# Eigenvalue for Densely Defined(S+)L Perturbations of Multivalued Maximal Monotone Operators in Reflexive Banach Spaces

Authors: Boubakari Ibrahimou
Journal: Journal of Mathematics
 (2013)

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## Abstract

LetX be a real reflexive Banach space and let X∗ be its dual. Let Ω⊂X be open and bounded such that 0∈Ω. Let T:X⊃D(T)→2X∗ be maximal monotone with 0∈D(T) and 0∈T(0). Using the topological degree theory developed by Kartsatos and Quarcoo we study the eigenvalue problem Tx+λPx∋0, where the  operator P:X⊃D(P)→X∗ is a single-valued of class (S+)L. The existence of continuous branches of eigenvectors of infinite length then could be easily extended to the case where the operator P:X→2X∗ is multivalued and is investigated.


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## Body

## 1. Preliminaries

In what follows we assume thatX is a real or complex Banach space and has been renormed such that it and its dual X* are locally uniformly convex. The normalized duality mapping is defined by
(1)J(x)={x*∈X:∥x*∥=∥x∥,〈x,x*〉=∥x∥2}.
The mapping T:X⊃D(T)→2X* is said to be “monotone” if for every x,y∈D(T), u∈Tx, and v∈Ty we have
(2)〈u-v,x-y〉≥0.
A monotone operator T is “maximal monotone” if G(T) is maximal in X×X*, when X×X* is partially ordered by inclusion. In our setting, a monotone operator T is maximal if and only if R(T+λJ)=X* for every λ>0. If T:X⊃D(T)→2X* is maximal monotone operator, the operator Tt≡(T-1+tJ-1)-1:X→X* is called the Yosida approximant of T and the following, which can be found in [1, page 102] is true.Lemma 1.
LetT:X⊃D(T)→2X* be a maximal monotone operator with 0∈D(T) and 0∈T(0). Then(i)
Tt is a bounded maximal monotone mapping with 0=Tt(0) for each t>0;(ii)
Ttx⇀T{0}x in X as t→0 for all x∈D(T), where T{0}x denotes the element y*∈Tx of minimum norm;(iii)
||Ttx||→∞ as t→0 for all x∉D(T)¯.
AlsoTtx∈TJtx, where Jt=I-tJ-1Tt:X→X and satisfies: if x∈convD(T)¯, then Jtx→x in X as t→0, where convM denote the convex hull of the set M.The proof of the next lemma can be found in Kartsatos and Skrypnik [2],Lemma 2.
LetT:X⊃D(T)→2X* be maximal monotone such that 0∈D(T) and 0∈T(0). Then the mapping (t,x)→Ttx is continuous on the set (0,∞)×X.An operatorT:X⊃D(T)→Y, with Y another Banach space, is bounded if it maps bounded subsets of D(T) onto bounded sets. It is compact if it is continuous and maps bounded subsets of D(T) onto relatively compact subsets of Y. It is demicontinuous if for every sequence {xn} such that xn→x0 we have Txn⇀Tx0.We say that the operatorT:X⊃D(T)→2X* satisfies condition (Sq) on a set A⊂D(T) if for every sequence {xn}⊂A such that xn⇀x0∈X and any yn*∈Txn, with yn*→ (some)y*∈X*, we have xn→x0. If A=D(T), then we say that T satisfies (Sq).Definition 3.
LetX be a separable reflexive Banach space and let L be a dense subspace of X. A mapping P:D(P)⊂X→X* is said to be of class (S+)L if, for any sequence of finite dimensional subspaces Fn of L with Un=1∞Fn¯=X, h∈X*, {xn}n=1∞⊂D(P) with xn⇀x0 and
(3)limsupn→∞〈Pxn-h,xn〉≤0,limn→∞〈Pxn-h,v〉=0
for all v∈Un=1∞Fn, we have xn→x0, x∈D(P) and Px0=h.
Ifh=0, then we call P a mapping of class (S+)0,L.Definition 4.
