Chapter 1

EXAMPLES OF FIXED POINT FREE MAPPINGS

Brailey Sims

Mathematics, School of Mathematical and Physical Sciences The University of Newcastle NSW, 2308, Australia

bsims@maths.newcastle.edu.au

1. Introduction

In this short chapter we collect together examples of fixed point free nonexpansive mappings defined on nonempty closed convex subsets of a Banach space X, that are bounded, weak compact, or, in the case when X = E^{∗}, weak^{∗} compact; that is, σ(X, E) compact. These example help delineate the class of spaces enjoying the fpp, the w-fpp, or the w^{∗}-fpp. Unfortunately, known examples are rather sparse. With the exception of Alspach’s example (or modifications of it, see section ???), the mappings concerned are adaptations of affine maps (indeed, modified shifts), or minor variants thereof. This dearth of examples is a major impediment to a fuller understanding of metric fixed point theory and the discovery of informative new examples would be an important step forward.

In the next section we will document examples that demonstrate failure of the fpp. Subsequent sections will deal with more specialized examples that demonstrate; failures of the w^{∗}-fpp in duals of certain Banach spaces and finally Alspach’s famous demonstration that the w-fpp fails in L_{1}[0, 1].

2. Examples on closed bounded convex sets

Example 2.1 c_{0} fails the fpp.

Let C = B_{c_{0}}^{+} := {(x_{n}) ∈ c_{0} : 0 ≤ x_{n} ≤ 1, all n} and define T by

T (x_{n}) := (1 − x_{1}, x_{1}, x_{2}, …).

Then for any x, y ∈ c_{0} we have ∥T x − T y∥_{∞} = ∥x − y∥_{∞}. So, T is nonexpansive, indeed a metric isometry, and T maps C into C. On the other hand the only possible fixed point for T is (\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, …) ∉ c_{0}.

To put the next example into context it is important to recall that c_{0} enjoys the w-fpp.

Example 2.2 c_{0} fails the fpp for a contraction and on a set which is compact in a topology only slightly coarser than the weak topology.

Let a = (a(n)) ∈ l_{∞} be a strictly decreasing sequence of ‘weights’satisfying α ≤ a(n) ≤ β, for some 0 < α ≤ β < ∞, and take

a_{0} := 0

a_{n} := (a(1), …, a(n), 0, 0, …) for n = 1, 2, 3, ….

Now consider the set

K := \overline{co}\{a_{n}\}_{n=0}^{∞}

in c_{0} consisting of vectors of the form

∑_{n=0}^{∞} λ_{n}a_{n} = (a(1)(1 − λ_{0}), a(2)(1 − (λ_{0} + λ_{1})), a(3)(1 − (λ_{0} + λ_{1} + λ_{2})), …),

where λ_{n} ≥ 0 and ∑_{n=0}^{∞} λ_{n} = 1, that admits fixed point free affine contraction T_{a} defined by T_{a}(a(1)(1−λ_{0}), a(2)(1−(λ_{0}+λ_{1})), …) := (a(1), a(2)(1−λ_{0}), a(3)(1−(λ_{0}+λ_{1})), …). The proof of these properties of T follows. Lemma 2.3 (i) T_{a} maps K into K, (ii) T_{a} is a contraction, (iii) T_{a} is fixed point free in K. Proof. To establish (i) it suffices to note that for λ_{n} ≥ 0 and ∑_{n=0}^{∞} λ_{n} = 1, we have

T ( ∑_{n=0}^{∞} λ_{n}a_{n}) = (a(1), a(2)(1 − λ_{0}), a(3)(1 − (λ_{0} + λ_{1})), …)

= ∑_{n=1}^{∞} λ_{n−1}a_{n} ∈ K.

(ii) follows, since for x = (a(1)(1 − λ_{0}), a(2)(1 − (λ_{0} + λ_{1})), dots) and y = (a(1)(1 − μ_{0}), a(2)(1 − (μ_{0} + μ_{1})), …) we have that

∥x − y∥ = sup{a(1)|μ_{0} − λ_{0}|, a(2)|μ_{0} − λ + μ_{1} − λ_{1}|, … }

and

∥T_{a}x − T_{a}y∥ = sup{a(2)|μ_{0} − λ_{0}|, a(3)|μ_{0} − λ + μ_{1} − λ_{1}|, … }.

Since a = (a(n)) is a strictly decreasing sequence T_{a} is a contraction. Notice that if instead a = (a(n)) were chosen to be merely decreasing then T_{a} would still be a nonexpansive map. Finally, if x = (a(1)(1 − λ_{0}), a(2)(1 − (λ_{0} + λ_{1})), …) were such that x = T x = (a(1), a(2)(1 − λ_{0}), a(3)(1 − (λ_{0} + λ_{1})), …) then we would have λ_{0} = 0, λ_{1} = 0, … contradicting the requirement that ∑_{n=0}^{∞} λ_{n} = 1, and so we have (iii). ■

K is convex, since for x = (a(1)(1 − λ_{0}), a(2)(1 − (λ_{0} + λ_{1})), …), y = (a(1)(1 − μ_{0}), a(2)(1 − (μ_{0} + μ_{1})), …) and α ∈ [0, 1]

αx + (1 − α)y

= α ∑_{n=0}^{∞} λ_{n}a_{n} + (1 − α) ∑_{n=0}^{∞} μ_{n}a_{n}

= ∑_{n=0}^{∞}(αλ_{n} + (1 − α)μ_{n})a_{n}

∈ K.

