Dual qualification conditions and their relations
Dual qualification conditions in purely algebraic setting
Let X, Y be l.c.H.t.v.s with its topological dual X ∗ , Y ∗ , respectively, f, g, h :
X → R ∪ {+∞} are proper functions, H : domH ⊂ X → Y is a mapping, and k :Y →R∪ {+∞} is a proper function Note that the functions f, g, h, λH (λ ∈Y ∗ ) and k are proper functions but not necessarily convex nor lsc Let
We start with the relations between the sets in consideration.
Lemma 2.1.1 With the previous notions, one gets
In particular, if A = epi(f+g+k◦H) ∗ then A =B =C =D =E =F.
The proof is straightforward, primarily utilizing (1.7) Initially, we note that epi g ∗ + [ λ∈dom k ∗ epi(λH−k ∗ (λ)) ∗ equals [ λ∈dom k ∗ epi g ∗ + epi(λH−k ∗ (λ)) ∗ By applying (1.7) with φ=g and ψ=λH−k ∗ (λ) for λ∈dom k ∗, we establish that A is a subset of B Furthermore, by applying (1.7) again with φ=f and ψ=g+λH−k ∗ (λ) for λ∈dom k ∗, we conclude that B is a subset of E Consequently, for any λ in dom k ∗ and x in dom H, we derive the necessary relationships.
−k ∗ (λ)≤(k◦H)(x)−(λH)(x), and so, λH−k ∗ (λ) ≤ k◦H, which yields f +g+λH −k ∗ (λ)≤ f +g +k◦H We thus have, for all λ∈domk ∗ , f +g +λH−k ∗ (λ)∗
, and therefore, it holds for allλ∈domk ∗ , epi f+g+λH−k ∗ (λ)∗
, which, in turn, shows that
⊂epi(f+g+k◦H) ∗ , and epi f ∗ + [ λ∈dom k ∗ epi(g+λH−k ∗ (λ)) ∗ ⊂epi f ∗ + epi (g+k◦H) ∗
We have just proved that
A ⊂ B ⊂ E ⊂ epi(f +g+k◦H) ∗ andB ⊂ F The inclusionF ⊂epi (f+g+k◦H) ∗ follows from (1.7) Other inclusions can be proved in the same way The last assertion is obvious.
We now introduce the following dual conditions:
(CA) A= epi(f+g+k◦H) ∗ , (CB) B= epi(f+g+k◦H) ∗ , (CC) C = epi(f +g+k◦H) ∗ , (CD) D= epi(f+g+k◦H) ∗ , (CE) E = epi(f+g+k◦H) ∗ , (CF) F = epi(f +g+k◦H) ∗ The relations between these conditions are given in the next theorem.
Theorem 2.1.1 The following implications hold. a a
Here (A) =⇒(B) means that condition (A) implies condition (B).
Proof We observe that if (CA) holds, i.e.,
According to Lemma 2.1.1, we establish that A, B, C, D, E, and F are all equal to epi(f + g + k ◦ H) ∗ This indicates that if condition (CA) is satisfied, then conditions (CB), (CC), (CD), (CE), and (CF) are also satisfied Furthermore, using the same reasoning, we conclude that if any of the conditions (CB), (CC), or (CD) are met, then (CE) must also hold, and if (CB) is true, then (CF) follows as well Thus, the proof is complete.
The conditions (CA) and (CB) were introduced in [32], where some Farkas-type results were established under (CA).
We now consider a special case LetC be a subset ofX,K be a convex cone in Y. Let g :=iC, k := i−K and A :=C∩H −1 (−K) Then it is easily seen that k ∗ = i K + , and hence, domk ∗ =K + Moreover, iA=iC+i−K◦H, where
Thus, the condition (CA) becomes epif ∗ + [ λ∈K + epi (λH) ∗ + epii ∗ C = epi (f+iA) ∗
By the same procedure, we get other modifications of other conditions (CB), (CC), (CD), (CE), and (CF) Specifically, we consider the following modifications of these conditions:
(CA 1 ) epif ∗ + [ λ∈K + epi (λH) ∗ + epii ∗ C = epi (f +i A ) ∗ ,
(CC1) epi (f +iC) ∗ + [ λ∈K + epi (λH) ∗ = epi (f +iA) ∗ ,
(CE 1 ) [ λ∈K + epi (f+λH+i C ) ∗ = epi (f +i A ) ∗ , (CF 1 ) epif ∗ + epii ∗ A = epi (f +i A ) ∗
Then as a direct consequence of Theorem 2.1.1, we get
Corollary 2.1.1 The following relations hold. a a
Dual qualification conditions in convex setting
This article presents essential dual qualification conditions for scenarios where the functions exhibit convexity and lower semi-continuity Additionally, it references previous work that outlines sufficient and necessary conditions for the validity of (CA) as discussed in [32].
In this subsection we assume that f, g ∈ Γ(X), k ∈ Γ(Y), λH ∈ Γ(X) for all λ∈domk ∗ , and dom(f+g+k◦H)6=∅.
Proposition 2.1.1 epi (f+g+k◦H) ∗ =C if and only if C is w ∗ -closed.
Proof Let us introduce the extended real-valued function φ defined on X ∗ by φ = (f+g) ∗ 2 inf λ∈dom k ∗ (λH−k ∗ (λ)) ∗
We firstly observe thatφis convex Indeed, let ψ : domk ∗ ×X ∗ −→R∪ {+∞}be the function defined by ψ(λ, u) := (λH−k ∗ (λ)) ∗ (u).
It is easy to verify thatψ is convex By Theorem 2.1.3(v) in [66], the function λ∈dominfk ∗ (λH −k ∗ (λ)) ∗ (.) = inf λ∈dom k ∗ ψ(λ, ) is convex and hence, φ is proper convex.
Now, from the assumption that f, g ∈ Γ(X), k ∈ Γ(Y), and λH ∈ Γ(X) for all λ∈domk ∗ , one gets, for any x∈X, φ ∗ (x) = sup x ∗ ∈X ∗ n hx ∗ , xi −
= sup x ∗ ∈X ∗ nhx ∗ , xi − inf λ∈dom k ∗ inf x ∗ =u+v
= sup λ∈dom k ∗ sup u,v∈X ∗ nhu, xi+hv, xi −(f+g) ∗ (u)−(λH) ∗ (v)−k ∗ (λ)o
= f(x) +g(x) + (k◦H)(x) and thus, epi (f+g+k◦H) ∗ = epiφ ∗∗ = epi clφ, (2.1) where clφ is the lower semi-continuous regularization of φ (which is defined via the equation epi clφ= cl epiφ).
On the other hand, by Theorem 2.2(e) in [63], we get epi clφ = epi cl (f+g) ∗ 2 inf λ∈dom k ∗ (λH−k ∗ (λ)) ∗
= cl epi (f +g) ∗ + epi [ inf λ∈dom k ∗ (λH −k ∗ (λ)) ∗ ]
= cl epi (f +g) ∗ + epi s [ inf λ∈dom k ∗ (λH−k ∗ (λ)) ∗ ]
Combining (2.1) and (2.2), we get epi (f+g+k◦H) ∗ = clC.The proof is complete.
In the same way, we get
Proposition 2.1.2 The following statements are true:
(i) epi (f +g+k◦H) ∗ =D if and only if D is w ∗ -closed,
(ii) epi (f +g+k◦H) ∗ =E if and only if E is w ∗ -closed,
(iii) epi (f +g+k◦H) ∗ =F if and only if F is w ∗ -closed.
Proof Similar to the proof of Proposition 2.1.1, using Theorem 2.2 in [63] and consid- ering the following functions φ 1 , φ 2 , andφ 3 for the case (i), (ii), and (iii), respectively: φ 1 = g ∗ 2 inf λ∈dom k ∗ (f +λH−k ∗ (λ)) ∗
Characterizations of dual conditions–Generalized Moreau-Rockafellar re-
Dual conditions characterizing generalized Moreau-Rockafellar
Assume that the functions f, g, k, and λH (for arbitrary λ ∈ Y ∗ ) are proper functions but not necessarily convex nor lower semicontinuous Assume further that dom(f+g+k◦H)6=∅ We start by establishing the characterizations of (CA).
Theorem 2.2.1 The following statements are equivalent:
(c) For all x¯∈dom(f +g+k◦H) and all ≥0,
Proof [(a) ⇒ (b)] Applying the Fenchel inequality to the conjugate functions f ∗ , g ∗ , (λH) ∗ , and k ∗ , it is easy to see that for all x ∗ ∈X ∗ ,
To proceed, letx ∗ ∈X ∗ If (f+g+k◦H) ∗ (x ∗ ) = +∞,then the conclusion arises trivially by (2.3) So we can considerx ∗ ∈dom(f+g+k◦H) ∗ Then x ∗ ,(f+g+k◦H) ∗ (x ∗ )
∈ epi(f +g+k◦H) ∗ By (a) there exist λ ∈ domk ∗ ,(u, r)∈ epif ∗ ,(v, s)∈ epig ∗ , and (w, t)∈epi(λH−k ∗ (λ)) ∗ such that x ∗ ,(f +g+k◦H) ∗ (x ∗ )
(f +g+k◦H) ∗ (x ∗ ) = r+s+t≥f ∗ (u) +g ∗ (v) + (λH) ∗ (x ∗ −u−v) +k ∗ (λ). Combining the last inequality and (2.3) we get
The equality in (b) is confirmed, indicating that the minimum is achieved The inclusion in (c) is derived directly from the definition of the subdifferential To establish the converse inclusion, consider \( x^* \in \partial(f+g+k \circ H)(\bar{x}) \) Given that \( \bar{x} \) is within the domain of \( f+g+k \circ H \), Lemma 1.0.2 applies.
By (b) there exist λ ∈ domk ∗ , u ∈ domf ∗ , v ∈ domg ∗ , w ∈ dom(λH) ∗ such that u+v +w=x ∗ and
Combining this and (2.4), we get
Note that (u, f ∗ (u)) ∈ epif ∗ , and so, if we set 1 = f ∗ (u)− hu,xi¯ +f(¯x) ≥ 0 then by Lemma 1.0.2, u ∈ ∂ 1 f(¯x) Similarly, if we set 2 = g ∗ (v)− hv,xi¯ +g(¯x) ≥ 0 and
Now, (2.5) together with the fact that x ∗ =u+v +w yields
1+2+3+k ∗ (λ) + (k◦H)(¯x) =+ (λH)(¯x), which shows that x ∗ =u+v+wbelongs to the right-hand side of the relation in (c). [(c)⇒(a)] By Lemma 2.1.1, it is sufficient to show that epi(f+g+k◦H) ∗ ⊂epif ∗ + epig ∗ + [ λ∈dom k ∗ epi(λH−k ∗ (λ)) ∗
Take (x ∗ , r)∈epi(f+g+k◦H) ∗ and ¯x∈dom(f+g+k◦H) By Lemma 1.0.2 there exists≥0 such thatx ∗ ∈∂ (f+g+k◦H)(¯x) andr=+hx ∗ ,xi −¯ (f+g+k◦H)(¯x).
