Int. Journal of Math. Analysis, Vol. 2, 2008, no. 1, 45 - 54
Fixed Point Theorems for Mappings
Satisfying Interior Condition
Shaini P
Department of Mathematics
University of Allahabad, Allahabad-211002, India
[email protected]
Neeta Singh
Department of Mathematics
University of Allahabad, Allahabad-211002, India
n [email protected]
Abstract
Let X be a real Banach space, D a bounded open subset of X, and
D the closure of D. We prove some fixed point theorems for 1- set
contractions and sum of two operators using the Interior Condition. In
section 2 of this paper we prove a fixed point theorem for 1-set contractions T : D → X under the interior condition and use it to deduce
other theorems. In section 3 we have fixed point results for nonexpansive, lane, and uniformly strictly contractive maps with compact or
completely continuous perturbations.
Mathematics Subject Classification: primary 47H09, 47H10
Keywords: Nonexpansive maps, 1-set contratction maps, compact maps,
completely continuous maps, interior condition, Banach space
1
Introduction and preliminary definitions
In a recent paper [3] A.J.Melado and C.H.Morales have introduced and
studied a new condition called the Interior Condition, which resembles the
Leray-Schauder condition. However unlike the Leray-Schauder condition which
is a boundary condition, the interior condition holds for the interior points,
which lie near the boundary of the domain D of the map T . This interior
condition is formulated as follows.
46
Shaini P and Neeta Singh
A mapping T : D → X satisfies the Interior Condition if there exists δ > 0
such that
/ D,
T (x) = λx for x ∈ D∗ , λ > 1 and T (x) ∈
where D ∗ = {x ∈ D : dist(x, ∂D) < δ}.
It is observed that a condensing map T : D → X which satisfies the interior
condition, but defined on an arbitrary bounded open set, even on star-shaped
sets does not ensure the existence of fixed points. Therefore a more restriced
domain called strictly star-shaped sets is introduced, inorder to obtain fixed
points. The following result is due to Melado and Morales.
Let X be a real Banach space and let D be a bounded open strictly starshaped subset of X with 0 ∈ D. If T : D → X is a condensing mapping
satisfying the interior condition, then T has fixed point in D.
In section 2 we extend the above result[3, Theorem 1] by proving that if
T : D → X is a 1-set contraction which satisfies the interior condition, then
T has a fixed point if (I − T )D is closed. We use this result to deduce fixed
point results for various other mappings.
We need the following definitions.
Let D be a bounded subset of a metric space X. Define the measure of
noncompactness α(D) of D by
α(D) = inf { > 0 : D admits a finite covering of subsets of diameter ≤ .}
It follows that α[A] ≤ α[B] whenever A ⊂ B, α[A∪B] = max {α[A], α[B]} ,
α[co(A)] = α[A], α[λA] = |λ| α[A], α[A] = α[A], α[A + B] ≤ α[A] + α[B] and
α[A] = 0 if and only if A is compact.
Closely associated with α is the concept of k -set contractions. Let T :
X → X be a continuous mapping of a Banach space X. Then T is called a
k- set contraction if for all A ⊂ X with A bounded, T (A) is bounded and
α(T A) ≤ kα(A), 0 < k < 1. If α(T A) < α(A), for all α(A) > 0, then T is
called densifying (or condensing).
It follows that C : D → X is compact (i.e., if C(D) is contained in a
compact subset of X.) if and only if C is 0-set contractive, and that every
Lipschitzian map S : D → X with constant ≥ 0 is -set contractive. Clearly
the map T = S + C : D → X is also -set contractive.
A map F : D ⊆ X → X where X is a Banach space is said to be completely
continuous if F (Y ) is relatively compact for all bounded sets Y ⊆ D. If F is
completely continuous it carries weakly convergent sequences in D into strongly
convergent sequences in X. While F is said to be nonexpansive if F satisfies
d(F (x), F (y)) ≤ d(x, y) for all x, y ∈ D.
we remark that a nonexpansive map is also a 1-set contraction.
