Continuation methods and condensing mappings

Nonlinear Analysis 102 (2014) 84–90

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Continuation methods and condensing mappings J.M. Soriano ∗ , M. Ordoñez Cabrera Departamento de Análisis Matemático, Facultad de Matemáticas, Universidad de Sevilla, Aptdo. 1160, Sevilla 41080, Spain

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Article history: Received 21 November 2013 Accepted 23 January 2014 Communicated by S. Carl MSC: primary 58C30 secondary 65H10

abstract Some sufficient conditions are provided to assert that a condensing perturbation of the identity defined on a Banach space over K reaches a fixed value. When conditions are strengthened, this mapping reaches at most finitely many times the fixed value throughout the entire space. The proof of the result is constructive and is based upon a continuation method. © 2014 Elsevier Ltd. All rights reserved.

Keywords: Fixed point Continuation method Surjective implicit function theorem C 1 -homotopy Proper mapping Compact mapping Condensing mapping Fredholm mapping

1. Preliminaries Let X , Y be two Banach spaces. If F : D(F ) ⊆ X → Y is a continuous mapping, then one way of solving the equation F (x) = k0

(1)

is to embed (1) in a continuum of problems H (x, t ) = k0

(0 ≤ t ≤ 1),

(2)

which can easily be resolved when t = 0. When t = 1, the problem (2) becomes (1). In the case when it is possible to continue the solution for all t in [0, 1] then (1) is solved. This method is called continuation with respect to a parameter. In this paper, some sufficient conditions are provided in order to guarantee that a C 1 -condensing perturbation of the identity defined on a Banach space over K reaches a fixed value. If these conditions are conveniently strengthened then this mapping reaches at most finitely many times the fixed value. Other conditions, sufficient to guarantee the existence of k0 -points, have been given by the first author in a finite-dimensional setting [1], and in an infinite-dimensional setting [2]. A continuation method was used in the proofs of these papers. The proofs supply the existence of implicitly defined continuous mappings whose ranges reach k0 -points. A continuation method is also used here. The key is the use of the surjective implicit function theorem [3], properties of Kuratowski noncompactness measure for bounded sets in metric spaces, properties of condensing weakly coercive C 1 -mappings, and the properties of proper and Fredholm C 1 -mappings [4]. Ascoli’s



Corresponding author. Tel.: +34 954557004; fax: +34 954557972. E-mail addresses: [email protected] (J.M. Soriano), [email protected] (M. Ordoñez Cabrera).

http://dx.doi.org/10.1016/j.na.2014.01.024 0362-546X/© 2014 Elsevier Ltd. All rights reserved.

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Theorem is used to prove that our example fulfills the required conditions. Sard–Smale’s Theorem tells us that regular values, here needed, are ‘‘non-rare’’. This theorem avoids frequently, to write explicitly the sentence ‘‘regular value’’ among the hypotheses of our propositions. The iinterested reader can find information on continuation and homotopy methods and the number of solutions of some non-linear equations in [5–10]. We briefly recall some theorems and concepts to be used. Definitions ([4,3]). Let M be a bounded set in a metric space (X , d). The Kuratowski noncompactness measure χ (M ) is defined as the infimum of the set of all numbers ε > 0 with the property that M can be covered by finitely many sets, each of whose diameters is less than or equal to ϵ . For ε > 0, a set of points x1 , x2 , . . . , xn ∈ M is called a finite ε -net for M if and only if min1≤i≤n d(x, xi ) < ε for all x ∈ M. If X is complete, then M is relatively compact if and only if for every ε > 0 there is a finite ε -net for M. Therefore χ (M ) = 0 is equivalent to relative compactness for M. Henceforth we will assume that X , Y are Banach spaces over K, where K = R or K = C. A mapping F : D(F ) ⊆ X → Y is called weakly coercive if and only if

