Extremal solutions for second-order fully discontinuous problems with nonlinear functional boundary conditions

Electronic Journal of Qualitative Theory of Differential Equations 2018, No. 29, 1–14; https://doi.org/10.14232/ejqtde.2018.1.29 www.math.u-szeged.hu/ejqtde/ Extremal solutions for second-order fully discontinuous problems with nonlinear functional boundary conditions Rubén Figueroa1,Rodrigo López Pouso1, 2 and Jorge Rodríguez–LópezB1, 2 1Departamento de Estatística, Análise Matemática e Optimización, Universidade de Santiago de Compostela, 15782, Facultade de Matemáticas, Campus Vida, Santiago, Spain 2Instituto de Matemáticas, Universidade de Santiago de Compostela Received 6 February 2018, appeared 21 May 2018 Communicated by Petru Jebelean Abstract. We provide new results concerning the existence of extremal solutions for a class of second-order problems with nonlinear functional boundary conditions where the nonlinearity considered may be discontinuous with respect to all of its variables. The main result relies on recent fixed point theorems for discontinuous operators and the lower and upper solution method. Keywords: discontinuous differential equations, upper and lower solutions, fixed point theorems. 2010 Mathematics Subject Classification: 34A36, 34B15, 47H10. 1 Introduction and preliminaries We study the existence of extremal solutions for the differential equation x00(t) = f(t,x,x0),t∈I= [a,b], (1.1) where the nonlinear term fmay be discontinuous in all the arguments. More specifically, we shall prove existence of extremal solutions to (1.1) coupled with nonlinear functional boundary conditions 0=L1(x(a),x(b),x0(a),x0(b),x), 0=L2(x(a),x(b)),(1.2) where L1∈ C R4× C(I),Ris nonincreasing in the third and in the fifth variables, and nondecreasing in the fourth one; and L2:R2→Ris a continuous function and it is nondecreasing with respect to its first argument. BCorresponding author. Email: jorger[email protected] 2R. Figueroa, R. L. Pouso and J. Rodríguez-López In particular, the nonlinear boundary conditions (1.2) contain Dirichlet boundary conditions x(a) = x(b) = 0, (1.3) and periodic conditions x(a) = x(b),x0(a) = x0(b). (1.4) Since fmay be discontinuous in all the arguments, we are forced to use new fixed point theorems (see [6,9]) combined with the lower and upper solutions method [3,5]. Similar fixed point methods were employed in [7] in the analysis of first-order differential problems with initial functional conditions. Let us start with some preliminary results and definitions. Let Kbe a nonempty closed convex subset of a normed space (X,k·k)and T:K−→ Kan operator, not necessarily continuous. Definition 