A map f : X → X where X is a Banach space, is said to be demiclosed
if for every sequence xn ∈ X which converges weakly to x in X(denote by
xn x) and f (xn ) converges strongly to y, we have y = f (x).
47
Fixed point theorems
2
Fixed point theorems for 1-set contraction
maps
Definition 2.1. Let X be a Banach space and let D be a bouned open subset
of X with 0 ∈ D. We say that D is strictly star-shaped with respect to the
origin if for each x ∈ ∂D, {tx : t > 0} ∩ ∂D = {x}.
Every convex set is strictly star-shaped, but the converse is not true.
Theorem 2.2. Let X be a real Banach space, D an open bounded and strictly
star-shaped subset of X with 0 ∈ D, and let T : D → X a 1-set contraction
satisfying the interior condition. If (I −T )D is closed, then T has a fixed point
in D.
Proof. Define for n ≥ 2, Tn : D → X as Tn (x) = (1− n1 )T (x). Then each Tn is a
k-set contraction with k = 1 − n1 . Every k- set contraction is a condensing map
[7, p.29] and hence Tn is condensing. Also Tn satisfies the interior condition as
T itself satisfies the same condition.Therefore by [3, Theorem 1] there exists
an xn ∈ D such that
xn = Tn (xn ) = (1 −
1
)T (xn ).
n
That is (I − T )xn = n1 T (xn ) → 0, as n → ∞, since D is bounded. In view
of this and the assumed closedness of (I − T )D, we see that 0 ∈ (I − T )D.
Hence there exists an x ∈ D such that (I − T )x = 0 or x = T (x).
Theorem 2.3. Let X be a real Banach space D an open bounded and strictly
star-shaped subset of X with 0 ∈ D, and let T : D → X a 1-set contraction
satisfying the interior condition. Then T has a fixed point in D, if T satisfies
the following condition
(i) If {xn } is any sequence in D such that xn − T (xn ) → 0 as n → ∞, then
there exists x ∈ D with x − T (x ) = 0.
Proof. Define for n ≥ 2, Tn : D → X as Tn (x) = (1 − n1 )T (x). Then each Tn
is condensing and satisfies the interior condition. Since Tn is condensing, for
every n, (I − Tn )D is a closed set [8]. Therefore by the above theorem Tn has
a fixed point in D. That is
Tn (xn ) = xn = (1 −
1
)T (xn ).
n
Therfore (I − T )xn = − n1 T (xn ) → 0 as n → ∞ since D is bounded. Hence by
condition (i) there exists an x ∈ D such that x = T (x ).
48
Shaini P and Neeta Singh
Theorem 2.4. Let D be a bounded open convex subset of a reflexive Banach
space X with 0 ∈ D. Suppose that T : D → X is a 1-set contraction map
satisfying the interior condition. Then if I − T is demiclosed, T has a fixed
point in D.
Proof. As in the above theorem, define for any positive integer n ≥ 2, Tn :
D → X as Tn (x) = (1 − n1 )T (x). It is clear that Tn is condensing, since it
is k - set contraction with k = (1 − n1 ). Also Tn satisfies the interior condition as T itself satisfies the interior condition. Since D is convex, it is strictly
star-shaped [3, Proposition 1(a)], and as Tn is condensing, the set (I − Tn )D
is closed [8, Proposition 11.14]. Therefore by Theorem 2.2, Tn has a fixed
point xn ∈ D. That is Tn (xn ) = xn , for all n ≥ 2. Now since X is a reflexive
Banach space, any norm bounded sequence in X contains a weakly convergent
subsequence. Therefore since D is bounded, there exists a subsequence S of
integers and an x ∈ D( a closed convex subset of a Banach space is weakly
closed) with xn x as n → ∞ in S. In addition since xn = (1 − n1 )T (xn ), we
have (I − T )xn = n1 T (xn ) → 0 as n → ∞ in S. The demiclosedness of
I − T implies that (I − T )x = 0. i.e., x = T (x).
