Existence and Uniqueness of Solutions for Linear Fredholm-Stieltjes Integral Equations Via Henstock-Kurzweil Integral

Existence and uniqueness of solutions for linear Fredholm-Stieltjes integral equations via Henstock-Kurzweil integral

M. Federson∗ and R. Bianconi†

Abstract We consider the linear Fredholm-Stieltjes integral equation

Z b x (t) − α (t, s) x (s) dg(s) = f (t) , t ∈ [a, b] , a in the frame of the Henstock-Kurzweil integral and we prove a Fredholm Alternative type result for this equation. The functions α, x and f are real-valued with x and f being Henstock-Kurzweil integrable.

Keywords and phrases: Linear integral equations, Fredholm integral equations, Fredholm Alternative, Henstock-Kurzweil integral. 2000 Mathematical Subject Classiﬁcation: 45A05, 45B05, 45C05, 26A39.

1 Introduction

The purpose of this paper is to prove a Fredholm Alternative type result for the following linear Fredholm-Stieltjes integral equation

Z b x (t) − α (t, s) x (s) dg(s) = f (t) , t ∈ [a, b] , (1) a in the frame of the Henstock-Kurzweil integral. Let g :[a, b] → R be an element of a certain subspace of the space of continuous functions from [a, b] to R. Let Kg([a, b], R) denote the space of all functions f :[a, b] → R such that ∗Instituto de CiênciasMatemáticase de Computa¸cão,Universidade de SãoPaulo, CP 688, SãoCarlos, SP 13560-970, Brazil. E-mail: [email protected] †Instituto de Matemática e Estatstica, Universidade de SãoPaulo, CP 66281, SãoPaulo, SP 05315-970, Brazil. E-mail: [email protected]

1 R b the integral a f(s)dg(s) exists in the Henstock-Kurzweil sense. It is known that even when g(s) = s, an element of Kg([a, b], R) can have not only many points of discontinuity, but it can also be of unbounded variation. The Henstock-Kurzweil integral encompasses the Lebesgue, Riemann and Newton integrals as well as its improper integrals. When (1) is considered in the sense of Henstock-Kurzweil, f can have singularities, for instance, and yet it may be possible to ﬁnd a solution xf in Kg([a, b], R). In [4], the authors considered the linear Fredholm integral equation

Z b x (t) − α (t, s) x (s) ds = f (t) , t ∈ [a, b] , a and established a Fredholm Alternative for such equation in the frame of the Henstock- Kurzweil integral (see [4], Theorem 3.10). On that occasion, the functions α, x and f were considered as being Banach-space valued and x and f were Kurzweil integrable. The latter could also be considered in any of the spaces of Henstock, Bochner or improper Riemann integrable functions (see [4], Remarks 3.2 and 3.3). Also, an example was given showing that, under the hypothesis we considered, the existing results were not enough. Suppose the function α satisﬁes certain conditions. In the present paper, we prove that:

• if for every f ∈ Kg([a, b], R), equation (1) has a unique solution xf ∈ Kg([a, b], R), then the application f 7→ xf is bicontinuous and equation (1) has a resolvent with similar integral representation; if besides α(t, s) = 0 for m-almost every s > t, then f 7→ xf is causal and the resolvent of (1) is given by the Neumann series (see Theorem 5.1 in the sequel).

We also state the following Fredholm Alternative for equation (1):

• for every f ∈ Kg([a, b], R), either equation (1) has a unique solution xf ∈ Kg([a, b], R) and (1) has a resolvent with similar integral representation, or the corresponding homogeneous equation admits non-trivial solutions in Kg([a, b], R) (see Theorem 5.2 in the sequel).

Corollary 5.1 in the sequel mixes the two results. Although the above results are proved in the case where α, x and f are real-valued, the auxiliary theory developed throughout the paper is placed in a general abstract context. As in [6], the reason for considering functions taking values in a ﬁnite dimensional space is because some results of the Kurzweil-Henstock integration theory either hold for one integral or for the other. For instance, in a Banach-space valued context, the spaces of Henstock and of Kurzweil integrals may not coincide; a Riemann integrable function may not be Henstock integrable; the Fundamental Theorem of Calculus may fail for the Kurzweil integral. See [7], [8] and [3] for examples. See also [18]. Another obstacle we encountered is the fact that the normed space of Henstock-Kurzweil integrable functions is not complete (it is ultrabornological though). Therefore one can not

2 apply usual fixed point theorems in order to get existence results, for instance. This difficulty was also faced by the authors in [4] and in [6] and by Hönig in [11], but it can be overcome when some ideas due to Hönigare applied. Such ideas concern applying representation theorems, an integration by parts formula and the fundamental theorem of calculus in order to transform integral equations in the sense of the Henstock-Kurzweil integral into integral equations in the sense of the Riemann-Stieltjes integral. As we did in [6] for a Volterra-Stieltjes type integral equation, we adapt, in the present paper, the ideas mentioned above to the Stieltjes case, that is, by means of representation theorems, integration by parts and substitution formulas and the fundamental theorem of calculus for Henstock-Kurzweil vector integrals, we transform the Fredholm-Stieltjes integral equation in the sense of Henstock-Kurzweil integral into an ordinary Fredholm-Stieltjes integral equation. In particular, we apply the Fredholm Alternative for the Riemann-Stieltjes integral (proved by the authors in [4], Theorems 2.4 and 2.5) to obtain a Fredholm Alter- native for our equation (1). This result is presented in Section 5. The other sections are organized as follows. Section 2 is devoted to the fundamental theory of the Riemann-Stieltjes integral in Banach spaces, where we present basic results, representation theorems and the Fredholm Alternative. In Section 3, we give some basic definitions of the Henstock-Kurzweil integration theory. In Section 4, we present auxiliary results for the Henstock-Kurzweil integral such as the fundamental theorem of calculus and a substitution formula.

2 The Riemann-Stieltjes integral in Banach spaces

In this section we introduce the deﬁnitions and terminology that we are going to use throughout the paper for Riemann-Stieltjes integrals and we mention some properties of these integrals. We start by deﬁning functions of bounded B-variation, of bounded semi-variation and of bounded variation and giving some results concerning these spaces.

2.1 Functions of bounded B-variation, of bounded semi-variation and of bounded variation A bilinear triple (we write BT ) is a set of three vector spaces E, F and G, where F and G are normed spaces with a bilinear mapping B : E × F → G. For x ∈ E and y ∈ F , we write xy = B (x, y) and we denote the BT by (E,F,G)B or simply by (E,F,G). A topological BT is a BT (E,F,G) where E is also a normed space and B is continuous. We suppose that kBk ≤ 1. We denote by L (E,F ) the space of all linear continuous functions from E to F , where E and F are normed spaces. We write L (E) = L (E,E) and E0 = L (E, R), where R denotes the real line. Throughout this paper, X, Y and Z will always denote Banach spaces.

3 Example 2.1. As an example of a BT we can consider E = L (X,Y ), F = L (Z,X), G = L (Z,Y ) and B (v, u) = v ◦ u. In particular, when Z = R, we have E = L (X,Y ), F = X, G = Y and B (u, x) = u (x); when X = R, we have E = Y , F = Y 0, G = R and B (y, y0) = hy, y0i; when X = Z = R, we have E = G = Y , F = R and B (y, λ) = λy.

Let [a, b] be a compact interval of R. Any ﬁnite set of closed non-overlapping subintervals [ti−1, ti] of [a, b] such that the union of all intervals [ti−1, ti] equals [a, b] is called a division of [a, b]. In this case we write d = (ti) ∈ D[a,b], where D[a,b] denotes the set of all divisions of [a, b]. By |d| we mean the number of subintervals in which [a, b] is divided through a given d ∈ D[a,b].

Given a BT (E,F,G)B and a function α :[a, b] → E, for every division d = (ti) ∈ D[a,b] we deﬁne   |d|  X  SBd (α) = SB[a,b],d (α) = sup [α (ti) − α (ti−1)] yi ; yi ∈ F, kyik ≤ 1

 i=1  and SB (α) = SB[a,b] (α) = sup SBd (α); d ∈ D[a,b] . Then SB (α) is the B-variation of α on [a, b]. We say that α is a function of bounded B-variation whenever SB (α) < ∞. When this is the case, we write α ∈ SB ([a, b] ,E). The following properties are not diﬃcult to prove:

(SB1) SB ([a, b] ,E) is a vector space and the mapping α ∈ SB ([a, b] ,E) 7→ SB (α) ∈ R+ is a seminorm;

(SB2) for α ∈ SB ([a, b] ,E), the function t ∈ [a, b] 7→ SB[a,t] (α) ∈ R+ is monotonically increasing;

(SB3) for α ∈ SB ([a, b] ,E) and c ∈ [a, b], SB[a,b] (α) ≤ SB[a,c] (α) + SB[c,b] (α)

Consider the BT (L (X,Y ) ,X,Y ). In this case, we write SV ([a, b] ,L (X,Y )) and SV (α) respectively by SB ([a, b] ,L (X,Y )) and SB (α), and any element of SV ([a, b] ,L (X,Y )) is called a function of bounded semi-variation. Given a function α :[a, b] → E, a normed space E and a division d = (ti) ∈ D[a,b], we deﬁne |d| X Vd (α) = Vd,[a,b] (α) = kα (ti) − α (ti−1)k i=1 and the variation of α is given by V (α) = V[a,b] (α) = sup Vd (α); d ∈ D[a,b] .

4 If V (f) < ∞, then α is called a function of bounded variation. In this case, we write α ∈ BV ([a, b] ,E). We also have

BV ([a, b] ,L (E,F )) ⊂ SV ([a, b] ,L (E,F )) and 0 SV ([a, b] ,L (E, R)) = BV ([a, b] ,E ) . Remark 2.1. Consider a BT (E,F,G). The deﬁnition of variation of a function α :[a, b] → E, where E is a normed space, can also be considered as a particular case of the B-variarion of α in two diﬀerent ways.

• Let E = F 0, G = R or G = C and B(x0, x) = hx, x0i. By the deﬁnition of the norm in E = F 0, we have

|d| X Vd (α) = kα (ti) − α (ti−1)k i=1   |d|  X  = sup hxi, α (ti) − α (ti−1)i ; xi ∈ F, kxik ≤ 1 = SBd(α).

 i=1 

Thus when we consider the BT (Y 0,Y, R), we write BV (α) and BV ([a, b] ,Y 0) instead of SB (α) and SB ([a, b] ,Y 0) respectively.