∥F (x)∥ → ∞ as ∥x∥ → ∞. A mapping F is said to be compact whenever it is continuous and the image F (B) is relatively compact (i.e. its closure F (B) is compact in Y ) for every bounded subset B ⊂ D(F ). A mapping F is said to be proper whenever the pre-image F −1 (K ) of every compact subset K ⊂ Y is also a compact subset of D(F ). A mapping is called bounded if and only if it takes bounded sets into bounded sets. A mapping T : D(T ) ⊆ X → X , is said to be k-contractive if and only if there exists k ∈ [0, 1) such that ∥Tx−Ty∥ ≤ k∥x−y∥ for all x, y ∈ X . A mapping T : D(T ) ⊆ X → X , is said to be condensing if and only if T is bounded and continuous, and χ (T (M )) < χ (M ) for all bounded sets M in D(T ) with χ (M ) > 0. A mapping T is called a k-set contraction if and only if it is bounded and continuous, and there is a number k ≥ 0 such that χ(T (M )) ≤ kχ (M ) for all bounded sets M in D(T ). The symbol dim means dimension, codim means codimension, ker means kernel, and R(L) stands for the range of the mapping L. That L : X → Y is a linear Fredholm mapping means that L is linear and continuous and both the numbers dim(ker(L)) and codim(R(L)) are finite, and therefore ker(L) = X1 is a Banach space and has topological complement X2 , since dim(X1 ) is finite. The integer number Ind(L) = dim(ker(L)) − codim(R(L)) is called the index of L. Let F : D(F ) ⊆ X → Y . If D(F ) is open, then F is said to be a Fredholm mapping whenever both F is a C 1 -mapping and F ′ (x) : X → Y is a Fredholm linear mapping for all x ∈ D(F ). If Ind(F ′ (x)) is constant with respect to x ∈ D(F ), then we call this number the index of F and write it as Ind(F ). X , Y are called isomorphic if and only if there is a linear homeomorphism (isomorphism) L : X → Y . Let F (X , Y ) denote the set of all linear Fredholm mappings A : X → Y . Let L(X , Y ) denote the set of all linear continuous mappings L : X → Y . Let Isom(X , Y ) denote the set of all isomorphisms L : X → Y . Let F : D(F ) ⊆ X → Y with D(F ) open be a C 1 -mapping. The point u ∈ X is called a regular point of F if and only if ′ F (u) ∈ L(X , Y ) maps X onto Y . The point v ∈ Y is called a regular value of F if and only if the pre-image F −1 (v) is empty or consists solely of regular points. If F (u) = v , we say u is a v -point to F . Theorem 1 (The Surjective Implicit Function Theorem [3]). Let X , Y , Z be Banach spaces over K = R or K = C, and let F : U (u0 , v0 ) ⊆ X × Y → Z be a C -mapping on an open neighborhood of the point (u0 , v0 ). Suppose that: 1

(i) F (u0 , v0 ) = 0, and (ii) Fv (u0 , v0 ) : Y → Z is surjective. Then the following is true: Let r > 0. There is a number ρ > 0 such that, for each given u ∈ X with ∥u − u0 ∥ < ρ , the equation F (u, v) = 0 has a solution v , such that ∥v − v0 ∥ < r. Theorem 2 ([4]). Let X be a Banach space over K = R or K = C. Then for all bounded subsets M , M1 , . . . , Mn and N of X the following hold:

χ(∅) = 0 χ(M ) = 0 if and only if M is relatively compact. 0 ≤ χ (M ) ≤ diam χ(M )

