R E S E A R C H
Open Access
Fixed points of nonexpansive potential
operators in Hilbert spaces
Biagio Ricceri
**Correspondence:
[email protected] Department of Mathematics, University of Catania, Viale A. Doria 6, Catania, 95125, Italy
Abstract
In this paper, we show the impact of certain general results by the author on the topic described in the title. Here is a sample:
Let (X,·,·) be a real Hilbert space and letT:X→Xbe a nonexpansive potential operator.
Then, the following alternative holds: eitherThas a fixed point, or, for each sphereS
centered at 0, the restriction toSof the functionalx→01T(sx),xdshas a unique global maximum towards which each maximizing sequence inSconverges.
MSC: 47H09; 47H10; 47J30; 47N10; 49K40; 90C31
Keywords: nonexpansive operator; potential operator; fixed point; well-posedness
1 Introduction
There is no doubt that fixed point theory for nonexpansive mappings is one of the central topics in modern analysis. Actually, since [, , ], such a theory has had (and contin-ues to have) a strong development, and several deep (often spectacular) results have been achieved within it in the settings such as abstract harmonic analysis (where the contribu-tions of Professor Lau are fundamental) and the geometry of Banach spaces.
On the other hand, another very important class of operators is that composed of poten-tial operators. That is to say, the operators that can be regarded as the Gâteaux derivative of a suitable functional. Actually, the variational methods to study linear and nonlinear equations are fully based on potential operators.
In the present paper, as the title says, we are interested in fixed point theory for the intersection of the two above classes of operators in the setting of Hilbert spaces. More precisely, we intend to show the impact of certain general results that the author has es-tablished in the last years on such a topic.
Referring to [] for a thorough introduction to potential operators (with several exam-ples related to them), we recall here a specific situation where one can easily appreciate the relationships between the two classes of operators we are dealing with. Namely, let (X,·,·) be a real Hilbert space,L:X→Xa continuous linear operator andz∈X. Then,
L+zis a potential operator if and only if
L(x),y=L(y),x
for allx,y∈X, while it is nonexpansive if and only ifLL(X)≤.
The following result subsumes very well the spirit of the ones that we will establish in Section :
Theorem . Let(X,·,·)be a real Hilbert space and let T:X→X be a nonexpansive potential operator. Then, the following alternative holds: either T has a fixed point, or, for each sphere S centered at, the restriction to S of the functional x→T(sx),xds has a unique global maximum towards which each maximizing sequence in S converges.
2 Preliminaries
From now on, (X,·,·) will be a real Hilbert space. For eachr> , we put
Br=
x∈X:x<r,
Br=
x∈X:x≤r
and
Sr=
x∈X:x=r.
T:X→Xwill be a nonexpansive operator,i.e.,
T(x) –T(y)≤ x–y
for all x,y∈X. We also assume that there exists a Gâteaux differentiable functional
J:X→R, withJ() = , such thatJ=T, whereJ is the Gâteaux derivative ofJ. This amounts to say that
lim
λ→+
J(x+λy) –J(x)
λ =
T(x),y
for allx,y∈X. It can easily be checked that
J(x) =
T(sx),xds
for allx∈X. We also put
Iλ(x) = x
–λJ(x)
for allx∈X,λ∈R. Forλ= we will simply use the symbolIinstead ofI.
The basic proposition which relates the fixed points ofλT(|λ| ≤) with the functional
Iλis as follows.
Proposition . The functional Iλis strictly convex and coercive for|λ|< , and convex for
Proof Let|λ| ≤. For eachx,y∈X, we have
x–λT(x) –y+λT(y),x–y=x–y–λT(x) –λT(y),x–y
≥ x–y–|λ|T(x) –T(y)x–y
≥ –|λ|x–y.
From this, it follows that the derivative of the functional Iλ (that is the operatorx→
x–λT(x)) is monotone and that it is uniformly monotone if|λ|< . Now, the conclusion
follows from classical results ([], pp.-).
Another very useful proposition [] is as follows.