1.1. The closed–convex envelope (or Krasovskij envelope [8]) of an operator T: K−→ Kis the multivalued mapping T:K−→ 2Kgiven by Tx=\ ε>0 co TBε(x)∩Kfor every x∈K, (1.5) where Bε(x)denotes the closed ball centered at xand radius ε, and co means closed convex hull. Remark 1.2. Note that Tis an upper semicontinuous multivalued mapping which assumes closed and convex values (see [2,9]) provided that T K is a relatively compact subset of X. Theorem 1.3 ([9, Theorem 3.1]).Let K be a nonempty, convex and compact subset of X. Any mapping T :K−→ K has at least one fixed point provided that for every x ∈K we have {x} ∩ Tx⊂ {Tx}, (1.6) where Tdenotes the closed–convex envelope of T. Remark 1.4. Condition (1.6) is equivalent to Fix(T)⊂Fix(T), where Fix(S)denotes the set of fixed points of the operator S. Theorem 1.5 ([6, Theorem 2.7]).Let K be a nonempty, closed and convex subset of X and T :K−→ K be a mapping such that T K is a relatively compact subset of X and it satisfies condition (1.6). Then T has a fixed point in K. 2 Existence of solution for discontinuous BVP with nonlinear boundary conditions We shall work in the Banach space X=C1(I)endowed with its usual norm kxkC1=kxk∞+ x0 ∞=max t∈I|x(t)|+max t∈Ix0(t). Following [5] and the review article [3] we shall use lower and upper solutions for obtaining an existence result for problem (1.1)–(1.2). In the proof of the main result we shall consider a modified problem in the line of [4]. Extremal solutions for second-order discontinuous problems 3 Definition 2.1. We say that α∈ C(I)is a lower solution for the differential problem (1.1)–(1.2) if it satisfies the following conditions. (i)For any t0∈(a,b), either D−α(t0)<D+α(t0), or there exists an open interval I0such that t0∈I0,α∈W2,1(I0)and α00(t)≥f(t,α(t),α0(t)) for a.a. t∈I0. (ii)D+α(a),D−α(b)∈Rand L1(α(a),α(b),D+α(a),D−α(b),α)≤0. (iii)L2(α(a),α(b))=0, and L2(α(a),·)is injective. Similarly β∈ C(I)is an upper solution for (1.1)–(1.2) if it satisfies the inequalities in the reverse order. Now we present a Nagumo condition which provides a priori bound on the first derivative of all possible solutions between the lower and upper solutions for the differential problem. Proposition 2.2. Let ¯ α,¯ β∈ C(I)be such that ¯ α≤¯ βand define r=max ¯ β(b)−¯ α(a),¯ β(a)−¯ α(b)/(b−a). Assume there exist a continuous function ¯ N:[0, ∞)→(0, ∞),¯ M∈L1(I)and R >r such that ZR r 1 ¯ N(s)ds >k¯ MkL1. Define E :=(t,x,y)∈I×R2:¯ α(t)≤x≤¯ β(t). Then, for every function f :E→Rsuch that for a.e. t ∈I and all (x,y)∈R2with (t,x,y)∈E, |f(t,x,y)|≤¯ M(t)¯ N(|y|), and for every solution x of (1.1)such that ¯ α≤x≤¯ β, we have  x0 ∞<R. Proof. Let xbe a solution of (1.1) and