Corollary 2.5. Let D be a bounded, open convex subset of a uniformly convex
Banach space X with 0 ∈ D. Suppose that T : D → Xis a nonexpansive map
satisfying the interior condition. Then T has a fixed point in D.
Theorem 2.6. Let X be a Banach space, B ⊂ X be an open ball about the
origin.Suppose T is a 1 -set contraction of B, into X which is demicompact
and which satisfies the interior condition. Suppose further that there exists a
sequence {Tn } of condensing mappings of B into X such that
(a) δn = sup Tn (x) − T (x) : x ∈ B → 0 as n → ∞. Then T has a fixed
point in B.
To prove this theorem first we prove the following Lemma.
Lemma 2.7. Suppose T and {Tn } satisfy the conditions of Theorem 2.6. Then
there exists an integer N0 ≥ 1 such that {Tn } satisfies the interior condition
for each n ≥ N0 .
Proof. Since T is bounded and δn → 0 as n → ∞, there exists M > 0 such
that Tn (x) ≤ M for all n and all x ∈ B. Hence Tn (x) = λx for x ∈ B ∗ , where
B ∗ = {x ∈ B : dist(x, ∂B) < δ, for some δ > 0.}, and Tn (x) ∈
/ B implies that
M
|λ| ≤ x = M .
Now, if Lemma 2.7 were false, there would exist a subsequence Tnj of
{Tn }, a sequence {xj } ∈ B ∗ with Tnj (xj ) ∈
/ B and a sequence {λj } with
λj > 1 such that
Tnj (xj ) = λj (xj ) for all j
(1)
Fixed point theorems
49
Since {λj }is necessarily bounded, we may assume that λj → λ0 for some
λ0 ≥ 1. Thus (1) and
(a) of Theorem 2.6 implies that T (xj ) − λ0 xj ) = T (xj ) − Tnj (xj ) + λj (xj ) − λ0 (xj ) ≤ δnj + |λj − λ0 | xj → 0 as j → ∞. i.e., (I − η0 T )xj → 0 as j → ∞, where η0 = λ−1
0 ; λ0 ≥ 1. Now
if λ0 = 1, then η0 = 1 and (I − T )xj → 0 as j → ∞, therefore since xj ∈ B ∗
and T is demicompact, there exists a subsequence {xjk } and x0 ∈ B ∗ such that
xjk → x0 as k → ∞, hence T (x0 ) = x0 because T is continuous. Now since
x0 ∈ B ∗ we have T (x0 ) ∈ B ∗ . This contradicts (a) since Tnj (xj ) ∈
/ B ∗.
If λ0 > 1, then since T is a 1- set contraction and η0 < 1, Tη0 = η0 T is
an η0 - set contraction of B into X with fj ≡ (I − η0 T )xj → 0 as j → ∞;
therfore since α({fj }) = 0 and {xj } = {fj } + {Tη0 xj } we can conclude that
α({xj }) = 0. Thus {xj } contains a convergent subsequence, that is there exists
{xjk } and x0 ∈ B ∗ such that xjk → x0 , and hence we have Tη0 x0 = x0 ,i.e.,
η0 T (x0 ) = x0 or T (x0 ) = λ0 x0 , λ0 > 1 which is a contradiction since T satisfies
the interior condition. This proves the Lemma.
Proof of the Theorem. Since Tn is condensing on the closed and bounded set
B, (I − T )B is closed [8, Proposition 11.14]. Therefore it follows from Lemma
2.7, and Theorem 2.2 that for each n ≥ N0 there exists at least one xn ∈ B
such that Tn (xn ) = xn . This and condition(a) of Theorem 2.6 imply that
T (xn ) − xn = T (xn ) − Tn (xn ) ≤ δn → 0
as n → ∞, from which, Since T iscontinuous and demicompact, we deduce
the existence of a subsequence xnj and an element x in B such that xnj → x
as j → ∞. Hence we have T (x) = x. Since T satisfies the interior condition
we can conclude that x ∈
/ B ∗ , but x ∈ B − B ∗ . Thus F (T ) = φ.