LetX be a separable reflexive Banach space and let L be a dense subspace of X. A multivalued mapping P:D(P)⊂X→2X* is said to be of class (S+)L if it satisfies the following conditions: (i)
Px is bounded closed and convex for each x∈D(P),(ii)
P is weakly upper semicontinuous in each finite dimensional space, that is, for each finite dimensional space F of L, F∩D(P)≠∅, P:F∩D(P)→2X* is upper semicontinuous in the weak topology,(iii)
if for any sequence of finite dimensional subspacesFn of L with L⊂Un=1∞Fn¯, h∈X*, {xn}n=1∞⊂D(P)∩L and xn⇀x0 such that
(4)limsupn→∞〈pn-h,xn〉≤0,limn→∞〈pn-h,v〉=0for all v∈Un=1∞Fn and some pn∈Pxn, then we have xn→x0, x∈D(P) and h∈Px0.
Ifh=0, then we call P a mapping of class (S+)0,L.We will need the following two conditions:(P1)
there exist a subspaceL of X such that L⊂D(P), L¯=X and that the operator P satisfies condition (S+)L;(P2)
there exist a functionϕ:ℝ+→ℝ+ which is nondecreasing such that
(5)〈p,x〉≥-ϕ(∥x∥),p∈Px,x∈D(P).The following lemma can be found in Zeidler [3, page 915].Lemma 5.
LetT:X⊃D(T)→2X* be maximal monotone. Thus the following are true:(i)
{xn}⊂D(T)xn→x0 and Txn∋yn⇀y0 imply x0∈D(T), and y0∈Tx0;(ii)
{xn}⊂D(T)xn⇀x0 and Txn∋yn→y0 imply x0∈D(T), and y0∈Tx0.We will need the following lemma from Adhikari and Kartsatos [4].Lemma 6.
Assume that the operatorsT:X⊃D(T)→2X* and T0:X⊃D(T0)→X* are maximal monotone, with 0∈D(T)∩D(T0) and 0∈T(0)∩T0(0). Assume further that T+T0 is maximal monotone. Assume that there is a positive sequence {tn} such that tn↓0, a sequence {xn}⊂D(T0) and a sequence wn∈T0xn such that xn⇀x0 and Ttnxn+wn⇀y0*∈X*. Then the following are true:(i)
the inequality(6)limn→∞〈Ttnxn+wn,xn-x0〉<0
is impossible;(ii)
if(7)limn→∞〈Ttnxn+wn,xn-x0〉=0,
then x0∈D(T+T0) and y*∈(T+T0)x0.The proof of the following lemma is given in [5] proof of Theorem 7 but we will repeat it here for completeness and future reference.Lemma 7.
LetT:X⊃D(T)→2X* be a quasibounded maximal monotone operator such that 0∈T(0). Let {tn}⊂(0,∞) and {tn}⊂X be such that
(8)∥un∥≤S,〈Ttnun,un〉≤S1,
where S,S1 are positive constants. Then there exists a number K>0 such that ||Ttnun||≤K for all n=1,2,….Proof.
Let(9)wn=Ttnun=(T-1+tnJ-1)-1un.
We have that
(10)wn∈TJtnun=Txn,tnwn=J(un-xn),
where xn=Jtnun. Thus,
(11)〈wn,xn〉=〈wn,un-tnJ-1wn〉=〈wn,un〉-tn〈wn,J-1wn〉=〈wn,un〉-tn∥wn∥2≤〈Ttnun,un〉≤S1.
From (11) we obtain
(12)tn∥wn∥2=〈wn,un〉-〈wn,xn〉.
Now since 0∈T(0) and wn∈Txn, we have that 〈wn,xn〉≥0, which implies tn||wn||2≤S1. We claim that {wn} is bounded, if not we may assume that ||wn||→∞ and ||wn||≤||wn||2 for all n. Thus tn||wn||≤S1 and
(13)tn∥wn∥=∥J(un-xn)∥=∥un-xn∥
which implies that {xn} is bounded. Now since T is strongly quasibounded, the boundedness of {xn} and {〈wn,xn〉} imply the boundedness of {wn}, that is, a contradiction. It follows that {Ttnun} is bounded and the proof is complete.We denote byJψ the duality mapping with gauge function ψ. The function ψ:ℝ+→ℝ+ is continuous, strictly increasing such that ψ(0)=0 and ψ(r)→∞ as r→∞. This mapping Jψ is continuous, bounded, surjective, strictly and maximal monotone, and satisfies condition (S+). Also,
(14)〈Jψx,x〉=ψ(∥x∥)∥x∥,∥Jψx∥=ψ(∥x∥),x∈X
for these facts we refer to Petryshyn [6, pages 32-33 and 132].


