The antipodal mapping theorem and difference equations in Banach spaces†

October 15, 2004 13:26 Journal of Difference Equations and Applications FOP

Journal of Difference Equations and Applications

Vol. , No. , , 1–13

The antipodal mapping theorem and difference equations in

Banach spaces †

Daniel Franco ‡, Donal O’Regan § and Juan Peran *‡

‡ Departamento de Matematica Aplicada, Universidad Nacional de Educacion aDistancia, Apartado 60149, Madrid 28080, Spain.

§ Department of Mathematics. National University of Ireland. Galway. Ireland.()

We employ the Borsuk-Krasnoselskii antipodal theorem to prove a new fixed point theorem inordered Banach spaces. Then, the applicability of the result is shown by presenting sufficientconditions for the existence of solutions to initial value problems for first-order difference equationsin Banach spaces. To prove that result we shall employ set valued analysis techniques.

Keywords: difference equations; antipodal theorem; set valued analysisAMS Subject Classification (2000): 39A05; 47H10

1 Introduction

Let E be a Banach space and f a compact map from B = {x ∈ E : ‖x‖ ≤ 1}to E .

Theorem 1.1 Borsuk-Krasnoselskii antipodal theorem. Suppose that f hasno fixed points on ∂B and that the antipodal condition

f(x)− x

‖f(x)− x‖6= f(−x) + x

‖f(−x) + x‖, x ∈ ∂B

is satisfied.Then f has a fixed point in B.

The above result has been taken from section 16.3 in [22] and it was first

† This research has been supported in part by Ministerio de Ciencia y Tecnologıa (Spain), projectMTM2004-06652-C03-03* Corresponding author. E-mail: [email protected]


2 Franco, O’Regan and Peran

proved by Borsuk in 1933 in the finite dimensional case [7] and this general-ization to the infinite dimensional setting is due to Krasnoselskii [16].

There is a vast literature on applications of the Borsuk-Krasnoselskii antipo-dal theorem and its variants and generalizations to different fields of mathe-matics (we recommend the interested reader to the nice survey paper [21]). Inparticular, it has been employed extensively to prove existence and multipleresults for nonlinear integral and differential equations (see [19, 20] and thereferences therein).

In the first part of this paper, using Borsuk-Krasnoselskii’s result we shallpresent a new fixed point theorem in Banach spaces with an order structuregiven by an normal order cone. More precisely, we shall introduce new abstractdefinitions of lower and upper solutions for the equation u = f(u) in theBanach space E . Then, we shall show the existence of maximal and minimalfixed points in a sector defined by the lower and the upper solution withoutassuming monotone conditions on the nonlinearity f .

The second part of the paper deals with the application of the previousresult to prove new sufficient conditions for the existence of solutions of initialand boundary value problems for first and higher order difference equationsin Banach spaces. Evidently, if the dimension of the Banach space is finite wejust have a difference equation or a system of difference equations (see [1, 15]for a general theory).

Existence of solutions for difference equations, difference equations in Ba-nach spaces or systems of difference equations have been considered by severalauthors using different techniques [2, 8, 9, 11, 13, 17]. We note that in [9] theauthors also introduce a cone in order to prove existence of single and multiplefixed sign solutions. In [9] the cone is needed so that a fixed point theoremof cone expansion and compression of norm type can be used. The approachin [9] is completely different from our approach. In [8,10,13,14] the lower andupper solution technique was employed.

In Section 3 we prove an existence result for first order initial value problemsin Banach spaces, assuming that there exist suitably defined upper an lowersolutions. Then, we study higher order initial and boundary problems by re-ducing these problems to initial value ones. Finally, we note that our techniquewas inspired by methods employed to solve the first order scalar case in [10].

2 Existence principle

Let E be an ordered Banach space whose order cone K ⊂ E is normal and hasnon-empty interior. That is,K is a closed, convex, proper ({−x, x} ∈ K ⇒ x =0}), normal (there is a number r0 > 0 such that 0 ≤ u ≤ v ⇒ ‖u‖ ≤ r0‖v‖)cone and int(K) 6= ∅.