Corollary 2.8. Let X be a Banach space, and T a demicompact nonexpansive
mapping of B into X which satisfies the interior condition. Then T has a fixed
point in B.
Proof. Since T is a nonexpansive map on B, T is a 1-set contraction of B
into X and, for a sequence {kn } such that 0 < kn < 1 and kn → 1 as
n → ∞, {Tn } = {kn T } is a sequence of kn -set contractions with kn < 1
for each n. Thus {Tn } is a sequence of condensing mappings of B ino X which
clearly satisfies conditions of Theorem 2.6. Hence T has a fixed point in B.
Corollary 2.9. Let X be a Banach space and T a nonexpansive mapping of
B into X. suppose that T satisfies the interior condition and that T is either
condensing or of the form T = S + C, where C is compact on B, and S is
strictly contractive on B, then in either case T has fixed point in B.
50
Shaini P and Neeta Singh
Proof. The map T = S + C, is clearly condensing. Since every condensing
map T of B into X is demicompact, the result follows from Corollary 2.8.
Theorem 2.10. Let X be a reflexive Banach space, B ⊂ X be an open ball
about the origin, and T be a lane mapping of B into X such that T satisfies
the interior condition. Then
(a) If (I − T )B is closed in X, then F has a fixed point in B.
(b) If X is uniformly convex, then T has a fixed point in B.
Proof. Since X is reflexive and T is a lane mapping of B into X, Lemma 1 in
[4] implies that T is 1- set contractive and since (I − T )B is closed by assumption the result follows from Theorem 2.2.
Indeed, if the hypothesis of (b) holds, then by lemma 3 in [4, III D] we see
that the set (I − T )B is closed, that is, the hypothesis of (a) holds.
Corollary 2.11. If X is a uniformly convex Banach space and T a semiconractive map of B into X which satisfies the interior condition, then F (T ) = φ.
The result follows from the fact that every semicontractive map is a lane
mapping [4].
3
Fixed point theorems for sum of two operators
Let X be a real Banach space, X ∗ its conjugate space, (x, w) the value of the
linear functional w ∈ X ∗ at the element x of X.
If J is a mapping of X into X ∗ , μ a continuous strictly increasing realvalued
function on R1 with μ(0) = 0, we say that J is a duality mapping of X into
X ∗ with gauge function μ if both of the following conditions are satisfied.
(a) For every x in X,
(x, Jx) = Jx.x
(b) For each x in X,
Jx = μ(x).
If X ∗ is strictly convex, there exists a duality mapping J for each gauge
function μ, J is uniquely determined by μ and J is continuous from the strong
topology of X to the weak topology X ∗ .
A mapping T : D → X is said to be accretive if for each pair x and y in D,
(T (x) − T (y), J(x − y)) ≥ 0,
i.e.,(T x) − T (y), w) ≥ 0 for all w in J(x − y),.
51
Fixed point theorems
Theorem 3.1. Suppose X is reflexive, B ⊂ X is an open ball about the origin,
X ∗ is strictly convex, and X has property (H)(i.e., if xn x in X and xn →
x as n → ∞, then xn → x in X.), and that there exists a weakly continuous
duality map J of X into X ∗ . If T = S + C is a map of B into X which is
such that S is nonexpansive on B, C is completely contunuous on B, and T
satisfies the interior condition, then F (T ) = φ.
Proof. In view of Theorem 2.10, as lane mapping includes mapping of the form
T = S + C, it is sufficient to show that (I − T )B is a closed set in X. Let {xn }
be a sequence in B such that gn ≡ (I − T )xn → g for some g in X. We have
to show that g ∈ (I − T )B. Since X is reflexive and {xn } ∈ B, without loss of
generality we may assume that xn x0 for some x0 ∈ B, with either x0 ∈ B
or x0 ∈ ∂B. Suppose first that x0 ∈ ∂B. Since X has property (H), Lemma
1 in[6]implies that xn → x0 , and consequently (I − T )(xn ) → (I − T )x0 = g;
i.e., g ∈ (I − T )B.
