By ilikeafrica in arXiv — 19 Nov 2025

EVERY NONREFLEXIVE SUBSPACE OF L1[0, 1]

ℓ1 space에 대한 연구로, T는 ℓ1 공간의 일부 부분 집합 D에서 쌍대적 인 계약 매핑이 되는 것을 증명합니다. 이러한 결과를 이용하여 모든 비 반사성의 L1[0, 1] 하위 공간은 FPP( Fixed Point Property)를 만족하지 못하는 것으로 증명했습니다.

연구의 주요 결과는 다음과 같습니다.

* T는 ℓ1 space에 부분 집합 D에서 쌍대적 인 계약 매핑이 됩니다.
* T가 D내에서 유한 차원으로 변환 가능한 것을 보입니다.
* 이러한 결과를 이용하여 모든 비 반사성의 L1[0, 1] 하위 공간은 FPP( Fixed Point Property)를 만족하지 못하는 것으로 증명했습니다.

연구는 ℓ1 space에 대한 중요한 이론적 성과로, 기존 연구에서 밝혀지지 않은 새로운 결과를 제공합니다.

EVERY NONREFLEXIVE SUBSPACE OF L1[0, 1]

arXiv:math/9302208v1 [math.FA] 4 Feb 19930EVERY NONREFLEXIVE SUBSPACE OF L1[0, 1]FAILS THE FIXED POINT PROPERTYP.N. Dowling and C.J.

LennardMiami University and University of PittsburghABSTRACTThe main result of this paper is that every non-reﬂexive subspace Y of L 1[0, 1] failsthe ﬁxed point property for closed, bounded, convex subsets C of Y and nonexpansive(or contractive) mappings on C. Combined with a theorem of Maurey we get that forsubspaces Y of L 1[0, 1], Y is reﬂexive if and only if Y has the ﬁxed point property.For general Banach spaces the question as to whether reﬂexivity implies the ﬁxed pointproperty and the converse question are both still open.Keywords and phrases. nonexpansive mapping; closed, bounded, convex set; ﬁxed point property; nonre-ﬂexive subspaces of L 1[0, 1]; normal structure

1EVERY NONREFLEXIVE SUBSPACE OF L1[0, 1]FAILS THE FIXED POINT PROPERTYP.N. Dowling and C.J.

LennardMiami University and University of PittsburghIntroductionWe ﬁrst give a new example of a renorming of ℓ1 such that the positive part of theusual closed unit ball supports a ﬁxed point free nonexpansive map. The mapping is thatof Lim [L], whose renorming of ℓ1 provided the ﬁrst such example.

This example leadsto a new ﬁxed point free nonexpansive map with respect to the usual norm. A variationon this theme produces a ﬁxed point free contraction on a closed, bounded, convex setin ℓ1 with its usual norm.

Using this example we show that every nonreﬂexive subspaceof L1[0, 1] fails the ﬁxed point property for nonexpansive mappings, proving the converseof a theorem of Maurey [M]. In particular, the Hardy space H1 on the unit circle mustfail to have the ﬁxed point property; which contrasts with Maurey’s result in [M] thatH1 has the weak (and weak-star) ﬁxed point property.

We next provide an example ofa reﬂexive subspace of L1[0, 1] that fails normal structure; which means that Maurey’sﬁrst above-mentioned theorem cannot be deduced from Kirk’s theorem. In passing, weobserve that a standard example shows that ℓ1 with its usual norm fails the weak-starﬁxed point property and the weak-star Kadec-Klee property with respect to the predualc of all convergent scalar sequences.

2We thank Brailey Sims, Mark Smith and Barry Turett for helpful discussions. Thesecond author acknowledges the support of a University of Pittsburgh F.A.S.

ReseachGrant.PreliminariesRecall that ℓ1 is the Banach space of all scalar sequences x = (xn)∞n=1 for which∥x∥1 := P∞n=1 |xn| < ∞. For each n ∈N, we let en denote the n-th vector in the usualunit vector basis of ℓ1.

L1[0, 1] is the usual space of Lebesgue integrable functions (wherealmost everywhere equal functions are identiﬁed), with its usual norm.Let (X, ∥·∥X) be a Banach space. We say that (X, ∥·∥X) has the ﬁxed point propertyif given any non-empty, closed, bounded and convex subset C of X, every nonexpansivemapping T : C →C has a ﬁxed point.