Antipodal theorem and difference equations 3

As usual, we will write u� v for v − u ∈ int(K) and

[u, v] = (u+K) ∩ (v −K) = {w ∈ E : u ≤ w ≤ v}

(u, v) = int([u, v]) = (u+ int(K)) ∩ (v − int(K)) = {w ∈ E : u� w � v}.

Notice that [u, v] is a bounded set, because for each w ∈ [u, v] one has

‖w‖ ≤ ‖w − u‖+ ‖u‖ ≤ r0‖v − u‖+ ‖u‖.

We consider on Em, with m ∈ N, the norm ‖x‖ =√‖x1‖2 + . . .+ ‖xm‖2

and the partial order relation correspondent to the cone Km (which is closed,convex, proper, normal and with non-empty interior), that is: x ≤ y if andonly if xk ≤ yk for all k ∈ {1, . . . ,m}.

A map is said to be compact if it is continuous and maps bounded sets intorelatively compact sets.

Remark 1 A compact subset S ⊂ E always has maximal and minimal elements.To see that, suppose J is a linearly ordered subset of S and consider the filterbase on S formed by the sets Fv = {u ∈ S : u ≥ v} with v ∈ J . Since S is acompact set, this filter base has a cluster point u0 ∈ S, that is u0 ∈

⋂v∈J Fv.

Therefore, u0 is an upper bound of J . By Zorn’s Lemma, S has a maximalelement. A slight change in the proof actually shows the minimal elementexistence.

Theorem 2.1 If a � b in E and f : [a, b] → E is a compact map such thata� f(a+ v); b� f(b− v) for all v ∈ ∂K with v ≤ b− a, then f has a fixedpoint in (a, b) but none in ∂[a, b].

Furthermore, the set of fixed points {u ∈ [a, b] : f(u) = u} is a non-emptycompact subset of (a, b), having maximal and minimal elements.

Proof First suppose that f has a fixed point u ∈ [a, b] such that

u ∈ ∂[a, b] = ((a+ ∂K) ∩ (b−K)) ∪ ((a+K) ∩ (b− ∂K)) .

Then one has u−a ∈ ∂K with u−a ≤ b−a or b−u ∈ ∂K with b−u ≤ b−a.If u − a ∈ ∂K with u − a ≤ b − a, then a � f(a + (u − a)) = f(u) = u, sou− a ∈ int(K), which is impossible. Similarly, if we assume b− u ∈ ∂K withb− u ≤ b− a, we have b− u ∈ int(K), once more a contradiction.

Now consider the Minkowski functional defined on E by

p(w) = inf{r > 0 :w

r∈ [a−b

2 , b−a2 ]}.



The functional p is a norm on E , since [a−b2 , b−a

2 ] is a symmetric, closed,bounded, convex set with 0 ∈]a−b

2 , b−a2 [. Furthermore, p and ‖·‖ are equivalent

norms. To see this notice

1r0‖ b−a

2 ‖‖w‖ ≤ p(w) ≤ 1

r1‖w‖,

where r0 comes from the normal cone condition and r1 is the radius of a ballwith centre 0 and contained in

(a−b2 , b−a

2

).

Since(

a−b2 , b−a

2

)= {w ∈ E : p(w) < 1}, the result follows from the antipodal

theorem if the antipodal condition

f ′(u)− u

‖f ′(u)− u‖6= f ′(−u) + u

‖f ′(−u) + u‖

holds for all u ∈ ∂[a−b2 , b−a

2 ]; here f ′(u) = f(u+ b+a

2

)− b+a

2 .Recall that

∂[a−b2 , b−a

2 ] =((a−b

2 + ∂K) ∩ ( b−a2 −K)

)∪((a−b

2 +K) ∩ ( b−a2 − ∂K)

).

Suppose the antipodal condition does not hold. Then there is a numberr > 0 and v ∈ ∂K with v ≤ b− a such that

f ′(

a−b2 + v

)−(

a−b2 + v

)= r

(f ′(

b−a2 − v

)−(

b−a2 − v

)),

that is

f (a+ v)− (a+ v) = r (f (b− v)− (b− v)) .