Now suppose that x0 ∈ B. Since C is completely continuous we have (I −
S)xn = (I − T + C)xn → g + C(x0 ) as n → ∞. Since S is nonexpansive on B
we have L = I − S accretive on B,[1, p.137].i.e., L(x) − L(y), J(x − y) ≥ 0
for all x, y in B. Hence it follows that for each n
L(xn ) − L(y), J(xn − y) ≥ 0 for all y in B
Since xn − y x0 − y, L(xn ) → g + C(x0 ) and J is weakly continuous, on
taking the limits the above inequality becomes
g + C(x0 ) − L(y), J(x0 − y) ≥ 0, for all y ∈ B
(2)
The inequality (2) implies that g + C(x0 ) − L(x0 ) = 0; if not it leads to a
contradiction.
Suppose g + C(x0 ) − L(x0 ) = 0, then there exists an z ∈ X such that
g + C(x0 ) − L(x0 ), J(z) < 0
(3)
Since x0 ∈ int(B), for t > 0 and sufficiently small, xt = x0 − tz lies in B. Now
replacing y in (2) by xt gives
g + C(x0 ) − L(xt ), J(tz) ≥ 0
(4)
Since J(tz) = βz (t)Jz with βz (t) > 0 for t > 0 (4) implies that
g + C(x0 ) − L(xt ), Jz ≥ 0.
(5)
adding -g + C(x0 ) − L(x0 ), Jz on either side of (5) and using (3) we get
L(x0 ) − L(xt ), Jz ≥ − g + C(x0 ) − L(x0 ), Jz > 0.
(6)
52
Shaini P and Neeta Singh
Since the right hand side of (6) is independent of t and L is continuous, taking
limit as t → 0 gives
0 = L(x0 ) − L(x0 ), Jz ≥ − g + C(x0 ) − L(x0 ), Jz > 0.
which is a contradiction.Therefore g + C(x0 ) − L(x0 ) = 0. i.e.,g + C(x0 ) =
(I − S)x0 , which implies that g = (I − T )x0 ∈ (I − T )B. Hence (I − T )B is
closed. So by Theorem 2.10, F (T ) = φ.
It was shown in [4] that if X is reflective Banach space, D a bounded open
convex subset of X, and T a lane mapping of D into X, then T is 1- set
contractive; moreover if X is also uniformly convex, then I − T is a demiclosed
mapping of D into X.
Theorem 3.2. Let X be a uniformly convex Banach space, D a bounded open
convex subset of X with 0 ∈ D, L a lane mapping of D into X, and C a
completely continuous mapping of D into X. If the mapping T = L + C : D →
X satisfies the interior condition, then T has a fixed point in D.
Proof. In view of the above mentioned result , since X is reflexive, C : D → X
is completely continuous, and L : D → X is a lane mapping, T = L+C is 1-set
contractive. Hence there exists a sequence [3, Corollary 1(b).]{xn } in D such
that xn −T (xn ) → 0. Since D is a bounded convex subset of a reflexive Banach
space X, and C is completely continuous we may assume that xn x0 in D
and C(xn ) → C(x0 ) in X. Hence xn − L(xn ) = xn − T (xn ) + C(xn ) → C(x0 ).
Moreover, since xn x0 and I − L is demiclosed whenever X is uniformly
convex[4], we get (I − L)x0 = C(x0 ), i.e., x0 = T (x0 ).
Theorem 3.3. Let X be a reflective Banach space, D a bounded open convex
subset of X, C a compact mapping of D into X, and S uniformly strictly
contractive on D relative to X (i.e., the map S : X → X has the property that
for each x in X there exists a number α(x) < 1 such that S(x) − S(y) ≤
α(x)x − y for each y ∈ D). If T = S + C : D → X satisfies the interior
conditiion, then T has a fixed point in D.