• Let F = E0, G = R or G = C and B(x, x0) = hx, x0i. By the Hahn-Banach Theorem, we have

0 0 0 0 kα (ti) − α (ti−1)k = sup {|hα (ti) − α (ti−1) , xii| ; xi ∈ E , kxik ≤ 1}

and hence

|d| X Vd (α) = kα (ti) − α (ti−1)k i=1   |d|  X 0 0 0 0  = sup hα (ti) − α (ti−1) , x i ; x ∈ E , kx k ≤ 1 = SBd(α). i i i  i=1 

Given c ∈ [a, b], we deﬁne the spaces

BVc ([a, b] ,X) = {f ∈ BV ([a, b] ,X); f (c) = 0} and SVc ([a, b] ,L (X,Y )) = {α ∈ SV ([a, b] ,L (X,Y )) ; α (c) = 0} .

5 Such spaces are complete when endowed, respectively, with the norm given by the variation V (f) and the norm given by the semi-variation SV (α). See [19], for instance. The next results are borrowed from [15]. We include the profs here, since this reference is not easily available. Lemmas 2.1 and 2.2 below are respectively Theorems I.2.7 and I.2.8 from [15].

Lemma 2.1. Let α ∈ BV ([a, b],X). Then

(i) For all t ∈ ]a, b], there exists α(t−) = lim α(t − ε). ε↓0 (ii) For all t ∈ [a, b[, there exists α(t+) = lim α(t + ε). ε↓0 Proof. We will prove (i). The proof of (ii) follows analogously. Consider a strictly increasing sequence {tn} in [a, t[ converging to t. Then

n X kα(ti) − α(ti−1)k ≤ V[a,t](α), for all n. i=1 Hence ∞ X kα(ti) − α(ti−1)k ≤ V[a,t](α). i=1

Then {α(tn)} is a Cauchy sequence, since

m X kα(tm) − α(tn)k ≤ kα(ti) − α(ti−1)k ≤ ε, i=n+1 for suﬃciently large m, n. The limit α(t−) of {α(tn)} is independent of the choice of {tn} and we ﬁnish the proof.

Lemma 2.2. Let α ∈ BV ([a, b],X). For every t ∈ [a, b], let v(t) = V[a,t](α). Then (i) v(t+) − v(t) = kα(t+) − α(t)k, t ∈ [a, b[.

(ii) v(t) − v(t−) = kα(t) − α(t−)k, t ∈ ]a, b].

Proof. By property (SB2), v is monotonically increasing and hence v(t+) and v(t−) exist. By Lemma 2.1, α(t+) and α(t−) also exist. We will prove (i). The proof of (ii) follows analogously. Suppose s > t. Then property (SB3) implies V[a,s] (α) ≤ V[a,t] (α) + V[ t,s] (α) . Therefore kα(s) − α(t)k ≤ V[t,s](α) = V[a,s](α) − V[a,t](α) and hence kα(t+) − α(t)k ≤ v(t+) − v(t).

6 Conversely, given d ∈ D[a,t], let vd(t) = Vd(α). Then for every ε > 0, there exists δ > 0 such that v(t + σ) − v(t+) ≤ ε and kα(t + σ) − α(t+)k ≤ ε and there exists d : a = t0 < t1 < . . . < tn = t < tn+1 = t + σ such that v(t + σ) − vd(t + σ) ≤ ε whenever 0 < σ ≤ δ. Thus

v(t + σ) − v(t) ≤ vd(t + σ) + ε − vd(t) = kα(t + σ) − α(t)k + ε ≤ kα(t+) − α(t)k + 2ε and hence v(t+) − v(t) ≤ kα(t+) − α(t)k. This completes the proof.

Let α ∈ BV ([a, b],X). Since kα(t)k ≤ kα(a)k + V[a,t], then Lemma 2.2 implies that the sets {t ∈ [a, b[; kα(t+) − α(t)k ≥ ε} and {t ∈ ]a, b]; kα(t) − α(t−)k ≥ ε} are finite for every ε > 0. Thus we have the next result which can be found in [15], for instance (Proposition I.2.10 there). Proposition 2.1. Let α ∈ BV ([a, b],X). Then the set of points of discontinuity of α is countable (and all discontinuities are of the first kind). Let us define

+ BVa ([a, b],X) = {α ∈ BVa([a, b],X); α(t+) = α(t), t ∈]a, b[}.

+ A proof that BVa ([a, b],X) with the variation norm is complete can be found in [15], The- orem I.2.11. We reproduce it next.

+ Theorem 2.1. BVa ([a, b],X) is a Banach space when endowed with the variation norm. Proof. For every t ∈ ]a, b[, we have kα(t)k ≤ kα(a)k + V (α) = V (α). Hence the mappings

Tt : α ∈ BVa([a, b],X) 7→ α(t) ∈ X and Tt+ : α ∈ BVa([a, b],X) 7→ α(t+) ∈ X + are continuous. Therefore BVa ([a, b],X) is a closed subspace of BVa([a, b],X), since it is given by the continuous mappings Tt and Tt+, t ∈ ]a, b[, and the result follows from the fact that BVa([a, b],X) is a Banach space with the variation norm. Given u ∈ L(X,Z0) and z ∈ Z, we denote an element of X0 by z ◦ u which is given by

hz ◦ u, xi = hz, u(x)i, x ∈ X.

We have |hz ◦ u, xi| = |hz, u(x)i| ≤ kzk ku(x)k ≤ kzk kuk kxk. We denote by u∗ ∈ L (Y 0,X0) the adjoint or transposed operator of u ∈ L(X,Y ) which is deﬁned by hx, u∗ (y0)i = hu (x) , y0i , x ∈ X, y0 ∈ Y 0.

7 Then y0 ◦ u = u∗(y0), y0 ∈ Y 0 since (y0 ◦ u)(x) = hy0, u(x)i = hu∗(y0), xi for every x ∈ X. Next we have [15], Proposition I.3.5 and the corollary that follows it. Proposition 2.2. Given a function α :[a, b] → L(X,Z0). Then (i) SV (α) = sup{V (z ◦ α); z ∈ Z, kzk ≤ 1};

(ii) α ∈ SV ([a, b],L(X,Z0)) if and only if z ◦ α ∈ BV ([a, b],X0), for every z ∈ Z.

Proof. In order to prove (i), it is enough to observe that, if d = (ti) ∈ D[a,b], then   |d|  X  SVd(α) = sup [α(ti) − α(ti−1)]xi ; xi ∈ X, kxik ≤ 1

 i=1    |d|  X  = sup sup hz, [α(ti) − α(ti−1)]xii ; xi ∈ X, kxik ≤ 1

kzk≤1 i=1   |d|  X  = sup sup hxi, z ◦ α(ti) − z ◦ α(ti−1)i; xi ∈ X, kxik ≤ 1 kzk≤1  i=1 

= sup {Vd(z ◦ α); z ∈ Z, kzk ≤ 1} .

Now we will prove (ii). By (i), if α ∈ SV ([a, b],L(X,Z0)), then z ◦ α ∈ BV ([a, b],X0), z ∈ Z. The converse follows by the uniform boundedness principle. Indeed. Let us deﬁne D = (d, x); d ∈ D[a,b], x = (x1, . . . , x|d|), xi ∈ X, kxik ≤ 1 . Then for each (d, x) ∈ D, we 0 deﬁne Fd,z ∈ Z by

|d| X Fd,z(z) = hz, [α(ti) − α(ti−1)]xii, z ∈ Z. i=1

0 Then the set {Fd,z;(d, z) ∈ D} ⊂ Z is simply bounded on Z, since |Fd,z(z)| ≤ Vd(z ◦ α) ≤ V (z ◦ α), for all (d, z) ∈ D, and all z ∈ Z. Therefore by uniform boundedness principle, there exists M ≥ 0 such that kFd,xk ≤ M,(d, z) ∈ D, that is, for all (d, z) ∈ D, we have sup {|Fd,z(z)|; z ∈ Z, kzk ≤ 1} ≤ M. Hence SV (α) ≤ M. Corollary 2.1. Suppose α ∈ SV ([a, b],L(X,Z0)). Then (i) For every t ∈ ]a, b], there exists α(t−) ∈ L(X,Z0) in the sense that

lim (z ◦ α)(t − ε) = z ◦ α(t−), z ∈ Z; ε→0+

8 (ii) For every t ∈ [a, b[, there exists α(t+) ∈ L(X,Z0) in the sense that

lim (z ◦ α)(t + ε) = z ◦ α(t+), z ∈ Z. ε→0+

Proof. We will prove (i). The proof of (ii) follows analogously. We may suppose, without loss of generality, that α(a) = 0. By Lemma 2.1 (i) and Proposition 2.2 (ii), given z ∈ Z, there exists

0 Tz = lim (z ◦ α)(t − ε) ∈ X . ε→0+

0 Then the mapping T : z ∈ Z 7→ Tz ∈ X is linear. It is also continuous, since kz ◦ α(t)k ≤ kz ◦ α(a)k + V[a,t](z ◦ α). Besides, Proposition 2.2 (i) implies k(z ◦ α)(t − ε)k ≤ V (z ◦ α) ≤ kzkSV (α). Hence kT k ≤ SV (α). Let tT be the transposed mapping of T , that is, tT : x ∈ X 7→ tT x ∈ Z0 is deﬁned by t t 0 t hz, T xi = hTz, xi, where z ∈ Z. Then T ∈ L(X,Z ) and k T k = kT k ≤ SV (α). Also tT = α(t−) in the sense of (i), since for every x ∈ X, we have

t t hz ◦ T, xi = hz, T xi = hTz, xi = h lim (z ◦ α)(t − ε), xi = lim h(z ◦ α)(t − ε), xi ε→0+ ε→0+ and we ﬁnished the proof. In view of Corollary 2.1, we deﬁne the space

+ 0 0 + 0 SVa ([a, b],L(X,Z )) = α ∈ SVa([a, b],L(X,Z )); z ◦ α ∈ BVa ([a, b],X ), z ∈ Z which is complete when equipped with the semi-variation norm. This result can be found in [15], Theorem I.3.7. We include it here.

+ 0 Theorem 2.2. SVa ([a, b],L(X,Z )) is a Banach space with the semi-variation norm. Proof. By Theorem 2.2 (i), for every z ∈ Z, the mapping

0 0 Fz : α ∈ SVa([a, b],L(X,Z )) 7→ z ◦ α ∈ BVa([a, b],X )

+ 0 0 is continuous. By Theorem 2.1, BVa ([a, b],X ) is a closed subspace of BVa([a, b],X ) and + 0 −1 0 therefore SVa ([a, b],L(X,Z )) = ∩ {(Fz) (BVa([a, b],X )); z ∈ Z} is a closed subspace of 0 the Banach space SVa([a, b],L(X,Z )) which implies the result.

2.2 Riemann-Stieltjes integration For the next results we need the concept of the Riemann-Stieltjes integral which we deﬁne by means of tagged divisions.