Here T is nonexpansive if ∥Tx −Ty∥X ≤∥x −y∥Xfor all x, y ∈C. Moreover, T is a contraction if ∥Tx−Ty∥X < ∥x−y∥X for every x, y ∈Cwith x ̸= y.

Also, (X, ∥· ∥X) has normal structure if every closed, bounded, convex subsetC of X containing at least two points has radius less than its diameter. For such a C andx ∈C, rad(x; C) equals supy∈C ∥x−y∥X, the radius of C is rad(C) := infx∈C rad(x; C) andthe diameter of C is diam(C) := supx∈C rad(x; C).

Kirk [Ki] showed that in a reﬂexiveBanach space, normal structure implies the ﬁxed point property. If X is a dual space,isometrically isomorphic to Y ∗for some Banach space Y , then (X, ∥· ∥X) has the weak-star ﬁxed point property (with respect to Y ) if given a non-empty, weak-star compact,convex set C in X, every nonexpansive mapping on C has a ﬁxed point.

Further, X hasthe weak-star Kadec-Klee property (with respect to Y ) if weak-star and norm convergenceof sequences coincides in the unit sphere of X. The weak Kadec-Klee property on a Banachspace X is deﬁned similarly.1.

A new exampleLet C := {x ∈ℓ+1 : P∞n=1 xn ≤1}. C is a closed, bounded and convex set in ℓ1.

3Deﬁne T : C →C byT : x 7−→ 1 −∞Xn=1xn, x1, x2, . .

. , xk, .

. .

!.T is clearly ﬁxed point free and P∞n=1(Tx)n = 1 for all x ∈C.Lim [L] renormed ℓ1 with the equivalent norm x 7−→∥x+∥1∨∥x−∥1. T is nonexpansivewith respect to Lim’s norm.Let us renorm ℓ1 in another way.

Deﬁne ∥· ∥on ℓ1 by setting∥x∥:=∞Xn=1xn +∞Xn=1|xn| , ∀x ∈ℓ1.Then for all x and y in C,Tx −Ty = ∞Xn=1(yn −xn), x1 −y1, x2 −y2, . .

. , xk −yk, .

. .

! ; and so∥Tx −Ty∥=∞Xn=1(yn −xn) +∞Xk=1(xk −yk) +∞Xn=1(yn −xn) +∞Xk=1|xk −yk| = ∥x −y∥.Thus, in summary, T is a ﬁxed point free isometry on the closed, bounded and convex setC, for the equivalent norm ∥· ∥on ℓ1.2.

A new nonexpansive map for the usual norm on ℓ1Deﬁne the linear mapping Q from (ℓ1, ∥· ∥) into (ℓ1, ∥· ∥)1 byx 7−→ ∞Xn=1xn, x1, x2, . .

. , xk, .

. .

!.Note that ∥Qx∥1 = ∥x∥for all x ∈ℓ1 and the range of Q is V := {y ∈ℓ1 : y1 = P∞n=2 yn}.Moreover, Q−1 : y 7−→(y2, y3, . .

. , yk, .

. .

).Now, let K := Q(C) = {y ∈ℓ+1 : y1 = P∞n=2 yn ≤1} and deﬁne the mappingS : K →K by S := QTQ−1. It is easy to check that for all y ∈K,Sy = (1, 1 −y1, y2, y3, .

. .

, yk, . .

. ) ;and that S is a ﬁxed point free isometry on the closed, bounded, convex set K in ℓ1 withrespect to the usual norm.

43. A variation on this themeFix a sequence (εk)∞k=1 in [0, 1) with P∞k=1 εk < ∞.

Fix δ ∈(0, 1]. Deﬁne T : ℓ1 →ℓ1byT : x 7−→ δ 1 −∞Xn=1xn!, (1 −ε1)x1, (1 −ε2)x2, .

. .

, (1 −εk)xk, . .

. !.We claim that T is ﬁxed point free.

Indeed, suppose not. Then there is an x ∈ℓ1 withTx = x. Thusδ 1 −∞Xn=1xn!= x1and(1 −εk)xk = xk+1 , ∀k ∈N .Consequently, for each k ∈N, xk+1 = x1Qkj=1(1−εj).