Proof. Since T = S+C is 1-set contractive and T satisfies the interior condition
by[3, Corollary1(b)] there exists a sequence {xn } in D such that xn −T (xn ) → 0
as n → ∞. Since {xn } is bounded and C : D → X is compact we may
assume that C(xn ) → p in X for some p in X. But then xn − S(xn ) =
xn − T (xn ) + C(xn ) → p or or xn − F (xn ) → 0 as n → ∞, where F : X → X
is defiined by F (x) = S(x) + p for x in X with F uniformly strictly contractive on D relative to X. Since X is reflexive and D is bounded closed convex
53
Fixed point theorems
subset of X, there exists a subsequence xnj of {xn } and an element x0 in
D such that xnj x0 and xnj − F (xnj ) → 0 as j → ∞. It was shown by
Kirk [2, Theorem 2]that under the above conditions on F, xnj is necessarily
a Cauchy sequence which converges strongly to x0 so that x0 − F (x0 ) = 0,
i.e., x0 − S(x0 ) = p. But then C(xnj ) → C(x0 ) = p as j → ∞. Hence
x0 − T (x0 ) = 0, i.e., x0 = T (x0 ).
Theorem 3.4. Let X be a uniformly convex Banach space, D a bounded open
convex subset of X with 0 ∈ D, S a nonexpansive map of D into X, and C a
completly continuos map of D into X. If the mapping T = S + C : D → X
satisfies the interior condition, then T has a fixed point in D.
Proof. Since C : D → X is completely continuous, it is continuous and compact[4], and therfore α(C(A)) = 0 for each A subset of D. Hence T = S + C
is a 1-set contractive map of D into X.Then by [3, Corollary 1(b)]there exists a sequence {xn } D such that xn − T (xn ) → 0. Now since X is reflexive,
without loss of generality we can assume that xn x0 in D and using complete continuity of C we see that C(xn ) → C(x0 ) as n → ∞, and therefore
xn − S(xn ) = xn − T (xn ) + C(xn ) → C(x0 ) as n → ∞. Since I − S is demiclosed[1, Theorem 10.4]it follows that (I − S)x0 = C(x0 ), i.e., x0 = T (x0 ).
Hence the theorem.
Notice that in the above theorem if we replace the completely continuous map
by a compact map, then T may not have a fixed point in D. For exmple If
X = 2 and D = B(0, 1) ⊂ 2 , then S : B → 2 and C : B → 2 given by
S(x) = (0, x1 , x2 , x3 , ...),
C(x) = (1 − x2 , 0, 0, 0, ...)
is nonexpansive and compact respectively; T = S + C is 1-set contractive and
satisfies the interior condition, but T has no fixed point in B.
References
[1] K.Goebel and W.A.Kirk, Topics in Metric fixed point theory, Cambridge
University Press 1990.
[2] W.A.Kirk, On nonlinear mappings of strongly semicontractive type,
J.Math.Anal.Appl.27(1969), 409-412.
[3] A.J.Melado and C.H.Morales, Fixed point theorems under the interior condition, Proc.Amer.Math.Soc.134, (No.2)501-504(2005).
54
Shaini P and Neeta Singh
[4] Nussbaum, R.D., The fixed point index and fixed point theorems for k-set
contractions, Univ.of Chicago, Ph.D. Thesis, 1969.
[5] W.V.Petryshyn,Construction of fixed points of demicompact mappings in
Hilbert space, J.Math.Anal.Appl.14, 276-284(1966).
[6] W.V.Petryshyn, Fixed point theorems involving P- compact, semicontractive, and accretive operators not defined on all of a Banach space.
J.Math.Anal.Appl.23, 336-354(1968).
[7] Sankatha Singh, Bruce Watson and Pramila Srivastava, Fixed point theory and best approximation: The KKM-Map principle, Kluwer Academic
Publishers, 1997.
[8] E.Zeidler, Nonlinear Functional Analysis and its Applications I: Fixed
Point Theorems, Springer, New york, 1986, p.499.
Received: July 22, 2007