It now follows from (c) that there are λ ∈ domk ∗ , 1, 2, 3 ≥ 0, u ∈ ∂ 1f(¯x), v ∈∂ 2g(¯x), and w∈∂ 3(λH)(¯x) such that x ∗ =u+v +w and
Lets= 1 +hu,xi −¯ f(¯x),t = 2 +hv,xi −¯ g(¯x), q= 3 +hw,xi −¯ (λH)(¯x) Again, by Lemma 1.0.2, (u, s)∈epif ∗ , (v, t)∈epig ∗ , and (w, q)∈epi(λH) ∗ Moreover, s+t+q = 1 + 2 + 3 +hx ∗ ,xi −¯ f(¯x)−g(¯x)−(λH)(¯x)
⊂ epif ∗ + epig ∗ + [ λ∈dom k ∗ epi(λH−k ∗ (λ)) ∗ The proof is complete.
In the same way, we get
Theorem 2.2.2 The following statements are equivalent:
(f) For all x¯∈dom(f +g+k◦H) and all ≥0,
Theorem 2.2.3 The following statements are equivalent:
(i) For all x¯∈dom(f +g+k◦H) and all ≥0,
Characterizations for (CC), (CD), and (CF) are easy modifications and will be ignored.
Remark 2.2.1 By Theorem 2.1.1, if the condition (CA) holds then all the others(CB), (CC), (CD), (CE), and(CF) do, too So Theorems 2.2.1–2.2.3 yield variants of representations of (f+g+k◦H) ∗ (x ∗ ) which extend Moreau-Rockafellar-type formulae(see [10]), and representations of ∂ (f +g+k◦H)(¯x).
Special cases
Theorems 2.2.1–2.2.3 also yield many extended versions of results known in the literature Let us take an example, when k ≡0 and H ≡0, Theorem 2.2.1 turns back to a recent result in [14], which is a nonasymptotic version of [66, Corollary 2.6.7], and is also a relaxed version of a result in [47, Section II] where it is assumed that at least one of the functions f, g is continuous at a point in the domain of the other, which is stronger than the condition proposed in this corollary [14].
It is also worth noting that when f is convex and = 0 then -subdifferential off,
∂ f(ã) reduces to the ∂f(ã) in the sense of convex analysis The following corollary is a direct consequence of Theorem 2.2.1.
Corollary 2.2.1 Assume that f, g ∈ Γ(X) and λH ∈ Γ(X) for all λ ∈ domk ∗ If (CA) holds then for each x¯∈dom(f+g+k◦H), one has
In particular, if epi (f+k◦H) ∗ = epif ∗ + S λ∈domk ∗ epi(λH−k ∗ (λ)) ∗ then
Proof This is a direct consequence of Theorem 2.2.1 by taking = 0 in (c) Indeed, if = 0 then from the equality
1+2+3+k ∗ (λ) + (kH)(¯x) =+ (λH)(¯x), the Fenchel inequalityk ∗ (λ)≥λH(¯x)−k(H(¯x)), and the fact that 1 , 2 , 3 ≥0 we get
1 = 2 = 3 = 0 andk ∗ (λ) =λH(¯x)−k(H(¯x)) The last equality yieldsλ∈∂k(H(¯x)). The second assertion is obvious by taking g ≡0.
We now consider the case where H is a continuous linear operator H = A ∈ L(X, Y), where L(X, Y) denotes the set of all continuous linear operators from X to
Y, we then again find the results in convex analysis which were established recently in
[5, Theorem 7.5] and [9, Theorem 5.4] under the assumptions thatf is lsc convex, k is lsc convex and K-increasing Here the mentioned assumptions are removed.
We denote A ∗ to be the adjoint operator of A and (A ∗ ×Id R )(epik ∗ ) to be the image of the set epik ∗ through the functionA ∗ ×Id R :Y ∗ ×R−→X ∗ ×R, defined by (A ∗ ×Id R )(y ∗ , r) = (A ∗ y ∗ , r).
Corollary 2.2.2 Assume that A∈ L(X, Y) Assume further that epi (f +k◦A) ∗ = epif ∗ + (A ∗ ×Id R )(epik ∗ ) (2.6) Then
(b) If in addition, f ∈Γ(X) and k ∈Γ(Y) then for each x¯∈domf ∩A −1 (domk),
Proof We firstly observe that if λ∈domk ∗ then λA−k ∗ (λ)∗
(x ∗ , r)∈(A ∗ ×Id R )(epik ∗ ) ⇐⇒ (∃λ∈domk ∗ : A ∗ λ=x ∗ ,and (λ, r)∈epik ∗ )
⇐⇒ (x ∗ , r)∈ [ λ∈domk ∗ epi(λA−k ∗ (λ)) ∗ , which, together with (2.6), shows that epi (f +k◦A) ∗ = epif ∗ + (A ∗ ×Id R )(epik ∗ ) = epif ∗ + [ λ∈domk ∗ epi(λA−k ∗ (λ)) ∗ , and hence, (d) in Theorem 2.2.2 holds (i.e., (CB) holds) for the case where g ≡0 and
Note that the left-hand side of (2.7) is finite and for u∈domf ∗ , λA∗
Combine this and (2.7), one gets
• Proof of (b) It is easy to see that (λA) ∗ (u) = 0 whenu=A ∗ λ and that u∈A ∗ ∂k(A(¯x)) ⇐⇒ (∃λ ∈∂k(Ax), u¯ =A ∗ λ)
⇐⇒ (∃λ ∈∂k(Ax),¯ (λA) ∗ (u) +hA ∗ λ,xi¯ =hu,xi)¯
∂(λA)(¯x) = A ∗ ∂k(A¯x) Assertion (b) now follows fromCorollary 2.2.1 The proof is complete.
Nonconvex Farkas-type results
Nonconvex Farkas-type results
In the previous section, we established that the functions f, g, k, and λH (for any λ in Y*) are proper functions; however, they may not be convex or lower semicontinuous, as long as the sum f + g + k ◦ H is proper We will now present various characterizations of the functional inequality (1).
Theorem 2.3.1 Assume that (CA) holds and h ∈ Γ(X) The following statements are equivalent:
(II) For any x ∗ ∈X ∗ , there exist λ ∈domk ∗ , u∈domf ∗ , v ∈domg ∗ such that h ∗ (x ∗ )≥f ∗ (u) +g ∗ (v) + (λH) ∗ (x ∗ −u−v) +k ∗ (λ);
(III) For any x ∗ ∈X ∗ , there exist λ ∈domk ∗ , u∈domf ∗ such that h ∗ (x ∗ )≥f ∗ (u) + (g+λH) ∗ (x ∗ −u) +k ∗ (λ);
(IV) For any x ∗ ∈X ∗ , there exists λ ∈domk ∗ such that h ∗ (x ∗ )≥(f +g+λH) ∗ (x ∗ ) +k ∗ (λ);
(V) For any x ∗ ∈X ∗ , there exist λ ∈domk ∗ , u∈domg ∗ such that h ∗ (x ∗ )≥g ∗ (u) + (f +λH) ∗ (x ∗ −u) +k ∗ (λ);
(VI) For any x ∗ ∈X ∗ , there exist λ ∈domk ∗ , u∈dom(f +g) ∗ such that h ∗ (x ∗ )≥(f +g) ∗ (u) + (λH) ∗ (x ∗ −u) +k ∗ (λ);
(VII) For any x ∗ ∈X ∗ , there exists u∈domf ∗ such that h ∗ (x ∗ )≥f ∗ (u) + (g+k◦H) ∗ (x ∗ −u).
Proof • [(I)=⇒(II)] Assume that (I) holds Then h ∗ ≥(f+g+k◦H) ∗ which yields epih ∗ ⊂epi(f+g+k◦H) ∗ Letx ∗ ∈X ∗ Ifh ∗ (x ∗ ) = +∞,then (II) holds trivially So we can considerx ∗ ∈domh ∗ , it is clear that x ∗ , h ∗ (x ∗ )
∈epih ∗ ⊂epi(f+g+k◦H) ∗ Since (CA) holds, there exist λ ∈ domk ∗ , (u, r) ∈ epif ∗ , (v, s) ∈ epig ∗ ,(w, t) ∈ epi(λH−k ∗ (λ)) ∗ such that x ∗ , h ∗ (x ∗ )
Thush ∗ (x ∗ ) =r+s+t ≥f ∗ (u) +g ∗ (v) + (λH) ∗ (x ∗ −u−v) +k ∗ (λ) Note that here we have u∈domf ∗ , v ∈domg ∗ , and w=x ∗ −u−v ∈dom(λH) ∗
Letx ∗ ∈X ∗ For u∈domf ∗ , v ∈domg ∗ and λ∈domk ∗ whose existence ensure from (II), one has
So, (III) follows from this and (II).
This assertion can be proved by the same argument as in that of (II) =⇒(III)
Take arbitrarily x∈X, x ∗ ∈X ∗ By (IV), there existsλ∈domk ∗ satisfying h ∗ (x ∗ )≥(f +g+λH) ∗ (x ∗ ) +k ∗ (λ), which gives rise to hx ∗ , xi −h ∗ (x ∗ ) ≤ hx ∗ , xi −(f +g+λH) ∗ (x ∗ )−k ∗ (λ)
(note that the inequality still holds even x ∗ ∈/ domh ∗ ) Since h ∈ Γ(X), taking the supremum over allx ∗ ∈X ∗ , we get h(x) = h ∗∗ (x)≤(f+g+k◦H)(x), which is (I).
The proofs of the implications
(II) =⇒(V) and (V) =⇒(IV) are similar to that of
, using the inequality concerning the conjugate of the sum and the convolution of two conjugates of functions (see (1.7)).
For any x ∗ ∈X ∗ ,u∈domg ∗ , v ∈domf ∗ and λ∈domk ∗ , one has
So, (V) follows from this and (II).
For any x ∗ ∈X ∗ ,u∈domg ∗ and λ∈domk ∗ , one has
(IV) follows from this and (V).
Assume that (I) holds Then h ∗ ≥(f +g+k◦H) ∗ and epi h ∗ ⊂ epi(f +g +k ◦H) ∗ Let x ∗ ∈ X ∗ If h ∗ (x ∗ ) = +∞, then (VI) holds trivially So we can consider x ∗ ∈ domh ∗ , then x ∗ , h ∗ (x ∗ )
∈ epi h ∗ ⊂ epi(f +g +k◦H) ∗ Since (CA) holds then (CC) also holds, there exist λ ∈ domk ∗ ,(u, r) ∈ epi(f +g) ∗ and (v, s)∈epi(λH−k ∗ (λ)) ∗ such that x ∗ , h ∗ (x ∗ )
Thush ∗ (x ∗ ) =r+s ≥(f+g) ∗ (u) + (λH) ∗ (x ∗ −u) +k ∗ (λ) Obviously,u∈dom(f+g) ∗ and v =x ∗ −u∈dom(λH) ∗
Letx ∗ ∈X ∗ For u∈dom(f +g) ∗ and λ ∈domk ∗ , one has (f +g+λH) ∗ (x ∗ )≤ (f +g) ∗ 2(λH) ∗
So, (IV) follows from this and (VI).