In addition we would like the compactness of K but the sequence {a_{n}} has no norm converent subsequence and hence K, although norm closed, is not compact. Similarily, {a_{n}} has no weak convergent subsequence in c_{0} and so fails to be compact in the weak topology. However we can ‘improve’ the weak topology on c_{0} so that K is compact. To define this topology, E_{a}, let

E_{a} := ker(a) = {(y(n)) ∈ l_{1} : ∑ y(n)a(n) = 0}.

E_{a} is closed with respect to the 1-norm, since if e_{k} → e in the 1-norm and ∑_{n} e_{k}(n)a(n) = 0 then ∑_{n} e(n)a(n) = 0, but not with respect to the w^{∗}-topology. By [Guerre-Delabriére, 1992, Lemma 1.1.11] E_{a} is a norming subspace for c_{0}, that is there exists a constant C > 0 such that for all x ∈ c_{0},

sup y ∈ E(x, y) > C∥x∥,

where (x, y) = ∑ x(k)y(k), as usual. Alernatively this is easily verified by direct calculation for any x ∈ c_{0}, where |x(l)| = ∥x∥_{∞}, by defining

y_{n} := \frac{α}{α + β} (\underbrace{0, …, 0, 1}_{l}, \underbrace{0, …, 0, \frac{a(l)}{∥a∥}_{n}, 0, …)} .

Clearly ∥y_{n}∥_{1} ≤ 1 and

|(x, y)| = \frac{α}{α + β} |x(l) − x(n + l)\frac{a(l)}{∥a∥}| → \frac{α}{α + β} |\frac{α}{α + β} x(l)| = ∥x∥_{∞}, as n → ∞.

So

\frac{α}{α + β} ∥x∥_{∞} ≤ sup\{(x, y) : y ∈ E_{a}, ∥y∥_{1} ≤ 1\},

but also when ∥y∥ = 1

∑ x(k)y(k) ≤ ∥x∥_{∞}y(k) ≤ ∥x∥_{∞},

and hence

\frac{α}{α + β} ∥x∥_{∞} ≤ sup\{(x, y) : y ∈ E_{a}, ∥y∥_{1} ≤ 1\} ≤ ∥x∥_{∞}.

E_{a} separates points of c_{0} since for x, z ∈ c_{0}, x ≠ z there exists y ∈ E such that

y(x − z) ≥ \frac{α}{α + β} ∥x − z∥_{∞} 0.

Therefore, by [Jameson, 1974, 27.3], the set E_{a} is dense in c_{0}^{∗} = l_{1} with respect to the w^{∗}topology. We consider c_{0} equipped with the topology E_{a} := σ(c_{0}, E_{a}). That is, E is the smallest locally convex linear topology on c_{0} making continuous all the elements of E_{a}(as linear functionals on c_{0}).

The topology E_{a} may be seen as ’slightly’ coarser than the weak topology on c_{0}, being induced by a norming codimension one subspace of c_{0}^{∗}. It displays some unusual, although not too pathological, properties. For example, the following propositions can be proved by more or less standard methods of locally convex space theory.

Proposition 2.4 The topology E_{a} consists of ∅, c_{0} with the weak topology, all finite intersections of the sets

{(x(n)) ∈ c_{0} : a < ∑ x(n)y(n) < b, ∑ a(n)y(n) = 0}

and all arbitrary unions of these finite intersections.

Proposition 2.5 A sequence (x_{n}) in c_{0} is E_{a} convergent to x ∈ c_{0} if and only if for every y ∈ E_{a},

(x_{n}, y) → (x, y).

As a consequence of the above result a_{n} ⟶^{E_{a}} a_{0} and does not have weakly null subsequences. This will be used to show K is E_{a}-compact. However, first we need the following lemma.

Lemma 2.6 K is E_{a}-closed.

Proof. For n = 1, 2, … let

x_{n} = ∑_{k=0}^{∞} λ_{k}^{(n)}d_{k} = (a(1)(1 − λ_{0}^{(n)}), a(2)(1 − (λ_{0}^{(n)} + λ_{1}^{(n)})), …),

where λ_{k}^{(n)} ≥ 0 and ∑_{k=0}^{∞} λ_{k}^{(n)} = 1, be such that x_{n} ⟶^{E_{a}} x = (μ_{1}a(1), μ_{2}a(2), …).

Choosing f := (\frac{1}{a(1)}, \frac{−1}{a(2)}, 0, 0, …) ∈ E_{a} we have

f(x_{n} − x) = (1 − λ_{0}^{(n)} − μ_{1}) − (1 − λ_{0}^{(n)} − λ_{1}^{(n)} − μ_{2}) → 0.

That is

λ_{1}^{(n)} → μ_{1} − μ_{2}.

Similarily, choosing f := (0, \frac{1}{a(2)}, \frac{−1}{a(3)}, 0, 0, …) we obtain

λ_{2}^{(n)} → μ_{2} − μ_{3},

and in general

λ_{k}^{(n)} → μ_{k} − μ_{k+1}.