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M ⊆ N implies χ (M ) ≤ χ (N ). χ(M + N ) ≤ χ (M ) + χ (N ). χ(β M ) ≤ |β|χ (M ) for all β ∈ K. χ(M ) = χ (M ). χ(M ) = χ (co(M )) = χ (co(M )), where co(M ) denotes the convex hull of M, and co(M ) denotes the closed convex hull of  M. χ( ni=1 Mi ) = max{χ (M1 ), . . . , χ (Mn )}. Theorem 3 ([4]). Let S ∈ F (X , Y ). The perturbed mapping S + C satisfies S + C ∈ F (X , Y ) and Ind(S + C ) = Ind(S ) if C ∈ L(X , Y ) and C is a compact mapping. Theorem 4 ([4]). Let g : D(g ) ⊂ X → Y be a compact mapping. Let a ∈ D(g ). If the derivative g ′ (a) exists, then g ′ (a) is also a compact mapping. Theorem 5 ([4]). The Sard–Smale Theorem. Let A : X → Y be a proper C k -Fredholm mapping with k > max {Ind(A), 0}. Then the set of regular values of A is open and dense in Y . 2. Continuation methods and condensing mappings Next, we establish our main result. Theorem 6. Let T : X → X be a C 1 -mapping, where X is Banach space over K = R or K = C. Let k0 ∈ X . We assume that: (i) T is a condensing mapping. (ii) For any fixed t ∈ [0, 1], the vector k0 of X is a regular value of the mapping H (x, t ) := I (x) − tT (x), where I is the identity mapping on X . (iii) H is a weakly coercive mapping. Then the following holds: (a) (I − T )−1 (k0 ) is not-empty, i.e., there is at least one x∗ ∈ X such that I (x∗ ) − T (x∗ ) = k0 . In particular, if k0 = 0, T has at least a fixed point x∗ ∈ X , x∗ = T (x∗ ). If condition (i) is strengthened as follows (iv) T is a compact mapping, then in addition we have (b) I − T has at most finitely many k0 -points on X , and at least one. (c) Hypothesis (ii) can be erased. Proof. (a1) We will prove here that H is a proper mapping on X × [0, 1], which will imply that H −1 (k0 ) is a compact set. Let K be any fixed compact subset of X , and let N := H −1 (K ). It suffices to show that N is a relatively compact set, since N is a closed set, which in turn holds because H is continuous. Denote by π1 : X × [0, 1] → X , π2 : X × [0, 1] → [0, 1] the natural projections on X and [0, 1], respectively. Then N ⊂ N1 × N2 , where Ni := πi (N )(i = 1, 2). We prove now that N is a relative compact set. Firstly, N is bounded, since it is the pre-image H −1 (K ) of the bounded set K by the weakly coercive mapping H. Next, consider any point (x, t ) ∈ N. Then H (x, t ) = y ∈ K , that is, I (x) − tT (x) = y, so I (x) = x = y + tT (x), which implies x ∈ K + tT (N1 ). This yields N1 ⊂ K + tT (N1 ). From Theorem 2,

χ (N1 ) ≤ χ (K + tT (N1 )) ≤ χ (K ) + χ (tT (N1 )) = χ (K ) + |t |χ (T (N1 )). Since K is a compact set and t ∈ [0, 1], then χ (K ) = 0, and χ (N1 ) ≤ χ (T (N1 )). Since T is a condensing mapping and N1 is a bounded set, if χ (N1 ) > 0, then χ (T (N1 )) < χ (N1 ), which implies a contradiction, therefore χ (N1 ) = 0. Since χ (N1 ) = 0, N1 is a relatively compact set in X . Since N2 is a bounded subset of a space of finite dimension, it is relatively compact. Therefore N1 × N2 is a relatively compact set in the product topology. Since N ⊂ N1 × N2 , we get that N is also relatively compact. We have also proved here that I − T is a proper mapping, since it is the particular case in which t = 1. (a2) Let us suppose that H (xa , ta ) − k0 = 0

(3)

for any fixed (xa , ta ) ∈ X × [0, 1]. We will prove that there is ρ > 0 such that for any t ∈ (ta − ρ, ta + ρ), there exists x = x(t ) such that H (x(t ), t ) = k0 . Since zero is a regular value for the mappings I (x) − tT (x) − k0 by hypothesis (ii), we get that Hx (x, t ) =

∂(H (x, t ) − k0 ) = I − tT ′ (x) ∈ L(X , X ) ∂x

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is surjective for every pair (x, t ) such that H (x, t ) = k0 . In particular, the mapping Hx (xa , ta ) = I − tT ′ (xa ) ∈ L(X , X ) is surjective, which together with identity (3) and Theorem 1 implies the existence of ρ > 0, r > 0 such that for any t ∈ (ta − ρ, ta + ρ) there is x(t ) with H (x(t ), t ) − k0 = 0 and ∥x(t ) − xa ∥ < r. (a3) We will prove that for every t in [0, 1] there exists x(t ) such that H (x(t ), t ) − k0 = 0. Let P denote the set of all t such that there is a solution x(t ) for the last equation. Since H (k0 , 0) − k0 = I (k0 ) − k0 = 0, this set is not empty. By (a2), the set P is relatively open in [0, 1]. Finally, since P is the projection of H −1 (k0 ) into the second component (i.e. the t component), and since it is the image of a compact set by the continuous function projection, it is therefore compact, and hence also closed. By the connection of [0, 1], P = [0, 1]. In particular H (x(1), 1) = I (x(1)) − T (x(1)) = k0 , which is conclusion (a) of the theorem, with x∗ := x(1). (b) We substitute the general hypothesis (i) by hypothesis (iv) in part (b). (b1) Since T is a compact mapping, T is a k-set contraction with k = 0. Since every k-set contraction mapping with 0 ≤ k < 1 is a condensing mapping, T is a condensing mapping. Then (iv) implies that T is a condensing mapping, and this together with (ii) implies that conclusion (a) holds. Furthermore the C 1 -mapping L : X → X , L := I − T = H (·, 1) is proper, which was proved in (a1). We will see that L is a Fredholm mapping of index zero: For each x ∈ X , Theorem 4 implies that the mapping T ′ (x) ∈ L(X , X ) is also a compact mapping. Since T ′ (x) is a linear compact mapping and given that I ′ (x) = I is a linear Fredholm mapping of index zero, Theorem 3 implies that L′ (x) = I ′ (x) − T ′ (x) ∈ L(X , X ),