9 A tagged division of [a, b] is any set of pairs (ξi, ti) such that (ti) ∈ D[a,b] and ξi ∈ [ti−1, ti] for every i. In this case we write d = (ξi, ti) ∈ TD[a,b], where TD[a,b] denotes the set of all tagged divisions of [a, b]. Any subset of a tagged division of [a, b] is a tagged partial division of [a, b] and, in this case, we write d ∈ TPD[a,b]. A gauge of a set E ⊂ [a, b] is any function δ : E → ]0, ∞[. Given a gauge δ of [a, b], we say that d = (ξi, ti) ∈ TPD[a,b] is δ-ﬁne, if [ti−1, ti] ⊂ {t ∈ [a, b]; |t − ξi| < δ (ξi)} for every i, that is, ξi ∈ [ti−1, ti] ⊂ ]ξi − δ(ξi), ξi + δ(ξi)[, i = 1, 2,... |d|.

Now we will deﬁne the Riemann-Stieltjes integrals by means of tagged divisions d = (ξi, ti) of [a, b] and constant gauges δ (i.e., there is a δ0 > 0 such that δ (ξ) = δ0 for every ξ ∈ [a, b]). Let (E,F,G) be a BT . Any function α :[a, b] → E is said to be Riemann integrable with respect to a function f :[a, b] → F if there exists an I ∈ G such that for every ε > 0, there is a constant gauge δ of [a, b] such that for every δ-ﬁne d = (ξi, ti) ∈ TD[a,b],

|d| X [α (ti) − α (ti−1)] f (ξi) − I < ε.

i=1

R b In this case, we write I = a dα (t) f (t) . By Rf ([a, b] ,E) we mean the space of all functions α :[a, b] → E which are Riemann integrable with respect to f :[a, b] → F . Analogously we say that f :[a, b] → F is Riemann integrable with respect to α :[a, b] → E if there exists an I ∈ G such that for every ε > 0, there is a constant gauge δ of [a, b] such that for every δ-ﬁne d = (ξi, ti) ∈ TD[a,b],

|d| X α (ξi)[f (ti) − f (ti−1)] − I < ε.

i=1

Then Rα ([a, b] ,F ) denotes the space of all functions f :[a, b] → F which are Riemann R b integrable with respect to a given α :[a, b] → E with integral I = a α (t) df (t) . R b R b The integrals a dα (t) f (t) and a α (t) df (t) deﬁned above are known as Riemann- Stieltjes integrals. Consider the BT (E,F,G) with E = Y 0 = L (Y, R), F = L (X,Y ), G = X0 = L (X, R) and B (y0, u) = y0 ◦ u, for all y0 ∈ Y 0 and all u ∈ L (X,Y ). We will use the identiﬁcation

Z d Z d dy (t) ◦ K (t, s) = K (t, s)∗ dy (t) , c c where y :[a, b] → Y 0, K :[c, d] × [a, b] → L (X,Y ) and K (t, s)∗ denotes the adjoint of K (t, s) ∈ L (X,Y ).

10 2.3 Some properties Let E be a normed space. By C ([a, b] ,E) we mean the space of all continuous functions from [a, b] to E endowed with the usual supremum norm which we denote by k·k∞. We deﬁne Ca ([a, b] ,E) = {f ∈ C ([a, b] ,E); f (a) = 0} . The next result is well-known. It gives the Integration by Parts Formula for the Riemann- Stieltjes integrals. For a proof of it, see for instance [14] or [8], Theorem 2.5. Theorem 2.3 (Integration by Parts). Let (E,F,G) be a BT . If either α ∈ SB ([a, b] ,E) and f ∈ C ([a, b] ,F ), or α ∈ C ([a, b] ,E) and f ∈ BV ([a, b] ,F ), then α ∈ Rf ([a, b],E), f ∈ Rα([a, b],F ) and the Integration by Parts Formula

Z b Z b dα (t) f (t) = α (b) f (b) − α (a) f (a) − α (t) df (t) a a holds. The assertions in the next remark follow by the Integration by Parts Formula and some easy computation. Remark 2.2. Suppose (E,F,G) is a BT and α ∈ SB ([a, b] ,E). If we deﬁne

Z b Fα(f) = dα(t)f(t), f ∈ C ([a, b] ,F ) , a then Fα ∈ L(C([a, b],F ),G) and kFαk ≤ SB(α). In particular we have

• If E = F 0 and G = C or R as in Remark 2.1, then given α ∈ BV ([a, b],F 0), there exists

|d| Z b X Fα(f) = hf(t), dα(t)i = lim hf(ξi), α(ti) − α(ti−1)i, f ∈ C([a, b],F ), ∆d→0 a i=1

where d = (ξi, ti) ∈ TD[a,b] and ∆d = max{ti − ti−1; i = 1, 2,..., |d|}. Also Fα ∈ 0 C([a, b],F ) and kFαk ≤ V (α). • If α ∈ SV ([a, b],L(X,Y )), then for every f ∈ C ([a, b] ,X) there exists the Riemann- R b R b Stieltjes integral a dα (t) f (t). Furthermore Fα (f) = a dα (t) f (t), f ∈ C ([a, b] ,X), is such that Fα ∈ L (C ([a, b] ,X) ,Y ) and kFαk ≤ SV (α). The next theorem says that all operators in L(C([a, b],X),Y )) can be represented by functions of bounded semi-variation. The version we present here is a special case of [14], Theorem I.5.1. In particular, it will be shown later that if Y = Z0, then L(C([a, b],X),Y )) can be represented by functions of bounded semi-variation which are right continuous.

11 Theorem 2.4. The mapping

α ∈ SVa([a, b],L(X,Y )) 7→ Fα ∈ L(C([a, b],X),Y )

R b where Fα(f) = a dα(t)f(t), for f ∈ C([a, b],Y ), is an isometry (i.e., kFαk = SV (α)) of the ﬁrst Banach space onto the second. We also have α(t)x = Fα(χ]a,t]x), x ∈ X. We proceed with the presentation of results borrowed from [15] with their proofs. In the sequel, we assume that c is some point in the interval [a, b]. Given α ∈ BV ([a, b],Y 0), let us deﬁne an auxiliary function α :[a, b] → Y 0 by   0, t = a, α(t) = α(t+) − α(a), t ∈ ]a, b[, (2)  α(b) − α(a), t = b.

The next result will be useful to prove that the operators of C([a, b],Y )0 can be represented 0 by elements of BVc([a, b],Y ). It can be found in [15], Theorem I.2.12. Lemma 2.3. Let α ∈ BV ([a, b],Y 0). Then

+ 0 (i) α ∈ BVc ([a, b],Y ) and V (α) ≤ V (α);

(ii) For every f ∈ C([a, b],Y ), Fα(f) = Fα(f). Proof. Let us prove (i). By the deﬁnition of α, α(a) = 0 and α is right continuous at t ∈ ]a, b[. + 0 Hence α ∈ BVc ([a, b],Y ). It remains to prove that V (α) ≤ V (α). We can suppose, without loss of generality, that α(a) = 0. Then given ε > 0 and d = (ti) ∈ D[a,b], there exist si ∈ ]ti−1, ti[, i = 1, 2,..., |d| − 1, such that α(si+) = α(si) (by Proposition 2.1) and kα(ti+) − α(si)k ≤ ε. Therefore

kα(ti+) − α(ti−1)k ≤ kα(ti+) − α(si)k + kα(si) − α(si−1)k + kα(ti−1+) − α(si−1)k

≤ kα(si) − α(si−1)k + 2ε.

0 If we consider d = (si) with a = s0 < s1 < . . . < s|d|−1 < s|d| = b, then

|d| |d| |d| X X X kα(ti+) − α(ti−1)k ≤ kα(si) − α(si−1)k + 2ε i=1 i=1 i=1 and hence Vd(α) ≤ Vd0 (α)+2|d|ε ≤ V (α)+2|d|ε. Thus Vd(α) ≤ V (α) (since ε is independent of the choice of d) and then V (α) ≤ V (α). Now we will prove (ii). By deﬁnition, we have

|d| Z b X Fα(f) = hf(t), dα(t)i = lim hf(ξi), α(ti) − α(ti−1)i, ∆d→0 a i=1

12 where ξi ∈ [ti−1, ti]. We may suppose, without loss of generality, that α(a) = 0 and then Proposition 2.1 implies α(t) = α(t), for all t ∈ [a, b] but a countable subset. Then if we take the points ti of the division d = (ti) ∈ D[a,b] in the complement of that countable subset, we obtain (ii). The next representation theorem can be found in [15], Theorem I.2.13.

Theorem 2.5. The mapping

0 0 α ∈ BVc([a, b],Y ) 7→ Fα ∈ C([a, b],Y ) is an isometry (i.e., kFαk = V (α)) of the ﬁrst Banach space onto the second.

Proof. It is clear that the mapping is linear and kFαk ≤ V (α). We will prove that the mapping is one-to-one, that is, α 6= 0 implies Fα 6= 0. It is enough to show that there exists f ∈ C([a, b],Y ) such that Fα(f) 6= 0. If α(b) 6= 0, then there exists y0 ∈ Y such that hy0, α(b)i = 1. If we take f(t) ≡ y0, then

Z b Fα(f) = hy0, dα(t)i = hy0, α(b)i = 1. a

If α(b) 6= 0, let t0 ∈ ]a, b[ be such that α(t0) 6= 0 and consider y0 ∈ Y such that hy0, α(t0)i = 1. Deﬁne   0, t ≤ t0  1 fn(t) = n(t − t0)y0, t0 ≤ t ≤ t0 + n  1 y0, t0 + n ≤ t ≤ b.