Proposition . Let Y be a nonempty set, f,g:Y→Rtwo functions, andλ,μtwo real numbers, withλ<μ. Letyˆλbe a global minimum of the function f +λg and letyˆμ be a
global minimum of the function f +μg. Then, one has
g(yˆμ)≤g(yˆλ).
If eitheryˆλoryˆμis strict andyˆλ=ˆyμ, then
g(yˆμ) <g(yˆλ).
LetSbe a topological space. As usual, given a functionf:S→Rand a setC⊆S, we say that the problem of minimizing (resp. maximizing)f overCis well posed if the following two conditions hold:
- the restriction off toChas a unique global minimum (resp. maximum), sayxˆ; - every sequence{xn}inCsuch thatlimn→∞f(xn) =infCf (resp.
limn→∞f(xn) =supCf), converges toxˆ.
A set of the type{x∈S:f(x)≤r}is said to be a sub-level set off.
Given two functionals,:X→R, for eachλ∈]–∞, +∞], we denote byM(,,λ) either the set of all global minima of+λor the empty set according to whetherλ∈R orλ= +∞. We adopt the conventionsinf∅= +∞,sup∅= –∞. We also set
α(,,λ) :=max
inf
X ,M(sup,,λ)
,
β(,,λ) :=min
sup
X
, inf
M(,,λ)
.
Note that, by Proposition ., ifa<b, one has
α(,,b)≤β(,,a).
In [], we established the following basic result:
Theorem . Let,:X→Rand let a,b∈]–∞, +∞], with a<b. Assume that
and that, for eachλ∈]a,b[, the functional+λhas weakly compact sub-level sets and admits a unique global minimum in X.
Then, for each r∈]α(,,b),β(,,a)[, the problem of minimizing over–(r)is
well posed with respect to the weak topology. More precisely, the unique global minimum of|–(r), sayxˆr, agrees with the unique global minimum of+λfor someλ∈]a,b[.
Moreover, the functions r→ ˆxrand r→(xˆr)are continuous in]α(,,a),β(,,b)[
with respect to the weak topology.
Finally, let us recall the result of M. Schechter and K. Tintarev [] that we will apply jointly with Theorem . in the next section.
Theorem . Assume that J is sequentially weakly continuous. For each r> , set
ψ(r) =sup
x∈Sr
J(x).
Moreover, let A⊆], +∞[be an open interval such that, for each r∈A,J has no local max-ima in Brand there exists a uniquexˆr∈Srsatisfying J(xˆr) =ψ(r).
Then, the following conclusions hold:
(i) the functionψ isCand increasing inA;
(ii) for eachr∈A, one has
T(xˆr) = ψ(r)xˆr.
3 Results
Our first result (inspired by []) shows the key role which a certain functionϕ: ], +∞[→ [, +∞[ plays in dealing with the fixed points ofT.
Theorem . For each r> , put
ϕ(r) = inf
x∈Br
supBrJ–J(x)
r–x .
If there is r> such thatϕ(r) <, then T has a fixed point which lies in Br.
If T()= , then one has
lim inf
r→+ ϕ(r)≥
.
In any case, one has
lim sup
r→+∞
ϕ(r)≤ .
Proof Letr> be such thatϕ(r) <. So, there existsx∈Br, such that
supBrJ–J(x)
r–x <
By Proposition .,Iis weakly lower semicontinuous, and so there existsxˆ∈Brsuch that
I(xˆ)≤I(x) (.)
for allx∈Br. We claim thatˆx<r. Actually, ifˆx=r, then, by (.), taking into account
thatsupBrJ=supBrJ, we would have
x
–J(x) <
r–supy∈BrJ(y)≤ ˆx
–J(xˆ)
against (.). As a consequence,xˆis a local minimum of the functionalI, and so it is a fixed point ofT, by Proposition . again.
Now, assume thatT()= . Arguing by contradiction, suppose that
lim inf
r→+ ϕ(r) <
.
Then, we could find a sequence of positive numbers{rn}converging to such thatϕ(rn) <
for alln∈N. But then, for eachn∈N, there would be a fixed pointxnofTlying inBrn. Hence,{xn}would converge to inXand so, by continuity, we would haveT() = .