∀x ∈ X

is a Fredholm linear mapping, and Ind(L (x)) = Ind(I ′ (x)) = 0, ∀x ∈ X . Therefore the nonlinear mapping L is a proper Fredholm C 1 -mapping of index zero. (b2) We will see that, if there exists x ∈ X verifying L(x) = k0 , then L is a local C 1 -diffeomorphism at x. Let x ∈ X exist such that L(x) = k0 . Since k0 is a regular value of L, the linear Fredholm mapping L′ (x) maps onto X , and since Ind(L′ (x)) = 0, dim(ker(L′ (x))) = 0. Thus L′ (x) ∈ Isom(X , X ). By the Local Inverse theorem [3], L is a local diffeomorphism at x. (b3) We will prove here that L has at most a finite number of k0 -points on X . Since L is proper, L−1 (k0 ) is a compact set. If there were an infinite sequence ′

(xn )n≥1 ⊂ X ,

with xn ̸= xm when n ̸= m

satisfying L(xn ) = k0 , ∀n ∈ N, there would be a subsequence (xnk )k≥1 ⊂ L−1 (k0 ), converging to a point x ∈ L−1 (k0 ), and x would not be a non-isolated k0 -point of L. However, since L is a local diffeomorphism at x, x is an isolated zero of L. This is a contradiction. Hence there is not an infinite number of k0 -points of L on X . Thus L has at most finitely many k0 -points in X , and at least one. (c) Let us construct the mapping F : X ×[0, 1] → X , F (x, t ) := tT (x), and consider any bounded sequence (xn , tn ). Since T is compact, there exists a subsequence (xnk ) of (xn ) such that T (xnk ) → y as k → ∞. Since (tnk ) is a sequence of real numbers in a bounded set, there is a subsequence (tmk ) such that tmk → t as k → ∞. Then the sequence (F (tn , xn )) = (tn T (xn )) has subsequence (tmk T (xmk )) such that tmk T (xmk ) → ty as k → ∞. Hence F is a compact mapping. Since I is a Fredholm mapping of index zero, Theorems 4 and 3 imply that H = I − F is a Fredholm mapping of index zero, and it is a proper C 1 -mapping, which was proved at (a1). Since H is a proper C 1 -Fredholm mapping, and since 1 > max {Ind(H ), 0} = 0, Theorem 5 implies that the set of regular values V ⊂ X of H is open and dense in X . Let us consider any value k0 ∈ X . Since V is dense in X , for any ε = 1n there is a kn ∈ V with ∥kn − k0 ∥ < 1n , and then kn − k0 → 0 as n → ∞. Since T is a compact mapping and H a weakly coercive mapping for any selected kn , conclusion (b) holds, and so finitely many kn -points xn exist for the mapping I − T . Since I − T is proper and (kn ) ∪ {k0 } a countable compact set where (I − T )−1 (kn ) is a finite set of kn -points, (I − T )−1 {(kn ) ∪ {k0 }} is a compact set which is countable. This set will be called again (xn ). Then there is subsequence (xmk ) such that xmk → x as mk → ∞. Since I (xmk ) − T (xmk ) = kmk and I − T is continuous, I (x) − T (x) = k0 . Then, if stronger hypothesis (iv) is satisfied instead of (i), hypothesis (ii) can be erased.  Corollary 7. Let us consider the equation

− Tx + x = y,

(4)

where the unknown is x ∈ X , y ∈ X is fixed, and T : X → X is a condensing C -mapping. 1

(a) If H (x, t ) := I (x) − tT (x) is a weakly coercive Fredholm C n -mapping of index m, with m < n and n ≥ 1, then there is an open dense subset V ⊂ X , whose elements are regular values of H and for any y ∈ V Eq. (4) has a solution.

