Slater conditions without interior points for programs in Lebesgue spaces with pointwise bounds and finitely many constraints

We consider optimization problems in Lebesgue spaces with pointwise box constraints and finitely many additional linear constraints. We prove that the existence of a Slater point which lies strictly between the pointwise bounds and which satisfies the linear constraints is sufficient for the existence of Lagrange multipliers. Surprisingly, the Slater point is also necessary for the existence of Lagrange multipliers in a certain sense. We also demonstrate how to handle additional finitely many nonlinear constraints.

In this problem, we have pointwise bounds x a , x b , and finitely many linear constraints.We are interested in the existence of Lagrange multipliers.This is a delicate question, since (under typical assumptions on the measure µ) the set has empty interior unless p = ∞.Moreover, if p = ∞, this would lead initially to multipliers in the space L ∞ (µ) ⋆ , but often their regularity can be improved.We mention that the above problem is also interesting from an application point of view, see, e.g., Bonnans, Zidani, 1999;Casas, Tröltzsch, 2002;Buchheim, Grütering, Meyer, 2022.We say that x ∈ L p (µ) is a Slater point, if the conditions x a < x < x b µ-a.e. on Ω, ⟨g i , x⟩ L p (µ) ≤ a i ∀i = 1, . . ., n, ⟨h j , x⟩ L p (µ) = b j ∀j = 1, . . ., m are satisfied.Note that we only require that x lies strictly between the pointwise bounds and this does not imply that x is an interior point of K. Our main result Theorem 3.3 shows the existence of Lagrange multipliers for locally optimal solutions x of the above problem, provided that a Slater point exists.Under certain circumstances, we are even able to prove that a Slater point is necessary for the existence of Lagrange multipliers (see Section 4), which is rather surprising.We also address the case of nonlinear inequality constraints in Section 5.
We briefly describe our idea to prove the existence of multipliers.We denote by P the set of points satisfying the finite-dimensional constraints, i.e., the feasible set of our problem is given by K ∩ P .For the existence of Lagrange multipliers, we have to characterize the normal cone of K ∩ P .By using the notion of n-polyhedricity from Wachsmuth, 2019, we automatically get the formula T K∩P (x) = T K (x) ∩ T P (x) for the tangent cone.By dualizing this formula, we get N K∩P (x) = N K (x) + N P (x) provided that the set on the right-hand side is closed (or weak-⋆ closed, see Lemma 2.7 below).This closedness can be shown by utilizing the Slater point x.
We mention that a similar result is obtained in the rather unknown work Tröltzsch, 1977.Therein, the existence of Lagrange multipliers is proved using a similar notion of Slater points.In contrast to our approach, the main tool is duality for linear programs.Under rather weak conditions, it is shown that no duality gap occurs and the Slater point is used to prove the existence of a dual solution.Moreover, the problem is addressed in more abstract spaces.However, our approach has certain advantages.We are able to discuss also p ∈ {1, ∞}, whereas Tröltzsch, 1977 requires reflexive spaces.We further show necessity of Slater points for the existence of multipliers.We address inequality constraints whereas the approach of Tröltzsch, 1977 is only given for equality constraints.

Preliminaries
Throughout the paper, we use the following setting.
We fix some notation.The set of nonnegative integers is denoted by N. The set of positive, real numbers is R + .By L q (µ) we denote the usual Lebesgue spaces.Recall that L q (µ) ⋆ = L q ′ (µ) for q ∈ [1, ∞), see Fremlin, 2003, 243G and 244K.
Finally, we mention that we use {x a = x} with functions x a , x : Ω → R as an abbreviation for the set {ω ∈ Ω | x a (ω) = x(ω)}.Similar abbreviations are also used for different relation symbols.