Then fn ∈ C([a, b],Y ) and kfnk = ky0k, n ∈ N, which implies

1 Z t0+ n Z b Fα(fn) = n(t − t0)hy0, dα(t)i + hy0, dα(t)i, 1 t0 t0+ n

1 where the second integral equals hy0, α(t0 + n )i which converges to hy0, α(t0+)i = 1 and the norm of the first integral is bounded by ky0kV 1 (α) which converges to 0. [t0,t0+ n ] Now we will prove that the mapping is onto and it is an isometry. It suffices to show 0 0 that for every F ∈ C([a, b],Y ) , there exists an α ∈ BV ([a, b],Y ) such that F = Fα and V (α) ≤ kF k. Then by Lemma 2.3, the same applies to α given by (2). Since C([a, b],Y ) is a subspace of the space B([a, b],Y ) of bounded functions from [a, b] to Y , then the Hahn-Banach Theorem implies F ∈ C([a, b],Y )0 admits a linear continuous extension F ∈ B([a, b],Y )0 such that kF k = kF k. If for all s ∈ [a, b], s 6= c, we define ψs(t) = 1, if a ≤ t ≤ s, ψs(t) = 0, if s < t ≤ b, and ψc = 0, then for every y ∈ Y and t ∈ [a, b], we have yψt ∈ B([a, b],Y ) and this function takes only two values: 0 and y. The mapping y ∈ Y 7→ F (yψt) ∈ R is linear and continuous

13 0 because |F (yψt)| ≤ kF k kyk). Therefore there exists one and only one element α(t) ∈ Y such that F (yψt) = hy, α(t)i, y ∈ Y . 0 We assert that α ∈ BV ([a, b],Y ) and V (α) ≤ kF k. Indeed, given d = (ti) ∈ D[a,b], we have |d| X Vd(α) = kα(ti) − α(ti−1k i=1   |d|  X  = sup hyi, α(ti) − α(ti−1i ; yi ∈ Y, kyik ≤ 1

 i=1    |d| X = sup F  yi(ψti − ψti−1 ) ≤ kF k,

kyik≤1 i=1 P|d| R b since k i=1 yi(ψti −ψti−1 )k ≤ 1. Also, given f ∈ C([a, b],Y ), we have F (f) = a hf(t), dα(t)i, that is, Fα = F . Indeed, because for ξi ∈ [ti−1, ti], we have

|d| R b X hf(t), dα(t)i = lim hf(ξi), α(ti) − α(ti−1)i a ∆d→0 i=1  |d|  X = lim F  f(ξi)(ψti − ψti−1 ) = F (f), ∆d→0 i=1

P|d| since lim∆d→0 i=1 f(ξi)(ψti − ψti−1 ) = f, where the limit is taken in B([a, b],Y ). Also

|d| |d| X X f − f(ξi)(ψti − ψti−1 ) = sup f(s) − f(ξi)[ψti (s) − ψti−1 (s)] ≤ ωd(f), a≤s≤b i=1 i=1 where ωd(f) is the oscillation of f with respect to d and converges to 0 as ∆d → 0.

0 0 Remark 2.3. Theorem 2.5 also holds for BVc([a, b],Y ), c ∈ [a, b], instead of BVa([a, b],Y ). The next two results are also borrowed from [15] (see respectively Theorem I.3.8 and Corollary I.3.9 there).

Theorem 2.6. The mapping

+ 0 0 α ∈ SVc ([a, b],L(X,Z )) 7→ Fα ∈ L(C([a, b],X),Z ) is a linear isometry (i.e., kFαk = SV (α)) of the ﬁrst Banach space onto the second.

14 Proof. The mapping is clearly linear. Also kFαk ≤ SV (α) by Remark 2.2. Let us prove that the mapping is injective. If α 6= 0, there exists t0 ∈ ]a, b] such that 0 α(t0) 6= 0. Hence there exists z ∈ Z such that z ◦ α(t0) 6= 0, where z ◦ α(t) ∈ X . Therefore + 0 z ◦ α ∈ BVc ([a, b],X ) and z ◦ α 6= 0. By Theorem 2.5, Fz◦α 6= 0, where Fz◦α is the element 0 of C([a, b],X) deﬁned by z ◦ α. Thus there exists f ∈ C([a, b],X) such that Fz◦α(f) 6= 0. On the other hand, Z b Z b Fz◦α(f) = hf(t), d(z ◦ α)(t)i = hz, dα(t) f(t)i = hz, Fα(f)i a a and hence Fα(f) 6= 0, that is, Fα 6= 0. 0 + 0 Now we will show that given F ∈ L(C([a, b],X),Z ), there exists α ∈ SVc ([a, b],L(X,Z )) such that F = Fα and SV (α) ≤ kF k. For every z ∈ Z, we have z ◦ F ∈ C([a, b],X)0 and kz ◦ F k ≤ kzk kF k. Then Theorem 2.5 + 0 implies that there is one and only one element αz ∈ BVc ([a, b],X ) such that z ◦ F = Fαz , that is, for every f ∈ C([a, b],X)0, we have

Z b (z ◦ F )(f) = hf(t), dαz(t)i a and V (αz) = kz ◦ F k.

We assert that αz1+z2 (t) = αz1 (t) + αz2 (t), t ∈ [a, b]. Indeed, we have (z1 + z2) ◦ F = 0 z1 ◦ F + z2 ◦ F and hence for every f ∈ C([a, b],X) , we have Z b Z b

hf(t), dαz1+z2 (t)i = hf(t), d(αz1 + αz2 )(t)i. a a Then the uniqueness of the representation in Theorem 2.5 implies the assertion. In a similar way, one proves that αλz(t) = λαz(t). We have α(t)x ∈ Z0. Indeed. Given t ∈ [a, b] and x ∈ X, if we deﬁne α(t)x ∈ RZ by (α(t)x)z = hx, αz(t)i, then the above computation implies that the mapping z ∈ Z 7→ (α(t)x)z = hx, αz(t)i ∈ R is linear. The mapping is also continuous because kα(t)xk = sup {|hx, αz(t)i|; z ∈ Z, kzk ≤ 1} and |hx, αz(t)i| ≤ kxk kαz(t)k ≤ kxkV (αz) = kxk kz◦F k ≤ kxk kF k kzk and hence kα(t)xk ≤ kxk kF k. We have α ∈ L(X,Z0), since the mapping x ∈ X 7→ α(t)x ∈ Z0 is linear and kα(t)k ≤ kF k (by the last inequality). + 0 We also assert that α ∈ SVc ([a, b],L(X,Z )) and SV (α) ≤ kF k. Indeed, by the deﬁnition + 0 of α, we have z ◦ α = αz ∈ BVc ([a, b],X ) for every z ∈ Z. Hence by Proposition 2.2 (i),

SV (α) = sup {V (z ◦ α); z ∈ Z, kzk ≤ 1} = sup {V (αz); z ∈ Z, kzk ≤ 1} ≤ kF k, since V (αz) = kz ◦ F k ≤ kzk kF k.

Finally F = Fα, since for every z ∈ Z, we have z ◦ F = Fαz = Fz◦α = z ◦ Fα and then z ◦ F = z ◦ Fα.

15 The next result is a consequence of Theorem 2.6 with Z = Y 0.

Corollary 2.2. For every Fα ∈ L(C([a, b],X),Y ), there exists one and only one α ∈ + 00 SVc ([a, b],L(X,Y )) such that F = Fα.

Remark 2.4. Under the hypotheses of Corollary 2.2, we write αF = α. Note that the 00 mapping F 7→ αF may not be onto if Y 6= Y . By Corollary 2.1, if α ∈ SV ([a, b],L(X,Z0)), then for every t ∈ [a, b[, there exists one and only one element α(t+) ∈ L(X,Z0) such that for every x ∈ X and every z ∈ Z, we have

limhα(t + ε)x, zi = hα(t+)x, zi. ε→0

If we deﬁne α+(t) = α(t+), a < t < b, and α+(a) = α(a), then α+ is a function of bounded semi-variation and we write α+ ∈ SV +(]a, b],L(X,Z0)). Moreover for every f ∈ C([a, b],X), R b + R b + we have a dα (t)f(t) = a dα(t)f(t) and kFαk = SV (α ). The next result follows from [2], Satz 10 and from Theorem 2.6.

Theorem 2.7. The mapping

+ 0 0 α ∈ SVc (]a, b],L(X,Z )) 7→ Fα ∈ L(C([a, b],X),Z ) is a linear isometry (i.e., kFαk = SV (α)) of the ﬁrst Banach space onto the second. Similarly as in Corollary 2.2, we have

Corollary 2.3. For every Fα ∈ L(C([a, b],X),Y ), there exists one and only one α ∈ + 00 SVc (]a, b],L(X,Y )) such that F = Fα. R R + ˙ + 00 Let SVb (]a, b],L(X, Y )) denote the space of functions α ∈ SVb (]a, b],L(X,Y )) such R t that α(a) ∈ L(X,Y ) and a α(s)xds ∈ Y , for all t ∈ [a, b] and all x ∈ X. Let χA be the characteristic function of a set A ⊂ [a, b]. The next theorem completes Corollary 2.3 and characterizes the image of the mapping F 7→ αF . It is borrowed from [11] (Theorem 1.4 there) and its proof can also be found in [6], Theorem 10.

Theorem 2.8. The mapping

+ R ˙ α ∈ SVb (]a, b],L(X, Y )) 7→ Fα ∈ L(C([a, b],X),Y ),

R b where Fα(f) = a dα(t)f(t), is an isometry (i.e., kFαk = SV (α)) of the ﬁrst Banach space R t onto the second. Furthermore a α(s)xds = −Fα(gt,x) and α(a)x = −Fα(χ[a,b]x), where for t ∈ [a, b] and x ∈ X, we deﬁne gt,x(s) = (s − a)x, if a ≤ s ≤ t, and gt,x(s) = (t − a)x, if t ≤ s ≤ b.

16 + R ˙ + R ˙ Let SVb ([a, b],L(X, Y )) = {α ∈ SVb (]a, b],L(X, Y )); α(a+) = α(a) instead of α(a) ∈ L(X,Y )}. The next two representation theorems are respectively Theorems 1.5 and 1.6 from [11]. Their proofs can also be found in [6], Theorems 11 and 12. Theorem 2.8 is used to prove Theorem 2.9 which, in turn, is applied in the proof of Theorem 2.10.

Theorem 2.9. The mapping

+ R ˙ α ∈ SVb ([a, b],L(X, Y )) 7→ Fα ∈ L(Ca([a, b],X),Y ) is an isometry of the ﬁrst Banach space onto the second.

The notation below is going to be used in the next theorem. 00 t Given a function α :[c, d] × [a, b] → L(X,Y ), we write α (s) = αs(t) = α(t, s) and we consider the following properties:

σ (C ) for a ≤ s ≤ b and x ∈ X, the function t ∈ [c, d] 7→ αs(t)x ∈ Y is continuous,

σ R s (Ce ) for a < s ≤ b and x ∈ X, the function t ∈ [c, d] 7→ a α(t, σ)xdσ ∈ Y is continuous, and for x ∈ X (and s = a), the function t ∈ [c, d] 7→ αa(t)x ∈ Y is continuous, (SV u) For all t ∈ [c, d], αt ∈ SV ([a, b],L(X,Y )) and SV u(αt) := sup SV (αt) < ∞. If c≤t≤d u u moreover α(t, b) = 0 for all t ∈ [c, d], then we write (SVb ) instead of (SV ). R R + u t + ˙ u t (SVb ) For all t ∈ [c, d], α ∈ SVb ([a, b],L(X, Y )) and SV (α) := sup SV (α ) < ∞. c≤t≤d We write α ∈ CσSV u([c, d] × [a, b],L(X,Y )) if α satisfies (Cσ) and (SV u). Analo- σ u σ u gously, α ∈ Ce SVb ([c, d] × [a, b],L(X,Y )) if α satisfies (Ce ) and (SVb ). We write α ∈ R R σ + u ˙ σ + u Ce SVb ([c, d] × [a, b],L(X, Y )) if α satisfies (Ce ) and (SVb ). Theorem 2.10. The mapping

R σ + u ˙ α ∈ Ce SVb ([c, d] × [a, b],L(X, Y )) 7→ Fα ∈ L(C([a, b],X),C([c, d],Y )),

R b u where (Fαf)(t) = dsα(t, s)f(s), c ≤ t ≤ d, is an isometry (i.e., kFαk = SV (α)) of the a R s ﬁrst Banach space onto the second. Besides, a α(t, σ)xdσ = −Fα(gs,x)(t), a ≤ s ≤ b and α(t, a)x = −Fα(χ[a,b]x)(t). We denote by K (X,Y ) the subspace of L (X,Y ) of compact linear operators. In particular, we write K (X) = K (X,X). We conclude this section of auxiliary results mentioning a Fredholm Alternative for Riemann-Siteltjes integrals. For a proof of it, see [13] and [4], Theorems 2.4 and 2.5.