## 2. The Eigenvalue Result

In this section we are using the topological degree developed by Kartsatos and Quarcoo in [7]. The following result will also improve Theorem 4 of Kartsatos and Skrypnik in [8] since we are no longer assuming that the perturbation is quasibounded.Theorem 8.
LetΩ be open and bounded with 0∈Ω. Assume that the operator T:X⊃D(T)→2X* is maximal monotone with 0∈D(T) and 0∈T(0). Assume that the operator P:X⊃D(P)→X*, with L⊂D(P), satisfies condition (S+)L and (P2). Let ϵ, ϵ0, and Λ be positive numbers. Assume that(𝒫)
there existsλ∈(0,Λ] such that
(15)Tx+λPx+ϵJψx∋0
has no solution in D(T+P)∩Ω.
Then(i)
there exists(λ0,x0)∈(0,Λ]×(D(T+P)∩∂Ω) such that
(16)Tx0+λ0Px0+ϵJψx0∋0;(ii)
if0∉T(D(T)∩∂Ω), T satisfies (Sq) on ∂Ω, and property (𝒫) is satisfied for every ϵ∈(0,ϵ0], then there exists (λ0,x0)∈(0,Λ]×(D(T+P)∩∂Ω) such that Tx0+λ0Px0∋0.Proof.
(i) Assume that(𝒫) is true and that (16) is false. We consider the homotopy inclusion
(17)H(t,x)≡Tx+tΛPx+ϵJψx∋0,t∈[0,1].
It is clear that this inclusion has no solution x∈D(H(t,·))∩∂Ω for t∈(0,1], because tΛ∈(0,Λ] and our assumption about (16). It is also true for t=0 because 0∈(T+ϵJψ)(0) and the operator T+ϵJψ is strictly monotone, hence one-to-one. We are going to show that H(t,x) is an admissible homotopy for the degree in [4]. To this end, we set Tt=T, Pt≡tΛP+Jψ and we recall the operators Tt,s=Ts≡(T-1+sJ-1)-1:X→X*, s>0, and we set Jt,s=Js=I-sJ-1Tt,s=I-sJ-1Ts:X→X. We have D(H(0,·))=D(T) and D(H(t,·))=D(T+P), t∈(0,1]. We also set Dt=D(tΛP)=D(tP). We have D0=X and Dt=D(P) for t∈(0,1]. Let Ω be an open and bounded subset of X. We know that the equation
(18)Ttx+Ptx∋0
has no solution x∈D(H(t,·))∩∂Ω for any t∈[0,1]. Now we consider the equation
(19)Tsx+tΛPx+ϵJψ=0,
and we show that there exists s1>0 such that
(20)0∉(Ts+tΛP+ϵJψ)(Dt∩∂Ω),(s,t)∈(0,s1]×[0,1].
Assume that this is not true, then there exist {sn}⊂(0,∞) with sn↓0, {tn}⊂[0,1], with tn→t0, {xn}⊂∂Ω with xn⇀x0, and
(21)Tsnxn+tnΛPxn+ϵJψxn=0.
Clearly we cannot have tn=0 for any n, since (Tsn+Jψ)(0)=0 and the operator Tsn+Jψ is strictly monotone, hence one-to-one. Thus tn>0, for all n. From (21) we have that
(22)〈Tsnxn,xn〉=-tnΛ〈Pxn,xn〉-ϵ〈Jψxn,xn〉≤tnΛϕ(∥xn∥)-ϵψ(∥xn∥)≤Λϕ(S1)+ϵψ(S1),
where S1 is the bound of {xn}. Thus we have the boundedness of 〈Tsnxn,xn〉. Using Lemma 7, we have the boundedness of Tsnxn. We may thus assume that Tsnxn⇀h1*. From (21) we also have that the sequence {Pxn} is bounded and we may assume that Pxn⇀h0*.