Tangent cones and normal cones
We recall some preliminary results concerning tangent cones and normal cones.Further information can be found in Bonnans, Shapiro, 2000.
Definition 2.2.Let C ⊂ X be a closed subset of the Banach space X.For x ∈ C, we define the radial cone and the (Bouligand) tangent cone In the case that C is convex, the tangent cone T C (x) coincides with the closure of the radial cone R C (x).
For box constraints in L p (µ), p ∈ [1, ∞), the following formula for the tangent cone is well known.We also provide estimates for the tangent cone in L ∞ (µ).
Lemma 2.3.Let measurable functions x a , x b : Ω → [−∞, ∞] be given such that is nonempty and we fix x ∈ K.The estimates hold for all ε > 0. For p ∈ [1, ∞), we even have Proof.The estimates are straightforward to verify.Now, let p < ∞ and let h belong to the right-hand side of (2.2).We define the sequence Then, it is easy to check that In case p = ∞, we do not have Thus, an analogue formula to (2.2) holds, if we define the tangent cone in L ∞ (µ) by using the weak-⋆ topology.
Our next goal is the definition of the normal cone which utilizes the concept of the polar cone.When we are working in the space L ∞ (µ), we do not want to arrive in the cumbersome space L ∞ (µ) ⋆ , but in the nice predual space L 1 (µ).This is already reflected in the definition of the polar cone.
Definition 2.4.Let C ⊂ X be a subset of a Banach space X.We define Another perspective is to equip the pair of spaces L p (µ) and L p ′ (µ) with compatible topologies.That is, we can use the norm topologies in L p (µ), p ∈ [1, ∞), but L ∞ (µ) has to be equipped with the weak-⋆ topology.In this sense, Definition 2.4 is compatible with, e.g., Bonnans, Shapiro, 2000, Section 2.1.4.As usual, the normal cone is defined as the polar of the tangent cone.Definition 2.5.Let C ⊂ X be a closed subset of a Banach space X.
Note that the polar of T C (x) is defined as in Definition 2.4.In particular, this is the usual definition of the normal cone, unless X = L ∞ (µ).In this case case, we define the normal cone to stay in the predual space L 1 (µ) of L ∞ (µ).
Lemma 2.6.Let K and x be given as in Lemma 2.3.Then, Proof.The inclusion "⊃" follows easily from the estimate (2.1b).Now, let ζ ∈ N K (x) be given.Let ε > 0 be given and let M be an arbitrary subset of For later reference, we provide the following result for the polar of an intersection of tangent cones.
Lemma 2.7.Let C, P ⊂ L p (µ) be closed and convex and fix x ∈ C ∩ P .Then, Here, cl ⋆ (A) for A ⊂ L p ′ (µ) denotes the closure of A w.r.t. the norm convergence in case p ∈ (1, ∞] and the closure of A w.r.t. the weak-⋆ topology of L ∞ (µ) in case p = 1.
Finally, we provide simple formulas for sets defined by linear restrictions.
given.We define For all x ∈ P we have is the active set.
Proof.The formula for the tangent cone follows from a simple calculation.The normal cone is given in Bonnans, Shapiro, 2000, Proposition 2.42.

Polyhedric sets
In this section, we briefly review some of the theory of polyhedric sets.More details can be found in Wachsmuth, 2019.
Polyhedric sets are a suitable generalization of polytopes to infinite dimensional spaces.In many regards, they seem to behave similarly to polytopes, in particular if they are in interaction with sets defined via finitely many continuous and linear inequalities.They were introduced in the seminal works Mignot, 1976;Haraux, 1977. In Wachsmuth, 2019 it was shown that "almost all" of the polyhedric sets are even n-polyhedric for all n ∈ N.
Definition 2.9.Let C ⊂ X be a closed, convex subset of the Banach space X and let n ∈ N be given.We call C n-polyhedric at x ∈ C, if The classical case of polyhedricity corresponds to the case n = 1 in Definition 2.9.
A main result of Wachsmuth, 2019 is that sets defined by lower and upper bounds are n-polyhedric for all n ∈ N if the underlying Banach space possesses a certain lattice structure.This, in particular, applies to our situation and yields the following result, cf.Wachsmuth, 2019, Example 4.21(1).
Theorem 2.10.The set K from Lemma 2.3 is n-polyhedric for all n ∈ N.
This n-polyhedricity yields an easy formula for the tangent cone and the normal cone of the intersection with finitely many linear restrictions.
Lemma 2.11.Let subsets K, P ⊂ L p (µ) be given.We assume that K is n-polyhedric for all n ∈ N and that P is defined by finitely many continuous and linear equalities and inequalities.Then, Proof.The formula for the tangent cone follows from Wachsmuth, 2019, Lemma 4.4.Consequently, Lemma 2.7 yields the identity for the normal cone.
Recall that T P (x) and N P (x) are given in Lemma 2.8.

