊤ k¯r! σGk(¯r)! 1 .
Passing to the limit, ¯d⊤¯r! 1; a contradiction to (37). Therefore ¯r ∈ V . " Now we assume the existence of an S-free set W containing V ; it satisfies in particular
W•⊂ W◦⊂ V◦= [0, 1]V•. (38) If W•⊂ V•, this W is of no use to disprove maximality of V (Proposition 4.8). We are therefore in the situation
W•̸⊂ V•, which implies from (38): 0 /∈ V•. (39) Thus, W• contains some points out of V•. The key argument for our analysis is that one of these points lies on an extreme ray of V∞◦ – which will be the dV of Lemma 5.3, crucial to construct the unbounded sequence{rk} of Figure 8.
Lemma 5.4 (Constructing an appropriate extreme ray) Let W ⊃ V satisfy (39). There is an extreme rayR+dV of V∞◦ such that the set Nk defined by (32) satisfies W◦∩ Nk=∅ for k large enough. Proof. From (39), we are in the framework of Corollary 3.10; Figure 10 is helpful to follow the proof. If $W◦⊂ V• then W• = conv%$W◦!⊂ V•, contradiction. So there is e∈ $W◦(hence σ
W(e) = 1) which does not lie in V•; because V ⊂ W , i.e. σV ! σW, this e satisfies σV(e) < 1 (otherwise σV(e) = 1, hence e∈ #V◦⊂ V•). $ W◦ e de ¯b V• B bj0
Figure 10: The extreme rayR+bj0 contains some point in V•\W•
Then construct de := σV1(e)e∈ #V◦ (remember (21): σV(e) > 0). For every e′ ∈ [0, e], the segment [e′, de] contains e. Being a convex set, V• cannot contain such an e′ (otherwise it would contain e as well). As a result, the compact convex sets V•and [0, e] can be separated: there is ℓ∈ Rq(appropriately scaled) such that
max&0, e⊤ℓ'< 1 < min d∈V•d
⊤ℓ . (40)
Observe that
1 > e⊤ℓ = σV(e)d⊤eℓ > 0 . (41) Now introduce the closed convex set
B :=&b∈ V∞◦ : b⊤ℓ = 1 '
.
Figure 5.10: The extreme rayR+bj0 contains some point inV•\W•
containe as well). As a result, the compact convex sets V•and[0, e] can be separated: there is `∈ Rq (appropriately scaled) such that
max0, e>` < 1 < min d∈V•d
>` . (5.40)
Observe that
1 > e>` = σV(e)d>e` > 0 . (5.41) Now introduce the closed convex set
B :=b∈ V∞◦ : b>` = 1 .
Clearly,R+B ⊂ V∞◦. Conversely, apply (5.28): every nonzerod∈ V∞◦ can be scaled to sometd∈ V•. By (5.40),td>` > 1, then d can be scaled again to td/(td>`), which lies in B. We have shown
R+B = V∞◦ . (5.42)
By (5.28), everyb ∈ B can be obtained by scaling some d ∈ bV◦: b = td; and t = d>1` ∈ ]0, 1[ by (5.40). This means that
B⊂ ]0, 1[ bV◦ ⊂ V◦; (5.43)
B is therefore bounded (and closed because V∞◦ is closed), hence compact.
Using (5.41), scalee to ¯b := e>1`e ∈ B and express ¯b = Pjαjbj as a convex combination of extreme pointsbj ofB (Minkowski’s Theorem). Then
σW(¯b) = 1
e>`σW(e) = 1 e>` > 1 .
By convexity of σW, there is some j0 such that σW(bj0) > 1 (we may have σW(bj0) = +∞). Altogether, we have exhibited
bj0 extreme inB and satisfying 1 < σW(bj0) .
Extremality ofbj0 inB implies extremality of the rayR+bj0 inR+B, i.e. in V∞◦ because of (5.42). The intersection ofW◦with this extreme ray is some[0, dW] (dW may be 0) which, by definition of a polar, does not containbj0. Sinceb>j0` = 1 (because bj0 ∈ B), d
>` < 1 for all d∈ [0, d
W]. Then, (5.40) shows that[0, dW] and [dV, tVdV] are separated.
As a result, the two compact setsW◦and[dV, tVdV] are disjoint. If there were dk ∈ W◦∩ Nk for allk, then the bounded sequence {dk} would have some cluster point d∗; butW◦ is closed: d∗
The setB constructed in the above proof is a so-called basis of the pointed cone V∞◦. The case σW(bj0) = +∞, dW = 0 corresponds to a W as in Figure 5.5; it occurs in Figure 5.10. This latter picture is still helpful to follow the next proof. Recall thatL is the lineality space of V .
Proposition 5.5.5. Assume0∈ S = conv (S). If a minimalCGFρ represents the S-free set V = V (ρ)
which is not maximal, thenVkexists as described by Lemma 5.5.3. There isrk∈ Vk∩S, decomposed asrk= `k+ ukwith`k∈ L and uk∈ L⊥, such that
for someK ⊂ N , lim k∈Kkr
kk = +∞ and lim k∈Kku
kk = +∞ .