Assume that (I) holds From the fact that h ∗ ≥ (f+g+k◦H) ∗ and epi h ∗ ⊂ epi(f +g +k ◦H) ∗ , for any x ∗ ∈ X ∗ , we can consider x ∗ ∈ domh ∗ , x ∗ , h ∗ (x ∗ )
∈ epi h ∗ ⊂ epi (f +g +k◦H) ∗ In this case, it follows from (CF) (as (CA) holds) that there exist (u, r)∈epif ∗ and (v, s)∈epi(g+k◦H) ∗ such that x ∗ , h ∗ (x ∗ )
Let x ∈ X It follows from (VII) that for any x ∗ ∈ X ∗ (there is u∈domf ∗ ), hx ∗ , xi −h ∗ (x ∗ ) ≤ hx ∗ , xi −f ∗ (u)−(g+k◦H) ∗ (x ∗ −u)
≤ hx ∗ , xi − hu, xi+f(x)− hx ∗ −u, xi+ (g+k◦H)(x)
(note that the inequality still holds even x ∗ ∈/ domh ∗ ) Since h ∈ Γ(X), taking the supremum over x ∗ ∈domh ∗ , we get h(x) =h ∗∗ (x)≤(f +g+k◦H)(x) which is (I) The proof is complete.
Remark 2.3.1 It is worth noting that one of the equivalent pairs between (I)and one among the left (i.e., (II)–(VII)), is a characterization of (1) or a Farkas-type result.
Theorem 2.3.1 presents six distinct versions of Farkas-type results related to the specified inequality system While it provides only sufficient conditions, this theorem broadens and encompasses various existing Farkas-type results found in the literature, as referenced in sources [8], [21], [30], [31], [41], and [37].
This section delves deeper into the Farkas-type results outlined in Theorem 2.3.1 In Theorem 2.3.2, we will present necessary and sufficient conditions for each of the equivalent pairs discussed previously.
Theorem 2.3.2 The following assertions hold:
The assertions in (i) and (ii) have been validated as demonstrated in [32] The proofs for these assertions follow a reasoning akin to that of Theorem 2.3.1 Therefore, we will elaborate only on the proof of assertion (iii), while the proofs for the other assertions will be summarized.
The (CC) condition leads to the equivalence of (I) and (VI) for all h in Γ(X) The proof that (I) implies (VI) under (CC) mirrors that of Theorem 2.3.1, while the converse, (VI) implies (I), holds unconditionally Specifically, for any x∗ in X∗, (VI) guarantees the existence of u in dom(f+g)∗ and λ in domk∗ such that h∗(x∗) is greater than or equal to (f+g)∗(u) plus (λH)∗(x∗−u) plus k∗(λ) Consequently, for any x in X, the inequality hx∗, xi - h∗(x∗) is less than or equal to hx∗, xi - hu, xi plus (f+g)(x) minus hx∗−u, xi plus hλ, H(x)i.
(note that the inequality still holds even x ∗ ∈/ domh ∗ ) Now, taking the supremum overx ∗ ∈X ∗ with the fact that h∈Γ(X), we get h(x) =h ∗∗ (x)≤(f +g+k◦H)(x) ∀x∈X, which shows that (I) holds.
• We now show that [∀h∈Γ(X), (I)⇐⇒ (VI)] ensures (CC) By Lemma 2.1.1, it is sufficient to show that epi(f +g+k◦H) ∗ ⊂ C.
Let (x ∗ , r) ∈ epi(f +g +k ◦H) ∗ Then (f +g +k◦H)(x) ≥ hx ∗ , xi −r for all x ∈ X In other words, (I) holds with h(ã) = hx ∗ ,ãi −r and hence, (VI) holds and consequently, there exist λ∈domk ∗ , u∈dom(f +g) ∗ such that
Note that here we have x ∗ −u ∈ dom(λH) ∗ = dom λH − k ∗ (λ)∗
+ epi(f +g) ∗ + epi(λH −k ∗ (λ)) ∗ ⊂ C.The proof is complete.
Remark 2.3.2 Characterizations of Farkas-type results given in Theorem 2.3.2 are very general In special cases, they produce new results, extend or cover many knownFarkas-type results for systems involving convex and DC functions as will be shown in the next section.
Special cases
This subsection highlights the importance of the general Farkas-type results by presenting special cases that yield new versions and extensions for both convex and nonconvex systems, including those involving DC functions Additionally, it provides necessary and sufficient conditions for achieving strong stable Lagrange duality in a general convex optimization problem, while maintaining the same assumptions regarding spaces, sets, and functions as outlined in Subsection 2.3.1.
The case when h is an affine function
Taking has an affine function, h(ã) = hx ∗ ,ãi+α whereα∈R and x ∗ ∈X ∗ , we get from Theorem 2.3.2 (v),
Proposition 2.3.1 Let C be a nonempty subset of X The following statements are equivalent:
(a) epi (g+i C +k◦H) ∗ = S λ∈domk ∗ epi (g+i C +λH−k ∗ (λ)) ∗ ; (b) For any α∈R and any x ∗ ∈X ∗ , g(x)+(k◦H)(x)≥ hx ∗ , xi+α∀x∈C
Proof [(a)⇒(b)] The implication follows from Theorem 2.3.2(v) when taking f =i C and h(x) =hx ∗ , xi+α for all x∈X (and hence, domh ∗ ={x ∗ }and h ∗ (x ∗ ) =−α).
[(b)⇒(a)] Assume (b) holds In order to show that (b) holds, by Lemma 2.1.1, it is sufficient to show that epi (g+i C +k◦H) ∗ ⊂ [ λ∈domk ∗ epi (g+i C +λH−k ∗ (λ)) ∗
If (x ∗ , r)∈epi (g+iC +k◦H) ∗ then g(x) + (k◦H)(x)≥ hx ∗ , xi −r, ∀x∈C.
It now follows from (b) that (−r plays the role ofα) there exists λ∈domk ∗ such that g(x) + (λH)(x)≥ hx ∗ , xi −r+k ∗ (λ), ∀x∈C.
This is equivalent to r≥(g+i C +λH−k ∗ (λ)) ∗ (x ∗ ), which means that
Now consider the case whereY is preordered by a nonempty convex coneK Letting k =i−K, we get from Proposition 2.3.1,
Corollary 2.3.1 Let C be a nonempty subset of X and let K be a nonempty convex cone in Y Assume that C∩dom (g+i−K◦H)6=∅ Then the following assertions are equivalent:
(a) epi (g+iC+i−K◦H) ∗ = S λ∈K + epi (g+iC +λH) ∗ ; (b) For any x ∗ ∈X ∗ and any α∈R,
Remark 2.3.3 Corollary 2.3.1 is an extension of generalized Farkas lemmas in [45, Theorem 3.1], [29, Corollary 5.2], [31, Corollary 6.2], and in [21, Theorem 2.2] (where withC=X) to nonconvex systems The extension is twofold: Firstly, it is a nonconvex version of results mentioned in the previous papers Namely, the closedness of C and the cone K, the convexity of the function g, K-convexity of the mapping H are all removed Secondly, Corollary 2.3.1 gives necessary and sufficient conditions for this Farkas lemma version while only the sufficient condition was proposed in the mentioned papers.
Characterization of Farkas-type results with composed functions in convex setting
In this article, we consider a nonempty closed convex subset C of a space X and a closed convex cone K in a space Y We examine a function f belonging to the class Γ(X) and a mapping H that is K-convex Additionally, we assume that for every positive scalar λ in K, the scaled mapping λH is also in Γ(X), and that the intersection of C with the preimage of -K under H is non-empty These conditions lead to important implications derived from Theorem 2.3.2.
Corollary 2.3.2 The following statements (a) and (b) are equivalent:
(b) For any h∈Γ(X), the following assertions are equivalent:
In the proof, we define Letk as i − K and g as i C, establishing that k belongs to Γ(Y) and g belongs to Γ(X) It is evident that the domain of k* is K+, and k*(λ) equals 0 for all λ in K+ This indicates that condition (CE) holds true Additionally, the statement (i) corresponds directly to (I) in this specific context, while (IV) transforms to h*(x*) being greater than or equal to (f + i C + λH)*(x*), which aligns with condition (ii) The corollary's conclusion is derived from Theorem 2.3.2(v).
Remark 2.3.4 Farkas-type result for systems involving DC function of the form (b) in the previous corollary was established in [29], [31] However, Corollary 2.3.2 is stronger than the one in [29], [31] in the sense that Corollary 2.3.2 gives a necessary and sufficient condition for (b) while only the sufficient condition was given in the mentioned papers Our condition(in (a)) is(CE1), which is weaker than the condition in the mentioned references which is equivalent to (CA1) The same comment applies to the next corollaries in comparison with their counterparts one in [29], [31], [37], [45].
In the case where h is an affine function we get the following from Corollary 2.3.2. Corollary 2.3.3 The following statements are equivalent:
(a) S λ∈K + epi (f+iC+λH) ∗ is weak ∗ -closed;
Proof It is clear that the implication [(a)⇒ (b)] is a direct consequence of Corollary 2.3.2 The proof of [(b)⇒(a)] is similar to that of Proposition 2.3.1.
Remark 2.3.5 Consider the convex optimization problem (P):
Then the equivalence in (b) of Corollary 2.3.3 is nothing but
In convex optimization, the relationship H(x)∈−K, x∈Cinf [f(x)− hx ∗ , xi] = sup λ∈K + x∈Cinf[f(x) + (λH)(x)− hx ∗ , xi] is recognized as stable strong Lagrange duality for problem (P) Corollary 2.3.3 outlines the necessary and sufficient conditions for this duality to hold, as established in [5, Theorem 8.3] and [32, Corollary 4], which is based on a specific assumption regarding H.
“K-epi closed” which is a bit weaker than our assumption that λH ∈ Γ(X) for all λ ∈ K + However, for the sake of simplicity, we do not consider this notion in this thesis.
Applications
Alternative-type theorems
In this subsection, Theorem 2.3.2 (and its consequences such as Proposition 2.3.1, Corollaries 2.3.1, 2.3.2, 2.3.3) will be used to derive alternative-type theorems.
Theorem 2.3.2 presents alternative theorems that do not require convexity or lower semi-continuity To illustrate this, we will focus on one specific example, Theorem 2.3.2(iii).
Theorem 2.4.1 Let f, g : X → R ∪ {+∞}, k : Y → R ∪ {+∞} be proper and
H : domH ⊂X →Y such that f+g+k◦H is proper Then, epi (f+g+k◦H) ∗ =C if and only if for any h∈Γ(X), precisely one of the following statements is true: (i) ∃x∈X, f(x) +g(x) +k[H(x)]< h(x);
(ii) For any x ∗ ∈domh ∗ , there exist λ∈domk ∗ , u∈dom (f +g) ∗ such that h ∗ (x ∗ )≥(f +g) ∗ (u) + (λH) ∗ (x ∗ −u) +k ∗ (λ).
Proof Observe that non-(i) is equivalent to (I) and (ii) is nothing but (VI) Then, the conclusion follows from Theorem 2.3.2(iii) yields the result.
Theorem 2.4.1 provides necessary and sufficient conditions for the alternative result, which is significant in understanding alternative characterizations Similar characterizations in nonconvex settings under various qualification conditions are discussed in reference [32].