Now, to prove the third assertion, assume thatThas a fixed point, sayx˜. Then, by Propo-sition ., we have
˜x
–J(x˜)≤ x
–J(x) (.)
for allx∈X. Fixr>˜x. From (.), we then obtain
sup
Br
J≤
r–˜x +J(x˜)
and so
ϕ(r)≤supBrJ–J(x˜)
r–˜x ≤ .
This clearly implies that
lim sup
r→+∞
ϕ(r)≤ .
Fixλ∈], [. So,λT has a (unique) fixed point. By the previous remark, we clearly infer that
lim sup
r→+∞ λϕ(r)≤
and so the conclusion is obtained passing to the limit forλtending to .
Corollary . If T has no fixed points, then
lim
r→+∞ϕ(r) =infr>ϕ(r) = .
The next two results come from Theorem .. Clearly,fix(T) (resp.fix(–T)) will denote the set of all fixed points ofT (resp. –T).
Theorem . Assume that T()= . Set
η=max
inf
X J,fix(–supT)J
, θ=min
sup
X
J, inf
fix(T)J
.
For eachλ∈] – , [, letyˆλbe the unique fixed point of the operatorλT .
Then, the following assertions hold:
(a) the functionλ→g(λ) :=J(yˆλ)is increasing in]–, [and its range is]η,θ[;
(a) for eachr∈]η,θ[, the pointxˆr:=yˆg–(r)is the unique point of minimal norm ofJ–(r) towards which every minimizing sequence inJ–(r), for the norm, converges;
(a) the functionr→ ˆxris continuous in]η,θ[.
Proof We apply Theorem . taking
a= –,
b= and
(x) = –J(x), (x) =
x
for allx∈X. With these choices, we have α(,,b) = –θ
and
β(,,a) = –η.
By Propositions . and ., the functiong is non-decreasing in ]–, [ andg(]–, [)⊆ [η,θ]. Now, letA⊂]–, [ be a non-degenerate interval. Ifgwas constant inA, then, by Proposition . again, the functionλ→ ˆyλwould be constant inA. Letyˆbe its unique value. Then, we would have
ˆ
y=λT(yˆ)
semicontinuous, coercive and with a unique global minimum (that isˆyλ), we are allowed to apply Theorem .. Accordingly, for eachr∈]η,θ[, there existsλr∈]–, [, withyˆλr ∈
J–(r), such that
ˆyλr<x
for allx∈J–(r)\ {ˆy
λr}, and each sequence{yn}inJ–(r) withlimn→∞yn=ˆyλr, weakly converges toyˆλr. SinceXis a Hilbert space, this implies that{yn}strongly converges toyˆλr. Likewise, we get the strong continuity in ]η,θ[ of the functionr→ ˆyλr from its weak continuity that is ensured by Theorem . too. Now, to get (a), (a), (a), it is enough to observe that
λr=g–(r)
for allr∈]η,θ[.
Theorem . Assume that T()= . Set
θ= inf
x∈fix(T)x .
For eachλ> , letuˆλbe the unique fixed point of the operator λT .
Then, the following assertions hold:
(b) the functionλ→h(λ) :=ˆuλis decreasing in], +∞[and its range is],θ[;
(b) for eachr∈],θ[, the pointvˆr:=uˆh–(r)is the unique global maximum ofJ|Sr towards
which every maximizing sequence forJ|Sr converges;
(b) the functionr→ ˆvris continuous in],θ[.
If, in addition, the functional J is sequentially weakly continuous and has no local max-ima in Bθ, then, withψdefined as in Theorem ., the following further assertions hold: (b) the functionψ isC, increasing and strictly concave in],θ[;
(b) one has
T(vˆr) = ψ(r)vˆr for allr∈],θ[;
(b) one has
ψ(r) = h
–(r)
for allr∈],θ[.
Proof This time, we apply Theorem . taking
a= ,
and
(x) = x
,
(x) = –J(x)
for allx∈X. With these choices, we have
α(,,b) =
and
β(,,a) = θ.