Thus, for k = 1, 2, …

λ_{k} := μ_{k} − μ_{k+1} = lim_{n} λ_{k}^{n} ≥ 0

and

x = (μ_{1}a(1), (μ_{1} − λ_{1})a(2), (μ_{1} − λ_{1} − λ_{2})a(3), …) ∈ c_{0}.

So we must have

μ_{1} = ∑_{k=1}^{∞} λ_{k} ≥ 0,

and then, provided μ_{1} ≤ 1,

x = ∑_{k=1}^{∞} λ_{k}d_{k} ∈ K.

But, given ϵ > 0 there exists N so that

μ_{1} = ∑_{k=1}^{∞} λ_{k} < ∑_{k=1}^{N} λ_{k} + ϵ/2,

and there exists n for which

|λ_{k} − λ_{k}^{(n)}| ≤ ϵ/2N, for k = 1, 2, …, N.

Thus,

μ_{1} ≤ ∑_{k=1}^{N} λ_{k}^{(n)} + ϵ ≤ 1 + ϵ, as ∑_{k=0}^{∞} λ_{k}^{(n)} = 1,

and so μ_{1} ≤ 1, as required. ■

Since d_{n} ⟶^{E} d_{0}, we have that {d_{n}}_{n=0}^{∞} is E-compact. The E-compactness of K then follows from Lemma 2.6, the definition of E, and the following general result.

Lemma 2.7 Let X be a separable Banach space and let M be a closed norming subspace of X^{∗}. If D ⊂ X is σ(X, M)-compact then co(D) is σ(X, M)-precompact.

Proof. Since M is closed and norming, D is bounded and, equipped with the relative σ(X, M) topology, is a compact Hausdorff space. Let C := C(D, σ(X, M)), the space of continuous real valued functions on D with this topology. Then V defined by

V (f)(m) := f(m|_{D}), for f ∈ C^{∗} and m ∈ M

is a w^{∗}to w^{∗}; that is, σ(C^{∗}, C) to σ(M^{∗}, M), continuous linear operator from C^{∗} to M^{∗}. Since M is norming, X may be identified with a closed subspace of m^{∗} (the space (X, ∥.∥′) is complete, where ∥x∥′ := sup{m(x) : m ∈ M, ∥m∥ ≤ 1}). It suffices to shown that V (C^{∗}) ⊂ X, as then V (B_{C^{∗}}) is a σ(X, M)-compact convex subset of X containing D (for d ∈ D consider the action of V on d regarded as a point measure in B_{C^{∗}}).

To establish that V (C^{∗}) ⊂ X we first note that if f ∈ C^{∗} then V (f) is σ(M, X) boundedly continuous. Indeed, since X is separable, bounded subsets of M are σ(M, X) metrizable. So, if (m_{n}) is a bounded sequence in M with m_{n} → m in the σ(M, X) topology then the Lesbegue dominated convergence theorem gives that f(m_{n}|_{D}) → f(m|_{D}), as required.

Now, suppose there is an f ∈ C^{∗} with g := V (f) ∉ X. Then there exists F ∈ M^{∗∗} with ∥F ∥ = 1, F (g) ≠ 0, and F |_{X} = 0. B_{M} is σ(M^{∗∗}, M^{∗}) dense in B_{M^{∗∗}}, so there is a net (m_{i}) ⊂ B_{M} with m̂_{i}(m^{∗}) → F (m^{∗}), for all m^{∗} ∈ M^{∗}. In particular m̂_{i}(x) → F (x) = 0, for all x ∈ X ≤ M^{∗∗}. That is, m_{i} → 0 in the σ(M, X) topology, and so since (m_{i}) is bounded g(m_{i}) → g(0) = 0. But, g ∈ M^{∗} so g(m_{i}) = m̂_{i}(g) → F (g) ≠ 0, a contradiction establishing the result. ■

Example 2.8 (Lim, 1980) l_{1} with the equivalent dual norm ∥f⃗∥′ := ∥f⃗^{+}∥_{1} ∨ ∥f⃗^{−}∥_{1} fails the w^{∗}-FPP. In showing this we find a nonexpansive map T with a w^{∗}-compact minimal invariant set C that does not contain a fixed point of T. We first show that ∥ · ∥′ is indeed an equivalent dual norm for l_{1}. To this end, for x⃗ ∈ c_{0} define

∥x⃗∥ := ∥x^{+}∥_{∞} + ∥x^{−}∥_{∞}

Then ∥·∥ is an equivalent norm on c_{0} satisfying ∥x⃗∥_{∞} ≤ ∥x⃗∥ ≤ 2∥x⃗∥_{∞} and so it suffices to show that for f⃗ ∈ l_{1} we have

∥f⃗∥′ = sup{f⃗(x⃗) : x⃗ ∈ c_{0}, ∥x⃗∥ ≤ 1}.