In the same way, for a special case where h is a constant function, we get from Proposition 2.3.1,
Corollary 2.4.1 Let g, k, H be as in Theorem 2.4.1 If epi(g +k◦H) ∗ = S λ∈dom k ∗ epi (g+λH−k ∗ (λ)) ∗ then for any α∈R, precisely one of the following statements is true
Turning to the convex setting, Corollary 2.3.2 gives
Corollary 2.4.2 Let C be a nonempty closed, convex subset of X, K be a closed, convex cone in Y, and let f ∈ Γ(X) Assume that λH ∈ Γ(X) for all λ ∈ K + and
C∩H −1 (−K)6=∅ Then, S λ∈K + epi (f+i C +λH) ∗ is weak ∗ -closed if and only if for any h ∈Γ(X), precisely one of the following statements is true
(ii) ∀x ∗ ∈domh ∗ ,∃λ∈K + such that f(x) +λH(x)≥ hx ∗ , xi −h ∗ (x ∗ ) ∀x∈C.
The alternative result stems from Corollary 2.3.3, which has been discussed in previous studies and is commonly referred to as the generalized Farkas lemma for systems constrained by convex cones, although it was presented under more stringent qualification conditions in sources such as [31] and [37].
Corollary 2.4.3 Let C be a nonempty closed, convex subset of X, K be a closed, convex cone in Y, and let f ∈ Γ(X) Assume that λH ∈ Γ(X) for all λ ∈ K + and
C∩H −1 (−K)6=∅ If S λ∈K + epi (f+i C +λH) ∗ is weak ∗ -closed then for any α∈ R, precisely one of the following statements is true
Set containments
Theorem 2.3.2 can also be used to derive characterizations (necessary and sufficient conditions) of set containment results in very general cases As illustration examples, we give here some characterization of containments of convex sets in a reverse convex set (see [65]) or the one of a convex set in a DC set (defined by a DC function) For other results of this type, see [40] and the references therein.
Assuming C is a closed and convex subset of X, and f belongs to the class of functions Γ(X), we consider the S-convex mapping H defined on its domain mapping to Y, with K being a closed convex cone in Y We define g as the indicator function of C, thus g also belongs to Γ(X) The sets [H ≤ K 0] and [f ≥ h] are defined as {x ∈ X | H(x) ∈ -K} and {x ∈ X | f(x) ≥ h(x)}, respectively For any function h in Γ(X), the inequality f + i_C + i_{-K}◦H ≥ h holds true.
The containment of a convex set within a DC set is represented as C∩[H ≤ K 0]⊂[f ≥ h], where the DC set is defined by the difference of two convex functions (specifically, [f − h ≥ 0]) When h is equal to zero, this relationship simplifies to the containment of a convex set within a reverse convex set.
This observation allows us to utilize Theorem 2.3.2 to establish set containments in a broad context For simplicity, we will concentrate exclusively on the previously mentioned convex case Assuming that the intersection of C and [H ≤ K 0] is not empty, we derive important results from Corollary 2.3.2.
Corollary 2.4.4 The set S λ∈K + epi (f+i C +λH) ∗ is weak ∗ -closed if and only if for any h ∈Γ(X) the following statements are equivalent:
By treating h as a constant function in Corollary 2.4.4, we derive a characterization for the containment of a convex set defined by a cone-constrained and a reverse convex set, which broadens the findings presented in references [40] and [31].
Corollary 2.4.5 Letα ∈R If the set S λ∈K + epi (f+iC+λH) ∗ is weak ∗ -closed then the following statements are equivalent:
Fenchel-Rockafellar duality formula
We now consider the case where f and k are proper functions, g ≡ 0, h ≡ 0, and
H = A ∈ L(X, Y), where L(X, Y) denotes the set of all continuous linear operators from X to Y Assume that dom (f+k◦A)6=∅ Let A ∗ denote the adjoint operator of A Consider the problem (FR) as follow
The classical Fenchel dual problem to (FR) is defined as the problem (DFR) below sup y ∗ ∈Y ∗
As a consequence of Theorem 2.3.2, we get
Proposition 2.4.1 [Fenchel-Rockafellar duality formula]Assume thatinf(F R)∈
R If the following qualification condition holds: epi (f+k◦A) ∗ = epif ∗ + (A ∗ ×Id R )(epik ∗ ) then the strong duality holds between (FR) and (DFR) and the dual problem (DFR) possesses at least one solution, i.e., x∈Xinf{f(x) +k(A(x))}= max y ∗ ∈Y ∗ {−f ∗ (A ∗ y ∗ )−k ∗ (−y ∗ )}.
In Corollary 2.2.2, it is demonstrated that if condition (2.6) is satisfied, then the epigraph of the function \( f + k \circ A \) can be expressed as the sum of the epigraph of \( f \) and a term involving the epigraph of \( k \), indicating that the qualification condition (CC) holds with \( g \equiv 0 \) and \( H = A \in L(X, Y) \) Furthermore, defining \( h(x) = \alpha \) for all \( x \in X \) leads to \( g^*(0) = 0 \) and \( h^*(0) = -\alpha \), while both \( g^*(x^*) \) and \( h^*(x^*) \) equal \( +\infty \) for \( x^* \neq 0 \).
In the context of Theorem 2.3.2, the condition inf (FR) = α ∈ R is equivalent to the inequality f(x) + k(A(x)) ≥ α for all x in X If the condition in equation (2.6) is satisfied, Theorem 2.3.2(iii) indicates that this inequality (2.9) is equivalent to (VI), which asserts the existence of y* in the domain of k* and u in the domain of f*.
The equation (y ∗ A) ∗ (−u) equals zero when u is equal to −A ∗ y ∗, while in all other scenarios, it results in +∞, as demonstrated in the proof of Corollary 2.2.2 This leads to the inequality −α being equivalent to f ∗ (−A ∗ y ∗ ) + k ∗ (y ∗ ) Consequently, we find that α equals the infimum of FR, which is less than or equal to −f ∗ (A ∗ y˜ ∗ ) − k ∗ (−˜y ∗), where ˜y ∗ represents −y ∗ The conclusion is supported by the principle of weak duality, which is inherently valid.
Remark 2.4.1 The Fenchel-Rockafellar duality formula (i.e., strong duality between(FR) and (DFR)) in the case where f and k are convex and lower semi-continuous was given in [58, Theorem 31.2] (for finite dimensional), in [66, Corollary 2.8.5], [10],and in [7] under different kinds of qualification conditions Interior-type condition was used in [58], [66] and the condition (2.6) was used in [10] and [7] The mentionedFenchel-Rockafellar duality formula was also established recently in [49] for the convex case (i.e., f, k are convex) without any assumptions on the lower semi-continuity, and under the same qualification condition (2.6) The same result was given in [5, Remark 7.1] in the absence of convexity but the lower semicontinuity of f and k is still needed. Proposition 2.4.1 extends the mentioned results to nonconvex case or to the case where the assumption on the lower semicontinuity is removed.
If we set Y = X and A = Id X, the condition (2.6) simplifies to the requirement that epi(f+k)* = epi f* + epi k*, leading to the conclusion of Proposition 2.4.1, which states that x ∈ X inf{f(x) + k(x)} = max y* ∈ X* {-f*(y*) - k*(-y*)} This represents the nonconvex version of the Fenchel theorem A similar qualification condition for convex, lower semi-continuous functions f and g has been established recently.
[15] Our result for this special case also generalizes the ones in [2] and [57].
New versions of Farkas lemma and Hahn-Banach theorem under
This chapter presents new formulations of the Farkas lemma for cone-convex systems and extended sublinear functions, under specific Slater-type constraints without relying on lower semi-continuity or closedness of functions and sets It introduces an extended Hahn-Banach-Lagrange theorem applicable in locally convex Hausdorff spaces, which does not require lower semi-continuity or closed constrained sets This theorem leads to enhanced versions of the Mazur-Orlicz theorem, the sandwich theorem, and the Hahn-Banach theorem for extended sublinear functions The chapter demonstrates the equivalence of the generalized Farkas lemma and the extended Hahn-Banach-Lagrange theorem, applying these results to derive duality and optimization conditions for composite problems involving sublinear-convex mappings Additionally, it explores three specific cases that yield a general formula for the conjugate of the supremum of a possibly infinite family of convex functions, a generalized Fenchel duality theorem, a separation theorem for convex sets in normed spaces, and optimality conditions for penalty problems in convex programming.
New versions of the Farkas lemma under Slater-type conditions
Farkas lemma for cone-convex systems
We present a novel version of the Farkas lemma applicable to convex systems that adhere to the Slater constraint qualification This proof builds on findings from previous research Notably, this result is innovative as it does not require the lower semi-continuity of functions or mappings, nor the closedness of the constrained set Additionally, it broadens the scope of the Farkas lemma previously established in earlier work.
Theorem 3.1.1 [Farkas lemma for cone-convex systems]LetX, Y be l.c.H.t.v.s.,
C be a nonempty convex subset of X, K be a closed convex cone in Y, f : X →
R∪ {+∞} be a proper convex function, g : X → Y • be a K-convex mapping, and β ∈R Assume that the Slater condition holds, i.e.,
(SC1) ∃¯x∈(domf)∩C such that g(¯x)∈ −intK.
Then the following statements are equivalent:
(ii) there exists y ∗ ∈K + such that f +y ∗ ◦g ≥β on C.
Proof Let us set A:=C∩(domg) and consider the perturbed function Φ :X×Y →
The convexity of function f and cone K, along with the K-convexity of function g, demonstrates that Φ is convex Additionally, for any points x∗ in X∗ and y∗ in Y∗, a straightforward calculation reveals that Φ∗(x∗, -y∗) equals the supremum of the expression {hx∗, xi + h - y∗, yi - Φ(x, y)}, where x is in set X and y is in set Y.
= sup{hx ∗ , xi − hy ∗ , yi −f(x) : x∈C, g(x)−y ∈ −K, y∈Y}
= sup{hx ∗ , xi − hy ∗ , g(x) +pi −f(x) : x∈A, p∈K}
= sup{hx ∗ , xi − hy ∗ , g(x)i −f(x) : x∈A}+ sup{−hy ∗ , pi : p∈K}, and hence, Φ ∗ (x ∗ ,−y ∗ ) ( sup{hx ∗ , xi −(y ∗ ◦g)(x)−f(x) : x∈A} if y ∗ ∈K + ,
Seth defines the function h(y) as the infimum of Φ(x, y) over the set X, indicating that h is convex due to the properties of Φ Furthermore, we assert that Φ(¯x, ) is continuous at the origin 0Y Given that condition (SC1) is satisfied, there exists a neighborhood V0 around 0Y such that the set −g(¯x) combined with V0 is contained within K Consequently, for any ε > 0, the relationship holds.
It now follows from Theorem 2.7.1 in [66] that either h(0 Y ) = −∞ or h(0 Y ) ∈ R and h|Y 0 is continuous at 0 Y Moreover, in both cases there exists ¯y ∗ ∈Y ∗ such that x∈Xinf Φ(x,0 Y ) = −Φ ∗ (0 X ∗ ,−¯y ∗ ) (3.2)
Note that (i) is equivalent to h(0 Y ) = inf X Φ(x,0 Y ) ≥ β and (ii) is nothing else but
The theorem concludes that Φ ∗ (0 X ∗ ,−¯y ∗ ) is minimized when infx∈C f(x) + (¯y ∗ ◦g)(x) is greater than or equal to β, given that ¯y ∗ belongs to K + This condition is satisfied if either (i) or (ii) is true, leading to Φ ∗ (0 X ∗ ,−¯y ∗ ) being less than +∞, which confirms that ¯y ∗ is indeed in K + according to (3.1) Thus, the proof is complete.