t∈Isuch that x0(t)>R. Notice that −r≤¯ α(b)−¯ β(a) b−a≤x(b)−x(a) b−a≤¯ β(b)−¯ α(a) b−a≤r, and then by Lagrange Theorem there exists τ∈Isuch that x0(τ)= x(b)−x(a) b−a ≤r. Thus we can choose t0<t1(or t1<t0) such that x0(t0) = r,x0(t1) = Rand r≤x0(s)≤Rin [t0,t1](or [t1,t0]). Therefore we have ZR r 1 ¯ N(s)ds =Zt1 t0 x00(s) ¯ N(x0(s))ds =Zt1 t0 f(s,x(s),x0(s)) ¯ N(x0(s)) ds ≤Zt1 t0 ¯ M(s)ds ≤k¯ MkL1, a contradiction, so we deduce that x0(t)<R. In the same way we prove that x0(t)>−R. We consider the differential problem (1.1)–(1.2), under weaker conditions about fthan the well-known Carathéodory’s conditions, and we look for solutions for this problem, namely functions x∈W2,1(I)satisfying (1.1)–(1.2). We shall allow fto be discontinuous in the second argument over countably many curves in the conditions of the following definition. They imply a ‘transversality’ condition whose geometrical idea recalls that of the discontinuity surfaces described in [8]. 4R. Figueroa, R. L. Pouso and J. Rodríguez-López Definition 2.3. An admissible discontinuity curve for the differential equation (1.1) is a W2,1 function γ:[c,d]⊂I−→ Rsatisfying one of the following conditions: either γ00(t) = f(t,γ(t),γ0(t)) for a.a. t∈[c,d](and we then say that γis viable for the differential equation), or there exist ε>0 and ψ∈L1(c,d),ψ(t)>0 for a.a. t∈[c,d], such that either γ00(t) + ψ(t)<f(t,y,z)for a.a. t∈[c,d], all y∈[γ(t)−ε,γ(t) + ε](2.1) and all z∈[γ0(t)−ε,γ0(t) + ε], or γ00(t)−ψ(t)>f(t,y,z)for a.a. t∈[c,d], all y∈[γ(t)−ε,γ(t) + ε](2.2) and all z∈[γ0(t)−ε,γ0(t) + ε]. We say that the admissible discontinuity curve γis inviable for the differential equation if it satisfies (2.1) or (2.2). Moreover, we shall allow fto be discontinuous in the third argument over some curves satisfying the conditions of the following definition, slightly different from the previous one. As far as the authors are aware, this is the first time that such discontinuity sets are considered. Definition 2.4. Given αand βlower and upper solutions for problem (1.1)–(1.2) such that α≤βon I, an inviable discontinuity curve for the derivative of the differential equation (1.1) is an absolutely continuous function Γ:[c,d]⊂I−→ Rsatisfying that there exist ε>0 and ψ∈L1(c,d),ψ(t)>0 for a.a. t∈[c,d], such that either Γ0(t) + ψ(t)<f(t,y,z)for a.a. t∈[c,d], all y∈[α(t),β(t)] (2.3) and all z∈[Γ(t)−ε,Γ(t) + ε]∪ {α0(t),β0(t)}, or Γ0(t)−ψ(t)>f(t,y,z)for a.a. t∈[c,d], all y∈[α(t),β(t)] (2.4) and all z∈[Γ(t)−ε,Γ(t) + ε]∪ {α0(t),β0(t)}. Now we state three technical results that we need in the proof of our main existence result of this section for (1.1)–(1.2). Their proofs can be lookep up in [9]. In the sequel