17 σ u Theorem 2.11. Suppose K ∈ C (SVa) ([a, b] × [a, b] ,L (X)). Given t ∈ [a, b], let

∗ 0 ∗ 0 K (t, s0) x = lim K (t, s) x s↓s0

0 0 for every s0 ∈ ]a, b[ and every x ∈ X . Suppose the mapping

♦ ♦ t K : t ∈ [a, b] 7→ K (t) = K ∈ SVa ([a, b] ,L (X)) belongs to C ([a, b] ,SVa ([a, b] , K (X))). Given λ ∈ R, λ 6= 0, consider the integral equations Z b λx (t) − dsK (t, s) x (s) = f (t) , t ∈ [a, b] , (3) a

Z b λu (t) − dsK (t, s) u (s) = 0, t ∈ [a, b] , (4) a Z b λy (s) − K (t, s)∗ dy (t) = g (s) , s ∈ [a, b] , (5) a Z b λz (s) − K (t, s)∗ dz (t) = 0, s ∈ [a, b] . (6) a Then the Fredholm Alternative holds for these equations, that is,

(i) either for every f ∈ C ([a, b] ,X), equation (3) has one and only one solution and the same applies to equation (5),

(ii) or equation (4) has non-trivial solutions and the same applies to equation (6).

R b If (i) holds, then equation (3) admits a solution if and only if a f (t) dz (t) = 0 for every solution z of equation (6). Similarly, if (i) holds, then equation (5) admits a solution if and only R b if a u (t) dg (t) = 0 for every solution u of equation (4). Also the space of solutions of (4) has ﬁnite dimension equal to that of the space of solutions of (6) which equals the codimension ∗ + 0 of (λI − FK )C([a, b],X) in C([a, b],X) and the codimension of (λI − (FK ) )BVb ([a, b],X ) + 0 in BVb ([a, b],X ).

3 Gauge integrals in Banach spaces

3.1 Deﬁnitions and terminology In this section, we consider functions α :[a, b] → L (X,Y ) and f :[a, b] → X.

18 We say that α is Kurzweil f-integrable or Kurzweil integrable with respect to f, if there exists I ∈ Y such that for every ε > 0, there is a gauge δ of [a, b] such that for every δ-ﬁne d = (ξi, ti) ∈ TD[a,b],

|d| X α (ξi)[f (ti) − f (ti−1)] − I < ε.

i=1 R b In this case, we write I = (K) a α (t) df (t) and α ∈ Kf ([a, b] ,L (X,Y )). Analogously, we say that f is Kurzweil α−integrable or Kurzweil integrable with respect to α, if there exists I ∈ Y such that given ε > 0, there is a gauge δ of [a, b] such that

|d| X [α (ti) − α (ti−1)] f (ξi) − I < ε,

i=1

R b whenever d = (ξi, ti) ∈ TD[a,b] is δ-fine. In this case, we write I = (K) a dα (t) f (t) and f ∈ Kα ([a, b] ,X). If the gauge δ in the definition of α ∈ Kf ([a, b] ,L (X,Y )) is a constant function, then R b we get the Riemann-Stieltjes integral a α (t) df (t) and we write α ∈ Rf ([a, b] ,L (X,Y )). Similarly, when we consider only constant gauges δ in the definition of f ∈ Kα ([a, b] ,X), R b α we obtain the Riemann-Stieltjes integral a dα (t) f (t) and we write f ∈ R ([a, b] ,X). The vector integral of Henstock is more restrictive than that of Kurzweil. We define it in the sequel. We say that α is Henstock f-integrable or Kurzweil integrable with respect to f, if there exists a function Af :[a, b] → Y (called the associate function of α) such that for every ε > 0, there is a gauge δ of [a, b] such that for every δ-fine d = (ξi, ti) ∈ TD[a,b],

|d| X kα (ξi)[f (ti) − f (ti−1)] − [Af (ti) − Af (ti−1)]k < ε. i=1

We write α ∈ Hf ([a, b] ,L (X,Y )) in this case. In an analogous way we define the Henstock α-integrability of f :[a, b] → X or Henstock integrability of f with respect to α and we write f ∈ Hα ([a, b] ,X) (see [7]). α α Clearly Hf ([a, b] ,L (X,Y )) ⊂ Kf ([a, b] ,L (X,Y )) and H ([a, b] ,X) ⊂ K ([a, b] ,X). If we identify the isomorphic spaces L(R, R) and R, then all the spaces Kf ([a, b] ,L (R)), Kf ([a, b] , R), Hf ([a, b] ,L (R)) and Hf ([a, b] , R) can also be identified, since Kf ([a, b] , R) = Hf ([a, b] , R) (see, for instance, [17]). Given f :[a, b] → X and α ∈ Kf ([a, b] ,L (X,Y )), we define the indefinite integral αef :[a, b] → Y of α with respect to f by Z t αef (t) = (K) α (s) df (s) , t ∈ [a, b] . a

19 If in addition α ∈ Hf ([a, b] ,L (X,Y )), then αef (t) = Af (t) − Af (a) for every t ∈ [a, b]. In an analogous way, given α :[a, b] → L (X,Y ), we define the indefinite integral feα : [a, b] → Y of f with respect to α by Z t feα (t) = (K) dα (s) f (s) , t ∈ [a, b] , a for every f ∈ Kα ([a, b] ,X). In particular, when α (t) = t, then instead of Kα ([a, b] ,X), Rα ([a, b] ,X), Hα ([a, b] ,X) and feα we write, respectively, K ([a, b] ,X), R ([a, b] ,X), H ([a, b] ,X) R t and fe, that is, fe(t) = (K) a f (s) ds, for every t ∈ [a, b]. We proceed as to define the equivalence classes of Kurzweil and of Henstock integrable functions. Let m denote the Lebesgue measure. A function f :[a, b] → X satisfies the Strong Lusin Condition and we write f ∈ SL ([a, b] ,X) if given ε > 0 and B ⊂ [a, b] with m (B) = 0, there is a gauge δ of B such that for every δ-fine d = (ξi, ti) ∈ TPD[a,b] with ξi ∈ B for all i, we have |d| X kf (ti) − f (ti−1)k < ε. i=1 If we denote by AC ([a, b] ,X) the space of all absolutely continuous functions from [a, b] to X, then we have AC ([a, b] ,X) ⊂ SL ([a, b] ,X) ⊂ C ([a, b] ,X) .

In SL ([a, b] ,X), we consider the usual supremum norm, k · k∞, induced from C ([a, b] ,X). Given f ∈ SL ([a, b] ,X) and α ∈ Hf ([a, b] ,L (X,Y )), let β :[a, b] → L (X,Y ) be such that β = α m-almost everywhere. Then β ∈ Hf ([a, b] ,L (X,Y )) and βef (t) = αef (t), for every t ∈ [a, b]. See [7] (the corollary after Theorem 5 there) for a proof of this fact. An analogous result holds when we replace Hf ([a, b] ,L (X,Y )) by Kf ([a, b] ,L (X,Y )). Suppose f ∈ SL ([a, b] ,X). Two functions β, α ∈ Kf ([a, b] ,L (X,Y )) are called equivalent if and only if βef = αef . We denote by Kf ([a, b] ,L (X,Y )) and by Hf ([a, b] ,L (X,Y )) respectively the spaces of all equivalence classes of functions of Kf ([a, b] ,L (X,Y )) and of Hf ([a, b] ,L (X,Y )) and we equip these spaces with the Alexiewicz norm Z t

kαkA,f = sup (K) α (s) df (s) ; t ∈ [a, b] = kαef k∞ . a

3.2 Some properties As it should be expected, the vector integrals of Kurzweil and of Henstock are linear, additive over non-overlapping intervals and invariant over sets of Lebesgue measure zero. In this section, me mention several other properties of these integrals. We start with the Fundamental Theorem of Calculus which holds for the Henstock integral. Its proof follows standard steps (see [17], p. 43, for instance) adapted to Banach space-valued functions. See also [18].

20 Theorem 3.1 (Fundamental Theorem of Calculus). If F ∈ C ([a, b] ,X) and there exists the derivative F 0 (t) = f (t), for every t ∈ [a, b], then f ∈ H ([a, b] ,X) and Z t (K) f (s) ds = F (t) − F (a) , t ∈ [a, b] . a The next two versions of the Fundamental Theorem of Calculus for the vector Henstock integral can be found in [7], Theorems 1 and 2 respectively. Theorem 3.2. If f ∈ SL ([a, b] ,X) and A ∈ SL ([a, b] ,Y ) are both diﬀerentiable and α : [a, b] → L (X,Y ) is such that A0 (t) = α (t) f 0 (t) for m-almost every t ∈ [a, b], then α ∈ Hf ([a, b] ,L (X,Y )) and A = αef .

Theorem 3.3. If f ∈ SL ([a, b] ,X) is diﬀerentiable and α ∈ Hf ([a, b] ,L (X,Y )) is bounded, 0 0 then αef ∈ SL ([a, b] ,Y ) and there exists the derivative (αef ) (t) = α (t) f (t) for m-almost every t ∈ [a, b]. Corollary 3.1. Suppose f ∈ SL ([a, b] ,X) is diﬀerentiable and non-constant on any non- degenerate subinterval of [a, b] and α ∈ Hf ([a, b] ,L (X,Y )) is bounded and such that αef = 0. Then α = 0 m-almost everywhere. The next result is a particular case of [8], Theorem 2.2.

Theorem 3.4. If f ∈ C ([a, b] ,X) and α ∈ Kf ([a, b] ,L (X,Y )), then αef ∈ C ([a, b] ,Y ). For the vector Henstock integral we have the following analogue which can also be found in [7], Theorem 7.