Now for x⃗ ∈ c_{0} with ∥x⃗∥ ≤ 1 let

y_{i} = \{ \begin{matrix}x_{i} & if f_{i}x_{i} > 0 \\ 0 & otherwise. \end{matrix}

Then ∥y⃗∥ ≤ ∥x⃗∥ ≤ 1 and

f⃗(x⃗) := ∑_{i=1}^{∞} f_{i}x_{i}

≤ ∑_{i=1}^{∞} f_{i}y_{i}

≤ ∥y^{+}∥_{∞}∥f⃗^{+}∥_{1} + ∥y^{−}∥_{∞}∥f⃗^{−}∥_{1}

= \underbrace{(\frac{∥y^{+}∥_{∞}}{∥y⃗∥} ∥f⃗^{+}∥_{1} + \frac{∥y^{−}∥_{∞}}{∥y⃗∥} ∥f⃗^{−}∥_{1})} ∥y⃗∥

≤ (∥f⃗^{+}∥_{1} ∨ ∥f⃗^{−}∥_{1})∥y⃗∥

≤ ∥f⃗∥′.

To see the reverse inequality note that ∥f⃗^{+}∥_{1} (or ∥f⃗^{−}∥_{1}) can be approximated arbitrarily well by f⃗(x⃗) where the x_{i} are a suitable choice of 0 or 1 (0 or −1) and so ∥x⃗∥ ≤ 1. Now let C = {f⃗ ∈ l_{1} : f_{i} ≥ 0, ∥f⃗∥′ ≤ 1} and define T by

T f⃗:= (1 − ∑_{i=1}^{∞} f_{i}, f_{1}, f_{2}, …).

C is closed and bounded with respect to ∥ · ∥_{1} and since the unit ball centred at 0 in the same norm is w^{∗}-compact we have C is a weak^{∗}-compact convex subset of l_{1}. It is readily verified that T is a fixed point free affine mapping of C into C. Futhermore C is a minimal invariant set for T. To see this note that for any f = (f_{m}) ∈ C we have as successive iterates

T f = (1 − ∑_{1}^{∞} f_{m}, f_{1}, f_{2}, …)

T^{2}f = (0, 1 − ∑_{1}^{∞} f_{m}, f_{1}, f_{2}, …)

T^{3}f = (0, 0, 1 − ∑_{1}^{∞} f_{m}, f_{1}, f_{2}, …)

etc. So T^{n}f ⇀^{w^{∗}} 0. Thus 0 belongs to any T invariant w^{∗}-compact convex subset K of C. Hence the n’th basis vector, e_{n} = T^{n}(0), is in K. It follows that C = \overline{co}\{e_{n}\} ⊆ K ⊆ C, so K = C. We conclude by showing that T is a metric isometry (hence certainly a nonexpansive mapping) on C. Given f⃗, g⃗ ∈ C let P := {i : f_{i} − g_{i} ≥ 0} and N := {i : f_{i} − g_{i} < 0}. In the case that

∑_{i∈P}(f_{i} − g_{i}) ≥ ∑_{i∈N}(g_{i} − f_{i}) we have

∥T f⃗ − T g⃗∥′ = ∥(∑_{i=1}^{∞}(g_{i} − f_{i}), f_{1} − g_{1}, f_{2} − g_{2}, …)∥′

= ∥(\underbrace{∑_{i∈N}(g_{i} − f_{i}) − ∑_{i∈P}(f_{i} − g_{i})}_{negative}, f_{1} − g_{1}, f_{2} − g_{2}, …))∥′

= Max{∑_{i∈P}(f_{i} − g_{i}), ∑_{i∈N}(g_{i} − f_{i})}

= ∥f⃗ − g⃗∥.

The equality follows similarly in the case when ∥f⃗ − g⃗∥′ = ∑_{i∈N}(g_{i} − f_{i}).

Example 2.9 l_{1} fails the w^{∗}-ffp for an affine contraction. To see this choose δ ∈ (0, 1) and a sequence {ϵ_{k}} ⊂ (0, 1) such that ∑ ϵ_{k} < ∞ and define

T (x) := (δ(1 − x_{1}) + ∑_{k=1}^{∞}(1 − ϵ_{k})x_{k+1}, δ(1 − x_{1}), (1 − ϵ_{1})x_{2}, (1 − ϵ_{2})x_{3}, ...)

acting on the subset

K := {(x_{n}) : 0 ≤ x_{n}, x_{1} = ∑_{k=2}^{∞} x_{k} ≤ 1}.

Infact K is invariant under T so T : K → K. To prove the T -invariance of K we need (T x)_{n} ≥ 0, (T x)_{1} = ∑_{n=1}^{∞}(T x)_{k} and (T x)_{1} ≤ 0. The first two are obvious, for the third:

(T x)_{1} = δ(1 − x_{1}) + ∑_{k=1}^{∞}(1 − ϵ_{k})x_{k+1} ≤ δ(1 − x_{1}) + x_{1} ≤ δ + (1 − δ)x_{1} ≤ δ + (1 − δ) = 1.

Now, K is a convex and w^{∗}-compact subset of l_{1}. The w^{∗}-compactness of K comes from considering (−1, 1, 1, ...) as a function acting on l_{1}. Since (−1, 1, 1, ...) is bounded it is w^{∗}-continuous. Hence C = ker(−1, 1, 1, ...) is w^{∗}-closed. Let B(0) denote the closed unit ball centred at 0 which is w^{∗}-closed. Now,

K = C ∩ {x ∈ l_{1} : x_{n} ≥ 0, n ≥ 0} ∩ B(0).

So K is the finite intersection of w^{∗}closed subsets of l_{1}, one of which is w^{∗}-compact. Therefore K is w^{∗}-compact.