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(b) In the particular case in which T is a compact weakly coercive mapping, the number of solutions of (4) is finite for any y ∈ X. Proof. We showed in (a1) that H (x, t ) := I (x) − tT (x) is a proper mapping. Since H : X × [0, 1] → X is a proper Fredholm C n -mapping, and since n > m = max {m, 0} = max {Ind(H ), 0}, Theorem 5 implies that the set of regular values V ⊂ X of H is open and dense in X . For any y ∈ V hypotheses (i)–(iii) of Theorem 6 are fulfilled, then Eq. (4) has a solution. If T is a compact weakly coercive mapping then conclusion (c) of Theorem 6 implies that the number of solutions is finite.  Corollary 8. Let T := C + S : X → X , where C , S : X → X are C 1 -mappings in the Banach space X , C is compact, S is k-contractive, y is a regular value of the mapping H (x, t ) := I (x) − tT (x), t ∈ [0, 1], and H is a weakly coercive mapping. Then the following hold: (a) The mapping T is condensing. (b) Eq. (4) has a solution. If the conditions are strengthened as follows: (vi) C is linear continuous and compact mapping, and I − S is bijective, then in addition we can deduce: (c) Eq. (4) has only a finite number of solutions. (d) I − T is a Fredholm mapping of index zero.

(i) (ii) (iii) (iv) (v)

Proof. Fix a bounded set M ⊂ X . Since C is a compact mapping, χ (C (M )) = 0. Let L1 , . . . , Ln be any finite covering of M. Then S (L1 ), . . . , S (Ln ) is a finite covering of S (M ), and since S is k-contractive diam S (Li ) ≤ k diam(Li ),

i = 1, . . . , n,

hence χ (S (M )) ≤ kχ (M ). Since T = C + S, Theorem 2 implies that

χ (T (M )) ≤ χ (C (M )) + χ (S (M )) ≤ kχ (M ). Then T is a bounded continuous k-set contraction with 0 ≤ k < 1, so T is a condensing mapping, and conclusion (a) holds. Since y is a regular value of the weakly coercive C 1 -mapping H (x, t ) = I (x) − tT (x), the conclusion (a) of Theorem 6 implies that L = I − T has a finite or infinite number of y-points, or equivalently Eq. (4) has a solution. In the particular case in which C is linear continuous compact mapping and I − S is bijective, since I − S is surjective, codim(R(I − S )) = dim(X /R(I − S )) = 0, and since I − S is injective, dim(ker(I − S )) = 0, then ind(I − S ) = 0. Theorem 3 implies that the mapping (I − S ) − C satisfies (I − S ) − C ∈ F (X , X ) and Ind(I − S − C ) = Ind(I − S ) = 0. Hence I − T is a Fredholm mapping of index zero. A proof similar to the one used in part (b) of Theorem 6 gives that I − T has finitely many y-points, or equivalently Eq. (4) has only a finite number of solutions.  Example. We show here an example in which Eq. (4) has finitely many solutions for either any value y ∈ V , where V is an open dense subset of X , or for any unrestricted y ∈ X . Let (E , A, µ) be a measure space. Let us consider the mapping T : C (A, K) → C (A, K),

(T (f ))(·) :=



K (·, y)f (y)dµ(y) + kf (·), A

where A is a compact subset of the normed space (E , ∥ · ∥), C (A, K) =: X is the Banach space of K-valued continuous functions defined on A with the norm ∥ · ∥∞ , K ∈ C (A × A, K), and so for fixed x ∈ A, K (x, ·) ∈ X , f ∈ X , and k is a real fixed number in [0, 1). We consider in E × E the product topology induced by the norm ∥ · ∥. We are interested in the knowledge of solutions of the following integral equation



K (·, y)f (y)dµ(y) + kf (·) − f (·) = g (·),

(5)

A

where the unknown function is f ∈ X , K and g are given functions in their respective spaces, with f A-measurable and K (x, ·) A-measurable for any fixed x ∈ A, A ∈ A with µ(A) = b ∈ K, and H : X × [0, 1] → X ,

H (f , t ) = tT (f ) − I (f )

is weakly coercive. For instance, K = R, the measure space is (R, Mλ∗ , λ), where λ is the Lebesgue measure, and A = [a, b].