There are two cases. If x1 = 0 thenxn = 0 for all n ∈N.

So, from the ﬁrst equation above, δ = 0; a contradiction. If x1 ̸= 0,then Q∞j=1(1 −εj) = limk→∞xk+1 = 0.

This is precisely equivalent to P∞j=1 εj = ∞; acontradiction.As before, we deﬁne C := {x ∈ℓ+1 : P∞n=1 xn ≤1}. Then T maps C into C. Indeed,each (Tx)n ≥0 and P∞n=1(Tx)n = δ (1 −P∞n=1 xn) + P∞k=1(1 −εk)xk ≤1.We verify that T is non-expansive with respect to the norm ∥· ∥.

Fix x, y ∈C.Tx −Ty =δ∞Xn=1(yn −xn), (1 −ε1)(x1 −y1), (1 −ε2)(x2 −y2), . .

. ,(1 −εk)(xk −yk), .

. .

5In summary, C is a closed, bounded, convex set in ℓ1 and T is a ﬁxed point freenonexpansive map on C for the equivalent norm ∥· ∥on ℓ1. Further, as in the previoussection, S := QTQ−1 is a ﬁxed point free nonexpansive map on the closed, bounded,convex set K := Q(C), for the usual norm ∥· ∥1 on ℓ1.Brieﬂy consider what happens when δ < 1 and 0 < εk ≤1 −δ for all k ∈N.

Then Tis not an isometry because 0 ∈C and ∥Tx −T0∥< ∥x∥for all x ∈C. Similarly, S is notan isometry.4.

Fixed point free contractions in ℓ1Fix δ ∈(0, 1) and a sequence (εk)∞k=1 in (0, 1 −δ) with P∞k=1 εk < ∞. Again deﬁneT : ℓ1 →ℓ1 byT : x 7−→ δ 1 −∞Xn=1xn!, (1 −ε1)x1, (1 −ε2)x2, .

. .

, (1 −εk)xk, . .

. !.From the previous section, we know that T is ﬁxed point free.Fix α ∈R and q > 0.

DeﬁneD :=x ∈ℓ1 :∞Xk=1(1 −δ −εk)xk = αand∞Xj=1|xj| ≤q.Note that given α we may always choose q so large that D is non-empty. Moreover, D isa norm closed, convex and bounded set in ℓ1.Let us specialize our discussion to the following case.

It is suﬃcient for our purposes,although other choices of parameters will also suﬃce. Let δ := 1/2 and ε1 := 1/4.

Letα = 1/2 and ﬁx q > 0 so large that D ̸= ∅. We introduce the auxilliary parameter η andgiven our previous choices and what we will require below of η, it turns out that η := 3/4is the best choice.

Finally, we deﬁne the sequence (εk)∞k=2 by the equations12 −εn+1(1 −εn) −18 = η12 −εn, ∀n ∈N .

6It is easy to inductively show that 0 < εn < 1/2 = 1 −δ for each n, and that theabove equations for εn, n ≥2 are equivalent to εn+1 =εn4(1 −εn), for all n ∈N. Thus0 < εn+1 < εn/2 for each n, and so P∞n=1 εn < ∞.

Consequently, δ and (εk)∞k=1 satisfythe hypotheses of the beginning of this section. In particular, the mapping T : ℓ1 →ℓ1 isﬁxed point free.4.1 Proposition.

T maps D into D.Proof: Fix x ∈D. Then∞Xk=112 −εkxk = 12and∞Xj=1|xj| ≤q .Tx = 12 1 −∞Xn=1xn!, (1 −ε1)x1, (1 −ε2)x2, .

. .

, (1 −εk)xk, . .

. !.Thus, we see that∞Xn=112 −εn(Tx)n =12 −ε1 12 1 −∞Xn=1xn!+∞Xn=112 −εn+1(1 −εn)xn= 18 −18∞Xn=1xn +∞Xn=112 −εn+1(1 −εn)xn= 18 +∞Xn=112 −εn+1(1 −εn) −18xn= 18 +∞Xn=13412 −εnxn = 18 + 3412= 12 ;

7and so we have P∞n=1 12 −εn(Tx)n = 12. Further,∞Xn=1|(Tx)n| =12 1 −∞Xn=1xn!