It is obvious that T is affine and we will now show that it is a contraction. First we prove that T is a nonexpansive map:

∥T x − T y∥

= |δ(y_{1} − x_{1}) + ∑_{k=1}^{∞}(1 − ϵ_{k}(x_{k+1} − y_{k+1})| + |δ(y_{1} − x_{1})| + ∑_{k=1}^{∞}(1 − ϵ_{k})|x_{k+1} − y_{k+1}|

= |(δ − 1)(y_{1} − x_{1}) + ∑_{k=1}^{∞} ϵ_{k}(x_{k+1} − y_{k+1})| + |δ(y_{1} − x_{1})| + ∑_{k=1}^{∞}(1 − ϵ_{k})|x_{k+1} − y_{k+1}|

≤ |(δ − 1)|y_{1} − x_{1}| + ∑_{k=1}^{∞} ϵ_{k}|x_{k+1} − y_{k+1}| + |δ(y_{1} − x_{1})| + ∑_{k=1}^{∞}(1 − ϵ_{k})|x_{k+1} − y_{k+1}|

= ∑_{k=1}^{∞} |x_{k} − y_{k}|

= ∥x − y∥.

If ∥T x − T y∥ = ∥x − y∥, then the above contains only equalities. Hence

|(1 − δ)(y_{1} − x_{1}) − ∑_{k=1}^{∞} ϵ_{k}(y_{k+1} − x_{k+1})|

= |(1 − δ)(y_{1} − x_{1})| + | ∑_{k=1}^{∞} ϵ_{k}(y_{k+1} − x_{k+1})|  ((1))

and

| ∑_{k=1}^{∞} ϵ_{k}(x_{k+1} − y_{k+1})| = ∑ ϵ_{k}|x_{k+1} − y_{k+1}|.  (2)

To satisfy (1) we must either have

(1 − δ)(y_{1} − x_{1}) ≥ 0

and

∑_{k=1}^{∞} ϵ_{k}(y_{k+1} − x_{k+1}) ≤ 0

or the reverse. Both cases follow a similar proof so we will prove the first case only. Since (2) gives us that the elements of the sum ∑_{k=1}^{∞} ϵ_{k}(x_{k+1} − y_{k+1}) are either all nonnegative or all nonpositive, we have

x_{k} ≥ y_{k}, fork ≥ 2.

Now, y_{1} ≥ x_{1}, and hence

x_{1} = ∑_{k=2}^{∞} x_{k} ≥ ∑_{k=2}^{∞} y_{k} = y_{1} ≥ x_{1}.

Thus x_{k} = y_{k} for k ≥ 1 and T is a contraction.

We now show that T is indeed fixed point free. Suppose T x = x. Then

T x_{n} = (1 − ϵ_{n−2})T x_{n−1} = (1 − ϵ_{n−2})(1 − ϵ_{n−3})...(1 − ϵ_{1})x_{2}.

If x_{2} = 0, then δ(1 − x_{1}) = 0 and hence x_{1} = 1. Also, x_{n} = 0 for n ≥ 2 which gives us the contradiction:

x_{1} = (T x)_{1} = δ(1 − x_{1}) + ∑_{k=1}^{∞}(1 − ϵ_{k})x_{k+1} = 0 ≠ x_{1}.

Thus x_{2} ≠ 0 and since x_{n} → 0 as n → ∞ we have

lim_{n→∞} \frac{x_{n}}{x_{2}} = 0.

This means ∏_{k=1}^{∞}(1 − ϵ_{k}) = 0 which contradicts ∑_{k=1}^{∞} ϵ_{k}.

Example 4. For conveniece, we take the dual action of (f_{n}) ∈ l_{1} on (x_{n}) ∈ c to be

(f_{n})(x_{n}) = f_{1}x_{1} + f_{2} lim x_{n} + f_{3}x_{2} + ...).

Consider the sequence x = (−1, 1, 1, ...) ∈ c as a continuous linear functional over l_{1}. Then

ker x = {f ∈ l_{1} : f_{1} = ∑_{i=2}^{∞} f_{i}}

is a w^{∗}−closed hyperplane and consequently the set

C = {f : f_{i} ≥ 0, f_{1} = ∑_{i=2}^{∞} f_{i} ≤ 1, }

being the intersection of ker x and w^{∗}−closed halfspaces is itself convex and w^{∗}−closed. Obviously C ⊂ 2B^{+}. So C is w^{∗}−compact. Choose δ ∈ (0, 1] and a sequence (ε_{k}) ⊂ [0, 1) such that ∑_{k=1}^{∞} ε_{k} < ∞ (consequently ∏_{k=1}^{∞} (1 − ε_{k}) > 0) and define a mapping by

T (f) = (δ (1 − f_{1}) + ∑_{k=1}^{∞} (1 − ε_{k}) f_{k+1}, δ (1 − f_{1}) , (1 − ε_{1}) f_{2}, (1 − ε_{2}) f_{3}, ...) .

There is some technicality involved in proving that T is a self mapping of C and that it is ∥·∥_{1} −nonexpansive. The summability condition imposed on the sequence (ε_{n}) implies that T is fixed point free.

A few further remarks about the above construction were made by M. A. Smyth [??]. In particular it was observed that if all ε_{k} > 0 and δ < 1, then T is contractive, in the sense that

f ≠ g ⇒ ∥T f − T g∥ < ∥f − g∥ .