Linear Constraints
In this section, we focus on a problem with linear constraints of the form Here, f : for all feasible points x of (3.1).
We denote by F ⊂ L p (µ) the feasible set of (3.1).By standard arguments, every local minimizer x satisfies ⟨f ′ (x), d⟩ L p (µ) ≥ 0 for all d ∈ T F (x) and, consequently, It remains to evaluate the normal cone.To this end, we write the feasible set as F = K ∩P , where Using the results on n-polyhedric sets recalled in Section 2.2, we get However, the set N K (x) + N P (x) is, in general, not (weak-⋆) closed, see Section 6.Thus, we have to utilize a Slater point.
are satisfied.We say that (3.1) satisfies the Slater condition if there exists a Slater point of (3.1).
In the next results, we use the formulas for N K (x) and N P (x) provided in Lemmas 2.6 and 2.8.
Lemma 3.2.Suppose that x is a Slater point of (3.1).Then, for every feasible point x of (3.1), the set Proof.We begin with the easier case p > 1. Suppose that the sequence (ξ k ) ⊂ N K (x) + N P (x) is convergent in L p ′ (µ), i.e., Due to the feasibility of the Slater point, we have Thus, all addends on the right-hand side of (3.3) are nonpositive and this yields This shows that the sequence ζ k is bounded in the space L 1 (μ).Since ξ k is bounded in L p ′ (µ), it is also bounded in L 1 (μ).Thus, the sequence βk,j h j for all k.By picking subsequences (without relabeling), we can assume that the sequences (α k,i ) k , ( βk,j ) k are convergent with limits α i ≥ 0, β j ∈ R.This shows The case p = 1 is slightly more difficult, since we have to work with the weak-⋆ topology in . Thus, we have to consider weak-⋆ convergent nets instead of sequences.By using the Krein-Šmulian theorem, Conway, 1985, Theorem V.12.1, it is sufficient to consider a bounded net (ξ k ) k∈K with where Here, K is a directed set.We can argue as above to arrive at (3.4).Now, we have to take an extra step, since convergent nets with values in R do not need to be bounded.However, there exists Thus, we consider subnets indexed by This is a directed set (with the order inherited from K) and it is easy to check that (ξ k ) k∈K ′ is a subnet of (ξ k ) k∈K .Now, we can continue as above with some obvious changes.
Together with Lemma 2.11, we get the remarkable formula if we require the existence of a Slater point of (3.1).In particular, by combining this result with (3.2) and with the formulas for the normal cones, we arrive at the following first-order condition.
Theorem 3.3.Let x ∈ F be a locally optimal solution of (3.1).Further, we assume the existence of a Slater point x of (3.1), cf.Definition 3.1.Then, there exist Note that the Slater point x does not need to satisfy the finitely many linear inequalities involving g i in a strict sense, since the associated normal cone is automatically closed, cf.Bonnans, Shapiro, 2000, Proposition 2.41.
Finally, we give a small remark which demonstrates that the Slater point can ignore all "real" inequalities, which cannot be reduced to equalities.
Remark 3.4.Let I ⊂ {1, . . ., n} be a subset of the inequality constraints which can be strictly satisfied, i.e., there exists x ∈ L p (µ) with Further, we require the existence of x ∈ L p (µ) with Then, it is easy to check that (1 − ε)x + εx is a Slater point for all ε > 0 small enough.Note that the point x ignores all inequalities from I.