mdenotes the Lebesgue measure in R. Lemma 2.5. Let a,b∈R,a<b, and let g,h∈L1(a,b),g≥0a.e., and h >0a.e. on (a,b). For every measurable set J ⊂(a,b)such that m(J)>0there is a measurable set J0⊂J such that m(J\J0)=0and for all τ0∈J0we have lim t→τ+ 0R[τ0,t]\Jg(s)ds Rt τ0h(s)ds =0=lim t→τ− 0R[t,τ0]\Jg(s)ds Rτ0 th(s)ds . Extremal solutions for second-order discontinuous problems 5 Corollary 2.6. Let a,b∈R,a<b, and let h ∈L1(a,b)be such that h >0a.e. on (a,b). For every measurable set J ⊂(a,b)such that m(J)>0there is a measurable set J0⊂J such that m(J\J0)=0and for all τ0∈J0we have lim t→τ+ 0R[τ0,t]∩Jh(s)ds Rt τ0h(s)ds =1=lim t→τ− 0R[t,τ0]∩Jh(s)ds Rτ0 th(s)ds . Corollary 2.7. Let a,b∈R,a<b, and let f ,fn:[a,b]→Rbe absolutely continuous functions on [a,b](n ∈N), such that fn→f uniformly on [a,b]and for a measurable set A ⊂[a,b]with m(A)>0we have lim n→∞f0 n(t) = g(t)for a.a. t ∈A. If there exists M ∈L1(a,b)such that |f0(t)|≤M(t)a.e. in [a,b]and also |f0 n(t)|≤M(t)a.e. in [a,b](n ∈N), then f 0(t) = g(t)for a.a. t ∈A. Now we present the main result in this paper. Theorem 2.8. Suppose that there exist α,β∈W1,∞((a,b)) lower and upper solutions to (1.1)–(1.2), respectively, such that α≤βon I. Let r=max {β(b)−α(a),β(a)−α(b)}/(b−a). Assume that for f :I×R2→Rthe following conditions hold: (C1)compositions t ∈I7→ f(t,x(t),y(t)) are measurable whenever x(t)and y(t)are measurable; (C2)there exist a continuous function N :[0, ∞)→(0, ∞)and M ∈L1(I)such that: (a) for a.a. t ∈I, all x ∈[α(t),β(t)] and all y ∈R, we have |f(t,x,y)|≤M(t)N(|y|); (b) there exists R >r such that ZR r 1 N(s)ds >kMkL1; (C3)there exist admissible discontinuity curves γn:In= [an,bn]−→ R(n∈N)such that α≤γn≤βon Inand their derivatives are uniformly bounded, and for all y ∈Rand for a.a. t∈I the function x 7→ f(t,x,y)is continuous on [α(t),β(t)] \S{n:t∈In}{γn(t)}; (C4)there exist inviable discontinuity curves for the derivative Γn:˜ In= [cn,dn]−→ R(n∈N) such that they are uniformly bounded and for a.a. t ∈I and all x ∈[α(t),β(t)], the mapping y7→ f(t,x,y)is continuous on [−R,R]\S{n:t∈˜ In}{Γn(t)}. Then problem (1.1)–(1.2)has at least a solution x ∈W2,1(I)between αand βsuch that kx0k∞<R. Proof. Without loss of generality, suppose that R>maxt∈I{|α0(t)|,|β0(t)|,|γ0 n(t)|,|Γn(t)|} for all n∈Nand define an integrable function ˜ M(t):=max s∈[0,R]{N(s)}M(t). Let us also define δR(z) = max {min {z,R},−R}for all z∈Rand f∗(t,x,y) = f(t,x,δR(y)) for all (t,x,y)∈I×R2. (2.5) 6R. Figueroa, R. L. Pouso and J. Rodríguez-López Consider the modified problem        x00(t) = f∗(t,ϕ(t,x(t)),(ϕ(t,x(t)))0)for a.a. t∈I, x(a) = L∗ 1(x(a),x(b),x0(a),x0(b),x), x(b) = L∗ 2(x(a),x(b)), (2.6) where ϕ(t,x) = max {min {x,β(t)},α(t)}for (t,x)∈I×R, (2.7) and L∗ 1(x,y,z,w,ξ)=ϕ(a,x−L1(x,y,z,w,ξ)) for all (x,y,z,w,ξ)∈R4× C(I)and L∗ 2(x,y)= ϕ(b,y+L2(x,y))for all (x,y)∈R2. We know from [10, Lemma 2] that if v,vn∈ C1(I)are such that vn→vin C1(I), then (a) (ϕ(t,v(t)))0exists for a.a. t∈I; (b) (ϕ(t,vn(t)))0→(ϕ(t,v(t)))0for a.a. t∈I. Now we consider the compact and convex subset of X=C1(I), K=x∈X:α(a)≤x(a)≤β(a),α(b)≤x(b)≤β(b), x0(t)−x0(s)≤Zt s ˜ M(r)dr (a≤s≤t≤b) and for each x∈Kdefine Tx(t) = L∗ 1(x) + t−a b−aL∗ 2(x)−L∗ 1(x)−Zb aZs af∗(r,ϕ(r,x(r)),(ϕ(r,x(r)))0)dr ds +Zt aZs af∗(r,ϕ(r,x(r)),(ϕ(r,x(r)))0)dr ds, where, for simplicity, we use the notations L∗ 1(x) = L∗ 1(x(a),x(b),x0(a),x0(b),x)and L∗ 2(x) = L∗ 2(x(a),x(b)). Observe that y=Tx is just the solution of (y00(t) = f∗(t,ϕ(t,x(t)),(ϕ(t,x(t)))0)for a.a. t∈I, y(a) = L∗ 1(x),y(b) = L∗ 2(x).(2.8) Conditions (C1)and (C2)guarantee that the operator Tis well defined. Moreover, Tmaps Kinto itself. Indeed, for any x∈Kand y=Tx we have, thanks to (C2) (a), that |y00(t)|=|f∗(t,ϕ(t,x(t)),(ϕ(t,x(t)))0)| ≤ M(t)N(|δR((ϕ(t,x(t)))0)|)≤˜ M(t), which, along with y(a) = L∗ 1(x)and y(b) = L∗ 2(x), imply that y∈K. Next we prove that the operator Tsatisfies condition (1.6) for all x∈Kand then Theorem 1.3 ensures the existence of a fixed point or, equivalently, a solution to the modified problem (2.6). This part of the proof follows the steps of that in [9, Theorem 4.4], but here some changes are necessary due to the use of lower and upper solutions and the derivative dependence in the ODE. We fix an arbitrary function x∈Kand we consider four different cases. Case 1: m({t∈In:x(t) = γn(t)} ∪ {t∈˜ In:x0(t) = Γn(t)}) = 0for all n ∈N. Let us prove that then Tis continuous at x. Extremal solutions for second-order discontinuous problems 7 The assumption implies that for a.a. t∈Ithe mapping f(t,·,·)is continuous at the point (ϕ(t,x(t)),(ϕ(t,x(t))0). Hence if xk→xin K, then f∗(t,ϕ(t,xk(t)),(ϕ(t,xk(t)))0)→f∗(t,ϕ(t,x(t)),(ϕ(t,x(t)))0)for a.a. t∈I, as one can easily check by considering all possible combinations of the cases x(t)∈[α(t),β(t)], x(t)>β(t)or x(t)<α(t), and |x0(t)| ≤ Ror |x0(t)|>R. Moreover, f∗(t,ϕ(t,x(t)),(ϕ(t,x(t)))0)≤˜ M(t)(2.9) for a.a. t∈I, hence Txk→Tx in C1(I). Case 2: m({t∈In:x(t) = γn(t)})>0for some n ∈Nsuch that γnis inviable. In this case we can prove that x6∈ Tx. First, we fix some notation. Let us assume that for some n∈Nwe have m({t∈In:x(t) = γn(t)})>0 and there exist ε>0 and ψ∈L1(In),ψ(t)>0 for a.a. t∈In, such that (2.2) holds with γreplaced by γn. (The proof is similar if we assume (2.1) instead of (2.2), so we omit it.) We denote J={t∈In:x(t) = γn(t)}, and we observe that m({t∈J:γn(t) = β(t)}) = 0. Indeed, if