# Implicit Vector Integral Equations Associated with Discontinuous Operators

Authors: Paolo Cubiotti; Jen-Chih Yao
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

LetI∶=[0,1]. We consider the vector integral equation h(u(t))=ft,∫Ig(t,z),u(z),dz for a.e. t∈I, where f:I×J→R,g:I×I→[0,+∞[, and h:X→R are given functions and X,J are suitable subsets of Rn. We prove an existence result for solutions u∈Ls(I,Rn), where the continuity of f with respect to the second variable is not assumed. More precisely, f is assumed to be a.e. equal (with respect to second variable) to a function f*:I×J→R which is almost everywhere continuous, where the involved null-measure sets should have a suitable geometry. It is easily seen that such a function f can be discontinuous at each point x∈J. Our result, based on a very recent selection theorem, extends a previous result, valid for scalar case n=1.


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## Body

## 1. Introduction

LetI:=[0,1]. Recently, in the paper [1], the following integral equation was studied: given λ>0, f:I×[0,λ]→R, g:I×I→[0,+∞[, h:]0,+∞[→R, and s>1; find u∈Ls(I) such that
(1)h(u(t))=f(t,∫I‍g(t,z)u(z)dz)fora.e.t∈I.
In [1], the following existence result was proved, where, unlike other results in the field (see, for instance, the papers [2–5] and references therein, to which we also refer for motivations for studying (1)), the continuity of f with respect to the second variable was not assumed.Theorem 1 (Theorem 1 of [1]).
Letλ>0, A⊆]0,+∞[ a closed interval, h:A→R a continuous function, f:I×[0,λ]→R, and g:I×I→[0,+∞[ two given functions. Let s∈]1,+∞], ϕ0∈Lj(I), with j>1 and j≥s′ (the conjugate exponent of s), ϕ1∈Ls′(I), and β∈Ls(I). Assume that (i)
there exists a functionf*:I×[0,λ]→R, two negligible sets E1,E2⊆[0,λ], with E2 closed, and a countable dense subset D of [0,λ], such that for all x∈D the function f*(·,x) is measurable, and for a.e. t∈I one has
(2){x∈[0,λ]:f(t,x)≠f*(t,x)}⊆E1,{x∈[0,λ]:f*(t,·)isdiscontinuousatx}⊆E2;(ii)
for allz∈int⁡(h(A)) (the interior of h(A)), one has int⁡h-1(z)=∅;(iii)
if one puts, for allt∈I,
(3)v(t):=essinfx∈[0,λ]f(t,x),z(t):=esssupx∈[0,λ]f(t,x),
then for a.e. t∈I one has
(4)[v(t),z(t)]⊆h(A),sup⁡h-1([v(t),z(t)])≤β(t);(iv)
one has(5)0<∥ϕ0∥Ls′(I)≤λ∥β∥Ls(I);(v)
for allt∈I, the function g(t,·) is measurable;(vi)
for a.e.z∈I, the function g(·,z) is continuous in I, differentiable in ]0,1[ and
(6)g(t,z)≤ϕ0(z),0<∂g∂t(t,z)≤ϕ1(z)∀t∈]0,1[. Then, there exists a solution u∈Ls(I) to (1).Of course, the main peculiarity of Theorem1 resides in the kind of discontinuity that is allowed for f. Indeed, it is easy to construct examples of functions f, g, and h satisfying the assumptions of Theorem 1 and such that for all t∈I the function f(t,·) is discontinuous at all points x∈[0,λ].Theorem1 extends a previous result (Theorem 1 of [6]), valid for the case where f does not depend on t explicitly. At this point, it is natural to ask if Theorem 1 above can be extended to the more general case where the function u takes its values in the space Rn. In this direction, we note that some results exist for the vector explicit equation
(7)u(t)=f(t,∫I‍g(t,z)u(z)dz)
(see [7, 8]), while for the implicit equation (1) the problem is still unsolved.The aim of this note is exactly to provide such an extension. In the following, ifn∈N and i∈{1,…,n}, we will denote by Pi:Rn→R the projection over the ith axis. Moreover, we will denote by mn the n-dimensional Lebesgue measure over Rn. If λ:=(λ1,…,λn)∈Rn, with λi>0 for all i=1,…,n, we will put Jλ:=∏i=1n‍[0,λi]. Finally, if n and λ are as above, we will denote by ℱn,λ the family of all subsets F⊆Jλ such that there exist sets F1,F2,…,Fn⊆Rn, with m1(Pi(Fi))=0 for all i=1…,n, such that F:=⋃i=1n‍Fi. The following is our main result (where Rn+ denotes the positive open orthant of Rn, and int⁡A(B) is the interior of B in A).Theorem 2.