+∞Xn=1(1 −εn)|xn|=12 −12∞Xn=1xn +∞Xn=1(1 −εn)|xn|=∞Xn=1−εnxn +∞Xn=1(1 −εn)|xn|≤∞Xn=1εn|xn| +∞Xn=1(1 −εn)|xn| ≤∞Xn=1|xn| ≤q .So, P∞n=1 |(Tx)n| ≤q. Thus Tx ∈D for all x ∈D.4.2 Proposition.

T is a contraction on D with respect to the norm ∥· ∥on ℓ1; andmoreover, for all x, y ∈D,∥Tx −Ty∥= 12∞Xn=1(yn −xn) +∞Xn=1(1 −εn)|xn −yn| .Proof: Fix x, y ∈D. ThenTx −Ty =12∞Xn=1(yn −xn), (1 −ε1)(x1 −y1), (1 −ε2)(x2 −y2), .

. .

,(1 −εk)(xk −yk), . .

8The following key result is now easy to verify. We omit the simple calculations in-volved.4.3 Theorem.

Let D and T be as in Propositions 4.1 and 4.2. Then S := QTQ−1 is aﬁxed point free contraction on the non-empty, closed, bounded, convex set L := Q(D), forthe usual norm ∥· ∥1 on ℓ1.

Moreover,L =(y ∈ℓ1 : y1 =∞Xn=2yn ,∞Xk=112 −εkyk+1 = 12 and∞Xn=2|yn| ≤q); whileSy =1, 12(1 −y1), (1 −ε1)y2, (1 −ε2)y3, . .

. , (1 −εk)yk+1, .

. .∀y ∈L ;and∥Sy −Sz∥1 = 12|z1 −y1| +∞Xn=1(1 −εn)|yn+1 −zn+1| ∀y, z ∈L .5.

All non-reﬂexive subspaces of L1[0, 1] fail the FPP5.1 Proposition. Let (εn)∞n=1 be precisely the scalar sequence from results 4.1, 4.2 and4.3 above.

Let (X, ∥· ∥X) be a Banach space and Y be a subspace of X such that thereexists a sequence (vn)∞n=1 in Y , a sequence (un)∞n=1 in X and a sequence (γn)∞n=1 in (0, ∞)with the following properties. (i)NXn=1tnunX=NXn=1|tn| , ∀scalar sequences t1, .

. .

tN ;(i.e. (un)∞n=1 is an isometric copy in X of the usual unit vector basis of ℓ1).

(ii)∥un −vn∥X < γn , ∀n ∈N ;where each γn is chosen so small that (vn)∞n=1 spans an isomorphic copy of ℓ1 in Y . (iii)γ1, γ2 < 1212andγn+1, γn+2 < 12εn ∀n ∈N .

9Then (Y, ∥· ∥X) fails the ﬁxed point property for closed, bounded, convex sets in Y andnonexpansive (or contractive) mappings on them.Proof: Let q be as in 4.1, 4.2 and 4.3 above. We will apply Theorem 4.3.

DeﬁneM := ∞Xn=1ynvn : y ∈ℓ1 ,y1 =∞Xn=2yn ,∞Xk=112 −εkyk+1 = 12 and∞Xn=2|yn| ≤q.M is a non-empty, closed, bounded, convex set in Y , by hypothesis (ii) and Theorem 4.3.Deﬁne R : M →M byR ∞Xn=1ynvn! :=∞Xn=1(Sy)nvn ,where S : L →L is as in Theorem 4.3.

Clearly, R maps M into M and R is ﬁxed pointfree.Moreover, for all σ = Pn ynvn and τ = Pn znvn in M, Theorem 4.3 gives us that∥Rσ −Rτ∥X ≤∞Xn=1((Sy)n −(Sz)n)unX+∞Xn=1((Sy)n −(Sz)n)(vn −un)X≤∞Xn=1|(Sy)n −(Sz)n| +∞Xn=1|(Sy)n −(Sz)n|γn≤∞Xn=1|(Sy)n −(Sz)n| +∞Xm=1|ym −zm|γm+1≤12|z1 −y1| +∞Xn=1(1 −εn)|yn+1 −zn+1|+ |y1 −z1|14 +∞Xm=2|ym −zm|12εm−1= 34|y1 −z1| +∞Xn=11 −εn2|yn+1 −zn+1| .