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This example is a particular case of the previous corollaries as we will see. Let S : X → X be S (f ) := (kf )(·), k ∈ [0, 1). The mapping S is linear and k-contractive, then S is linear and continuous, hence differentiable. Since S ′ (f ) = S, S is a kcontractive C 1 -mapping. Let C be (C (f ))(·) := A K (·, y)f (y)dµ(y). Considering Eq. (6), we know that C : X → X . We will first show that C is a linear continuous compact mapping. The linearity is trivial. Since for all f ∈ X and all x ∈ A

|(C (f ))(x)| ≤



sup |K (x, y)f (y)|d|µ|(y) ≤ ∥K ∥∞ ∥f ∥∞ A x∈A



d|µ|(y) A

we get ∥Cf ∥∞ ≤ (|b| ∥K ∥∞ )∥f ∥∞ ,

= ∥K ∥∞ ∥f ∥∞ |b|,

so C is bounded and then continuous being linear. Consequently, C is differentiable with C ′ (f ) = C , which is also continuous, and so C is a C 1 -mapping. Now we will prove that C is a compact mapping. To this aim, we use Ascoli’s Theorem. Let B be any bounded set in X . We will show that C (B) is a relatively compact set in X . We have that supf ∈B ∥f ∥∞ =: c < +∞. For any fixed x ∈ A,

     sup |(C (f ))(x)| = sup  K (·, y)f (y)d|µ|(y) ≤ (|b| ∥K ∥∞ )c . f ∈B f ∈B A

Hence {Cf (x) : f ∈ B} is a bounded set in K with the sup norm, so {(C (f ))(x) : f ∈ B} is a relatively compact set for any fixed x ∈ A. On the other hand, we have

|(C (f ))(x) − (C (f ))(z )| ≤



|K (x, y) − K (z , y)| |f (y)|d|µ|(y)  ≤ c |K (x, y) − K (z , y)|d|µ|(y). A

A

Since K is continuous, K is uniformly continuous in the compact set A × A, and then ∀ε > 0, ∃δ > 0 such that |K (x, y) − K (z , y)| < c |εb| , if ∥(x, y) − (z , y)∥ < δ . Then

|(C (f ))(x) − (C (f ))(z )| ≤ ε,

(6)

for all f ∈ B, and all x, z in A with ∥x − z ∥ < δ . Then the set C (B) is equicontinuous. From Ascoli’s Theorem, C (B) is a relatively compact subset of X . Hence the operator C is compact. Since S and C are C 1 -mappings, S is k-contractive and C is compact, we have that if g is a regular value of the weakly coercive C 1 -mapping H : X × [0, 1] → X , H (f , t ) = I (f ) − tT (f ), where T := S + C then Corollary 8 implies that: (a) The mapping T is condensing, and (b) Eq. (5) has a solution. Furthermore, since I − S is bijective, and C is linear continuous and compact, Corollary 8 implies that: (c) Eq. (5) has finitely many solutions. (d) I − T is a Fredholm mapping of index zero. Since the conditions of Corollary 7 part (a) are satisfied, there is an open dense subset V ⊂ X whose elements are regular values of H and for any g ∈ V Eq. (5) has a solution. Obviously the requirement that g is a regular value is satisfied on V . In the particular case in which k = 0, the conditions required in Corollary 7 part (b) are satisfied, then Eq. (5) has finitely many solutions for any value g ∈ X , and the requirement of a regular value can be dropped. Since H is linear in the f -component, we get that if tT ̸= I

∥tT (f ) − I (f )∥∞

       f f   = ∥f ∥∞ tT −I  ∥f ∥ ∥f ∥ ∞



→ ∞ as ∥f ∥∞ → ∞.



Then, a sufficient condition for H to be weakly coercive is T ̸= λI for any λ ≥ 1.

Acknowledgments The authors are grateful to the referee for carefully reading the manuscript and for offering extremely valuable comments and suggestions which enabled them to substantially improve the paper. The authors have been partially supported by the Plan Andaluz de Investigación de la Junta de Andalucía FQM-127 and Grant P08-FQM-03543, and by MEC Grant MTM2012-34847-C02-01.

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