Remark 3.1.1 It is worth noting that the above version of the Farkas lemma (i.e., the equivalence between (i) and (ii) in Theorem 3.1.1) was proved in several other works(see the recent survey [20] and references therein) under some constraint qualification condition such as the “closedness condition” in [22], [31] However, for these mentioned versions, the closedness of the convex set C, the lower semi-continuity of f and y ∗ ◦g for all y ∗ ∈ K + are always required Theorem 3.1.1 assumes a stronger constraint qualification condition to get the benefit of removing all the mentioned assumptions.
Farkas lemma for sublinear-convex systems
We apply the technique from [22] to adapt Theorem 3.1.1 into a revised version of the Farkas lemma for sublinear-convex (S-convex) systems This new iteration parallels Theorem 3.4 in [22], offering relaxed assumptions on functions, mappings, and constrained sets, while imposing a more stringent constraint qualification.
Theorem 3.1.2 Let X, Y be l.c.H.t.v.s., C be a nonempty convex subset of X, S :
Y → R∪ {+∞} be an lsc sublinear function, g : X → Y • be an S-convex mapping, and let f : X → R∪ {+∞} be a proper convex function, ψ : R → R∪ {+∞} be a proper convex function Assume that the following Slater-type constraint qualification condition holds:
(SC2) ∃¯a∈C∩(domf), ∃¯α∈R such that (¯α,+∞)∩(domψ)6=∅ and g(¯a)∈int{y∈Y : S(y)≤α}.¯ Then the following statements are equivalent:
(b) there exist γ ≥0 and y ∗ ∈Y ∗ such that y ∗ ≤γS on Y and f +y ∗ ◦g ≥ψ ∗ (γ) on C (3.3)
The proof mirrors the approach taken in Theorem 3.4 from [22] The central concept is to revert to the framework established in Theorem 3.1.1, demonstrating that the assumption (SC2) ensures (SC1) within this context, thereby allowing the application of Theorem 3.1.1.
[(a) =⇒ (b)] Assume that (a) holds Let us set Xe := X ×R, Ye = Y ×R,
The equations Ye = Y • × R and Ce = C × R define the set Se: Ye → R ∪ {+∞}, where S(y, λ) = e S(y) − λ for all (y, λ) ∈ Ye, indicating that Se is a lower semicontinuous (lsc) extended sublinear mapping Additionally, the mappings eg: Xe → Ye • and fe: Xe → R ∪ {+∞} are established, with eg(x, α) = (g(x), α) for all (x, α) ∈ Xe and f(x, α) = e f(x) + ψ(α) for all (x, α) ∈ Xe.
Since f, ψ are proper and convex, so is fe Similarly, as g is S-convex, eg is S-convex.e Moreover,geisK-convex withe Ke being the closed convex cone defined byKe :={(y, λ)∈
We now try to apply Theorem 3.1.1 with X,e Ye, C,e eg, fe, and Ke playing the roles of X, Y,C, g, f, and K, respectively, and with β = 0.
In the context of the Slater condition (SC1) for the function f, it is established that any value of α satisfying the condition in (SC2) leads to the inclusion of the set {y∈Y : S(y)≤α} × [α,+∞) within the negative cone -Ke Consequently, there exists a value α greater than α¯ (with α belonging to the domain of ψ) such that the pair (a, α) is part of the intersection of Ce and the domain of fe, and the expression eg(a, α) equals (g(a), α) within the interior of the negative cone -intK,e.
Secondly, from the definitions of f,e eg, and C, assertion (a) becomese
In the context of Theorem 3.1.1, we observe that for any point (x, α) in the set C, if the condition eg(x, α) = (g(x), α) belongs to the negative cone Ke, it follows that the function f(x, α) must be greater than or equal to zero Consequently, there exists a solution ye ∗ = (y ∗ ,−γ) within the positive cone Ke +, ensuring that the inequality fe + ye ∗ ◦ eg remains non-negative across the set C This leads us to the conclusion that the expression f(x) + ψ(α) + (y ∗ ◦ g)(x) - γα must be non-negative for all (x, α) in C, or alternatively, that f(x) + (y ∗ ◦ g)(x) is greater than or equal to γα - ψ(α) for all (x, α) in Ce.
(which holds even when α6∈ domψ) Taking the supremum over all α ∈R in the last inequality, (3.4), we get f(x) + (y ∗ ◦g)(x)≥ψ ∗ (γ), ∀x∈C.
Now, since S(0e Y,1) = S(0Y)−1 = −1 < 0, we have (0Y,1) ∈ −K, and hence,e ey ∗ (0Y,1) =y ∗ (0Y)−γ =−γ ≤0 as ye ∗ = (y ∗ ,−γ)∈Ke + Consequently, γ ≥0.
For every element y in the domain of S, the equation S(y, S(y)) equals zero, which implies that (y, S(y)) is included in the negative cone K This leads to the conclusion that y*(y) minus γS(y) is less than or equal to zero, indicating that y*(y) is less than or equal to γS(y) for all y within the domain of S Notably, this inequality also applies to all elements in Y that are outside the domain of S, resulting in the conclusion that y* is less than or equal to γS across the entire set Y.
[(b) =⇒ (a)] Assume that there are γ ≥ 0 and y ∗ ∈Y ∗ such that y ∗ ≤γS on Y and (3.3) holds Then for all x∈C, α∈R, f(x) + (y ∗ ◦g)(x)≥ψ ∗ (γ)≥γα−ψ(α) (3.5)
Since y ∗ ≤γS onY, (y ∗ ◦g)(x)≤ γ(S◦g)(x) for all x∈ X So, if x∈ C and α ∈ R satisfy (S◦g)(x)≤αthen (y ∗ ◦g)(x)≤γ(S◦g)(x)≤γα asγ ≥0, and it follows from (3.5) that f(x) +γα≥γα−ψ(α), or, f(x) +ψ(α)≥0, which means that (a) holds The proof is complete.
Remark 3.1.2 Theorem 3.1.2 can be considered a topological extension of the algebraic version of Lemma 5.3 in [60] whereS is a real-valued function andg is a mapping with its values inY and only the implication [(a)⇒(b)] is established To see this relation, we first recall the concept of (S, g)-compatible introduced in [60].
Definition 3.1.1 [60] Let Y be a nontrivial vector space, C be a nonempty convex subset of a vector space Let further S :Y →R, g :C →Y A proper convex function ψ :R→R∪{+∞}is said to be(S, g)-compatibleif(domψ)∩S a∈C((S◦g)(a),+∞)6∅.
This condition means that there exista∈Candα∈domψ such that (S◦g)(a)< α.
It is also worth noting that if ψ is a real-valued convex function then ψ is (S, g)- compatible whatever the values ofS and g are.
Now, the result in [60, Lemma 5.3] comes as a consequence of Theorem 3.1.2 Again, the proof follows the same line as in [22].
Corollary 3.1.1 [60] Let Y be a nontrivial vector space, S : Y → R be a sublinear function, C be a nonempty convex subset of a vector space X, g : C → Y be an S- convex mapping, f :C→R be a convex function, and ψ :R→R∪ {+∞} be a proper convex function Suppose that ψ is (S, g)-compatible Then the following statements are equivalent:
(ii) there exist γ ≥0 and a linear functional L on Y such that L≤γS on Y and f +L◦g ≥ψ ∗ (γ) on C (3.6)
Proof Let us equip X, Y with the finest locally convex Hausdorff topologies τ X , τ Y , respectively (i.e., the topologies defined by the set of all semi-norms on the corre- sponding spaces) [12] Define p:Y →R [22] p(y) := max{S(y), S(−y)}, ∀y∈Y.
In this context, p is identified as a semi-norm, which implies its continuity with respect to the topology τ Y This continuity indicates that p remains bounded above in a neighborhood surrounding 0 Y, and consequently, S is also bounded above in this vicinity since S is less than or equal to p As a result, S is confirmed to be continuous within the interior of its domain, specifically int(domS) = Y Additionally, the functions ¯f: X → R ∪ {+∞} and ¯g: X → Y • are defined by f(x) := ¯.
Then ¯f is proper convex and ¯g is S-convex.
Since ψ is (S, g)-compatible, there exist values ¯a in C and ¯β in domψ such that (S◦g)(¯a) is less than β Due to the density of real numbers, there is a value ¯α in R satisfying (S◦g)(¯a) < α < ¯β This implies that the intersection of (¯α, +∞) and (domψ) is non-empty, with ¯α being greater than the infimum of S over Y Additionally, as S is a continuous sublinear function, it follows that ¯g(¯a) belongs to the interior of the set {y in Y : S(y) ≤ α}.
The condition (SC2) is satisfied for the functions ¯g and ¯f, indicating that when x is in set C and α is a real number, the relationship (S ◦¯g)(x) = (S ◦g)(x) ≤ α leads to ¯f(x) + ψ(α) = f(x) + ψ(α) ≥ 0, which reflects statement (a) of Theorem 3.1.2 According to this theorem, it is equivalent to the existence of γ ≥ 0 and L ∈ Y ∗ such that L ≤ γS on Y, thereby confirming that the inequality (3.6) in statement (ii) holds true The proof is thus complete.
New versions of the Hahn-Banach theorem under Slater-type conditions 38
Extended Hahn-Banach-Lagrange theorem
In this article, we assume that X and Y are locally convex Hausdorff topological vector spaces, while C represents a nonempty convex subset of X, which may not be closed This leads us to important conclusions derived from Theorem 3.1.2.
Theorem 3.2.1 [Extended Hahn-Banach-Lagrange theorem] Let S : Y →
R∪ {+∞} be an lsc sublinear function, g :X →Y • be an S-convex mapping, and let f :X →R∪ {+∞} be a proper convex function Assume that the following condition holds:
Then the following statements are equivalent:
∈R, (ii) there exists y ∗ ∈Y ∗ such that y ∗ ≤S on Y and infC f+y ∗ ◦g
Proof Let ψ(λ) = λ for all λ ∈ R Note that ψ ∗ (γ) = 0 if γ = 1 and ψ ∗ (γ) = +∞ if γ 6= 1.
It is clear that only the implication [(i) =⇒ (ii)] needs to prove The converse one is straightforward Assume that (i) holds, i.e., β := inf C f +S◦g
∈ R Then f+S◦g ≥β onC It follows that ifx∈C, α∈R,(S◦g)(x)≤α then f(x) +ψ(α) = f(x) +α≥f(x) + (S◦g)(x)≥β.
In the context of Theorem 3.1.2, by defining \( fe := f - \beta \) and ensuring \( x \in C \) and \( \alpha \in \mathbb{R} \), we establish that if \( (S \circ g)(x) \leq \alpha \), then \( f(x) + e \psi(\alpha) \geq 0 \) This indicates that the conditions of Theorem 3.1.2 are satisfied with \( fe \) substituting for \( f \), and the condition (SC2) is also met, as defined by \( \psi \) Consequently, Theorem 3.1.2 guarantees the existence of \( \gamma \geq 0 \) and \( y^* \in Y^* \) such that \( y^* \leq \gamma S \) on \( Y \) and \( fe + y^* \circ g \geq \psi^*(\gamma) \) on \( C \).