Since /∈fix(T) andfix(T) is closed, we haveθ> . Of course, for eachλ> , the func-tionalx→λx–J(x) is weakly lower semicontinuous, coercive and with a unique global minimum (that isuˆλ), and so we can derive (b), (b), (b) from Theorem ., reasoning as in the proof of Theorem .. Under the additional assumptions onJ, (b), (b) follow directly from Theorem ., takingA= ],θ[. Finally, (b) is a consequence of (b) and of
the fact thatT(vˆr) =h–(r)ˆvr.
Remark . Of course, Theorem . is a by-product of Theorem ., as, ifThas no fixed points, we haveθ= +∞. On the other hand, if, for somer> , the problem of maximizing
JoverSris not well posed, thenThas a fixed point lying inBr. Indeed, from (b) it follows
thatr≥θ. But,fix(T) is a closed and convex set. So, it admits a point of minimal norm. By the above inequality, such a point lies inBrand we are done.
Now, we want to present the form that Theorem . assumes whenTis an affine oper-ator.
As usual, for a linear operatorL:X→X, we say that
- Lis compact if, for each bounded setC⊂X, the setL(C)is compact; - Lis symmetric if
L(x),u=L(u),x
for allx,u∈X.
Theorem . Let L:X→X be a symmetric continuous linear operator, with norm, and let z∈X\ {}.
For eachλ> , letwˆλbe the unique fixed point of the operatorλ(L–z). Moreover, set
θ= inf
x∈fix(L–z)x
and, for each r> ,
δ(r) =sup
x∈Sr
where
H(x) =
L(x) –z,x
.
Then, the following assertions hold:
(c) the functionλ→k(λ) :=wλis decreasing in], +∞[and its range is],θ[;
(c) for eachr∈],θ[, the pointωˆr:=wˆk–(r)is the unique global maximum ofH|Srtowards which every maximizing sequence forH|Sr converges;
(c) the functionr→ ˆωris continuous in],θ[.
If, in addition, T is compact, then the following further assertions hold:
(c) the functionδisC, increasing and strictly concave in],θ[;
(c) one has
L(ωˆr) – δ(r)ωˆr=z for allr∈],θ[;
(c) one has
δ(r) = k
–(r)
for allr∈],θ[.
Before giving the proof of Theorem ., we establish the following
Proposition . Let L:X→X be a symmetric continuous linear operator and let H be defined as in Theorem ..
Then, forx˜∈X, the following are equivalent:
(j) x˜is a local maximum ofH. (jj) x˜is a global maximum ofH. (jjj) L(x˜) =zandsupx∈XL(x),x ≤.
Proof First, observe that the symmetry ofLis equivalent to the fact that the functionalH
is Gâteaux differentiable with derivative given by
H(x) =L(x) –z
for all x∈X ([], p.). By the symmetry ofLagain, it is easy to check that, for each
x∈X, the inequality
H(x˜+x)≤H(x˜) (.)
is equivalent to
L(x˜) –z+ L(x),x
Now, if (j) holds, thenH(x˜) = (that isL(x˜) =z) and there isr> such that (.) holds for allx∈Br. So, from (.), we haveL(x),x ≤ for allx∈Brand then, by linearity, for all
x∈X, and this shows (jjj).Vice versa, if (jjj) holds, then (.) is satisfied for allx∈Xand
so by (.),x˜is a global maximum ofH, and the proof is complete.
Proof of Theorem . As we observed above, the symmetry ofLis equivalent to the fact thatLagrees with the derivative ofH. So, sincez= , we can apply Theorem . taking
T =L–z. In such a way, we derive (c)-(c) directly from (b)-(b). Now, assume thatL
is also compact. Then, this implies thatHis sequentially weakly continuous ([], Corol-lary .). Suppose thatH has a local maximum, sayx˜. Then, by Proposition .,x˜ is a global maximum ofH. In particular, this implies that the functionalx→x–H(x) is coercive and hence, by sequential weak lower semicontinuity, it has a global minimum. That is,fix(L–z)=∅, by Proposition .. So, by Proposition ., it clearly follows that
˜x≥θ.
In other words,Hhas no local maxima inBθ. At this point, (c)-(c) come directly from
(b)-(b).
Some remarks on Theorem . are now in order.
Remark . Note that the compactness ofLserves only to guarantee that the functional
x→ L(x),xis sequentially weakly continuous. So, Theorem . actually holds under such a weaker condition.