Theorem 3.5. If f ∈ SL ([a, b] ,X) and α ∈ Hf ([a, b] ,L (X,Y )), then αef ∈ SL ([a, b] ,Y ). The next result follows from Theorem 3.5. A proof of it can be found in [6], Theorem 5. Theorem 3.6. Suppose f ∈ SL ([a, b] ,X) is non-constant on any non-degenerate subinterval of [a, b]. Then the mapping

α ∈ Hf ([a, b] ,L (X,Y )) 7→ αef ∈ Ca ([a, b] ,X) is an isometry (i.e., kαef k∞ = kαkA,f ) onto a dense subspace of Ca ([a, b] ,X). The next result gives a substitution formula for the vector Kurzweil integral. It also holds for the Riemann-Stieltjes integral instead. For a proof of it, see [5], Theorem 11.

Theorem 3.7. Let α ∈ SV ([a, b] ,L (X,Y )), f :[a, b] → Z, β ∈ Kf ([a, b] ,L (Z,X)) ˜ R t and g(t) = βf (t) = a β(s)df(s), t ∈ [a, b]. Then α ∈ Kg([a, b],L(X,Y ) if and only if αβ ∈ Kf ([a, b] ,L (Z,Y )). In this case, we have Z b Z b (K) α (t) β (t) df (t) = α (t) dg (t) (7) a a

21 and Z b

(K) α (t) β (t) df (t) ≤ [SV (α) + kα (a)k] kβkA,f . (8) a Using Theorems 3.4 and 2.3, we have the next corollary. See [5], Corollary 8.

Corollary 3.2. If α ∈ SV ([a, b] ,L (X,Y )), f ∈ C ([a, b] ,W ), β ∈ Kf ([a, b] ,L (W, X)) and ˜ R t g(t) = βf (t) = a β(s)df(s), t ∈ [a, b], then αβ ∈ Kf ([a, b] ,L (W, Y )) and (7) and (8) hold. By E ([a, b] ,L (X,Y )) we mean the space of all step functions from [a, b] to L (X,Y ), that is, α belongs to E ([a, b] ,L (X,Y )) if and only if α is bounded, there is a division P|d| d = (ti) ∈ D[a,b] and there are numbers α1, α2, . . . , α|d| such that α(t) = i=1 αiχ[ti−1,ti[(t), for every t ∈ [a, b[. For a proof of the next proposition, see [6], Theorem 8.

Proposition 3.1. Let f ∈ SL([a, b],X) be diﬀerentiable and non-constant on any non- degenerate subinterval of [a, b]. Then the spaces C ([a, b] ,L (X,Y )) and E ([a, b] ,L (X,Y )) are dense in Kf ([a, b] ,L (X,Y )) in the Alexiewicz norm k ·kA,f .

4 Auxiliary results

In this section we prove auxiliary results concerning vector gauge integrals which will be useful in the next section. We start by giving a representation theorem which says that 0 given a certain g ∈ SL([a, b], R), the elements of Kg([a, b],L(R,X)) can be represented by functions of bounded variation which are continuous to the right.

Theorem 4.1. Given g ∈ SL([a, b], R) diﬀerentiable and non-constant in any non-degenerate subinterval of [a, b], then the mapping

+ 0 0 α ∈ BVb ([a, b],X ) 7→ Hα,g ∈ Kg([a, b],L(R,X)) , R b where Hα,g(f) = (K) a α(s)f(s)dg(s), is an isometry (i.e., kHα,gk = V (α)) onto and, for R t every t ∈ [a, b] and every x ∈ L(R,X), we have a α(s)xdg(s) = Hα,g(χ[a,t]x). Proof. The mapping is clearly linear. We assert that the mapping is one-to-one. Indeed. Given ρ ∈ [a, b] and x ∈ L(R,X), we deﬁne a function fρ,x :[a, b] → L(R,X) by fρ,x(s) = x if s ∈ ]a, ρ], and by fρ,x(s) = 0 other- R ρ wise. Then Theorem 2.3 implies fρ,x ∈ Rg([a, b],L(R,X)). Also Hα,g(fρ,x) = a α(s)xdg(s). Since for each x ∈ L(R,X), the function α(·)x :[a, b] → L(R) is such that α(·)x ∈ Rg([a, b],L(R)) ⊂ Kg([a, b],L(R)) = Hg([a, b],L(R)), then by the Fundamental Theorem R ρ 0 of Calculus (Theorem 3.2), there exists dρ (K) a α(s)xdg(s) = α(ρ)xg (ρ) for m-almost every ρ ∈ [a, b].

22 If α 6= 0, then there exist ρ ∈ [a, b] and x ∈ L(R,X) such that α(ρ)x 6= 0. Besides, we can suppose without loss of generality that g0(ρ) 6= 0 by the invariance of the integral over R ρ sets of Lebesgue measure zero. Indeed. From [7], Theorem 9 we have (K) a α(s)xdg(s) = R ρ 0 (K) a α(s)xg (s), for every ρ ∈ [a, b]. In fact the integral above is in the Riemann-Stieltjes R ρ R ρ 0 sense. In particular, a α(s)xdg(s) = a α(s)xg (s) and hence Z ρ 0 dρ α(s)xdg(s) = α(ρ)xg (ρ) 6= 0. a R ρ Thus Hα,g(fρ,x) = a α(s)xdg(s) is non-constant and hence Hα,g(fρ,x) 6= 0 and the mapping is one-to-one. 0 Since SV ([a, b],L(X, R)) = BV ([a, b],X ), it follows from Corollary 3.2 that kHα,gk ≤ V (α). 0 Let f ∈ Kg([a, b],L(R,X)) and H ∈ Kg([a, b],L(R,X)) and deﬁne Hb(feg) = −H(f). By Theorem 3.6, there is a unique continuous extension of Hb to Ca([a, b],X) which we still 0 denote by Hb. This new operator, Hb, belongs to Ca([a, b],X) . If α represents Hb, then R b Theorem 2.9 implies Hb(feg) = a dα(s)feg(s) and kHbk = V (α). Moreover Z b Z b Z b H(f) = −Hb(feg) = − α(s)dfeg(s) = dα(s)feg(s) = (K) α(s)f(s)dg(s), a a a where we applied Theorem 2.3 and Corollary 3.2 to obtain the last two equalities. Since by deﬁnition kfkg = kfegk, then kHk = kHbk = V (α) and the result follows. Let g ∈ SL([a, b],Z). Given a function α :[c, d] × [a, b] → L(X,Y 00), we consider the following properties:

σ R s (Ceg ) for a < s ≤ b and x ∈ L(Z,X), the function t ∈ [c, d] 7→ a α(t, σ)xdg(σ) ∈ Y is continuous, and for x ∈ L(Z,X) (and s = a), the function t ∈ [c, d] 7→ α(t, a)xg0(a) ∈ Y is continuous, and, again, we consider the following property mentioned earlier: R R + u t + ˙ u t (SVb ) for all t ∈ [c, d], α ∈ SVb ([a, b],L(X, Y )) and SV (α) := sup SV (α ) < ∞. c≤t≤d R R σ + u ˙ σ + u Then we write α ∈ Ceg SVb ([c, d] × [a, b],L(X, Y )) if α satisﬁes (Ceg ) and (SVb ). If, in + R u addition, we consider functions of bounded variation in property (SVb ), then we write α ∈ R R σ + u ˙ σ + u Ceg BVb ([c, d] × [a, b],L(X, Y )). Notice that when Y = R, the spaces Ceg BVb ([c, d] × ˙ σ u 00 [a, b],L(X, Y )) and Ceg BVb ([c, d] × [a, b],L(X,Y )) can be identiﬁed and hence we write σ u 0 simply Ceg BVb ([c, d] × [a, b],X ) in this case. The proof of the next result follows the steps of the proof of Theorem 4.1, however instead of Theorem 2.9, one should apply Theorem 2.10. See [6], Theorem 13.

23 Theorem 4.2. Let g ∈ SL([a, b], R) be diﬀerentiable and non-constant in any non-degenerate subinterval of [a, b]. Then the mapping

σ +u 0 α ∈ Ceg BVb ([c, d] × [a, b],X ) 7→ Hα,g ∈ L(Kg([a, b],L(R,X)),C([c, d], R)),

R b where (Hα,g(β))(t) = (K) a α(t, s)β(s)dg(s) for each t ∈ [c, d], is an isometry (i.e., kHα,gk = V (α)) onto and, for every s ∈ [a, b], every t ∈ [a, b] and every x ∈ L(R,X), we have R s a α(t, σ)xdg(σ) = (Hα,g(χ[a,s]x))(t). Using Proposition 3.1, the next result follows easily.

Corollary 4.1. Under the hypotheses of Theorem 4.2, for every s ∈ [a, b], every t ∈ [a, b] R s and every β ∈ Kg([a, b],L(R,X)), we have a α(t, σ)β(σ)dg(σ) = (Hα,g(χ[a,s]β))(t).

Let g ∈ SL(a, b], R). We say that an operator H ∈ L(Kg([a, b],L(R)),C([a, b], R)) is ∼ causal if given f ∈ Kg([a, b],L(R)) = Kg([a, b], R) and t ∈ [a, b], then f|[a,t] = 0 implies H(f)|[a,t] = 0, where h|A denotes the restriction of a function h to a subset A of its domain. We proceed as to show that in fact the isometry in Theorem 4.2 is onto the space of causal operators, provided α(t, s) = 0 for m-almost every s > t. The next result can be found in [6], Theorem 14.

Theorem 4.3. Let α :[a, b] × [a, b] → L(R) be such that α(t, s) = 0, for m-almost every s > t, and let g ∈ SL([a, b], R) be diﬀerentiable and non-constant in any non-degenerate subinterval of [a, b]. Then the mapping

σ +u α ∈ Ceg BVb ([a, b] × [a, b],L(R)) 7→ Hα,g ∈ L(Kg([a, b],L(R)),C([a, b], R)),

R b where (Hα,g(f))(t) = (K) a α(t, s)f(s)dg(s) for each t ∈ [c, d], is an isometry (i.e., kHα,gk = V (α)) onto the subspace of causal operators.

For a proof of the next lemma, see [9] for instance.

Lemma 4.1 (Straddle Lemma). Suppose f, F :[a, b] → X are such that F 0 (ξ) = f (ξ), for all ξ ∈ [a, b]. Then given ε > 0, there exists δ (ξ) > 0 such that

kF (t) − F (s) − f (ξ)(t − s)k < ε (t − s) , whenever ξ − δ (ξ) < s < ξ < t < ξ + δ (ξ).

When g(s) = s, Theorem 4.4 below is a particular case of [4], Theorems 3.6 and 3.7. The proof is a straightforward adaptation of such results to cope with the introduction of the function g. The version we give next completes [6], Lemma 4 whose proof is included here.