Letλ=(λ1,…,λn)∈R+n, and let X⊆R+n be a nonempty, closed, connected, and locally connected subset of Rn, with inf⁡Pi(X)>0 for all i=1,…,n. Let h:X→R be a continuous function and f:I×Jλ→R and g:I×I→[0,+∞[ two given functions. Let s∈]1,+∞], ϕ0∈Lj(I), with j>1 and j≥s′, ϕ1∈Ls′(I), and β∈Ls(I,Rn). Finally, let D be a countable dense subset of Jλ.
Assume that there exists a functionf*:I×Jλ→R and two sets E,F∈ℱn,λ, with F closed, such that(i)
for allx∈D the function f*(·,x) is measurable;(ii)
for a.e.t∈I one has
(8){x∈Jλ:f(t,x)≠f*(t,x)}⊆E,{x∈Jλ:f*(t,·)isdiscontinuousatx}⊆F.Moreover, assume that(iii)
int⁡X(h-1(t))=∅, for all t∈int⁡R(h(X));(iv)
if one puts, for allt∈I;
(9)v(t):=essinfx∈Jλf(t,x),z(t):=esssup⁡x∈Jλf(t,x),
then for a.e. t∈I and all i=1,…,n one has
(10)[v(t),z(t)]⊆h(X),sup⁡Pi(h-1([v(t),z(t)]))≤βi(t),
(where βi:I→R denotes the ith component of the function β);(v)
one has(11)0<∥ϕ0∥Ls′(I)≤min⁡1≤i≤nλi∥βi∥Ls(I);(vi)
for allt∈I, the function g(t,·) is measurable;(vii)
for a.e.z∈I, the function g(·,z) is continuous in I, differentiable in ]0,1[ and
(12)g(t,z)≤ϕ0(z),0<∂g∂t(t,z)≤ϕ1(z)∀t∈]0,1[. Then, there exists u∈Ls(I,Rn) such that
(13)h(u(t))=f(t,∫I‍g(t,z)u(z)dz)fora.e.t∈I.Theorem2 will be proved as an application of the following selection theorem, recently proved in [9], which we now state for the reader’s convenience (in the following, if S is a topological space, we will denote by ℬ(S) the Borel family of S).Theorem 3 (Theorem 2.2 of [9]).
LetT and X1,X2,…Xk be complete separable metric spaces, with k∈N, and let X:=∏j=1k‍Xj (endowed with the product topology). Let μ,ψ1,…,ψk be positive regular Borel measures over T,X1,X2,…Xk, respectively, with μ finite and ψ1,…,ψkσ-finite.
LetS be a separable metric space, and let F:T×X→2S be a multifunction with nonempty complete values. Finally, let E⊆X be a given set, and, for each i∈{1,…,k}, let P*,i:X→Xi be the projection over Xi. Assume that(i)
the multifunctionF is 𝒯μ⊗ℬ(X1)⊗⋯⊗ℬ(Xk)-measurable (where 𝒯μ denotes the completion of the Borel σ-algebra ℬ(T) of T with respect to the measure μ);(ii)
for a.e.t∈T, one has
(14){x:=(x1,…,xk)∈X:F(t,·)iiiisnotlowersemicontinuousatx}⊆E.Then, there exist sets Q1,…,Qk, with Qi∈ℬ(Xi) and ψi(Qi)=0 for all i=1,…,k, and a function ϕ:T×X→S such that(a)
ϕ(t,x)∈F(t,x) for all (t,x)∈T×X;(b)
for allx:=(x1,x2,…,xk)∈X∖[(⋃i=1k‍P*,i-1(Qi))∪E], the function ϕ(·,x) is 𝒯μ-measurable over T;(c)
for a.e.t∈T, one has
(15){x:=(x1,x2,…,xk)∈X:ϕ(t,·)iiiiisdiscontinuousatx(x1,x2,…,xk)∈X}⊆E∪[⋃i=1k‍P*,i-1(Qi)].The proof of Theorem2 will be given in Section 2. Further, we will point out some counterexamples to possible improvements of Theorem 2.


## 2. Proof of Theorem2

Before giving the proof of Theorem2, we fix some notations. If n∈N, the space Rn will be considered with its Euclidean norm ∥·∥n. Moreover, if x∈Rn and r>0, we put
(16)B(x,r):={v∈Rn:∥v-x∥n<r},B¯(x,r):={v∈Rn:∥v-x∥n≤r}.