We do not know if C is the minimal invariant set for T in these cases. However, when ε_{k} = 0, for k = 1, 2, ..., we observe that this is not the case. There is a smaller w^{∗}−closed convex T − invariant set; namely,

C′ = {f ∈ C : f_{1} = 1}

and a slightly more subtle variant of the argument used in example 2.8 shows that C′ is in fact the unique minimal invariant set for the nonexpansive map T : C → C. We would like to have simple examples of non-affine, nonexpansive maps that are f.p.free on non-weakly compact c.b.c. subsets of c_{0} and ℓ_{1} with their usual norms... Here are two such maps (that I told you about in ’94 or ’95).

(1) Consider X = c_{0} with its usual norm.

C := {x ∈ c_{0}^{+} : each x_{j} ≤ 1}.

Fix (p_{n}) , a real sequence that strictly decreases to 1. Define T : C → C by

T x := (1, \frac{x_{1}^{p_{1}}}{p_{1}} , …, \frac{x_{k}^{p_{k}}}{p_{k}} , …) .

Then T is a non-affine contraction on C. Furthermore, if p := ∏_{n=1}^{∞} p_{n} is finite, then T is fixed point free.

(2) X = ℓ_{1} with its usual norm.

K := {y ∈ ℓ_{1}^{+} : y_{1} = ∑_{n=2}^{∞} y_{n} ≤ 1}.

Fix (p_{n}) , a real sequence that strictly decreases to 1. Define U : K → K by

Uy := (1, 1 − ∑_{j=1}^{∞} \frac{y_{j+1}^{p_{j}}}{p_{j}} , \frac{y_{2}^{p_{1}}}{p_{1}} , …, \frac{y_{k+1}^{p_{k}}}{p_{k}} , …) .

Then U is a non-affine contraction on K. Furthermore, if p := ∏_{n=1}^{∞} p_{n} is finite, then U is fixed point free.

Example 2.10 L_{1}(μ) fails the w-FPP. Although the question had been raised more than twenty years earlier it was not until 1981 that Dale Alspach gave an example, drawn from ergodic theory, showing that not all Banach spaces enjoy the w-FPP.

(a) Alspach’s example [Alspach, 1981] Here we take C to be the set

C := \{f ∈ L_{1}[0, 1] : 0 ≤ f ≤ 1, ∫_{0}^{1} f dx = \frac{1}{2}\}

As the intersection of an order interval with a hyperplane in an order continuous Banach lattice, C is weak compact. The mapping T is essentially the baker transform of ergodic theory illustrated below.

Formally, for f ∈ C

T f(t) := \{ \begin{matrix}(2f(2t)) ∧ 1 & for 0 ≤ t ≤ \frac{1}{2}; \\ (2f(2t − 1) − 1) ∨ 0 & for \frac{1}{2} < t ≤ 1. \end{matrix}

It is clear from the above description that T is an isometry on C.

We now show that T is fixed point free.

Intuitively the idea is simple. First observe that the successive iterates of any point in C under T assume values closer to 0 or 1. Hence any fixed point for T must be a function which assumes only the values 0 or 1. By the “ergodic” nature of T it then follows that such a function must be either constantly 0 or constantly 1, and neither of these functions lie in C. The details follow.

For any f ∈ C we have T f(t) = 1 if and only if either

0 ≤ t ≤ \frac{1}{2} and \frac{1}{2} ≤ f(2t) ≤ 1

or

\frac{1}{2} < t ≤ 1 and f(2t − 1) = 1.

Futhermore if \frac{1}{2} ≤ t ≤ 1 and T f(t) = 1, then T f(t − \frac{1}{2}) = 1.

Now, suppose f is a fixed point for T then

A := {t : f(t) = 1}
= {t : T f(t) = 1}
= \{t : 0 ≤ t ≤ \frac{1}{2} and \frac{1}{2} ≤ f(2t) ≤ 1\} ∪ \{t : \frac{1}{2} < t ≤ 1 and f(2t − 1) = 1\}
= \{\frac{1}{2}t : \frac{1}{2} ≤ f(t) ≤ 1\} ∪ \{\frac{1}{2} + \frac{1}{2}t : f(t) = 1\}
= \frac{1}{2}\{t : \frac{1}{2} ≤ f(t) < 1\} ∪ \frac{1}{2}A ∪ (\frac{1}{2} + \frac{1}{2}A)

Since the three sets in the above union are mutually disjoint and each of the last two sets has measure one half that of A it follows that:

B_{1} := \{t : \frac{1}{2} ≤ f(t) < 1\}

is a null set. But, then

B_{1} = \{t : \frac{1}{2} ≤ T f(t) < 1\}
⊃ \{\frac{t}{2} : \frac{1}{4} ≤ f(t) < \frac{1}{2}\}

and so B_{2} := \{t : \frac{1}{4} ≤ f(t) < \frac{1}{2}\} is also a null set. Continuing in this way we have

B_{n} := \{t : \frac{1}{2^{n}} ≤ f(t) < \frac{1}{2^{n−1}} \}

is a null set for n = 1, 2, …, hence

{t : 0 < f(t) < 1} = ⋃_{n=1}^{∞} B_{n}

is null and

f ≡ χ_{A} (where meas(A) = ∫_{0}^{1} χ_{A} = \frac{1}{2}).