This gives

−v = a− (a+ v) � f (a+ v)− (a+ v) = r (f (b− v)− (b− v)) �� r (b− (b− v)) = rv,

and then (r + 1)v ∈ int(K), which is impossible because v ∈ ∂K.Therefore S = {u ∈ [a, b] : f(u) = u} is a non-empty compact set. To see

the compactness notice that f continuous implies S is closed, and also sinceS is bounded we have that S = f(S) is a relatively compact set. Finally, theminimal and maximal existence follows from Remark 1. �

In view of the last Theorem we make the following definition.



Definition 2.2 We say that a� b in E are a pair of lower and upper solutionfor equation u = f(u) iff a � f(a + v); b � f(b − v) for all v ∈ ∂K withv ≤ b− a.

3 Applications to difference equations

Before presenting our results in this section we need to recall some definitionsand results of the theory of set valued analysis [4].

LetX be a non-empty set and let P(X) denote the set of all subsets ofX. Wedefine a multifunction F from X to an arbitrary set Y , written F : X ⇒ Y , tobe a map F : P(X) → P(Y ) such that F (A) =

⋃a∈A F ({a}). Of course, given

F ({x}) for each x ∈ X, the multifunction F is determined. We denote F ({x})by F (x). The map F− : P(Y ) → P(X) defined by F−(B) = {x ∈ X : F (x) ∩B 6= ∅} is a multifunction, called the inverse multifunction of F . It is easilyproved that the composition of two multifunctions F : X ⇒ Y,G : Y ⇒ Z is amultifunction G ◦ F : X ⇒ Z such that

G ◦ F (x) =⋃

y∈F (x)

G(y).

As usual, we identify each map f : X → Y with the multifunction A ⊂ X →f(A). Obviously, the map usually denoted as f−1 : P(Y ) → P(X) coincideswith the multifunction f−.

LetX and Y be Hausdorff spaces. A multifunction F : X ⇒ Y is called uppersemicontinuous (u.s.c.) if F−(C) is a closed subset of X for each closed subsetC of Y . An u.s.c. multifunction with non-empty compact values is called anusco map.

Lemma 3.1 Let D ⊂ E be a closed bounded set, f : D → E a compact mapand h : D → E defined by h(u) = u − f(u). Then h is a closed map, that is,h−1 : E ⇒ D is an u.s.c. multifunction.

Proof Let C ⊂ D be a closed set, (yn) → y0 a convergent sequence with yn ∈h(C) and xn ∈ C such that yn = h(xn). Since cl(f(C)) is a compact set, thereexists a convergent subsequence (f(xni

)) of (f(xn)) with limit w0 ∈ cl(f(C)).Now,

limxni= lim f(xni

) + limh(xni) = w0 + y0

so w0 + y0 ∈ C and

f(w0 + y0) = f(limxni) = lim f(xni

) = w0.



Finally,

h(w0 + y0) = w0 + y0 − f(w0 + y0) = w0 + y0 − w0 = y0,

so y0 ∈ h(C) and h(C) =(h−1

)− (C) is a closed subset of E . �

3.1 First order difference equations.

Consider the following difference equation:

∆yk = fk(yk+1), k ∈ {1, . . . ,m− 1} (1)

where y = (y1, . . . , ym) ∈ Em, ∆yk = yk+1 − yk and fk : E → E is a compactmap for each k ∈ {1, . . . ,m− 1}.

Define hk : E → E by hk(u) = u−fk(u) for k ∈ {1, . . . ,m−1}. Then rewrite(1) as

yk = hk(yk+1), k ∈ {1, . . . ,m− 1}. (2)

We define a continuous map H = (H1, . . . ,Hm) : E → Em as follows:

Hm(u) = u (3)

Hk(u) = hk(Hk+1(u)), k ∈ {1, . . . ,m− 1}. (4)

It is easy to see that y = H(u) is a solution of (1) for each u ∈ E , andconversely, for each solution y of (1) there is a u ∈ E with y = H(u).

Definition 3.2 We say that α ≤ β in Em are a pair of lower and uppersolutions for equation (1) if for each k = 1, . . . ,m − 1 and v ∈ ∂K such thatv ≤ βk+1 − αk+1 one has ∆αk ≤ fk(αk+1 + v) and ∆βk ≥ fk(βk+1 − v).