If p∈[1,+∞], the space Lp(I,Rn) will be considered with the usual norm
(17)∥u∥Lp(I,Rn):=(∫I‍∥u(t)∥npdt)1/pifp<+∞,∥u∥L∞(I,Rn):=esssup⁡t∈I∥u(t)∥nifp=+∞.
As usual, we put Lp(I):=Lp(I,R). For the basic definitions and facts about multifunctions, we refer to [10].Proof of Theorem2.
Without loss of generality, we can assume that (8) and (10) hold for all t∈I. Moreover, we can suppose that j<+∞. Firstly, we prove that the functions v and z are measurable. Observe that, by assumption (ii), for all t∈I, one has
(18)v(t)=inf⁡x∈Jλ∖Ff*(t,x),z(t)=sup⁡x∈Jλ∖Ff*(t,x).
To see this, fix t∈I, and let ψ(t):=sup⁡x∈Jλ∖Ff*(t,x). Since mn(E∪F)=0 we get
(19)z(t)≤sup⁡x∈Jλ∖(E∪F)f(t,x)=sup⁡x∈Jλ∖(E∪F)f*(t,x)≤ψ(t).
Now, assume that z(t)<ψ(t). Hence, there is x*∈Jλ∖F such that f*(t,x*)>z(t). Since the function f*(t,·) is continuous at x*, there exist δ,ε>0 such that
(20)f*(t,x)>z(t)+ε∀x∈Jλ∩B(x*,δ).
Since mn(Jλ∩B(x*,δ))>0, we get
(21)z(t):=esssup⁡x∈Jλf(t,x)=esssup⁡x∈Jλf*(t,x)≥z(t)+ε,
which is absurd. Therefore, the second equality in (18) is proved. The first one can be checked in analogous way.
SinceF is closed, it can be easily checked that the set D∩(Jλ∖F) is nonempty, countable, and dense in (Jλ∖F). Consequently, by Lemma at page 198 of [11], the function f*|I×(Jλ∖F) is ℒ(I)⊗ℬ(Jλ∖F)-measurable (where ℒ(I) denotes the family of all Lebesgue-measurable subsets of I). By (18) and Lemma III.39 of [12], it follows that the functions v and z are measurable over I, as claimed.
By assumption (iii) and Theorem 2.4 di [13], there exists a set Y⊆X such that h(Y)=h(X) and the function h|Y is open (it carries open subsets of Y onto open subsets of h(X)=h(Y)). Consequently, the multifunction T:h(X)→2Y defined by putting, for each s∈h(X),
(22)T(s):=h-1(s)∩Y,
is lower semicontinuous in h(X) with nonempty values. Let f0:I×Jλ→R be defined by putting, for all (t,x)∈I×Jλ,
(23)f0(t,x)={f*(t,x)ifx∉Fz(t)ifx∈F.
Clearly, the function f0 is ℒ(I)⊗ℬ(Jλ)-measurable and, by (18), one has
(24)v(t)≤f0(t,x)≤z(t)∀(t,x)∈I×Jλ.
Moreover, assumption (ii) and the closedness of F imply that for all t∈I one has
(25){x∈Jλ:f0(t,·)isdiscontinuousatx}⊆F.
Let G:I×Jλ→2Y be the multifunction defined by setting, for each (t,x)∈I×Jλ,
(26)G(t,x):=T(f0(t,x))=h-1(f0(t,x))∩Y.
Observe that G is well-defined since for all (t,x)∈I×Jλ one has
(27)f0(t,x)∈[v(t),z(t)]⊆h(X).
Moreover, by the lower semicontinuity of T and by (25), for all t∈I, we get
(28){x∈Jλ:G(t,·)isnotlowersemicontinuousatx}⊆F.
Let Ψ:I×Jλ→2Rn (more precisely, Ψ:I×Jλ→2Y¯) be the multifunction defined by putting, for each (t,x)∈I×Jλ, Ψ(t,x):=G(t,x)¯. By (28), for all t∈[0,1], we get
(29){x∈Jλ:Ψ(t,·)isnotlowersemicontinuousatx}⊆F.
Moreover, the values of Ψ are closed (in Rn) subsets of X.
Sincef0 is ℒ(I)⊗ℬ(Jλ)-measurable and T is lower semicontinuous, by Proposition 13.2.1 of [10] the multifunction G is ℒ(I)⊗ℬ(Jλ)-weakly measurable. That is, for each set Ω⊆Y, with Ω open in the relative topology of Y, the set
(30)G-(Ω):={(t,x)∈I×Jλ:G(t,x)∩Ω≠∅}
belongs to ℒ(I)⊗ℬ(Jλ). By Proposition 2.6 and Theorem 3.5 of [14], the multifunction Ψ is ℒ(I)⊗ℬ(Jλ)-measurable.