It follows from (SC3) that (domf)∩(domg)∩C 6= ∅, and hence there exists ¯a ∈ C such that fe(¯a) + (y ∗ ◦g)(¯a) S(eg(\bar{a})) \), which implies \( \bar{\alpha} > \inf_{y \in Y} S(y) \) Given that \( S \) is a continuous sublinear function, it follows that \( eg(\bar{a}) \) belongs to the interior of the set where \( S(y) < \alpha \), thus satisfying condition (SC3) The conclusion is drawn from Theorem 3.2.1, where \( f \) and \( eeg \) correspond to the roles of \( f \) and \( g \), respectively, completing the proof.
3.2.2 Extension of the Hahn-Banach theorem, the sandwich theorem, and the Mazur-Orlicz theorem
The Hahn-Banach theorem is recognized for its limitations when the sublinear function has extended real values However, Theorem 3.2.1 provides an extension of this theorem to address such cases, given that a specific interior-type qualification condition is met.
Corollary 3.2.3 [Extended Hahn-Banach theorem] Let X be an l.c.H.t.v.s.,
S : X → R∪ {+∞} be an lsc extended sublinear function, M be a subspace of X, φ : M → R be a linear function satisfying φ ≤ S on M Assume that the following condition holds
Then there exists L∈X ∗ such that L≤S on X and L| M =φ, where L| M denotes the restriction of L to the subspace M.
Proof The conclusion of the corollary follows directly from Theorem 3.2.1 Indeed, let
Y =X,C =M,g :X →Xdefined byg(x) :=xfor allx∈X, andf :X→R∪{+∞} defined by f(x) :( −φ(x) if x∈M , +∞ else.
In this context, g is identified as an S-convex mapping, whereas f is recognized as a proper, convex function Notably, the condition (SC3) is satisfied due to the implications of equation (3.8) Additionally, the relationship expressed in (3.8), combined with the condition that φ is less than or equal to S on M, leads to the conclusion that the infimum of C f + S◦g is established.
∈R,which is (i) in Theorem 3.2.1 By this theorem, there exists L ∈ X ∗ such that L ≤ S on X and infM
≥0 (as φ ≤S on M), which means thatφ ≤LonM AsM is subspace andφ, Lare linear we haveL| M =φ. The proof is complete.
It is now easy to see that both the topological and algebraic versions of the cele- brated Hahn-Banach theorem are consequences of Corollary 3.2.3.
Corollary 3.2.4 [38](Topological Hahn-Banach theorem)LetX be an l.c.H.t.v.s. and M be a subspace of X Let further S :X →R be a continuous sublinear function and φ:M →R be a linear function such that φ ≤S on M Then there exists L∈X ∗ such that L≤S on X and L| M =φ.
Since S is a real-valued function, there exists a point ¯a in M and a real number ¯α such that S(¯a) is less than α, leading to ¯α being greater than the infimum of S(x) over X Additionally, because S is a continuous sublinear and convex function, the set {x in X: S(x) < α} is non-empty and is equal to the interior of the set {x in X: S(x) ≤ α} Therefore, ¯a belongs to both M and the interior of the set where S(x) is less than or equal to α, confirming that the condition in (3.8) is satisfied, and the conclusion is derived directly from Corollary 3.2.3.
Corollary 3.2.5 [38](Algebraic Hahn-Banach theorem) Let X be a nontrivial vector space, S : X → R be sublinear, M be a linear subspace of X, φ : M → R be linear andφ ≤S on M Then there exists a linear functionalL on X such that L≤S on X and L| M =φ.
The equivalence between extended versions of the Farkas lemma
lemma and the Hahn-Banach-Lagrange theorem
It is worth mentioning that the two versions of extended Farkas lemmas: Theorems 3.1.1, 3.1.2, and the extended Hahn-Banach-Lagrange theorem, Theorem 3.2.1, are equivalent to each other.
To show this, we first observe that the implication Theorem 3.1.1 =⇒Theorem 3.1.2 was proved in Section 3.1 while the implication Theorem 3.1.2 =⇒Theorem 3.2.1 has been just shown in Section 3.2 Consequently, we have
To complete the circle of equivalences, it is sufficient to prove the implication [Theorem 3.2.1 =⇒ Theorem 3.1.1].
Let X, Y, C, K, f, and g be defined in Theorem 3.1.1, and assume the Slater condition (SC1) is satisfied Define S as i−K, which is an lsc, extended sublinear function due to K being a closed convex cone Furthermore, it can be concluded that condition (SC3) is a consequence of (SC1).
On the other hand, if (i) in Theorem 3.1.1 holds, i.e., x∈C, g(x)∈ −K ⇒f(x)≥β, then f(x) + (S◦g)(x) =f(x)≥β,∀x∈C, or equivalently, inf C f+S◦g
≥β.Now, Theorem 3.2.1 yields the existence ofy ∗ ∈Y ∗ such thaty ∗ ≤S on Y and inf
≥β, which is equivalent to f+y ∗ ◦g ≥β onC.
It remains to prove that y ∗ ∈K + but this follows from the fact that y ∗ ≤S on Y, as one has hy ∗ , yi ≤S(y) =i−K(y) = 0, ∀y∈ −K.
This ensures that y ∗ ∈K + and the implication (i) =⇒ (ii) in Theorem 3.1.1 has been proved The converse implication is trivial.
Applications to optimization and convex analysis
Generalized optimization problems involving sublinear-convex map-
Let X, Y be l.c.H.t.v.s., C be a nonempty convex subset of X, and f : X →
R∪ {+∞} be a proper convex function Let further S : Y → R∪ {+∞} be an lsc sublinear function,g :X →Y • be an S-convex mapping and ψ :R→R∪ {+∞}be a proper convex function Assume that there are x∈ C∩(domf) and α∈ domψ such that (S◦g)(x)≤α.
Consider two optimization problems of the following forms:
The model (PS1) encompasses a variety of problems stemming from the diverse selections of the sublinear function S and the mapping g, including the general cone-convex problem.
(CP) inf f(x) s.t g(x)∈ −K, x∈C, whereK is a closed convex cone inY andg isK-convex To see this, just takeS =i−K and ψ ≡0.
The class of problems represented by model (PS2) encompasses several well-known issues, including the best approximation problem in normed spaces and penalty problems frequently encountered in numerical methods for convex problems Additionally, it includes special problem classes that lead to significant concepts such as the Fenchel duality theorem, the Fenchel-Moreau point, and the convex separation theorem, which will be explored in detail later in this section.
The concept of S-convex mappings encompasses various function classes, including affine, convex, and concave functions when Y = R Notably, certain "convexity" properties of S-convex mappings are preserved through compositions, which is demonstrated in the subsequent lemma This preservation is instrumental in establishing the duality results and optimality conditions for (PS1) The following results will maintain the initial assumptions regarding spaces and functions as outlined at the beginning of this section.
Lemma 3.3.1 Let S : Y → R∪ {+∞} be an (extended) lsc sublinear function and g :X →Y • be an S-convex mapping The following assertions are true:
(ii) If y ∗ ∈Y ∗ and y ∗ ≤γS for some γ ∈R + then y ∗ ◦g is convex,
(iii) If κ : Y → R is convex and S-increasing (i.e., y1, y2 ∈ Y, y1 ≤S y2 implies κ(y1)≤κ(y2)) then κ◦g is convex.
Proof (i) Since g is S-convex, for x 1 , x 2 ∈X, à 1 , à 2 >0, à 1 +à 2 = 1, one has g(à 1 x 1 +à 2 x 2 )≤ S à 1 g(x 1 ) +à 2 g(x 2 ), (3.12) which means that
Combining the last inequality and the sublinearity of S, one gets
=à 1 S(g(x 1 )) +à 2 S(g(x 2 )), which shows that S◦g is convex.
(ii) First note that if y 1 , y 2 ∈ Y and y 1 ≤ S y 2 (i.e., S(y 1 − y 2 ) ≤ 0) then by assumption, one has hy ∗ , y 1 −y 2 i ≤γS(y 1 −y 2 )≤0, which yieldshy ∗ , y 1 i ≤ hy ∗ , y 2 i, and hence,y ∗ isS-increasing.
Now, if x 1 , x 2 ∈X, à 1 , à 2 >0, à 1 +à 2 = 1, then (3.12) holds This inequality and the fact that y ∗ is S-increasing yield
(y ∗ ◦g)(à 1 x 1 +à 2 x 2 )≤ hy ∗ , à 1 g(x 1 ) +à 2 g(x 2 )i=à 1 (y ∗ ◦g)(x 1 ) +à 2 (y ∗ ◦g)(x 2 ), which shows that y ∗ ◦g is convex.
(iii) The proof is similar to that of (ii).
Remark 3.3.1 It is worth emphasizing that the condition “y ∗ ∈Y ∗ and y ∗ ≤γS for some γ ∈R+” is essential for the convexity of y ∗ ◦g In general, the convexity of the last function does not hold as shown in the next example (the mapping g was taken from [5, Example 3.3]) This is one of the reasons why one can not apply methods for composite convex problems to the models (PS1) or (PS2) (see [24], [32] and references therein).
Example 3.3.1 Let X =R, Y =R 2 and let K =R 2 + We add to Y =R 2 a greatest element with respect to the order defined byR 2 +, denoted by∞ R 2
+} be the functions defined respectively by S(y1, y2) =y2 for all (y1, y2)∈R 2 , and g(x)
Note thatS is a sublinear function and can be extended toY • =R 2 ∪ {∞ R 2
+) = +∞ Then this function, in turn, defines a binary relation “≤S” as shown in ((1.9)and (1.11)) We claim that g isS-convex and there is y ∗ ∈Y ∗ =R 2 such that y ∗ ◦g is not convex.
Firstly, we show that g is an S-convex mapping Indeed, take any λ1, λ2 > 0, λ1+λ2 = 1 and any x1, x2 ∈X =R.
+, i.e., λ 1 x 1 +λ 2 x 2 ≤0 then at least one of the two x 1 and x 2 is nonpositive, i.e., g(x 1 ) = ∞ R 2
+, or both Then (3.13) also holds Thus, g is S-convex.
Now let y ∗ = (1,−1)∈Y ∗ =R 2 Then hy ∗ ,(y 1 , y 2 )i=y 1 −y 2 for all (y 1 , y 2 )∈R 2 According to the convention made in Section 2, we get
It is clear thaty ∗ ◦g is not a convex function It is also worth mentioning here that in this example, y ∗ ◦g is not lsc either (see also [5, Example 3.3]).
The revised version of Farkas lemma for S-convex systems, outlined in Theorem 3.1.2, serves as a crucial tool for demonstrating strong duality and optimality conditions in problems (PS1) and (PS2).
We define the dual problem (DPS) of (PS1) as:
Recall that in the case that the dual problem (DPS) has at least one solution, we will write max (DPS) for the value of (DPS) instead of sup (DPS).