Remark 1 If α ≤ β are a pair of lower and upper solutions, we observe byconsidering v = 0 above that α is a subsolution and β is a supersolution for(1), in accordance with standard definitions (see for example [10]). But anarbitrary subsolution with an arbitrary supersolution, do not necessarily forma pair of lower and upper solutions in the sense of Definition 3.2. Finally, noticethat y ∈ Em is a solution of (1) if and only if it forms with itself a pair of lowerand upper solutions.

Remark 2 Observe that α ≤ β are a pair of lower and upper solutions if andonly if, for all k = 1, . . . ,m− 1 and v, v′ ∈ ∂K such that v, v′ ≤ βk+1 − αk+1,

hk(αk+1 + v)− v ≤ αk ≤ βk ≤ hk(βk+1 − v′) + v′ (5)



Lemma 3.3 Let α � β be a pair of lower and upper solutions for (1) andwk ∈ (αk, βk) for each k ∈ {1, . . . ,m− 1}. Then, the set

Sk(wk) = h−1k (wk) ∩ [αk+1, βk+1]

is a non-empty compact set contained in (αk+1, βk+1).Furthermore, the multifunction Sk : (αk, βk) ⇒ (αk+1, βk+1) is an usco map.

Proof For each v ∈ ∂K with v ≤ βk+1 − αk+1 the compact map f from[αk+1, βk+1] to E defined by

f(u) = fk(u) + wk

satisfies ∆αk ≤ f(αk+1 + v)− wk and ∆βk ≥ f(βk+1 − v)− wk. This gives

αk+1 � αk+1 + (wk − αk) ≤ f(αk+1 + v)

and

βk+1 � βk+1 + (wk − βk) ≥ f(βk+1 − v).

By Theorem 2.1, the set {u ∈ [αk+1, βk+1] : f(u) = u} is a non-empty compactsubset of (αk+1, βk+1). But this is the desired conclusion, since from hk(u) =u− fk(u) = u− f(u) + wk we have

Sk(wk) = {u ∈ [αk+1, βk+1] : f(u) = u}.

It remains to be proved that Sk is an u.s.c. multifunction. The set D =[αk+1, βk+1] is closed and bounded, the map fk : D → E is compact and, byLemma 3.1, hk : D → E is a closed map. Suppose B ⊂ int(D) = (αk+1, βk+1)is closed for the subspace topology induced on int(D), that is, there is a closed(in E) B′ ⊂ D such that B = B′ ∩ int(D). Then S−k (B) = S−k (B′) = hk(B′) ∩(αk, βk) is a closed subset of (αk, βk) for the induced topology on this set. �

Theorem 3.4 Let α � β be pair of lower and upper solutions for (1). Foreach w ∈ (α1, β1), denote by S(w) the subset of [α, β] formed by the solutionsy ∈ Em of (1) satisfying the initial value condition y1 = w.

Then S(w) is a non-empty compact subset of (α, β) having maximal andminimal elements. Furthermore, the multifunction S : (α1, β1) ⇒ (α, β) is anusco map satisfying S = H ◦ Sm−1 ◦ Sm−2 ◦ · · · ◦ S1.

Proof The proof falls in four parts:



a) We claim Sk ◦Sk−1 ◦ · · · ◦S1(w) is a non-empty subset of (αk+1, βk+1) fork ∈ {1, . . . ,m − 1}. For k = 1 it follows directly from Lemma 3.3. Assumingthe assertion holds for k−1, apply Lemma 3.3 for each wk ∈ Sk−1 ◦· · ·◦S1(w).

b) We claim S(w) ⊂ H ◦Sm−1 ◦Sm−2 ◦ · · · ◦S1(w). If y ∈ S(w), then y1 = wand yk = hk(yk+1) for k ∈ {1, . . . ,m− 1}. This gives y = H(ym) and

ym ∈ Sm−1(ym−1) ⊂ Sm−1 ◦ Sm−2(ym−2) ⊂ . . . ⊂ Sm−1 ◦ Sm−2 ◦ · · · ◦ S1(w).