Remark . A natural question is: if assertions (c)-(c) hold, must the operatorLbe sym-metric and the functionalx→ L(x),xsequentially weakly continuous?
Remark . Note that ifL, besides being compact and symmetric, is also positive (i.e.,
infx∈XL(x),x ≥), then, by classical results, the operatorx→L(x) –xis not surjective,
and so there arez∈Xfor which the conclusion of Theorem . holds withθ= +∞. In the previous results, the essential assumption is thatT()= . In the next (and last) result, to the contrary, we highlight a remarkable uniqueness property occurring when
supXJ= (and soT() = ). Actually, in such a case, is the unique fixed point ofλTfor eachλ∈], [.
More precisely, for each real Hilbert space (Y,·,·), we denote byAY the class of all
nonexpansive potential operatorsP:Y→Ysuch that
sup
x∈Y
P(sx),xds= .
Set
γY= inf P∈AY
infλ> :x=λP(x) for somex= .
Theorem . For any real Hilbert space(Y,·,·), with Y={}, one has
γY= .
We first prove
Proposition . One has
γR= .
Proof LetP∈AR. Fixλ∈], [. Let us prove that is the only fixed point ofλP. Arguing by contradiction, assume thatxis a non-zero fixed point ofλP. It is not restrictive to assume thatx> (otherwise, we would work withP(–x)). Consider now the function χ:R→Rdefined by
χ(x) =
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
–x ifx<x
,
x
– xx
+
x
if
x
≤x≤x, –x
+ xx
– x
ifx>x. Clearly,χ∈C(R). Letx> . If <x≤x
, we have
χ(x) = –x≤P(x). Ifx>x
, we have
χ(x) =x
–|x–x|= λP(x)
–|x–x|<P(x) –|x–x| ≤P(x). So, in particular, we get
x
P
x s
ds=
x
P(x)dx>
x
χ(x)dx=χ
x
=
which contradicts the factP∈AR. From what we have just proven, it clearly follows that ≤γR.
Now, fix anyμ> . Continue to consider the functionχdefined above (for a fixedx> ). Clearly, the function μχbelongs toARand
x= μ χ(x)
μ .
Of course, from this we infer that
γR≤μ,
Proof of Theorem . First, let us prove that
γY≤. (.)
To this end, fix anyQ∈ARand anyλ> such that
ˆ
t=λQ(ˆt)
for someˆt= . Fix alsou∈Y, withu= , and consider the operatorP:Y→Ydefined by
P(x) =Q u,xu
for allx∈Y. Clearly,P∈AY. Finally, set ˆ
x=λQ(ˆt)u.
Of course,xˆ= . Since
u,xˆ=λQ(ˆt)
we also have
u,xˆ=tˆ
and so
ˆ
x=λP(xˆ).
From this, it clearly follows that
γY≤γR,
and so (.) follows now from Proposition .. Now, let us prove that
≤γY. (.)
To this end, fixP∈AY,λ> andx˜∈Y\ {}such that ˜
x=λP(x˜). (.)
Then, consider the functionQ:R→Rdefined by
Q(t) =
P
tx˜
˜x
, x˜
˜x
for allt∈R. Clearly, the primitive ofQvanishing at is non-positive inR. Moreover, for eacht,s∈R, we have
Q(t) –Q(s)=
P
tx
x
–P
sx
x
, x
x
≤P
tx
x
–P
sx
x
≤ |t–s|.
This shows thatQis nonexpansive, and soQ∈AR. Now, from (.), we get
˜x=λ
P(x˜), x˜
˜x
that is
˜x=λQ˜x.
From this, we infer that
γR≤γY.
So (.) follows from Proposition ., and the proof is complete.
For specific consequences of Theorem . concerning nonlinear elliptic equations, we refer to the very interesting papers [] and [] where a problem asked in [] was solved.
Competing interests
The author declares that he has no competing interest.
Acknowledgement
Dedicated to Professor Anthony To-Ming Lau, with esteem and friendship.
Received: 4 April 2012 Accepted: 10 July 2012 Published: 24 July 2012
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