Necessity of Slater points
In this section, we are going to show that the existence of a Slater point is necessary for the existence of multipliers in certain situations.We continue to use the setting of Section 3.
First, we recall the important result that a finite linear system of equations and inequalities can be reformulated equivalently such that MFCQ holds.(iii) we have Proof.The conversion of our linear system can be performed in two steps.First, we convert all linear inequalities which cannot be strictly satisfied by feasible points into equalities.This ensures the existence of x satisfying (iii).Second, we reduce the system of linear equalities to a maximal linear independent subset.This yields (ii).Finally, it is easy to see that both steps do not change the set of feasible points, thus (i) holds.
Lemma 4.2.Suppose that x a < x b holds µ-a.e.Then, a dense subset of K is given by the set K : Proof.In the case that K is empty, there is nothing to show.Otherwise, we choose some x ∈ K.We first show that K is nonempty.Since µ is assumed to be σ-finite, it is easy to check that there exists a function w ∈ L p (µ) with w > 0.Then, a pointwise discussion shows that lies strictly between x a and x b and x ∈ L p (µ) follows from x, w ∈ L p (µ).Thus, x ∈ K.
Finally, for an arbitrary x ∈ K, the sequence ((1 − 2 −n )x + 2 −n x) n∈N belongs to K and converges towards x in L p (µ).
Lemma 4.3.We suppose that x a < x b and that the system (3.1)does not possess a Slater point.Then we have N P (x) ∩ −N K (x) ̸ = {0} for all feasible points x of (3.1).
Proof.In the proof, we use the modified linear system from Lemma 4.1.By we denote the active set of x and by ñ(x) we denote the cardinality of Ã(x).The assumption implies that is violated for all x ∈ L p (µ) with x a < x < x b , since otherwise (1 − t)x + tx would be a Slater point for t ∈ (0, 1) small enough.
Thus, the sets Since v 1 can be arbitrarily small, this gives λ ≥ 0. Now, we observe that the vector j=1,..., m belongs to A. Thus, the above inequality yields for all x ∈ L p (µ) with x a < x < x b .Due to Lemma 4.2, the above inequality continues to hold for all x ∈ K and, thus, Since the modified linear system from Lemma 4.1 still describes the polytope P , the sign condition on λ and Lemma 2.8 show that ζ ∈ N P (x).It remains to check ζ ̸ = 0.In case λ ̸ = 0, we use x from Lemma 4.1 to obtain Proof.First, we note that −ζ ∈ N K (x) yields x = x b on {ζ < 0} and x = x a on {ζ > 0}, see Lemma 2.6.
Let ξ ∈ L p ′ (µ) with ξ = 0 on {ζ = 0} be given.We define the sequence (ξ k ) k∈N ⊂ L p ′ (µ) Slater conditions without interior points Gerd Wachsmuth Via a distinction by cases, we get , see again Lemma 2.6.Thus, Finally, the definition of ξ k directly yields the pointwise µ-a.e.convergence of ξ k towards ξ.Moreover, |ξ k | ≤ |ξ| µ-a.e.Thus, we can invoke Lebesgue's dominated convergence theorem to show that ξ k converges strongly (in case p ′ < ∞) or weak-⋆ (in case p ′ = ∞) towards ξ.This yields the claim.Now, we are in position to prove the main theorem of this section.
Theorem 4.5.We suppose that x a < x b and that the system (3.1)does not possess a Slater point.Further, we assume that p > 1 (i.e., p ′ < ∞) and that the measure µ is nonatomic.Then, N K (x) + N P (x) is not closed in L p ′ (µ) for all feasible points x of (3.1).
We define an auxiliary function ϑ := n i=1 |g i | + m j=1 |h j | and we note that every function from N K (x) + N P (x) is bounded from above by a scalar multiple of ϑ on {ζ > 0}, since the functions in N K (x) are nonpositive on {ζ > 0}.Since ϑ is real valued, we can find a subset M ⊂ {ζ > 0} of positive measure such that ϑ is bounded from above on M .Consequently, all functions in N K (x) + N P (x) are bounded from above on M .However, since M has positive measure, since p ′ < ∞ and since µ is nonatomic, we can construct a nonnegative function ξ ∈ L p ′ (µ) \ L ∞ (µ), which vanishes outside of M .Now, Lemma 4.4 implies that ξ ∈ cl ⋆ (N K (x) + N P (x)), while the boundedness of ϑ on M and ξ ̸ ∈ L ∞ (µ) gives ξ ̸ ∈ N K (x) + N P (x).Thus, N K (x) + N P (x) cannot be closed.
The absence of Slater points also implies the nonexistence of multipliers.
Corollary 4.6.We suppose that x a < x b and that the system (3.1)does not possess a Slater point.Further, we assume that p > 1 (i.e., p ′ < ∞) and that the measure µ is nonatomic.Then, for each feasible point x of (3.1), there exists a linear functional F ∈ L p ′ (µ) such that x minimizes F on K ∩ P but no Lagrange multiplier exists.
The contrapositive of this corollary is also interesting.It yields that, if we fix a feasible point x and if we obtain Lagrange multipliers for all functionals which are minimized by x on K ∩ P , then there exists a Slater point.

Nonlinear Constraints
In this section we consider the situation with additional nonlinear constraints.For simplicity, we restrict ourselves to inequality constraints only.That is, we study the problem Minimize f (x) (5.1) Here, the data is as in Section 3 and, additionally, N ∈ N and G i : L p (µ) → R are continuously Fréchet differentiable.Similarly to Section 3, we require G ′ (x) ∈ L p ′ (µ) = L 1 (µ) in the case p = ∞ for all feasible points x of (3.1).In addition to the notation from Section 3, we define and the feasible set of (5.1) is denoted by In order to linearize the constraints, we utilize the constraint qualification by Robinson-Zowe-Kurcyusz (RZKCQ).
Lemma 5.1.Let x ∈ E be a feasible point and we assume the existence of a point x ∈ L p (µ) such that is the active set of the nonlinear constraints.Then, Proof.In order to apply the RZKCQ, we write the constraints in (5.1) as where G : L p (µ) → R N is the vector function with components G i and R = (−∞, 0] N .Since the interior of R is nonempty, the RZKCQ can be formulated as see Bonnans, Shapiro, 2000, (2.196).We can check that this condition is satisfied with x ′ = (1 − ε)x + εx for ε > 0 small enough.Thus, Bonnans, Shapiro, 2000, Corollary 2.91 yields the claim.
We mention that RZKCQ also yields a formula for the normal cone of E if we define this cone in the dual space of L p (µ).In case p = ∞, however, this is not compatible with our Definitions 2.4 and 2.5.Therefore, we provide the following lemma.
Note that the result Schirotzek, 2007, Proposition 2.4.3 (which does not need the assumption concerning finite dimensionality) is not applicable here, since, therein, the polars are defined to be subsets of L ∞ (µ) ⋆ .