Thus y ∈ H ◦Sm−1 ◦Sm−2 ◦ · · · ◦S1(w) and so S(w) ⊂ H ◦Sm−1 ◦Sm−2 ◦ · · · ◦S1(w).

c) We claim H ◦Sm−1 ◦Sm−2 ◦ · · · ◦S1(w) ⊂ S(w)∩ (α, β). If y ∈ H ◦Sm−1 ◦Sm−2 ◦ · · · ◦ S1(w), then y = H(ym), with ym ∈ Sm−1 ◦ Sm−2 ◦ · · · ◦ S1(w),which means that y is a solution of (1). Since ym ∈ Sm−1 ◦ Sm−2 ◦ · · · ◦ S1(w),there exist w2, w3 . . . , wm−1 such that

w2 ∈ S1(w) ⊂ (α2, β2) , . . . , wk+1 ∈ Sk(wk) ⊂ (αk+1, βk+1) , . . .

and

ym ∈ Sm−1(wm−1) ⊂ (αm, βm) .

But, by the definition of Sk, one has

wm−1 = hm−1(ym) = ym−1, wm−2 = hm−2(ym−1) = ym−2, . . .

. . . w2 = h2(y3) = y2, w = h1(y2) = y1.

Therefore, y ∈ S(w) ∩ (α, β).d) A continuous map, considered as a single-valued multifunction, is an usco

map and finite compositions of usco maps are usco maps (see §6.2 of [6]). ByLemma 3.3 and the result above, S is an usco map. The maximal and minimalexistence is a consequence of Remark 1. �

Remark 3 As we have mentioned in section 1, some authors have proposedvarious definitions of upper and lower solutions notions in the scalar case andhave developed related existence results. Notice Theorem 3.4 is stated for themultidimensional case. The crucial, but simple, idea is that one can reduce theorder of a problem by increasing its dimension (see Hartman’s paper [12] forthe linear case). Therefore, Definition 3.2 and Theorem 3.4 have a wide rangeof applications, as the following sections show. Observe that order reduction



and Definition 3.2 gives new and useful notions of upper and lower solutionsfor higher order problems.

3.2 Higher order difference equations.

Let p > n be natural numbers and consider the following n-th order initialvalue problem on an ordered Banach space F whose order cone is normal andhas non-empty interior:

∆nzk = ϕk(zk+1,∆zk+1,∆2zk+1, . . . ,∆n−1zk+1), k ∈ {1, . . . , p− n} (6)

z1 = w1, . . . , zn = wn, (7)

here ∆nzk = ∆(∆n−1zk

).

As we have seen, this problem can be written as

zk = ψk(zk+1, zk+2, zk+3, . . . , zk+n), k ∈ {1, . . . , p− n} (8)

z1 = w1, . . . , zn = wn. (9)

Assume p be a multiple of n, p = mn (if p wasn’t a multiple of n, we woulddefine ψk(u1, . . . , un) = un for k ∈ {p − n + 1, . . . ,mn}, where mn is thesmallest multiple of n greater than p).

Consider the first order initial value problem in E = Fn:

yk = hk(yk+1), k ∈ {1, . . . ,m− 1} (10)

y1 = (w1, . . . , wn), (11)

where the maps hk : E → E are defined as follows: the n-th component of hk

is defined by (hk)n (u) = ψnk(u). Assuming that (hk)n , (hk)n−1 , . . . , (hk)j+1

have been defined, let (hk)j be

(hk)j (u) = ψn(k−1)+j

((hk)j+1 (u), (hk)j+2 (u), . . . , (hk)n (u), u1, u2, . . . , uj

).

It is easily checked that, for each solution z1, z2, . . . , zp of (8)-(9), we obtain asolution of (10)-(11) by

yk = (zn(k−1)+1, zn(k−1)+2, . . . , znk)

with k ∈ {1, 2, . . . ,m} and vice versa. Now Section 3.1 can be applied toproblem (10)-(11).