Theorem 3.3.1 (Strong duality for (PS1)) If the condition (SC2) holds, then the strong duality holds for (PS1), i.e., inf (PS1) = max (DPS).
Proof • For the weak duality, take arbitrary x ∈ C, α ∈ domψ with (S◦g)(x) ≤ α, γ ≥0 and y ∗ ∈Y ∗ with y ∗ ≤γS onY Then one has f(x) +ψ(α) = f(x) +ψ(α)−γα+γα ≥f(x) +ψ(α)−γα+γ(S◦g)(x)
≥ f(x) +ψ(α)−γα+ (y ∗ ◦g)(x), and hence, inf (PS1) ≥ inf x∈C f(x) + inf α∈domψ[ψ(α)−γα] + (y ∗ ◦g)(x)
= inf x∈C f(x) + (y ∗ ◦g)(x)−ψ ∗ (γ) ,which gives inf (PS1)≥sup (DPS).
• By assumption, we get inf (PS1) < +∞ If inf (PS1) = −∞ then by the weak duality one gets inf (PS1) = sup (DPS) = −∞ and, in this case, any pair (y ∗ , γ) ∈
Y ∗ ×R+ such that y ∗ ≤ γS is a solution of (DPS), for instance, (0 Y ∗ ,0) is such a solution.
We now can assume that eβ := inf (PS1)∈R By definition of eβ, one gets x∈C, α∈domψ,(S◦g)(x)≤α
Theorem 3.1.2 confirms the validity of statement (a) by substituting fe with f−β Given that condition (SC2) is satisfied for f, the theorem guarantees the existence of eγ ≥ 0 and ey ∗ in Y ∗, with ey ∗ ≤ eγS on Y This leads to the equation f(x) + (e ye ∗ ◦ g)(x) = f(x) − eβ + (ye ∗ ◦ g)(x) ≥ ψ ∗ (eγ) for all x in C.
The inequality f(¯a)−eβ + (ye ∗ ◦g)(¯a) < +∞, where ¯a ∈ C, indicates that ψ ∗ (eγ) is finite From equation (3.14), we derive that for x ∈ C, the condition inf {f(x) + (ye ∗ ◦g)(x)−ψ ∗ (eγ)} ≥ eβ holds, establishing that sup (DPS) is at least eβ, which equals inf (PS1) This leads to the conclusion that inf (PS1) = sup (DPS) due to weak duality Furthermore, it is evident that (ye ∗, eγ) serves as a solution to the dual problem (DPS), thus completing the proof.
Corollary 3.3.1 Assume that the constraint qualification condition (SC2) holds and that ψ is non-decreasing on S a∈C
Then it holds inf (PS2) = inf (PS1) = max (DPS).
Proof By Theorem 3.3.1, we only need to prove that inf (PS2) = inf (PS1) under the extra assumption that ψ is non-decreasing on S a∈C
Indeed, inf (PS2) = inf x∈C∩(g −1 (domS))
Since the condition (SC2) holds, there exists (¯a,eα)∈C×domψ such that (S◦g)(¯a)≤eα and f(¯a) +ψ(α)e −∞, it follows from (3.29) thatψ ∗ (γ) = 0 andγ =c An argument similar to the one in the previous case and (3.29) gives us inf (PEN) = max λ=(λ i )∈ R m+p λ≤cS x∈Ωinf {f(x) +hλ, g(x)i}
We now establish optimality conditions for the penalty problem (PEN).
Theorem 3.3.4 (Optimality condition for(PEN))Letx¯∈Ω.Then x¯is a solution of (PEN) if and only if there exists λ¯ = (¯λ i )∈R m+p with max
Proof It is known that the penalty problem (PEN) is a special case of the problem
In the context of Theorem 3.3.3, we define parameters X, Y, C, ψ, S, and g, where S is identified as a continuous sublinear function, and g is recognized as an S-convex mapping Additionally, ψ is characterized as a convex and non-decreasing function on R, satisfying the condition (SC2) According to Corollary 3.3.3, the point ¯x is considered a solution to the problem (PEN) if and only if there exist a non-negative γ and a vector ¯λ = (¯λ i) in Y∗ = R^(m+p) such that ¯λ is less than or equal to γS on Y = R^(m+p).
On the one hand, it follows from (3.34) that for any i = 1, , m, one has ¯λ i hλ,¯ (0,ã ã ã ,0,1,0,ã ã ã ,0)i ≤cS(0,ã ã ã ,0,1,0,ã ã ã ,0) =c,and−λ¯ i =hλ,¯ (0,ã ã ã ,0,−1,0,ã ã ã,0)i ≤ cS(0,ã ã ã ,0,−1,0,ã ã ã ,0) =c, and hence,
The initial inclusion of (3.35) guarantees that γ equals c, leading to the equality and the expression in (3.33) Additionally, the continuity of the functions f, h i, and k j (where i ranges from 1 to m and j from 1 to p) combined with the final inclusion of (3.35) results in (3.32) Thus, the proof is concluded.
Generalized Fenchel duality theorem and a separation theorem 56
In a recent study, S Simons introduced the Hahn-Banach-Lagrange theorem, which he utilized to establish a generalized version of the Fenchel duality theorem This new approach eliminates the need for the lower semi-continuity assumption on the functions involved, expanding the theorem's applicability.
Inspired by previous results, we explore optimization problems involving sublinear-convex mappings, leading to a strong duality result that generalizes the Fenchel duality theorem This extension builds upon a recent version established by S Simons, accommodating extended sublinear functions Additionally, we examine a specific case in normed spaces, where the duality theorem directly results in the convex separation theorem.
In a locally convex Hausdorff topological vector space E, with its topological dual E∗, consider two nonempty convex subsets D and F of E Let S: E → R ∪ {+∞} be an extended lower semicontinuous (lsc) sublinear function, and h, k: E → R ∪ {+∞} be proper convex functions It is assumed that the intersection of the domains of h and k with the domain of S is non-empty, specifically (dom h - dom k) ∩ dom S ≠ ∅.
It is worth noting that the problem (Q) is a special case of (PS2) when taking
X := E ×E, Y := E, C := D×F, f(x, y) := h(x) +k(y), g(x, y) := x−y for all (x, y)∈X, and ψ(α) = αfor allα∈R.It is easy to see thatg is anS-convex mapping and that ψ ∗ (γ) = 0 if γ = 1 and ψ ∗ (γ) = +∞ if γ 6= 1.
We define the dual problem of (Q), denoted by (DQ), as follows:
Note that for any z ∗ ∈ E ∗ such that z ∗ ≤S on E and for any x ∈ D, y ∈ F, one has
≤ h(x) +k(y) +S(x−y), which shows that sup (DQ)≤inf (Q), i.e., the weak duality holds.
Theorem 3.3.5 (Generalized Fenchel duality theorem)Assume that the follow- ing constraint qualification condition holds:
(SC4) ∃¯a∈D∩(domh),∃¯b∈F ∩(domk) and ∃¯α∈R such that ¯ a−¯b ∈int{v ∈E : S(v)≤α}.¯
If inf (Q)∈R then the strong duality between (Q) and (DQ) holds, i.e., x∈Dinf y∈F h(x) +k(y) +S(x−y)
The problem (Q) is identified as a specific instance of (PS2) by defining Y as E, X as E×E, C as D×F, f(x, y) as h(x) + k(y), g(x, y) as x−y for all (x, y) in X, and ψ(α) as α for all α in R It is evident that the function f is a proper and convex function, while g represents an S-convex mapping, and ψ is characterized as a proper, convex, and non-decreasing function on R.
As (SC4) holds, (SC2) holds for the new setting, i.e., with X, Y, C, f, g, and ψ defined as above Corollary 3.3.1 is then applied to get inf (Q) = max z ∗ ∈E ∗ , γ≥0 z ∗ ≤γS x∈D, y∈Finf {f(x, y) + (z ∗ ◦g)(x, y)−ψ ∗ (γ)}.
This means that there exist ¯γ ≥0, z ∗ ∈E ∗ such thatz ∗ ≤γS, and¯ inf (Q) = inf x∈D, y∈F{f(x, y) + (z ∗ ◦g)(x, y)−ψ ∗ (¯γ)} (3.36)
Since inf (Q)∈ R, it follows from (3.36) that ψ ∗ (¯γ) = 0 and ¯γ = 1 Therefore, (3.36) collapses to inf (Q) = inf x∈D, y∈F{h(x) +k(y) +hz ∗ , x−yi}
This together with the weak duality completes the proof.
Remark 3.3.2 It is worth noting that in the case where D = F = E, then (DQ) collapses to sup z ∗ ∈E ∗ ,z ∗ ≤S
If S is a continuous sublinear function on E, the condition (SC4) is automatically satisfied For any points ¯a in the domain of h and ¯b in the domain of k, it follows that S(¯a−¯b) is finite By selecting a real number ¯α such that S(¯a−¯b) is less than ¯α (noting that ¯α exceeds the infimum of S), it can be concluded that ¯a−¯b belongs to the interior of the set where S(y) is less than or equal to ¯α This leads to the final conclusion.
As a consequence of Theorem 3.3.5, we get a version of the Fenchel duality theorem which was recently established in [62].
Corollary 3.3.4 [62, Theorem 4.1] Let E be an l.c.H.t.v.s with its topological dual
E ∗ and h, k :E →R∪ {+∞} be proper convex functions Then
∃z ∗ ∈E ∗ such that h(−z ∗ ) +k ∗ (z ∗ )≤0 (3.37) if and only if
≥0, (3.38) where S(E) stands for the family of all continuous seminorms onE.
Proof We shall apply Theorem 3.3.5 with D =F =E Note that for any S ∈ S(E), by Remark 3.3.2, condition (SC4) holds automatically.
Assume that (3.38) holds, i.e., there exists S ∈ S(E) such that x, y∈Einf h(x) +k(y) +S(x−y)
Since h, k are proper functions, it follows that x, y∈Einf h(x) +k(y) +S(x−y)
∈R. Applying Theorem 3.3.5, we obtain z ∗ ∈Emax ∗ ,z ∗ ≤S {−h ∗ (−z ∗ )−k ∗ (z ∗ )} ≥0, which means that (3.37) is satisfied.
Conversely, assume that (3.37) holds Then for any x, y ∈E one has h−z ∗ , xi −h(x) +hz ∗ , yi −k(y)≤h ∗ (−z ∗ ) +k ∗ (z ∗ )≤0, and hence, h(x) +k(y) +|hz ∗ , x−yi| ≥h(x) +k(y) +hz ∗ , x−yi ≥0.
Then we obtain the final conclusion by taking S(z) =|hz ∗ , zi| for all z ∈E.
As a consequence of Corollary 3.3.3 we get the following optimality condition for (Q).
Theorem 3.3.6 (Optimality condition for(Q))Assume that the condition (SC4) holds Let x ∈D∩(domh), y ∈ F ∩(domk) Then (x, y) is a solution of the problem (Q) if and only if there exists z ∗ ∈E ∗ such that z ∗ ≤S on E, and z ∗ ∈ −∂(h+i D )(x)
The relationship between (Q) and (PS2) is established by defining X as E×E, C as D×F, Y as E, with f(x, y) represented as h(x) + k(y) and g(x, y) as x − y for all (x, y) in X, while ψ(α) equals α for all α in R Additionally, f is identified as a proper, convex function, g is classified as an S-convex mapping, and ψ is recognized as a non-decreasing function on R+ Furthermore, the satisfaction of (SC4) implies that (SC2) is also fulfilled in this new context.