From the definition of T we have

T (χ_{A}) = (χ_{\frac{1}{2} A} + χ_{( \frac{1}{2} + \frac{1}{2} A)})

so, up to sets of measure zero,

A = \frac{1}{2}A ∪ (\frac{1}{2} + \frac{1}{2}A).

Continuing to iterate under T yields

A = \frac{1}{4}A ∪ (\frac{1}{4} + \frac{1}{4}A) ∪ (\frac{1}{2} + \frac{1}{4}A) ∪ (\frac{3}{4} + \frac{1}{4}A)
A = \frac{1}{8}A ∪ (\frac{1}{8} + \frac{1}{8}A) ∪ (\frac{1}{4} + \frac{1}{8}A) ∪ …

et hoc genus omne.

Thus the intersection of A with any dyadic interval (and hence any interval) has measure one half that of the interval, an impossibility for a set which is not of full measure. Notice that, unlike the previous example, the domain C of the baker transform T is not a minimal invariant set. This follows since

diam(C) = 1,

as

1 ≥ diam(C) ≥ ∥χ_{[0, \frac{1}{2} ]} − χ_{[ \frac{1}{2} ,1]}∥_{1} = 1),

while for any f ∈ C we have −\frac{1}{2} ≤ f − \frac{1}{2} ≤ \frac{1}{2} hence

∥f − \frac{1}{2}χ_{[0,1]}∥_{1} = ∫_{0}^{1} |f − \frac{1}{2}| ≤ \frac{1}{2}.

Thus, C is not diametral and therefore not a minimal invariant set. Indeed there seems to be no known explicit example of a non-trivial minimal invariant set for a nonexpansive map on a weak compact convex set.

(b) Sine’s modification of the Alspach example Robert Sine [1981] gave the following modification to the example of (0.5.1) which allows us to take as the domain C of our fixed point free nonexpansive mapping the whole order interval of 0 ≤ f ≤ 1.

For f ∈ C := {g : 0 ≤ g ≤ 1} let Sf := χ_{[0,1]} − f,

then S defines a mapping of C onto C with ∥Sf − Sg∥ = ∥f − g∥ for all f, g ∈ C. An argument similar to that for Alspach’s example shows that the composition ST, where T is the baker transform of (a), is an isometry on C with χ_{A} where A = [0, 1] or 0 the only possible fixed points. However, the action of ST is to map each of these functions onto the other, hence ST is fixed point free on the order interval 0 ≤ f ≤ 1.

(c) Schechtman’s construction. Gideon Schechtman [1982] gave a construction which leads to a greater variety of examples and is somewhat simpler than that of Alspach. Suppose (Ω, Σ, μ) is a measure space for which there exists a measure preserving transformation τ : Ω → Ω × [0, 1]; that is, for any measurable S ⊆ Ω × [0, 1] we have μ(τ^{−1}S) = meas(S) [Halmos, 1956]. Then if C is the weak compact convex set

C := \{f ∈ L_{1}(μ) : 0 ≤ f ≤ 1 and ∫_{Ω} f = \frac{1}{2}\}

we can define a mapping T : C → C by

T f := χτ^{−1}{(ω, t) : 0 ≤ t ≤ f(ω)}

Clearly T is an isometry on C and f ∈ C is a fixed point for T if and only if f = χ_{A} where A ∈ Σ is such that μ(A) = \frac{1}{2} and τ̂(A) := τ^{−1}(A × [0, 1]) = A a.e. Thus if τ is further chosen so that τ̂ is “ergodic”; that is τ̂(A) = A a.e. if and only if A = Ω or A = ϕ, then T is an example of a fixed point free nonexpansive mapping on C. Perhaps the simplest example of an (Ω, Σ, μ) and τ suitable for the above construction is the following. Let Ω = [0, 1]^{ℵ_{0}} with product Lebesgue measure and define τ by

τ^{−1}((ω_{1}, ω_{2}, …), t) := (t, ω_{1}, ω_{2}, …) .

Clearly τ is measure preserving, further if A ≠ ϕ and τ̂(A) = A, then for any (ω_{1}, ω_{2}, …) ∈ A we see that (t, ω_{1}, ω_{2}, …) ∈ A for any t ∈ [0, 1]. Iterating under τ̂ gives (t_{1}, t_{2}, …, t_{n}, ω_{1}, ω_{2}, …) ∈ A for any n ∈ ℕ and t_{1}, t_{2}, …, t_{n} ∈ [0, 1], and so we have A = Ω. An alternative example with Ω = [0, 1] is obtained by taking

τ^{−1}( ∑_{n=1}^{∞} \frac{ε_{n}}{2^{n}} , ∑_{n=1}^{∞} \frac{δ_{n}}{2^{n}} ) := \frac{δ_{1}}{2} + \frac{ε_{1}}{2^{2}} + \frac{δ_{2}}{2^{3}} + \frac{ε_{2}}{2^{4}} + ⋯,

where ε_{n}, δ_{n} ∈ {0, 1} for n = 1, 2, … A good way to view this example is via the correspondence

[0, 1] ⟷ \{0, 1\}^{ℵ_{0}} : ∑_{n=1}^{∞} \frac{ε_{n}}{2^{n}} ⟷ (ε_{1}, ε_{2}, …).