Ift0=0, then from
(23)ϵψ(∥xn∥)∥xn∥=〈ϵJψxn,xn〉≤-tnΛ〈Pxn,xn〉→0
we obtain xn→0∈∂Ω, which is a contradiction to 0∈Ω. Hence it follows that t0>0. Since {Jψxn} is bounded, we may assume that Jψxn⇀h2* and we have t0Λh0*=-h1*-ϵh2*.
From (21) it follows that
(24)〈Tsnxn+tnΛPxn+ϵJψxn,x〉=0,∀x∈Un=1∞Fn.
Thus
(25)〈Tsnxn+tnΛPxn+ϵJψxn,xn〉=0,limn→∞〈Tsnxn+tnΛPxn+ϵJψxn,v〉=0,∀v∈Un=1∞Fn.
Now
(26)〈tnΛPxn+h1*+ϵh2*,xn〉=〈Tsnxn+tnΛPxn+ϵJψxn,xn〉-〈Tsnxn+ϵJψxn,xn-x0〉-〈Tsnxn+ϵJψxn,x0〉+〈h1*+ϵh2*,xn〉,
hence
(27)limsupn→∞〈tnΛPxn+h1*+ϵh2*,xn〉≤limsupn→∞〈Tsnxn+ϵJψxn+tnΛPxn,xn〉--liminfn→∞〈Tsnxn+ϵJψxn,xn-x0〉--〈h1*+ϵh2*,x0〉+〈h1*+ϵh2*,x0〉≤-liminfn→∞〈Tsnxn+ϵJψxn,xn-x0〉≤0,
where the last inequality follows from Lemma 6. Now since P is of class (S+)L, it follows that xn→x0, x0∈D(P) and t0ΛPx0=-h1*-ϵh2*. Also we have that
(28)〈Tsnxn,xn-x0〉=0
and using Lemma 6 we get that x0∈D(T) and Tx0∋h1*=-t0ΛPx0-ϵJψx0. This is a contradiction since xn→x0, we have x0∈∂Ω. We have shown that H(t,x) is an admissible homotopy for our degree. We can now work as in Theorem 3 in [8] in order to show that d(H(t,·),Ω,0)=const.
Thus(29)d(H(t,·),Ω,0)=d(Ts+ϵJψ,Ω,0)=1,
where the last equality follows from Theorem 3, (i) of [9].
Consequently, the inclusionH(t,x)∋0 has a solution in Ω for each t∈[0,1]. In particular, this says that Tx+λPx+ϵJψx∋0 has a solution in Ω for every λ>0. This of course contradicts condition (𝒫) and finishes the proof of (i).
(ii) Letλn∈(0,Λ], xn∈D(P)∩∂Ω be such that, for some un*∈Txn,
(30)un*+λnPxn+(1n)Jψxn=0
again, λn=0 for any n is not possible. Since λn>0, we have by property (P2) that
(31)〈un*,xn〉=-λn〈Pxn,xn〉-1n〈Jψxn,xn〉≤λnϕ(∥xn∥)-1nψ(∥xn∥)∥xn∥≤Λϕ(S1)+ψ(S1)S1,
where S1 is the bound of {xn}. This show that 〈un*,xn〉 is bounded and further we obtain the boundedness of {un*} by Lemma 2. We may thus assume that λn→λ0, xn⇀x0, un*⇀h0*, then λnPxn⇀-h0*.
Ifλ0=0, then (30) implies that
(32)limn→∞un*=limn→∞[-λnPxn-(1n)Jψxn]=0.
Since T satisfies (Sq), it follows that xn→x0∈∂Ω. Now, by the demiclosedness of T (see Lemma 6) we obtain that x0∈D(T), 0∈Tx0 and this contradicts 0∉T(D(T)∩∂Ω). Hence, λ0>0. Repeating the proof of (i) starting from (21), we get again that
(33)limsupn→∞〈λnPxn+h0*,xn〉≤0
and since P is of class (S+)L, we have xn→x0, -h0*=λ0Px0. Using the demiclosedness of T, we obtain x0∈D(T)∩∂Ω, h0∈Tx0 and Tx0+λ0Px0∋0. The proof is now complete.