Every non-reﬂexive subspace Y of L1[0, 1], with its usual norm, failsthe ﬁxed point property for closed, bounded, convex sets in Y and nonexpansive (orcontractive) mappings on them.Proof: By the proof of the Kadec-Pe lczynski theorem [K-P] (or see [D], Chapter VII), forX := L1[0, 1] with its usual norm, sequences (vn)∞n=1 in Y , (un)∞n=1 in X and (γn)∞n=1 in(0, ∞) exist that satisfy all the hypotheses of Proposition 5.1 above.Combining 5.2 with Maurey’s theorem [M] allows us to state the fact below.5.3 Theorem. Let Y be a subspace of L1[0, 1] with its usual norm.

Then the followingare equivalent. (i) Y is reﬂexive.

(ii) Y has the ﬁxed point property.6. There is a reﬂexive subspace of L1[0, 1] failing NS

11Let us deﬁne the sequence of Rademacher functions (rn)∞n=1 on the real line R. r0is the characteristic function of [0, 1], r1 := χ(0,1/2) −χ(1/2,1) and rn(t) := rn−1(2t) +rn−1(2t −1), for all t ∈R and for each n ≥2. Henceforth we will restrict the domain ofeach rn to [0, 1].We deﬁne the subspace X of L1[0, 1] to be the closed linear span of the sequence(rn)∞n=0; which is isomorphic to ℓ2 by Khinchine’s inequalities (see, for example, Diestel[D]).

In particular, X is reﬂexive. We will denote the usual norm on L1[0, 1] by ∥· ∥1.Let C be the closed convex hull in X of the sequence (xn)∞n=1, where each xn := rn+r0.

(The translation factor of r0 is introduced to show that we may arrange for C to consistof all non-negative functions, and to link in with a remark below concerning the weakKadec-Klee property). The set C is closed, bounded and convex.We now show that x 7−→rad(x; C) is constant on C; and so X with the norm ∥· ∥1fails normal structure.Indeed, ﬁx x, y ∈C.We may suppose, without loss of gen-erality that x, y ∈D := the convex hull of (xn)∞n=1.So there exist non-negative realnumbers α1, .

. .

, αN and β1, . .

. , βN, with PNn=1 αn = 1 and PNn=1 βn = 1, such thatx = PNn=1 αnxn and y = PNn=1 βnxn.

Then∥x −y∥1 =NXn,m=1αnβm(rn −rm)1≤max1≤n,m≤N ∥rn −rm∥1 = 1 .On the other hand, xN+1 ∈C and it is straightforward to calculate that∥x −xN+1∥1 =NXn=1αnrn −rN+11=12N+1Xε∈{−1,1}N+1|ε1α1 + · · · + εNαN + εN+1|=12N+1 2Xε∈{−1,1}N(1 + ε1α1 + · · · + εNαN)=12N+1 2Xε∈{−1,1}N(1) = 1 .

12We remark that the above Banach space X is a rather natural example of a reﬂexivespace that has the ﬁxed point property but fails normal structure. Karlovitz [Ka2] showedthat another equivalent renorming of ℓ2 (due to R.C.

James) has the ﬁxed point property,while it fails normal structure.Also note that (xn)∞n=1 provides an example which shows that X fails the weak Kadec-Klee property.7. An old example revisitedFor each x ∈ℓ1, we let x0 = P∞n=1 xn.

One of the preduals of ℓ1 is c, the spaceof all convergent scalar sequences λ = (λn)∞n=1 with the supremum norm ∥λ∥∞. We letλ0 = limn→∞λn.

The duality is given by (see, for example, Banach [B]) :⟨x, λ⟩=∞Xn=1xn+1(λn −λ0) + x0λ0 =∞Xn=0xn+1λn.Consider the following well-known example. Let W equal {x ∈ℓ+1 : P∞n=1 xn = 1}and F : W →W be given by Fx := (0, x1, x2, .

. .

, xk, . .

. ).

Then F is a ﬁxed point freeisometry on the closed, bounded, convex set W in ℓ1. Let λn := 1 for all n ∈N.

Thenλ ∈c and ⟨x, λ⟩= P∞n=1 xn. So, W is weak-star compact with respect to the predual cof ℓ1.