Proof. If all of the S-free sets W containing V satisfy W• ⊂ V•, thenV is maximal (Proposition 5.4.8). Thus, there is anS-free set W ⊃ V satisfying (5.39) and we can construct dV as in Lemma 5.5.4.
IfR+dV = V∞◦, then bV◦ = V• ={dV} and V◦ = [0, dV] (Proposition 5.3.7): the S-free set V , represented byσV◦, is the half-space{r : d>Vr 6 1}, which separates 0 from S; this is ruled out by assumption.
Otherwise,R+dV ( V∞◦: we can apply Lemma 5.5.3 and construct the sequence of neighborhoods Vk. By minimality ofµ
V,Vkcannot beS-free (Lemma 5.5.3(i) and Theorem 5.4.10): there exists rk lying
inint Vk, hence from (5.12)
1 > σGk(rk) , (5.44)
and inS, hence rk∈ int W : σ/ W•(rk)> 1; since W•is compact,
∃ek∈ W• such that e>krk> 1 . (5.45) Now we claim that there isδ > 0 such that
tkek∈ V•∩ Nk, for sometk> 1 + δ and all k large enough . (5.46) Using (5.28), scaleek(nonzero from its definition) totkek ∈ V•; and note from (5.38) thattk > 1. Then (5.45) implies thattkek∈ G/ k: otherwise
16 e>krk 6 tke>krk6 σGk(rk)
by definition of a support function; this contradicts (5.44). It follows thattkek ∈ V•∩ Nk, which is far fromW•(Lemma 5.5.4); (5.46) is proved.
Now we can conclude. First, let ¯d ∈ [dV, tVdV] be a cluster point of the bounded sequence {tkek}. Next, use (5.46), (5.45), (5.44) to write for all d ∈ Gk
1 + δ6 tk6 tke>krk= (tkek− d)>rk+ d>rk < (tkek− d)>rk+ 1 . This holds in particular ford = dkstated in (5.34):
δ < (tkek− dk)>rk. (5.47) Then we obtain with the Cauchy-Schwarz inequality
δ <ktkek− ¯d + ¯d− dkk krkk 6 ktkek− ¯dk + 1 k krkk .
Furthermore, decomposerk= `k+ ukin (5.47) and observe that bothe>k`kandd>k`kare 0 (`k ∈ L whileekanddklie inV∞◦ ⊂ L⊥). So (5.47) gives also
δ < (tkek− dk)>uk6 ktkek− ¯dk + 1 k kukk .
As suggested in the beginning of this section, proving Theorem 5.5.1 is now easy. AnS-free set represented by a minimalCGFwill be automatically maximal under any assumption contradicting the existence of our unbounded sequences.
Proof. of Theorem 5.5.1 Construct the sequences{rk} and {uk} of Proposition 5.5.5. Case (i): Extract a cluster pointˆr of the normalized subsequence{rk}
k∈K: for someK0 ⊂ K,
lim k∈K0
rk krkk = ˆr .
Then take an arbitraryM > 0. We know that M/krkk 6 1 if k is large enough in K0 so, because both 0 andrklie inVk∩ S,
M krkkr
k∈ Vk∩ S , for large enough k ∈ K0.
By closedness, this impliesM ˆr ∈ S, hence ˆr ∈ S∞ because M is arbitrary. The same argument using Lemma 5.5.3(ii) givesrˆ∈ V∞.
Let us sum up. IfV is not maximal, then V∞∩ S∞contains a vector r of norm 1; this contra-ˆ dicts(i).
Case (ii): Writeuk = rk− `k∈ Vk− L = Vk+ L⊂ Vk+ V
∞⊂ Vk. Then proceed as in Case(i): extract a cluster pointu ofˆ uk
kukk
K and argue that M
kukkuk∈ Vk∩ L⊥to exhibit ˆ
u∈ V∞∩ L⊥ and kˆuk = 1 . (5.48)
Besides,ukis the projection ontoL⊥(a linear operator) ofrk ∈ S ⊂ U + S
∞; hence uk ∈ ProjL⊥U + ProjL⊥S∞.
By(ii)1,ProjL⊥U is a bounded set, so our cluster direction ˆu lies in ProjL⊥S∞: ˆ
u = ˆs− ˆ`, for some ˆs ∈ S∞and ˆ`∈ L . Use (5.48):
S∞3 ˆs = ˆu + ˆ` ∈ V∞+ L = V∞; then use(ii)2:
ˆ
s∈ V∞∩ S∞= L∩ S∞.
As a result,u = ˆˆ s− ˆ`lies in L; use (5.48) again: ˆu ∈ L ∩ L⊥cannot have norm 1.
Thus, in this case also,V has to be maximal.
Let us insist once more: the core of our proof is Proposition 5.5.5. Then (i) and (ii) appear as ad hoc assumptions to contradict the existence of the stated unbounded sequences; other similar assumptions might be designed.