24 σ +u 0 Theorem 4.4. Let α ∈ Ceg BVb ([c, d] × [a, b],X ), g ∈ SL([a, b], R) be diﬀerentiable and 0 non-constant in any non-degenerate subinterval of [a, b], and Kg :[c, d] × [a, b] → X be such R s σ u 0 that Kg(t, s)x = α(t, σ)xdg(σ), for each x ∈ L(R,X). Then Kg ∈ C SVb ([a, b]×[a, b],X ) a ∼ and for all t ∈ [c, d] and all y ∈ L(R) = R, the mapping

∗ 0 s ∈ [a, b] 7→ (Kg(t, s)) y ∈ X

∗ is continuous, where (Kg(t, s)) denotes the adjoint of Kg(t, s). Besides, we have

Z b Z b dsKg(t, s)β(s) = α(t, s)β(s)dg(s) a a for every function β ∈ C([a, b],L(R,X)) and all t ∈ [c, d]. Suppose, in addition, that αt ∈ SV ([a, b], K(X, R)), for every t ∈ [c, d]. Then the mapping

0 Kg : t ∈ [c, d] 7→ Kg(t, ·) ∈ SVa([a, b],X ) is continuous.

σ σ Proof. Since α fulﬁlls (Ceg ), then Kg(·, s) fulﬁlls (C ), for each s ∈ [a, b]. Let x ∈ L(R,X) and d = (si) ∈ D[a,b]. Then

|d| |d| Z si Z b X t t X K (si) − K (si−1) x = α(t, ρ)xdg(ρ) = α(t, ρ)xdg(ρ) ≤ g g i=1 i=1 si−1 a

≤ V (αt)kxk kg(b) − g(a)k t 0 and hence Kg = Kg(t, ·) ∈ SV ([a, b],X ), for each t ∈ [c, d]. t For every t ∈ [c, d] and every x ∈ L(R,X), the function Kg(·)x is continuous on [a, b], since it is an indeﬁnite integral (see Theorem 3.4). Therefore given t ∈ [c, d] and y ∈ L(R) ≡ R, t ∗ the function Kg(·)) y is continuous on [a, b]. 0 We assert that Kg ∈ C([c, d],SVa([a, b],X )). Indeed. Let t0 ∈ [c, d] and d = (si) ∈ D[a,b]. Then

|d| t t0 t t0 X t t0 t t0 K − K = SV K − K = sup K − K (si) − K − K (si−1) x = g g g g g g g g d, kxk≤1 i=1

|d| Z si Z si X = sup α(t, ρ)xdg(ρ) − α(t0, ρ)xdg(ρ) =

d, kxk≤1 i=1 si−1 si−1 Z b Z b

= sup α(t, ρ)xdg(ρ) − α(t0, ρ)xdg(ρ) = sup kKg(t, b)x − Kg(t0, b)xk d, kxk≤1 a a d, kxk≤1

25 σ 0 which tends to zero as t → t0, since Kg(·, s) ∈ C ([c, d],X ), for each s ∈ [a, b] and, in σ 0 particular, Kg(·, b) ∈ C ([c, d],X ). Let β ∈ C([a, b],L(R,X)), t ∈ [c, d] and γ = βeg. In accordance with Corollary 3.2, t α β ∈ Kg([a, b],L(R)) with Z b Z b Z b t t t (K) α (s)β(s)dg(s) = α (s)dγ(s) = − ds α (s) γ(s), (9) a a a

R s t where we applied Theorem 2.3 to obtain the last equality. Since Kg(t, s)x = a α (ρ)xdg(ρ) 0 with Kg(t, ·) ∈ SV ([a, b],X ) for each t ∈ [c, d], then Theorem 2.3 again implies the Riemann- R b Stieltjes integral a dsKg(t, s)β(s) exists for each t ∈ [c, d] and each β ∈ C([a, b],L(R,X)). We also assert that Z b Z b t dsKg(t, s)β(s) = (K) α (s)β(s)dg(s), (10) a a for every t ∈ [c, d] and every β ∈ C([a, b],L(R,X)). Indeed. It is enough to prove that (10) holds when β is a step function. Hence we need to show that for every t ∈ [c, d] and every x ∈ L(R,X), Z b Z b t dsKg(t, s)x = (K) α (s)xdg(s), (11) a a t Since α (·)x ∈ Rg([a, b],L(R)) by Theorem 2.3 and Rg([a, b],L(R)) ⊂ Kg([a, b],L(R)) = Hg([a, b],L(R)), it follows from the Fundamental Theorem of Calculus (Theorem 3.3) that R s t t 0 there exists ds a α (ρ)xdg(ρ) = α (s)xg (s) m-almost everywhere on [a, b]. Therefore given R s t t ∈ [c, d] and x ∈ L(R,X), we have dsKg(t, s)x = ds(Kg(t, s)x) = ds a α (ρ)xdg(ρ) = αt(s)xg0(s) for m-almost every s ∈ [a, b]. Then from the invariance of the integral over sets of m-measure zero, we obtain

Z b Z b t 0 dsKg(t, s)x = (K) α (s)xg (s)ds. (12) a a

R b t 0 R b t Now we will prove that a α (s)xg (s)ds = a α (s)xdg(s) which, together with (12), implies (11). Given ε > 0, t ∈ [c, d] and x ∈ L(R,X), let δ1 and δ2 be constant gauges of R b t R b t 0 [a, b] from the deﬁnitions of a α (s)xdg(s) and a α (s)xg (s)ds respectively. Given ξ ∈ [a, b], let δ3(ξ) > 0 be such that

|g(v) − g(s) − g0(ξ)(v − s)| < ε(v − s) (13) whenever ξ − δ3(ξ) < s < ξ < v < ξ + δ3(ξ) (see Lemma 4.1) and let δ(ξ) = min{δi(ξ); i = 1, 2, 3}. Then for every δ-ﬁne d = (ξi, si) ∈ TD[a,b], we have

Z b Z b t t 0 α (s)xdg(s) − α (s)xg (s)ds ≤ a a

Z b |d| t X t ≤ α (s)xdg(s) − α (ξi)x [g(si) − g(si−1)] +

a i=1

|d| |d| X t X t 0 + α (ξi)x [g(si) − g(si−1)] − α (ξi)xg (ξi)(si − si−1) +

i=1 i=1

|d| Z b X t 0 t 0 + α (ξi)xg (ξi)(si − si−1) − α (s)xg (s)ds <

i=1 a

|d| X t t < ε + V (α )kxk |g(si) − g(si−1) − g(ξi)(si − si−1)| + ε < 2ε + V (α )kxkε(b − a), i=1 where we applied by the integrability with respect to g of αt(·)x, the integrability of αt(·)xg0(·) and (13). The result follows easily. The next result can be found in [1] or in [12], Theorem 3.9.

Theorem 4.5. Given K ∈ CσSV u([a, b] × [a, b],L(X)), suppose there is a division d = (si) ∈ D[a,b] such that

t sup SV[si−1,t](K ); t ∈ [si−1, si] < 1, i = 1, 2,..., |d|.

Then K has resolvent given by the Neumann series.

Theorem 4.6 below is borrowed from [11], Theorem 3.1. See also [6], Lemma 7.

Theorem 4.6. Let E be a normed space and F be a Banach space such that F ⊂ R with continuous immersion. Let H ∈ L(E,F ) be such that for every f ∈ E, the equation x−Hx = f admits one and only one solution xf ∈ E. Then the mapping f ∈ R 7→ xf ∈ R is bicontinuous. If in addition the Neumann series I + H + H2 + H3 + ... = (I − H)−1 converges in L(F ), then it also converges in L(E).

5 The Fredholm Alternative for the Kurzweil-Henstock- Stieltjes integral

σ +u In this section, we write BV ([a, b], R) instead of BV ([a, b],L(R)) and Ceg BVb ([a, b] × σ +u [a, b], R) instead of Ceg BVb ([a, b] × [a, b],L(R)).

27 Theorem 5.1. Given g ∈ SL([a, b], R) diﬀerentiable and non-constant in any non-degenerate σ +u subinterval of [a, b] and H ∈ L(Kg([a, b], R),C([a, b], R)), let α ∈ Ceg BVb ([a, b] × [a, b], R) be the corresponding kernel by Theorem 4.2 (i.e., H = Hα,g). Suppose H is such that for each f ∈ Kg([a, b], R), the linear Fredholm-Henstock-Kurzweil-Stieltjes integral equation Z b x (t) − (K) α (t, s) x (s) dg(s) = f (t) , t ∈ [a, b] , (14) a

(i.e., the equation x − H(x) = f) admits a unique solution xf ∈ Kg([a, b], R). Then there σ +u exists a unique kernel ρ ∈ Ceg BVb ([a, b] × [a, b], R) and, for each f ∈ Kg([a, b], R),

Z b xf (t) = f(t) − (K) ρ (t, s) f (s) dg(s), t ∈ [a, b] . (15) a If moreover α(t, s) = 0, for m-almost every s > t, then H is causal as well as the bijection f 7→ xf given by (15) with ρ(t, s) = 0, for m-almost every s > t, and the Neumann series 2 3 I − Hρ,g = I + Hα,g + (Hα,g) + (Hα,g) + ... converges in L(Kg([a, b], R)). Proof. The ﬁrst part of the proof follows as in [11], Theorem 3.2 adapted for the Stieltjes −1 case. If in Theorem 4.6 we take E = Kg([a, b], R) and F = C([a, b], R), then (I − H) ∈ −1 −1 L(Kg([a, b], R)). If we deﬁne I − R = (I − H) , then R = H(I − H) belongs to σ +u L(Kg([a, b], R),C([a, b], R)) and it can be represented by a kernel ρ ∈ Ceg BVb ([a, b] × [a, b], R). Now we will prove the second part of the theorem which follows as in [6], Theorems 15 and 16. Suppose α(t, s) = 0 for m-almost every s > t. Let y = x − f and h(t) = R b (K) a α(t, s)f(s)dg(s). Both functions y and h are continuous, since h = Hα,g(χ[a,t]f) and y = Hα,g(χ[a,t]x) (see Corollary 4.1) with χ[a,t]f, χ[a,t]x ∈ Kg([a, b], R) and Hα,g ∈ L(Kg([a, b], R),C([a, b], R)) (see Theorem 4.3). Then equation (14) is equivalent to the following equation

Z b y (t) − (K) α (t, s) y (s) dg(s) = h (t) , t ∈ [a, b] . (16) a

R s t Let Kg :[a, b] × [a, b] → R be such that Kg(t, s)x = t α (σ)xdg(σ), for every x ∈ R. Then by Theorem 4.4, we have

Z b Z b t dsKg(t, s)y(s) = (K) α (s)y(s)dg(s). a a Thus equation (16) is equivalent to the following equation

Z b y (t) − dsKg(t, s)y(s) = h (t) , t ∈ [a, b] . (17) a

28 σ u Again by Theorem 4.4, Kg ∈ C BV ([a, b] × [a, b], R) and since α(t, s) = 0 for m-almost every s > t, then Kg(t, s) = 0 for s > t. Thus if (17) admits a unique solution, then the operator h 7→ yh is causal. Now we will prove that there exists d = (si) ∈ D[a,b] such that t sup BV[si−1,t]((Kg) ); t ∈ [si−1, si] < 1, i = 1, 2,..., |d|.