m({t∈J:γn(t) = β(t)})>0, then from (2.2) it follows that β00(t)−ψ(t)> f(t,β(t),β0(t)) on a set of positive measure, which is a contradiction with the definition of upper solution. Now we distinguish between two sub-cases. Case 2.1: m({t∈J:x(t) = γn(t) = α(t)})>0.Since m({t∈J:γn(t) = β(t)}) = 0, we deduce that m({t∈J:x(t) = α(t)6=β(t)})>0, so there exists n0∈Nsuch that mt∈J:x(t) = α(t),x(t)<β(t)−1 n0>0. We denote A={t∈J:x(t) = α(t),x(t)<β(t)−1/n0}and we deduce from Lemma 2.5 that there is a measurable set J0⊂Awith m(J0) = m(A)>0 such that for all τ0∈J0we have lim t→τ+ 0 2R[τ0,t]\A˜ M(s)ds (1/4)Rt τ0ψ(s)ds =0=lim t→τ− 0 2R[t,τ0]\A˜ M(s)ds (1/4)Rτ0 tψ(s)ds. (2.10) By Corollary 2.6 there exists J1⊂J0with m(J0\J1) = 0 such that for all τ0∈J1we have lim t→τ+ 0R[τ0,t]∩J0ψ(s)ds Rt τ0ψ(s)ds =1=lim t→τ− 0R[t,τ0]∩J0ψ(s)ds Rτ0 tψ(s)ds . (2.11) Let us now fix a point τ0∈J1. From (2.10) and (2.11) we deduce that there exist t−<τ0 and t+>τ0,t±sufficiently close to τ0so that the following inequalities are satisfied: 2Z[τ0,t+]\A ˜ M(s)ds <1 4Zt+ τ0 ψ(s)ds, (2.12) Z[τ0,t+]∩Aψ(s)ds ≥Z[τ0,t+]∩J0 ψ(s)ds >1 2Zt+ τ0 ψ(s)ds, (2.13) 2Z[t−,τ0]\A ˜ M(s)ds <1 4Zτ0 t− ψ(s)ds, (2.14) Z[t−,τ0]∩Aψ(s)ds >1 2Zτ0 t− ψ(s)ds. (2.15) 8R. Figueroa, R. L. Pouso and J. Rodríguez-López Finally, we define a positive number ρ=min 1 4Zτ0 t− ψ(s)ds,1 4Zt+ τ0 ψ(s)ds, (2.16) and we are now in a position to prove that x6∈ Tx. It is sufficient to prove the following claim: Claim – Let ˜ ε>0be defined as ˜ ε=min{ε, 1/n0}, where εis given by our assumptions over γnand 1/n0by the definition of the set A, and let ρbe as in (2.16). For every finite family xi∈B˜ ε(x)∩K and λi∈[0, 1](i =1, 2, . . . , m), with ∑λi=1, we have kx−∑λiTxikC1≥ρ. Let xiand λibe as in the Claim and, for simplicity, denote y=∑λiTxi. For a.a. t∈J= {t∈In:x(t) = γn(t)}we have y00(t) = m ∑ i=1 λi(Txi)00(t) = m ∑ i=1 λif∗(t,ϕ(t,xi(t)),(ϕ(t,xi(t)))0). (2.17) On the other hand, for every i∈ {1, 2, . . . , m}and for a.a. t∈Jwe have |xi(t)−γn(t)|+x0 i(t)−γ0 n(t)=|xi(t)−x(t)|+x0 i(t)−x0(t)<ε, (2.18) but by continuity x0(t) = γ0 n(t)for all t∈J, so (2.18) holds for every t∈J. Since γn(t)∈ [α(t),β(t)], for a.a. t∈Awe have |ϕ(t,xi(t)) −γn(t)|≤|xi(t)−γn(t)|, and (ϕ(t,xi(t)))0−γ0 n(t)≤x0 i(t)−γ0 n(t) because if xi(t)<α(t), then (ϕ(t,xi(t)))0=α0(t) = γ0 n(t). Hence, from (2.2) it follows that γ00 n(t)−ψ(t)>f(t,ϕ(t,xi(t)),(ϕ(t,xi(t)))0) for a.a. t∈Aand for all xi(t)satisfying (2.18). Moreover, since for a.a. t∈Awe have |γ0 n(t)|<Rand |x0 i(t)−γ0 n(t)|<ε, without loss of generality we can suppose |(ϕ(t,xi(t)))0|≤Rand thus γ00 n(t)−ψ(t)>f∗(t,ϕ(t,xi(t)),(ϕ(t,xi(t)))0)(2.19) for a.a. t∈A. Therefore the assumptions on γnensure that