# Fixed-Point Theorem for Isometric Self-Mappings

Authors: Joseph Frank Gordon
Journal: International Journal of Mathematics and Mathematical Sciences
 (2020)

---

## Abstract

In this paper, we derive a fixed-point theorem for self-mappings. That is, it is shown that every isometric self-mapping on a weakly compact convex subset of a strictly convex Banach space has a fixed point.


---

## Body

## 1. Introduction

LetX be a Banach space and C be a closed convex subset X. Let T:C⟶C be a self-mapping of C. Recall that T is said to be nonexpansive if(1)Tx−Ty≤x−y,for all x,y∈C. The fixed-point set of T is FixT:=x∈C:Tx=x. We say that the subset C of X is said to have an approximate fixed-point sequence for a nonexpansive self-map T if(2)limn⟶∞xn−Txn=0,for any sequence xnn≥1⊂C. When the closed convex subset C is bounded, then such a sequence always exists; indeed, by letting εn∈0,1, for all n≥1, be a null sequence and defining the maps Tnx=εnx0+1−εnTx where arbitrarily x0∈C, one can see that Tny−Tnx≤1−εnTy−Tx≤1−εny−x, implying that Tn is a contraction mapping with contraction constant 1−εn. By the Banach contraction mapping theorem, it follows that there exists a unique xn∈C such that Tnxn=xn, which implies that(3)xn=εnx0+1−εnTxn,from which we get xn−Txn=εnx0−Txn≤εnsupx,y∈Cx−y. Given that εn⟶0 and supx,y∈Cx−y<∞, it follows that xnn≥1 is an approximate fixed-point sequence.Another way of constructing an approximate fixed-point sequence is to require or assume thatFixT is nonempty. Now due to the assumption that FixT≠∅, the sequence xnn≥1⊂C is bounded (indeed, xn−p≤x0−p for all p∈FixT and x0 taken arbitrarily in C). Hence,(4)xn−Txn=εnx0−Txn⟶0,and xnn≥1 is an approximating fixed point for T.A mappingT:C⟶C of a set C in a Banach space X is called isometric if(5)Tx−Ty=x−y,for all x,y∈C. Note that an isometric mapping is just a nonexpansive mapping in which the inequality is always an equality. A well-known result of Brodskii and Milman [1] asserts that if C is a weakly compact convex subset of X and X has normal structure, then X has the fixed-point property for isometric mappings. In particular, any compact convex subset of X has the fixed-point property (see [2]).References [3–10] can be consulted for fixed-point problems on isometric mappings.Definition 1 (strictly convex Banach space).
A strictly convex Banach space is a Banach space such that wheneverx≠0≠y, then x+y=x+y if and only if x=λy for some λ>0.
An example of a strictly convex Banach space is a Hilbert space.Definition 2 (convex linear).
LetX be a Banach space and C⊆X be a closed convex subset of X. Then, the map T:C⟶X is said to be a convex linear if(6)Tax+1−ay=aTx+1−ay,for all x,y∈C and a∈0,1.
An example of a convex linear is a linear map.


## 2. Preliminaries

We introduce the following useful theorems that will be used in the proof of our main result.Theorem 1 (Mazur’s theorem).
Every nonempty convex subset of a Banach space is strongly closed if and only if it is weakly closed.Theorem 2 (Cantor’s intersection theorem).
LetX be a topological space. A decreasing nested sequence of nonempty compact closed subset of X has nonempty intersection. In other words, suppose that Ck is a sequence of nonempty compact closed subset of X satisfying C0⊃C1⊃,…,⊃Cn⊃Cn+1⊃,…, and it follows that(7)∩kCk≠∅.