## 3. Continuous Branches of Eigenvectors

In this section we are interested in showing that the results obtained in previous sections could be used in order to obtain the existence of continuous branches of eigenvectors. We need the following definition.Definition 9.
LetT:X⊃D(T)→2X*, P:X⊃D(P)→X* be given and consider the problem
(34)Tx+λPx∋0.
An eigenvector x is solution of (34) for some eigenvalue λ with x∈(D(T)∩D(P)). We say that the nonzero eigenvectors of the problem (34) form a continuous branch of infinite length if there exists r0>0 such that, for every r≥r0, the sphere ∂Br(0) contains at least one nonzero eigenvector of (34).Theorem 10.
Assume that the operatorT:X⊃D(T)→2X* is maximal monotone with 0∈D(T) and 0∈T(0). Assume that the operator P:X⊃D(P)→X* with L⊂D(P) is of class (S+)L and satisfies (P2). Let Λ be a positive number. Assume that Tx∋0 implies x=0, T satisfies (Sq) and(𝒫1) there exists α>0 and λ∈(0,Λ] such that
(35)|Tx+λPx|≥α,x∈D(T+P).Then the nonzero eigenvectors of problem (34) form a continuous branch of infinite length with corresponding eigenvalues λ∈(0,Λ].Proof.
Letr0>0 be given. Let ϵ0>0 be so small that ϵ0r0<α. Then
(36)|Tx+λPx+ϵJx|≥|Tx+λPx|-|ϵJx|≥α-ϵ∥x∥≥α-ϵ0r0,x∈D(T+P), which implies that the inclusion
(37)Tx+λPx+ϵJx∋0
has no solution x∈D(T+P)∩Br(0) for any ϵ∈(0,ϵ0]. Since 0∉T(∂Br0(0)) and T is of type (Sq), Theorem 8 implies the existence of a solution xλ0∈D(T+P)∩∂Br0(0), for some λ0∈(0,Λ]. The same argument can be repeated for any r>r0. This complete the proof.Remark 11.
This result is also true whenP:X⊃D(P)→2X* is assumed to be multivalued. We have the following theorem without proof.Theorem 12.
Assume that the operatorT:X⊃D(T)→2X* is maximal monotone with 0∈D(T) and 0∈T(0). Assume that the operator P:X⊃D(P)→2X* with L⊂D(P) is of class (S+)L and satisfies (P2). Let Λ be a positive number. Assume that Tx∋0 implies x=0, T satisfies (Sq), and
(
𝒫
1
) there exists α>0 and λ∈(0,Λ] such that
(38)|Tx+λPx|≥α,x∈D(T+P).
Then the nonzero eigenvectors of problem (34) form a continuous branch of infinite length with corresponding eigenvalues λ∈(0,Λ].In the following result, we assume that the operatorT is defined and bounded on all of X. In this case we demonstrate the fact that the assumption |Tx+Px|≥0 may be replaced by the assumption ||Px||≥0 on D(P).Theorem 13.
Assume that the operatorT:X⊃D(T)→2X* is maximal monotone and bounded with 0∈D(T) and 0∈T(0). Assume that the operator P:X⊃D(P)→X*, with L⊂D(P) satisfies condition (S+)L and (P2). Assume that Tx∋0 implies x=0, T satisfies (Sq) and there exist α>0 such that