• Necessity Assume thatx∈D∩(domh), y ∈F ∩(domk), and (x, y) is a solution of the Problem (Q) Then by Corollary 3.3.3 that there exist γ ≥0 and z ∗ ∈E ∗ such that z ∗ ≤γS onE, and
The expressions (3.39)-(3.40) now follow from (3.41) by some elementary calculation.
• Sufficiency It can be checked directly.
Let (E,k.k) be a normed space with its topological dual E ∗ and letcbe a positive number Consider the problem
It is clear that (Q1) is a special case of the problem (Q) on the normed space E with S(ã) = ck ã k, and hence, the dual problem of (Q1), denoted by (DQ1), is:
Since S(ã) = ck ã k is continuous, the constraint qualification condition (SC4) is satisfied Consequently, the duality result for (Q1), outlined in Corollary 3.3.5, directly follows from Theorem 3.3.5 Moreover, by setting D = E = F, this leads to Corollary 3.3.6, which corresponds to Corollary 6.1 in [62], representing a version of the Fenchel duality result for normed spaces.
Corollary 3.3.5 (Strong duality for (Q1)) If inf (Q1)∈R then the strong duality between (Q1) and (DQ1) holds, i.e., inf (Q1) = max z ∗ ∈E ∗ ,kz ∗ k E ∗ ≤c{−(h+i D ) ∗ (−z ∗ )−(k+i F ) ∗ (z ∗ )}. Corollary 3.3.6 [62, Corollary 6.1] Let (E,k.k) be a nontrivial normed space and h, k :E →R∪ {+∞} be proper convex functions Then
∃z ∗ ∈E ∗ such that h(−z ∗ ) +k ∗ (z ∗ )≤0 (3.42) if and only if
∃c≥0 such that inf x, y∈E h(x) +k(y) +ckx−yk
Theorem 3.3.6 provides necessary and sufficient conditions for optimality in problem (Q1), with the conditions mirroring those in Theorem 3.3.6, substituting S with ck.k Rather than delving deeper into this topic, we will focus on applying these results to a specific scenario, which yields a version of the separation theorem for convex sets in normed spaces.
We focus on two nonempty convex subsets, D and F, within a normed space (E, k.k) The distance between these sets is defined as dist(D, F) = inf{kx - yk : x ∈ D, y ∈ F}.
The challenge of determining the distance between two specific sets, D and F, is a particular instance of problem (Q1) where c = 1, h ≡ 0, and k ≡ 0 For broader discussions on the distances between convex sets, refer to [13] This leads us to the implications of Corollary 3.3.5.
Corollary 3.3.7 Let D and F be two nonempty convex subsets of a normed space (E,k.k) Then dist(D, F) = max z ∗ ∈E ∗ kz ∗ k E ∗ ≤1 z∈Dinfhz ∗ , zi −sup z∈F hz ∗ , zi
Remark 3.3.3 It is worth noting that the two sets D and F do not intersect if dist(D, F) > 0 Therefore, from Corollary 3.3.7 one can conclude that: If the two convex sets D and F satisfy dist(D, F)>0 then there exists z ∗ ∈E ∗ with kz ∗ k E ∗ ≤1 satisfying z∈Dinfhz ∗ , zi > sup z∈F hz ∗ , zi, which means that these two sets can be strictly separated by a continuous linear func- tional z ∗ ∈E ∗ with kz ∗ k E ∗ ≤1.
Corollary 3.3.7 presents a version of the separation theorem for convex sets in normed spaces, analogous to the one outlined in [66, p.6] This theorem emphasizes the significance of the closure of set A in E, known as hereclA, in establishing the relationship between convex sets.
Corollary 3.3.8 Let D and F be two nonempty convex subsets in a normed space (E,k.k) Then
∃¯z ∗ ∈E ∗ :k¯z ∗ k E ∗ ≤1 and sup z∈F h¯z ∗ , zi< inf z∈Dh¯z ∗ , zi
A conjugate formula for the supremum of a family of convex
The duality result for the (PS2) problem, outlined in Corollary 3.3.1, suggests potential extensions in convex analysis through various mappings of g One such mapping illustrates a formula for the conjugate of the supremum of a potentially infinite collection of convex functions, which may not be lower semi-continuous, within locally topological vector spaces.
Let X be a locally convex Hausdorff topological vector space, T be an arbitrary (possibly infinite) index set, andg t :X →R∪ {+∞}be proper convex (not necessarily lsc) function for allt∈T.We consider the product spaceR T endowed with the product topology and denoted by R (T ) the space of real tuples λ = (λ t )t∈T with only finitely manyλ t 6= 0.Note thatR (T ) is the topological dual ofR T (see [34], [55]) We represent byR (T + ) the positive cone inR (T ) , that is
Note thatR (T + ) is also the dual cone of the positive cone
R T + :={(γ t )t∈T ∈R T :γ t ≥0 for all t ∈T} in the product spaceR T The supporting set of λ∈R (T ) is suppλ :={t ∈T :λ t 6= 0}, and λ(u) :=X t∈T λ t u t = X t∈suppλ λ t u t ∀u= (u t )t∈T ∈R T ,∀λ= (λ t )∈R (T )
The following formula for the conjugate of the supremum function sup t∈T g t comes as a consequence of Corollary 3.3.1 This result can be considered a counterpart of the one in [5] (see Remark 3.3.4).
Proposition 3.3.1 Let g t : X →R∪ {+∞} be a proper convex (not necessarily lsc) function for all t∈T Assume that the following condition holds:
Then for any x ∗ ∈X ∗ with sup t∈T g t ∗
+} be the mapping defined by g(x)
DefineS :R T →R∪ {+∞} byS(y) := sup t∈T y t for all y= (y t ) t∈T ∈R T
We assert that S is an extended sublinear and lower semicontinuous (lsc) function It is evident that S qualifies as an extended sublinear function For any vector y = (yt)t∈T in R^T, S(y) is defined as the supremum of yt across all t in T, which can be expressed as S(y) = sup t∈T yt sup t∈T pt(y) Here, pt: R^T → R is the canonical function defined by pt(y) = yt, and it is continuous with respect to the product topology on R^T Consequently, we conclude that S is indeed an lsc function We further extend S to the set Y := R^T ∪ {∞ R^T}.
• g is a proper and S-convex mapping Since (3.44) holds, T t∈T domg t 6= ∅ and hence, g is proper Moreover, g is an S-convex mapping Indeed, for any à 1 , à 2 >
0, à 1 +à 2 = 1 and any x 1 , x 2 ∈X we will verify that g λ 1 x 1 +λ 2 x 2
≤ S λ 1 g(x 1 ) +λ 2 g(x 2 ), (3.46) where “≤ S ” is the binary relation associated with the extended sublinear function S defined by (1.9) We consider the following cases:
If x 1 , x 2 ∈ T t∈T domg t , then by the convexity of g t for all t ∈ T and the fact that T t∈T domg t is a convex set, one has
If x 1 ∈/ T t∈T domg t orx 2 ∈/ T t∈T domg t , then (3.46) holds (note that y≤ S ∞ R T
If λ 1 x 1 +λ 2 x 2 ∈/ T t∈T domg t , then at least one of the two x 1 and x 2 does not belong to T t∈T domg t (as T t∈T domg t is a convex set), i.e.,g(x 1 ) =∞ R T
+,or both. Hence, (3.46) also holds Thus g is S-convex.
•We now apply Corollary 3.3.1 to the function S, the mapping g defined as above,
C = X, and the function f : X → R with f(x) := −hx ∗ , xi for all x ∈ X, and ψ(α) :=α for all α∈R It is clear that ψ is proper convex lsc and non-decreasing on
R Note that the condition (3.44) ensures that (SC2) holds.
It now follows from Corollary 3.3.1 that x∈Xinf {f(x) + (S◦g)(x)}= max λ∈X ∗ , γ≥0 λ≤γS x∈Xinf {f(x) + (λ◦g)(x)−ψ ∗ (γ)}. This means that there exist ¯γ ≥0, ¯λ∈E ∗ such that ¯λ ≤¯γS, and x∈Xinf {f(x) + (S◦g)(x)}= inf x∈X f(x) + (¯λ◦g)(x)−ψ ∗ (¯γ) (3.48)
(x ∗ )∈ R, it follows from (3.47) and (3.48) that ψ ∗ (¯γ) = 0 and ¯γ = 1. Then one gets x∈Xinf {f(x) + (S◦g)(x)} = inf x∈X f(x) + (¯λ◦g)(x)
≤ max λ∈X ∗ λ(.)≤S(.) x∈Xinf {f(x) + (λ◦g)(x)} (3.49) Moreover, it is easy to that x∈Xinf {f(x) + (S◦g)(x)} ≥ sup λ∈X ∗ λ(.)≤S(.) x∈Xinf {f(x) + (λ◦g)(x)}.
This and (3.49) give rise to x∈Xinf {f(x) + (S◦g)(x)}= max λ∈X ∗ λ(.)≤S(.) x∈Xinf {f(x) + (λ◦g)(x)}.
To finalize the proof, we set λ = (λt)t∈T ∈ R (T) such that λ(y) ≤ S(y) = sup t yt for all y = (yt)t∈T ∈ R T We will demonstrate that λt ≥ 0 for all t ∈ T and that the sum of λt across T equals 1 For any k ∈ T, let u k = (u t)t∈T where u t = -1 if t = k and u t = 0 for t ≠ k The condition λ(u k) ≤ S(u k) implies λ k ≥ 0 for all k ∈ T Additionally, let ¯u = (¯u t)t∈T and u ∗ = (u ∗ t)t∈T ∈ R T, where ¯u t = 1 and u ∗ t = -1 for all t ∈ T The inequalities λ(¯u) ≤ S(¯u) and λ(u ∗) ≤ S(u ∗) confirm that P t∈T λ t = 1 Thus, the equality (3.45) is established from (3.49), completing the proof.
Remark 3.3.4 The conjugate formula (3.45) can be considered a counterpart of the one in [5, page 78] where it was proved by another technique and under a so-called closedness qualification condition and an extra assumption that g t is lsc for all t ∈ T. Here, in Proposition 3.3.1, we assume the condition (3.44), which is stronger than the closedness qualification condition in [5] However, in this proposition the assumption on the lower semi-continuity of g t for all t ∈T is removed.
This chapter presents qualification conditions known as closedness conditions for systems that are convex with respect to a convex cone K (termed K-convex systems) and those that are convex with respect to an extended sublinear function S (referred to as S-convex systems) We demonstrate that these closedness conditions serve as both necessary and sufficient criteria for two new extended versions of the Farkas lemma applicable to these systems.
The extended Farkas lemmas are equivalent to a new extended version of the Hahn-Banach-Lagrange theorem, leading to both analytic and algebraic extensions of the Hahn-Banach theorem, an extended Mazur-Orlicz theorem, and an extended sandwich theorem for sublinear functions Notably, these results are derived under necessary and sufficient conditions, unlike the findings in Chapter 3, which were based solely on sufficient conditions.