## 3. Main Result

We give the proof of the main result of this paper, which is accomplished in Theorem3 below. The following lemma, corollary, and proposition shall aid us in arriving at the conclusion of the main result.Lemma 1.
LetX be a strictly convex Banach space and C⊆X be a closed convex subset and T:C⟶C be an isometric mapping. Then, T is convex linear on C. That is, Tax+1−ay=aTx+1−aTy for all x,y∈C and a∈0,1.Proof.
Letw=ax+1−ay and a∈0,1. Without loss of generality, assume x≠y. Then, w−x=1−ay−x implies that(8)w−x=1−ay−x.
Similarly,w−y=ax−y which also implies that(9)w−y=ax−y.
First, we show thatTx≠Ty≠Tw. To see this, we observe that if Tx=Tw, then from (9), we have(10)w−y=ax−y=aTx−Ty=aTw−Ty=aw−y,leading to the contradiction that a=1.
Similarly, ifTy=Tw, then from (8), we have(11)w−x=1−ay−x=1−aTy−Tx=1−aTw−Tx=1−aw−x,leading to the contradiction that a=0.
Now sinceT is an isometry, it follows that(12)Tx−Ty=Tx−Tw+Tw−Ty≤Tx−Tw+Tw−Ty=x−w+w−y=1−ax−y+ax−y=x−y=Tx−Ty,which implies that Tx−Tw+Tw−Ty=Tx−Ty. Since X is strictly convex and Tx≠Tw≠Ty implies that there exist λ>0 such that(13)Tx−Tw=λTw−Ty,1+λTw=Tx+λTy,Tw=11+λTx+λ1+λTy,we obtain(14)Tw=βTx+1−βTy,where β∈0,1 and β=1/1+λ. We finally show that β=a.
From (14), we have Tw−Tx=1−βTy−Tx which implies that(15)Tw−Tx=1−βTy−Tx.
Also,Tw−Ty=βTx−Ty implies that(16)Tw−Ty=βTx−Ty.
From (16), βTx−Ty=Tw−Ty=w−y=ax−y=aTx−Ty which implies that β=a. Hence, from (14), we have shown that Tax+1−ay=aTx+1−aTy which establishes the convex linearity of T.Corollary 1.
LetX be a strictly convex Banach space, C⊆X be a closed convex subset, and T:C⟶C be an isometric mapping. Then, the function f:C⟶ℝ,x⟶I−Tx is a continuous convex function.Proof.
From Lemma1, T is a convex linear. Hence, subtracting the term ax+1−ay from both sides of Tax+1−ay=aTx+1−aTy for all x,y∈C and a∈0,1, we have(17)I−Tax+1−ay=aI−Tx+1−aI−Ty.
We have the following evaluation:(18)fax+1−ay=1−Tax+1−ay=a1−Tx+1−a1−Ty≤a1−Tx+1−a1−Ty=afx+1−afy.
Thus,f is a convex function and continuous (because T is continuous).Proposition 1.
Letf:C⟶ℝ be continuous convex function on a weakly compact convex subset C of any Banach space X. Then, f attains its minimum on C. That is, there exist x¯∈C such that(19)fx¯=minfx:x∈C.Proof.
Letm=inffx:x∈C. We show that m>−∞ and that fx¯=m for some x¯∈C.
Suppose thatm=−∞, and for each n∈ℕ, define Cn=x∈C:fx≤−n. For each n∈ℕ, the set Cn is closed (and weakly closed by Theorem 1), convex, and nonempty (since m=−∞). Therefore, Cnn=1∞ forms a nested decreasing sequence of weakly compact nonempty sets. By Theorem 2, ∩n=1∞Cn≠∅. But this implies that there is some x¯∈C such that fx¯≤−n for all n∈ℕ, an impossibility. Consequently, m>−∞.
So definen∈ℕ a sequence of sets Cn′=x∈C:fx≤m+1/n for all n belonging to N (natural numbers). As before, Cn′n=1∞ is a nested sequence of weakly compact nonempty sets and so ∩n=1∞Cn′≠∅. If x¯∈∩n=1∞Cn′, then x¯∈C and fx¯=m as required.Theorem 3 (main result).
Every isometric self-mappingT:C⟶C on a weakly compact convex subset C of a strictly convex Banach Space X has a fixed point.Proof.
We know from Corollary1 that fx:=I−Tx is a continuous convex function. So, by Proposition 1, f attains its minimum on C, say x¯. By the approximate fixed point of T, it is always possible to choose a sequence xnn≥1⊂C such that fxn=xn−Txn≤1/n for all n≥1. Hence, I−Tx¯=infx∈CI−Tx=0 and so Tx¯=x¯.Example 1.
LetC=0,1 and T:0,1⟶0,1,Tx=1−x for all x∈0,1. Then, T is an isometry, and by Theorem 3, T has a fixed point.
As an application of Theorem3, it is desired to solve the linear problem(20)x−v=λTx−v,λ∈0,1,for some nonzero scalar λ and constant vector v, where T:C⟶C is linear and C is closed and convex. The following theorem gives a solution.Theorem 4.
LetC be a closed convex subset of a Banach space X and let T:C⟶C be linear and continuous. Then, equation (20) has a solution if the following holds:(1)
TBX,X=λ−1(2)
C is weakly compact(3)
X is strictly convexProof.
By defining the auxiliary mapping(21)T′:C⟶C,x⟼λTx+1−λv,it is clearly seen that T′ is well defined and invariant on C since C is convex and λ∈0,1. Since T′y−T′x=λTy−Tx, then by taking norms gives us(22)T′y−T′xX=λTy−TxX.
SinceT is bounded, then(23)T′y−T′xX≤λTBX,Xy−xX=λ⋅λ−1y−xX=y−xX.
SinceT′ is an isometry, then by Theorem 3, one can find a solution x¯=T′x¯=λTx¯+1−λv.



---

*Source: 3640539-2020-10-13.xml*