SinceF∈ℱn,λ, there exist sets F1,F2,…,Fn⊆Rn such that F=⋃i=1n‍Fi and m1(Pi(Fi))=0 for all i=1…,n. By Theorem 3, there exist sets Q1,Q2,…,Qn⊆R, with Qi∈ℬ([0,λi]) and m1(Qi)=0 for all i=1,…,n, and a function ψ:[0,1]×Jλ→Rn such that(a)
for all(t,x)∈I×Jλ one has ψ(t,x)∈Ψ(t,x);(b)
for allx∈Jλ∖[⋃i=1n‍(Pi-1(Qi)∪Fi)], the function ψ(·,x) is ℒ(I)-measurable;(c)
for a.e.t∈I, one has
(31){x∈Jλ:ψ(t,·)isdiscontinuousatx}⊆[⋃i=1n(Pi-1(Qi)∪Fi)]∩Jλ.
SinceX is closed and h is continuous, for all (t,x)∈I×Jλ, the set h-1(f0(t,x)) is closed in Rn. Consequently, for all (t,x)∈I×Jλ, we get
(32)ψ(t,x)∈Ψ(t,x)=h-1(f0(t,x))∩Y¯⊆h-1(f0(t,x))¯iiiiiiiiiiiiiiiiiiiiiiiiii=h-1(f0(t,x)).
Now, let
(33)α:=min⁡1≤i≤n⁡inf⁡Pi(X)>0.
By (27), (32), and assumption (iv) we get
(34)ψ(t,x)∈∏i=1n‍[α,βi(t)]∀(t,x)∈I×Jλ.
Let ψ1:I×Rn→Rn be defined by putting
(35)ψ1(t,x)={ψ(t,x)ifx∈Jλβ(t)ifx∈Rn∖Jλ.
By (34) and (35) we easily get
(36)ψ1(t,x)∈∏i=1n‍[α,βi(t)]∀(t,x)∈I×Rn.
Let
(37)Λ:=[⋃i=1n‍(Pi-1(Qi)∪Fi)]∩Jλ,
and let D0 be any countable dense subset of Jλ∖Λ. Since mn(Λ)=0, it is easily seen that D0 is dense in Jλ. Let D1 be any countable dense subset of Rn∖Jλ. Then, the set D2:=D0∪D1 is countable and dense in Rn, and for all x∈D2 the function ψ1(·,x) is measurable by the above construction. Moreover, taking into account (31), for all t∈I, one has
(38){x∈Rn:ψ1(t,·)isdiscontinuousatx}⊆[⋃i=1n‍(Pi-1(Qi)∪Pi-1({0,λi})∪Fi)]∩Jλ.
Let H:[0,1]×Rn→2Rn be defined by putting, for all (t,x)∈[0,1]×Rn,
(39)H(t,x)=⋂m∈N‍conv¯(⋃y∈D2∥x-y∥n≤1/m{ψ1(t,y)})¯,
where “conv¯” stands for “closed convex hull.” By Proposition 2 of [8], taking into account (36) and (38), we have that(a)
H has nonempty closed convex values;(b)
for allx∈Rn, the multifunction H(·,x) is measurable;(c)
for allt∈I, the multifunction H(t,·) has closed graph;(d)
for allt∈I, one has
(40)H(t,x)={ψ1(t,x)}iiiiiiiiiiiiiiiiiiiiiiiii∀x∈Rn∖([⋃i=1n(Pi-1(Qi)∪Pi-1({0,λi})iiiiiiiiiiiiiiiiiiiIIIIIIIIIIIIIIiiiiiiiiiiiii∪Fi(Qi))⋃i=1n‍]∩Jλ).Moreover, by (36), we have
(41)H(t,x)⊆∏i=1n‍[α,βi(t)]∀(t,x)∈I×Rn.
Now we want to apply Theorem 1 of [15], choosing T=[0,1], X=Y=Rn, p=s, q=j′, V=Ls(I,Rn), Ψ(u)=u, r=∥β∥Ls(I,Rn), φ≡+∞, F=H, and
(42)Φ(u)(t)=∫I‍g(t,z)u(z)dz.