for a.a. t∈Awe have y00(t) = m ∑ i=1 λif∗(t,ϕ(t,xi(t)),(ϕ(t,xi(t)))0)< m ∑ i=1 λi(γ00 n(t)−ψ(t)) = x00(t)−ψ(t). (2.20) Now we compute y0(τ0)−y0(t−) = Zτ0 t− y00(s)ds =Z[t−,τ0]∩Ay00(s)ds +Z[t−,τ0]\Ay00(s)ds <Z[t−,τ0]∩Ax00(s)ds −Z[t−,τ0]∩Aψ(s)ds +Z[t−,τ0]\A ˜ M(s)ds (by (2.20), (2.17) and (2.9)) =x0(τ0)−x0(t−)−Z[t−,τ0]\Ax00(s)ds −Z[t−,τ0]∩Aψ(s)ds +Z[t−,τ0]\A ˜ M(s)ds ≤x0(τ0)−x0(t−)−Z[t−,τ0]∩Aψ(s)ds +2Z[t−,τ0]\A ˜ M(s)ds <x0(τ0)−x0(t−)−1 4Zτ0 t− ψ(s)ds (by (2.14) and (2.15)), Extremal solutions for second-order discontinuous problems 9 hence kx−ykC1≥y0(t−)−x0(t−)≥ρprovided that y0(τ0)≥x0(τ0). Similar computations with t+instead of t−show that if y0(τ0)≤x0(τ0)then we also have kx−ykC1≥ρ. The claim is proven. Case 2.2: m({t∈J:γn(t)∈(α(t),β(t))})>0. The set {t∈J:γn(t)∈(α(t),β(t))}can be written as the following countable union [ n∈Nt∈J:α(t) + 1 n<x(t)<β(t)−1 n, so there exists some n0∈Nsuch that m({t∈J:α(t) + 1/n0<x(t)<β(t)−1/n0})>0. Now we denote A={t∈J:α(t) + 1/n0<x(t)<β(t)−1/n0}. Since Ais a set of positive measure we can argue as in Case 2.1 for obtaining inequalities (2.12)–(2.15) and we are in position to prove the Claim again. Let xiand λibe as in the Claim and, for simplicity, denote y=∑λiTxi. Then for every i∈ {1, 2, . . . , m}and all t∈Awe have xi(t)∈(α(t),β(t)), so ϕ(t,xi(t)) = xi(t)and (ϕ(t,xi(t)))0=x0 i(t)and thus |ϕ(t,xi(t)) −γn(t)|+(ϕ(t,xi(t)))0−γ0 n(t)=|xi(t)−x(t)|+x0 i(t)−x0(t)<ε, (2.21) for all t∈A. Hence, from (2.2) it follows that γ00 n(t)−ψ(t)>f(t,ϕ(t,xi(t)),(ϕ(t,xi(t)))0) for a.a. t∈Aand all xi∈B˜ ε(x). Now the proof of the Claim follows exactly as in Case 2.1. Case 3: m({t∈˜ In:x0(t) = Γn(t)})>0for some n ∈Nsuch that Γnis an inviable discontinuity curve for the derivative. In this case, we can prove again that x6∈ Tx. As before, let us assume that for some n∈Nwe have m({t∈˜ In:x0(t) = Γn(t)})>0 and there exist ε>0 and ψ∈L1(˜ In),ψ(t)>0 for a.a. t∈˜ In, such that (2.4) holds with Γreplaced by Γn. Similarly, we can define ρas in (2.16) and we shall prove the Claim. Let xiand λibe as in the Claim and, for simplicity, denote y=∑λiTxi. For a.a. t∈J= {t∈˜ In:x0(t) = Γn(t)}we have (2.17). On the other hand, for every i∈ {1, 2, . . . , m}and for every t∈Jwe have |x0 i(t)−Γn(t)|=|x0 i(t)−x0(t)|<ε. Moreover, from (2.4) it follows that Γ0 n(t)−ψ(t)>f∗(t,ϕ(t,xi(t)),(ϕ(t,xi(t)))0) for a.a. t∈Inand for all xi(t)since ϕ(t,xi(t))∈[α(t),β(t)] and (ϕ(t,xi(t)))0∈{x0 i(t),α0(t),β0(t)}. Therefore the assumptions on Γnensure that for a.a. t∈Jwe have y00(t) = m ∑ i=1 λif∗(t,ϕ(t,xi(t)),(ϕ(t,xi(t)))0)< m ∑ i=1 λi(Γ0 n(t)−ψ(t)) = x00(t)−ψ(t), and the proof of Case 3 follows as in Case 2.1, but now the set Jplays the role of the set A there.