(39)∥Px∥≥α,x∈D(P).
Then the nonzero eigenvectors of the problem (34) form a continuous branch of infinite length.Proof.
We show that the problem (34) has eigenvectors on the set ∂Br(0) for every r>0. To this end, let fix r>0, ϵ>0 and show the existence of λ~>0 such that
(40)d(T+λ~P+ϵJ,Br(0),0)=0.
If this is not true, then there exist a sequence of {λn}⊂(0,∞) such that λn→∞ and one of the following is true:(i)
the degreed(T+λnP+ϵJ,Br(0),0) is not well-defined;(ii)
d(T+λnP+ϵJ,Br(0),0)≠0.
In the case (i) there exist eigenvectorsxn∈∂Br(0) such that
(41)Txn+λnPxn+ϵJxn∋0.
In the case (ii) there exist eigenvectors xn∈Br(0) such that (41) holds. Thus in either case, there exist a sequence {xn}⊂Br(0)¯ such that (41) is true. But this leads to a contradiction because ||λnPxn+ϵJxn||≥αλn-ϵr→∞, while the set Txn lie in a bounded set. Thus, (40) is true for some λ~>0. We consider the homotopy
(42)H(t,x)≡Tx+tλ~Px+ϵJx,t∈[0,1],x∈D(H(t,·)),
either there exist t0∈[0,1] and xt0∈∂Br(0) such that
(43)Txt0+t0λ~Pxt0+ϵJxt0∋0,
or
(44)d(H(t,·))=d(H(1,·))=0=d(H(0,·))=1,t∈[0,1]
that is a contradiction. The last equality in (44) follows from Theorem 3, (i) in [2]. It follows that (43) is true. Naturally, we must have t0≠0 in (43) because otherwise 0∈(T+ϵJ)(∂Br(0)). This cannot happen because we already have 0∈(T+ϵJ)(0) and T+ϵJ is one-to-one. From (43) we obtain sequences λn∈(0,∞), {xn}⊂∂Br(0) such that
(45)Txn+λnPxn+(1n)Jxn∋0.
We may assume that xn⇀x0. Again, {λn} cannot contain a subsequence {λnk} such that λnk→∞ as k→∞ because the sequence Txnk+(1/nk)Jxnk lies in a bounded set and λnk||Pxnk||→∞ as k→∞. Also we have that the sequence {Pxn} is bounded. Thus we may assume that Pxn⇀p*∈X*. Since the sequence {λn} is bounded, we may assume that λn→λ0. Again, λ0≠0 otherwise λnPxn+(1/n)Jxn→0 and the (Sq) property of T would imply that xn→x0∈∂Ω. Since T is demiclosed, Lemma 6 would imply Tx0∋0. This is a contradiction to our assumption that Tx∋0 implies x=0. It follows that λ0>0 and we may also assume that λn>0 for all n.
Now proceeding as in the proof of Theorem8, it is easy to see that
(46)limsupn→∞〈λnPxn+λ0p*,xn〉≤0,limn→∞〈λnPxn+λ0p*,v〉=0,∀v∈Un=1∞Fn.
Now since P is of class (S+)L, we conclude that xn→x0, x0∈D(P) and Px0=p*. Thus since also yn*⇀-λ0Px0, xn→x0 and the demiclosedness of T, we obtain x0∈D(P)∩∂Br(0) and Tx0+λ0Px0∋0. Since r>0 is arbitrary, the nonzero eigenvectors of problem (34) form a continuous branch of infinite length.



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