A set specified by prescribing precisely m of the ε_{n}’s has measure 1/2^{m}. From this it is clear that the product of two such sets has measure 1/2^{m_{1}+m_{2}} where m_{1} + m_{2} is also the number of digits prescribed for points in the τ^{−1} image of the product. It follows that τ is measure preserving. The ergodicity is estabished by iterating under τ̂ and an argument similar to that used for the conclusion of Alspach’s example.

Remarks.

Schechtman’s construction is both simpler and more versatile than that of Alspach and is of course also amenable to Sine’s modification. None-the-less, the Alspach example has some advantages. The relatively simple action of the baker transform permits detailed calculations. For example, it is possible to determine the orbit f_{0}, T f_{0}, T^{2}f_{0}, T^{3}f_{0}, …, of certain starting functions f_{0} under T. If f_{0} = \frac{1}{2}χ_{[0,1]} we obtain the iterates depicted below.

1 \frac{1}{2} 0\begin{aligned}0 & & 1 \\ & f_{0} \end{aligned}  \begin{aligned}0 & & 1 \\ & T f_{0} \end{aligned}  \begin{aligned}0 & & 1 \\ & T^{2}f_{0} \end{aligned}  \begin{aligned}0 & & 1 \\ & T^{3}f_{0} \end{aligned}

Here we see that the sequence T^{n}f_{0} = \frac{1}{2}(r_{n} + 1) is an orbit under T, where r_{n} is the n’th Rademacher function. This may be combined with a result of Maurey, that reflexive subspaces of L_{1}[0, 1] have the weak fixed point property, to show that the closed convex hull of an orbit of a nonexpansive mapping on a weakly compact convex set need not be invariant. Indeed, define D to be \overline{co}\{T^{n}(f_{0}) : n ∈ N\}. Since the closed linear span of the Rademacher functions is isomorphic to L_{2}[0, 1] [Lindenstrauss, Joram and Tzafriri, Lior, 1973], D can not be invariant. Indeed, were it invariant Maurey’s result would imply that T possessed a fixed point in D and, a fortiori, in C.

3. Notes and Remarks

Example 2.9 is essentially due to C. Lennards, the observstion that it is infact a contraction was made by M. Smyth who also extended it to the following broader result :

Theorem 3.1 (Smyth, M.A., 1994) Let Ω be an infinite compact Hausdorff topological space. Then C(Ω)^{∗} fails the w^{∗}-fpp with an affine contraction.

In our example Ω is the one point compactification of N, where′∞′ is the extra point. So we can write n ∈ Ω in the form n = (1, ∞, 2, 3, ...). Now, if for z = (z_{1}, z_{2}, z_{3}, ...) ∈ c we write

z = (z_{1}, lim_{n→∞} z_{n}, z_{2}, z_{3}, ...)

z is a continuous function on Ω. This is because

lim_{n→∞} z(n) = lim_{n→∞} z_{n} = z(∞).

So c=C(Ω). If we let l_{1} act on c by

x(z) = x_{1}z_{1} + x_{2} lim_{n→∞} +x_{3}z_{2} + ..., ∀x ∈ l_{1}, z ∈ c

then l_{1} = c^{∗} = C(Ω)^{∗} and T is the affine contraction for which the w^{∗}-fpp fails.

Chris Lennard has given a wavelet construction of a fixed point free isometry, similar to that of Alspach, and also on the order interval [0 ≤ f ≤ 1].

PROBLEMS. These examples indicate an intimate connection between fixed point free isometries and ergodic transformations of the underlying measure space. In the true tradition of ergodic theory, is the set of fixed point free isometries on the order interval [0 ≤ f ≤ 1] residual in an appropriate sense, at least among isometries which map into the set of 0, 1–valued functions? Clearly any space containing an isometric copy of L_{1}(μ) also fails the w-FPP.What do the examples look like when translated into l_{∞}, C[0, 1]? Examples 2.9 and ?? also suggest the following question. If a space X fails the (w, w^{∗})-FPP does it necessarily fail with an isometry?

References

[1] Alspach, Dale E., A fixed point free nonexpansive map, Proc. A.M.S., 82 (1981), 423-424.

[2] van Dulst, D. and Pach, A.J., On flatness and some ergodic super-properties of Banach spaces, Indagationes Mathematical, 43 (1981), 153-164.

[3] Goebel K., Kirk W. A., Topics in metric fixed point theory, Cambridge University Press, 1990 [4] Halmos Paul R., Measure Theory, Van Nostrand, 1950.

[5] Lim, T.-C., Asymptotic centers and nonexpansive mappings in conjugate Banach spaces, Pacific J. of Math., 90 (1980), 135-143.

[6] Sine, R., Remarks on the example of Alspach, Nonlinear Anal. and Appl., Marcel Decker, (1981), 237-241.

[7] Lindenstrauss, Joram and Tzafriri, Lior, Classical Banach Spaces, Springer-Verlag, Lecture Notes in Mathematics, 338 (1973).

[8] Schechtman, G., On commuting families of nonexpansive operators, Proc. Amer. Math. Soc., 84 (1982), 373-376.

[9] Smyth, M.A., [1994], Remarks on the weak star fixed point property in the dual of C(Ω), J. Math. Anal. and App., 195 (1994), 294-306.