Proof.
From the generalized open mapping theorem Zowe, Kurcyusz, 1979, Theorem 2.1, we get the existence of C > 0, such that for all f ∈ L ∞ (µ), there exist g This bound holds for all f with ∥f ∥ L ∞ (µ) ≤ 1, thus we get ∥ξ k ∥ L 1 (µ) ≤ CM .By assumption, (ξ k ) belongs to a finite-dimensional subspace, hence there exists a subsequence (without relabeling) such that ξ By combining the previous lemmas, we obtain a formula for the normal cone.
Lemma 5.3.Under the assumptions of Lemma 5.1, we have where Proof.We are going to apply Lemma 5.2 with For any f ∈ L ∞ (µ), we have see Bonnans, Shapiro, 2000, Proposition 2.42, and from Lemma 5.2.
Finally, it remains to utilize Lemmas 2.11 and 3.2 to obtain the formula (3.5) for N K∩P (x).Thus, we also need a Slater point as in Definition 3.1.
Theorem 5.4.Let x ∈ E be a feasible point and we assume the existence of x ∈ L p (µ) satisfying Now, assume additionally that x is a local minimizer of (5.1).Then, there exist ζ ∈ L p ′ (µ), α i ≥ 0, i ∈ A(x), β j ∈ R, j = 1, . . ., m and γ i ≥ 0, i ∈ B(x) such that The multiplier ζ can be split as in Theorem 3.3.
Proof.The result follows from Lemmas 2.11, 3.2 and 5.3.
On the first glance, one might think that it would be a weaker condition to require the existence of points x, x ∈ L p (µ) satisfying Then, one can use x in Lemma 5.3 to get N E (x) = N K∩P (x) + N Q (x) and afterwards x to get the representation of N K∩P (x).However, similar to Remark 3.4, we can set x = (1 − ε)x + εx.If ε > 0 small enough, x satisfies the system in Theorem 5.4.
The conditions posed on the point x from Theorem 5.4 are typically called linearized Slater conditions.As known from finite-dimensional optimization, we need a strict inequality for nonlinear constraints and a nonstrict inequality suffices for the (finitedimensional) linear constraints.Moreover, the point x is required to strictly satisfy the linear (infinite-dimensional) box constraints.

A counterexample
For illustration, we provide an example which does not possess a Slater point.We consider the problem (3.1) with the following setting.As a measure space, we use Ω = (0, 1) equipped with the Lebesgue measure and we take p = p ′ = 2.We use the simple pointwise bounds x a = 0, x b = ∞, a single inequality constraint (n = 1) g 1 ≡ 1, a 1 = 0, and no equality constraints (m = 0).In this situation, it is easy to check that x = 0 is the only feasible point and the Slater condition w.r.t.(K, P ) is violated.Here, K and P are as defined in Section 3.
For ζ ∈ L 2 (µ), it is easy to check that ζ ∈ N K (x) + N P (x) if and only if the positive part max(ζ, 0) belongs to L ∞ (µ).Thus, the closure of N K (x) + N P (x) coincides with L 2 (µ) and, in particular, N K (x) + N P (x) is not closed.This is in accordance with Theorem 4.5, which says that the assertion of Lemma 3.2 fails in absence of a Slater point and Finally, we choose a function z ∈ L 2 (µ) which is (essentially) unbounded from below and we set f (x) := ⟨z, x⟩ L 2 (µ) .Then, x minimizes (3.1), but the KKT conditions from Theorem 3.3 cannot be satisfied.The existence of such an objective f is expected from Corollary 4.6.
set and, hence, ζ ≤ 0 µ-a.e. on {x a = x < x b }.The other sign conditions on ζ can be shown analogously.
).Now, we can invoke Bonnans, Shapiro, 2000, Proposition 2.41 and this implies the existence of bounded sequences (α k,i