To this aim, we can argue as in [8]. In particular, observe the following.(a)
Φ(Ls(I,Rn))⊆C0(I,Rn). This follows easily from our assumptions (vi) and (vii) and the classical Lebesgue’s dominated convergence theorem.(b)
Ifv∈Ls(I,Rn) and {vk} is a sequence in Ls(I,Rn), weakly convergent to v in Lj′(I,Rn), then the sequence {Φ(vk)} converges to Φ(v) strongly in L1(I,Rn). This follows by Theorem 2 at page 359 of [16], since g is jth power summable in I×I (note that g is measurable on I×I by the classical Scorza-Dragoni theorem; see [17] or also [11]).(c)
By (41), the function
(43)h0:t∈I⟶sup⁡x∈Rninf⁡y∈H(t,x)∥y∥nbelongs to Ls(I) and ∥h0∥Ls(I)≤∥β∥Ls(I,Rn).
Therefore, taking into account the above construction, all the assumptions of Theorem 1 of [15] are satisfied. Consequently, there exists a function u*∈Ls(I,Rn) and a set K1⊆I, with m1(K)=0, such that
(44)u*(t)∈H(t,Φ(u*)(t))⊆∏i=1n‍[α,βi(t)]∀t∈I∖K1.
That is,
(45)u*(t)∈H(t,∫I‍g(t,z)u*(z)dz)⊆∏i=1n‍[α,βi(t)]iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii∀t∈I∖K1.
We now prove that the function u* satisfies our conclusion. To this aim, observe that, since E∈ℱn,λ, there exist sets E1,E2,…,En⊆Rn, with m1(Pi(Ei))=0 for all i=1,…,n, such that E:=⋃i=1n‍Ei.
Fixi∈{1,…,n}. Let γi:I→R be the function
(46)γi(t):=Pi(Φ(u*)(t))=∫I‍g(t,z)ui*(z)dz.
By (45) we get
(47)ui*(t)∈[α,βi(t)]∀t∈I∖K1;
hence for all t∈I we get the inequality
(48)0≤γi(t)≤∥ϕ0∥Ls′(I)∥u*∥Ls(I)ii≤λi∥βi∥Ls(I)·∥βi∥s=λi,
hence γi(I)⊆[0,λi]. By (vi), (vii), and (45) we have that γi is strictly increasing. Moreover, by Lemma 2.2. at page 226 of [18] we get
(49)γi′(t)=∫I‍∂g∂t(t,z)ui*(z)dz>0∀t∈]0,1[.
Consequently, by Theorem 2 of [19], the function γi-1 is absolutely continuous. By Theorem 18.25 of [20], the set
(50)Wi:=γi-1[(Pi(Ei∪Fi)∪Qi∪{0,λi})∩γi(I)]
has null Lebesgue measure. Now, put
(51)Ω:=(⋃i=1n‍Wi)∪K1.
Of course, m1(Ω)=0. Choose any point t*∈I∖Ω. Since t*∉K1, by (45) we get
(52)u*(t*)∈H(t*,Φ(u*)(t*)).
For each i∈{1,…,n}, since t*∉Wi, taking into account (48), we have
(53)γi(t*)∈[0,λi]∖[Pi(Ei∪Fi)∪{0,λi}∪Qi].
Therefore, Φ(u*)(t*)∈Jλ and for all i∈{1,…,n} we have
(54)Φ(u*)(t*)∉[(Ei∪Fi)∪Pi-1({0,λi})∪Pi-1(Qi)].
Consequently, we get
(55)Φ(u*)(t*)∈Jλ∖⋃i=1n‍[(Ei∪Fi)∪Pi-1({0,λi})∪Pi-1(Qi)].
By (40) we get
(56)u*(t*)∈H(t*,Φ(u*)(t*))={ψ1(t*,Φ(u*)(t*))}iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii={ψ(t*,Φ(u*)(t*))}.
By (32) and (56) we get
(57)u*(t*)∈h-1(f0(t*,Φ(u*)(t*)),
hence
(58)h(u*(t*))=f0(t*,Φ(u*)(t*)).
Since Φ(u*)(t*)∉E∪F, we get
(59)h(u*(t*))=f*(t*,Φ(u*)(t*))iiiiiiiiiiiiiii=f(t*,Φ(u*)(t*))iiiiiiiiiiiiiii=f(t*,∫i‍g(t*,z)u*(z)dz).
This completes the proof.Remark 4.
Of course, a functionf satisfying the assumptions of Theorem 2 can be discontinuous at each point x∈Jλ. The Example at the end of [8] shows that, in the statement of Theorem 2, none of the sets E and F can be assumed to depend on t. Moreover, the Example at the end of [6] shows that the second inequality in assumption (vii) cannot be weakened by assuming that
(60)0≤∂g∂t(t,z)≤ϕ1(z).



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