Let us consider the gauge δ of [a, b] defined as above and let d = (ξi, si) ∈ TD[a,b] be δ-fine. j Given i ∈ {1, 2,..., |d|} and t ∈ [si−1, si], let di = (r ) be a division of [si−1, t]. Then Z rj Z t X t j t j−1 X t t (Kg) (r ) − (Kg) (r ) xj = α (σ)xjdg(σ) = α (σ)xjdg(σ) ≤ j−1 j j r s−i |x | ≤ V u(α)|x | |g(t) − g(s )| < V u(α)|x |ε < j . j i−1 j 2 t Hence BV[si−1,t]((Kg) ) ≤ 1/2. Thus the assertion about the Neumann series for the resolvent of equation (17) in L(C([a, b], R)) follows from Theorem 4.5. Besides, the operator FKg given R t n −1 by FKg y (t) = a dsKg(t, s)y(s) is causal as well as FKg and hence I − FKg . By Theorem 4.6, the same applies to the resolvent of equation (17) in L(Kg([a, b], R)). Thus from −1 the fact that FKg = Hα,g (see Theorems 4.4 and 4.3), the operator I − Hρ,g = (I − Hα,g) is causal. Next we give a Fredholm Altenative type result for our Fredholm-Henstock-Kurzweil- Stieltjes integral equation. Theorem 5.2. Let g ∈ SL([a, b], R) be differentiable and non-constant in any non-degenerate σ +u subinterval of [a, b], f ∈ Kg([a, b], R) and α ∈ Ceg BVb ([a, b] × [a, b], R). Consider the linear Fredholm-Henstock-Kurzweil-Stieltjes integral equation Z b x (t) − (K) α (t, s) x (s) dg(s) = f (t) , t ∈ [a, b] , (18) a and its corresponding homogeneous equation Z b u (t) − (K) α (t, s) u (s) dg(s) = 0, t ∈ [a, b] . (19) a Consider also the following integral equations Z s Z b y (s) − α (t, σ) dy (t) dg(σ) = w (s) , s ∈ [a, b] , (20) a a and Z s Z b z (s) − α (t, σ) dz (t) dg(σ) = 0, s ∈ [a, b] . (21) a a Then

29 (i) either for each f ∈ Kg([a, b], R), equation (18) admits one and only one solution σ +u xf ∈ Kg([a, b], R) and there exists a unique kernel ρ ∈ Ceg BVb ([a, b] × [a, b], R) such that for each f ∈ Kg([a, b], R), Z xf (t) = f(t) − (K) ρ (t, s) f (s) dg(s), t ∈ [a, b] , [a,b]

and for each w ∈ C([a, b], R), equation (20) admits one and only one solution yg ∈ C([a, b], R);

(ii) or equation (19) admits non-trivial solutions in Kg([a, b], R). In this case, equation (18) admits a solution if and only if for each solution z ∈ C([a, b], R) of (21), Z b Z b (K) α(t, s)f(s)dg(s) dz(t) = 0 a a Z b Z b resp. (K) α(t, s)u(s)dg(s) dw(t) = 0 . a a Analogously, equation (20) admits a solution if and only if for each solution u ∈ Kg([a, b], R) of (19), we have Z b Z b (K) α(t, s)u(s)dg(s) dw(t) = 0. a a

R b R b Proof. Let h(t) = (K) a α(t, s)f(s)dg(s) and v = x−f. Then v(t) = (K) a α(t, s)x(s)dg(s) and, by Theorem 4.2, h, v ∈ C([a, b], R). Thus equation (18) is equivalent to equation Z b v (t) − (K) α (t, s) v (s) dg(s) = h (t) , t ∈ [a, b] . (22) a

R s t Let Kg :[a, b] × [a, b] → R be such that Kg(t, s)x = a α (σ)xdg(σ) for every x ∈ R. σ u According to Theorem 4.4, Kg satisﬁes (Ce ) and (BVa ) and for each t ∈ [a, b] and each y ∈ R, the mapping s ∈ ]a, b[ 7→ (Kg(t, s)) ∗ y ∈ R, where (Kg(t, s))∗ denotes the adjoint or transposed operator of Kg(t, s), is right continuous. Moreover the mapping

Kg : t ∈ [a, b] 7→ Kg(t, ·) ∈ BVa([a, b], R)) is continuous and we have Z b Z b t dsKg(t, s)v(s) = (K) α (s)v(s)dg(s). a a

30 Therefore (22) is equivalent to Z b v(t) − dsKg(t, s)v(s) = h(t), t ∈ [a, b], (23) a and by the Fredholm Alternative for the Riemann-Stieltjes integral (Theorem 2.11), either h ∈ C([a, b], R) and equation (23) admits one and only one solution which implies that for each f ∈ Kg([a, b], R), equation (18) admits a unique solution, or the homogeneous equation corresponding to (23) (respectively (18)) admits non-trivial solutions. In the second case, ∗ the adjoint (Kg) equals Kg. Then equation (18) (respectively equation (20)) admits a solution (not necessarily in Kg([a, b], R)) if and only if for each solution z ∈ C([a, b], R) of R b h R b i (21), a (K) a α(t, s)f(s)dg(s) dz(t) = 0 (respectively for each solution u ∈ Kg([a, b], R) R b h R b i of (19), we have a (K) a α(t, s)u(s)dg(s) dw(t) = 0). The missing part of assertion (i) follows easily from Theorem 5.1. By Theorems 5.1 and 5.2 we have the following corollary. Corollary 5.1. Suppose g ∈ SL([a, b], R) is diﬀerentiable and non-constant in any non- σ +u degenerate subinterval of [a, b], f ∈ Kg([a, b], R) and α ∈ Ceg BVb ([a, b] × [a, b], R) with α(t, s) = 0, for m-almost every s > t. Then (i) either the linear Fredholm-Henstock-Kurzweil-Stieltjes integral equation Z x (t) − (K) α (t, s) x (s) dg(s) = f (t) , t ∈ [a, b] , [a,b]

has a unique solution xf ∈ Kg([a, b], R); the operator Hα,g ∈ L(Kg([a, b], R),C([a, b], R)) R b given by (Hα,gf)(t) = (K) a α(t, s)f(s)dg(s), for each t ∈ [a, b], is causal as well as the bijection f 7→ xf which can be written as Z b xf (t) = f(t) − (K) ρ(t, s)f(s)dg(s), t ∈ [a, b], a σ +u where ρ ∈ Ceg BVb ([c, d] × [a, b], R) and ρ(t, s) = 0 for m-almost every s > t, and the 2 3 Neumann series I−Hρ,g = I+Hα,g+(Hα,g) +(Hα,g) +... converges in L(Kg([a, b], R));

(ii) or the corresponding homogeneous equation admits non-trivial solutions in Kg([a, b], R). Remark 5.1. Given g ∈ SL([a, b], R) differentiable and non-constant in any non-degenerate 2 subinterval of [a, b], let Rg([a, b], R)A be the space, endowed with the Alexiewicz norm, of the equivalence classes of functions f :[a, b] → R which are improper Riemann-Stieltjes integrable with respect to g and have a finite number of singularities. Also, let L1,g([a, b], R)A be the space, endowed with the Alexiewicz norm, of all equivalence classes of functions f : [a, b] → R which are Lebesgue-Stieltjes integrable with respect to g with finite integral. Then 2 Theorems 5.1 and 5.2 and Corollary 5.1 hold for Rg([a, b], R)A or L1,g([a, b], R)A instead of Kg([a, b], R).

31 References

[1] S. E. Arbex, Equa¸cõesIntegrais de Volterra-Stieltjes com Núcleos Descont´ınuos, Doctor Thesis, Instituto de Matemáticae Estat´ıstica, Universidade de São Paulo, 1976. (in Portuguese)

[2] J. Batt and H. K¨onig, Darstellung linearer transformationen durch vektorwertige Riemann-Stieltjes-integrale, Arch. Math., X, (1959), 273-287.

[3] M. Federson, Some peculiarities of the Henstock and Kurzweil integrals of Banach space- valued functions, Real Anal. Exchange, 29(1), (2003-2004), 439-460.

[4] M. Federson and R. Bianconi, Linear Fredholm integral equations and the integral of Kurzweil, J. of Applied Analysis, 8(1), (2002), 83-110.

[5] M. Federson, Substitution Formulas for the vector integrals of Kurzweil and of Henstock, Math. Bohem., 127(1), (2002), 15-26.

[6] M. Federson and R. Bianconi, Linear Volterra-Stieltjes integral equations in the sense of the Kurzweil-Henstock integral, Arch. Math., 37(4), (2001), 307-328.

[7] M. Federson, The Fundamental Theorem of Calculus for multidimensional Banach space-valued Henstock vector integrals, Real Analysis Exchange, 25(1), (1999-2000), 469-480.

[8] M. Federson and R. Bianconi, Linear integral equations of Volterra concerning the integral of Henstock, Real Anal. Exchange, 25(1), (1999-2000), 389-417.

[9] R. Henstock, A Riemann-type integral of Lebesgue power, Canad. J. of Math., 20 (1968), 79-87.

[10] R. Henstock, Deﬁnitions of Riemann type of the variational integrals, Proc. London Math. Soc., 11(3), (1961), 402-418.

[11] C. S. Hönig, On linear Kurzweil-Henstock integral equations, SeminárioBrasileiro de Análise, 32, (1990), 283-298.

[12] C. S. Hönig,Equations intégrales généraliséeset applications, Publ. Marh. dOrsay, 83- 01, exposé5, 1-50.

[13] C. S. H¨onig, Fredholm-Stieltjes integral equations I, in the Proceedings of the IV Escuela Latinoamericana de Matem´atica, (1978), 126-160.

[14] C. S. H¨onig, Volterra-Stieltjes integral equations, Math. Studies, 16, North-Holland Publ. Comp., Amsterdam, 1975.

32 [15] C. S. Hönig, The Abstract Riemann-Stieltjes Integral and its Applications to Linear Differential Equations with generalized boundary conditions, Notas do Instituto de Matemática e Estat´ıstica da Universidade de SãoPaulo, SérieMatemática,1, 1973.

[16] J. Kurzweil, Generalized ordinary diﬀerential equations and continuous dependence on a parameter, Czech. Math. J., 7 (82), (1957), 418-448.

[17] R. M. Mc Leod, The Generalized Riemann Integral, Carus Math. Monog., 20, The Math Ass. of America, 1980.

[18] S. Schwabik and Guoju Ye, Topics in Banach space integration. Series in Real Analysis, 10. World Scientiﬁc Publishing Co. Pte. Ltd., Hackensack, NJ, 2005.

[19] S. Schwabik, Abstract Perron-Stieltjes integral, Math. Bohem., 121 (4), (1996), 425-447.