Example 3.5 The discrete analogous of the autonomous boundary valueproblem z′′ + F (z) = 0; z(0) = z′(0) = 0 can be formulated as

zk = ψ(zk+1, zk+2), k ∈ {1, . . . , 2m− 2} (12)

z1 = z2 = 0 (13)

where m ∈ N, ψ(u1, u2) = g(u1)− u2 and g(x) = 2x− F (x)4m2 . Setting as above

yk = (z2k−1, z2k) for k ∈ {1, 2, . . . ,m}, the equivalent bidimensional first orderequation is

∆yk = f(yk+1), k ∈ {1, . . . ,m− 1} (14)

where f(u) = L(u) + γ(u), with

γ(u1, u2) =1

4m2(2F (u1) + F (g(u1)− u2), F (u1))

and L(u1, u2) =(−2 2−2 2

)(u1

u2

).

Consider the order relation � given by K = {(x2, x2) ∈ R2 : x2 ≥ x1 ≥ 0}and suppose F is such that there exist dm ≥ 0 with − dm

m2wm � γ(u) � dm

m2wm,where wm =

(2m1/m, 3m1/m − 1

). Remark 2 gives us the condition for the

upper and lower solutions to be inductively constructed from greater to smallerindexes. In such a way, we obtain

αk = αk+1 − L(αk+1) +dm

m2wm

βk = αk +dmckm2

wm,

where ck = mk/m + 2m1/m−1 . Observe that ck+1 − ck − 2 = m1/m−1

m1/m ck+1 andthat m1/m−1

m1/m wm � L(wm). It is obvious that βk − αk ∈ int(K). Furthermore,



for all v ∈ ∂K, 0 � v � (βk+1)− (αk+1), since L(v) � (0, 0), one has

∆αk = L(αk+1)−dm

m2wm � L(αk+1 + v) + γ(αk+1 + v) = f(αk+1 + v)

∆βk = ∆αk +dm(ck+1 − ck)

m2wm = L(αk+1)−

dm

m2wm +

dm(ck+1 − ck)m2

wm =

= L(αk+1) +dm

m2wm +

dm(ck+1 − ck − 2)m2

wm =

= L(αk+1) +dm

m2wm +

dmck+1(m1/m − 1)m2m1/m

wm �

� L(αk+1)− L(v) +dm

m2wm +

dmck+1

m2L(wm) �

� L(βk+1 − v) + γ(βk+1 − v) = f(βk+1 − v).

If we begin with αm = −dm

m2

(m1/mm+m+ 2m1/m − 1,m1/mm+m+ 3m1/m − 1

),

we obtain

α1 = (I − L)m−1(αm) +dm

m2

(m−2∑k=0

(I − L)k

)(wm) =

=−dm

m2

(2m− 1 −2m+ 22m− 2 −2m+ 3

)(m1/mm+m+ 2m1/m − 1m1/mm+m+ 3m1/m − 1

)+

+dm

m2

(m− 1 2−mm− 2 3−m

)(2m1/m

3m1/m − 1

)≺≺ (0, 0) ≺≺ α1 +

dmc1m2

wm = β1

We conclude by Theorem 3.4 that there exists a solution satisfying y1 = (0, 0),thus problem 12-13 has a solution located by α and β.

3.3 Other boundary conditions.

Some boundary value problems for discrete equations can be changed intoinitial value problems with a greater order. For instance, consider the followingfirst order problem with a periodic boundary condition:

ck = φk(wk+1), k ∈ {1, . . . ,m− 1} (15)

w1 = wm, (16)



with φk : E → E . Let the maps ψk : E2 → E , k ∈ {1, . . . , 2m − 1}, be definedas follows:

ψ1(u) = 0,

ψ2(u) = φ1(u2)− u1,

ψ2j−1(u) = u2, j ∈ {2, . . . ,m− 1}

ψ2j(u) = φj(u2), j ∈ {2, . . . ,m− 1}

ψ2m−1(u) = u1.

It can be easily shown that if z1, . . . , z2m, z2m+1 is a solution for the initialvalue problem

zk = ψk(zk+1, zk+2), k ∈ {1, . . . , 2m− 1} (17)

z1 = z2 = 0, (18)

then a solution for the problem (15)-(16) is obtained by setting w1 = z2m, ck =z2k for k ∈ {2, . . . ,m}. Observe that (17)-(18) can be handle like (8)-(9). Noticethat in order for ψk to be a compact map, E must be a finite dimensional space.

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The antipodal mapping theorem and difference equations in Banach spaces†

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