Thus ℓ1 with its usual norm fails the weak-star ﬁxed point property with respect toits predual c.For comparison, we remark that using the map T and set C described in section1, Lim [L] showed that ℓ1 with Lim’s norm fails the weak-star ﬁxed point property fornonexpansive maps with respect to the isometric predual c0 (suitably renormed). On theother hand, Karlovitz [Ka1] demonstrated that ℓ1 with its usual norm has the weak-starﬁxed point property with respect to c0.

Moreover, Soardi [So] showed that every Banachspace Y that is isomorphic to ℓ1, with Banach-Mazur distance from ℓ1 less than 2, has theweak-star ﬁxed point property.

13In a similar manner to that in section 6, one can show that W is diametral, i.e. x 7−→rad(x; C) is constant on C. Moreover, the extreme points of W provide us with an exampleshowing that ℓ1 fails the weak-star Kadec-Klee property with respect to the predual c. Letx(k) := ek+1 for each integer k ≥0.

Fix λ ∈c. For all n ∈N,⟨x(n), λ⟩=∞Xm=0x(n)m+1λm = λn −→nλ0 = ⟨x(0), λ⟩.So x(n) −→nx(0) weak-star with respect to c, each ∥x(k)∥1 = 1, yet ∥x(n) −x(0)∥1 = 2 forall n ≥1.We contrast the above with the fact that ℓ1 (with its usual norm) has the weak-star uniform Kadec-Klee property with respect to the predual c0 (see, for example, [D-S]and [Si]); and therefore has both the weak-star ﬁxed point property and the weak-starKadec-Klee property with respect to c0.Finally, we remark that Michael Smyth [Sm] has recently shown, by a variation of thetechniques in section 3 and 4 above, that ℓ1 fails the weak-star ﬁxed point property withrespect to its predual c with a contractive map.

By modifying our set D in section 4, heis also able to slightly simplify our mapping T.

14References[B] S. Banach, Th´eorie des Op´erations Lin´eaires, Chelsea Publishing Co., New York1978. [D] J. Diestel, Sequences and Series in Banach Spaces, Springer-Verlag New York Inc.1984.

[D-S] D. van Dulst and B. Sims, Fixed points of nonexpansive mappings and Cheby-shev centers in Banach spaces with norms of type (KK), in: Banach Space Theory and itsApplications, Proc. Bucharest 1981, Lecture Notes in Math.

991, Springer-Verlag 1983,35-43. [K-P] M.I.

Kadec and A. Pe lczynski, Bases, lacunary sequences and complementedsubspaces in the spaces Lp, Studia Math. 21 (1962), 161-176.

[Ka1] L.A. Karlovitz, On nonexpansive mappings, Proc. Amer.

Math. Soc.

55(2)(1976), 321-325. [Ka2] L.A. Karlovitz, Existence of ﬁxed points of nonexpansive mappings in a spacewithout normal structure, Paciﬁc J.

Math. 66(1) (1976), 153-159.

[Ki] W.A. Kirk, A ﬁxed point theorem for mappings which do not increase distances,Amer.

Math. Monthly 72 (1965), 1004-1006.

[L] T.C. Lim, Asymptotic centers and nonexpansive mappings in conjugate Banachspaces, Paciﬁc J.

Math. 90(1) (1980), 135-143.

[M] B. Maurey, Points ﬁxes des contractions de certains faiblement compacts deL1, Seminaire d’Analyse Fonctionelle, Expos´e no. VIII, ´Ecole Polytechnique, Centre deMath´ematiques (1980-1981).

[Si] B. Sims, The existence question for ﬁxed points of nonexpansive maps, LectureNotes, Kent State Univ. 1986.

[Sm] M. Smyth, preprint, University of Newcastle, Australia.

15[So] P.M. Soardi, Schauder bases and ﬁxed points of nonexpansive mappings, PaciﬁcJ. Math.

101(1) (1982), 193-198.P.N. DowlingMathematics and Statistics DepartmentMiami UniversityOxford, Ohio 45056U.S.A.C.J.

LennardMathematics and Statistics DepartmentUniversity of PittsburghPittsburgh, Pennsylvania 15260U.S.A.

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EVERY NONREFLEXIVE SUBSPACE OF L1[0, 1]

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