DUAL KADEC-KLEE PROPERTY AND FIXED POINTS

JEAN SAINT RAYMOND

Abstract. A Banach space X is said to have the dual Kadec-Klee property iff on the unit sphere of the dual space X^{∗} the weak*-topology coincides with the norm-topology. A. Amini-Harandi and M. Fakhar extended a theorem of B. Ricceri concerning the Kadec-Klee property by showing that if X has the dual Kadec-Klee property then for every compact mapping φ : B_{X^{∗}} → X \ {0} there exists some f in the unit sphere of X^{∗} such that ⟨f, φ(f)⟩ = ∥φ(f)∥ and conjectured that this yields a necessary and sufficient condition for X to have the dual Kadec-Klee property. We prove here that it is almost the case : we introduce a weakening of the dual Kadec-Klee property, here called the NAKK* property, and show that this property is equivalent to the above property about compact mappings from B_{X^{∗}} to X \ {0}.

1. Introduction

Recall that a Banach space X is said to have the Kadec-Klee property if the weak topology and the norm-topology of X agree on the unit sphere S_{X} of X. This property, which is weaker than being locally uniformly rotund, was introduced by M. Kadec in his proof that every separable Banach space can be equipped with an equivalent l.u.r. norm. Recall also that a function f : X → Y is said to be compact if it is continuous and its range f(X) is relatively compact in Y.

In a recent paper [5], B. Ricceri was interested in the following problem of fixed point. Let X be an infinite-dimensional reflexive Banach space and φ : S_{X} → X^{∗} \ {0} a continuous function ; is there some x ∈ S_{X} such that ⟨φ(x), x⟩ = ∥φ(x)∥ ? He proved that there are counterexamples in every infinite-dimensional Banach space if φ is not assumed to be compact. Nevertheless he has shown the following :

Theorem. Let X be a reflexive infinite dimensional Banach space having the Kadec-Klee property. If φ : S_{X} → X^{∗} \ {0} is a compact function then there exists some x̂ ∈ S_{X} such that ⟨φ(x̂), x̂⟩ = ∥φ(x̂)∥.

In fact since there exists a continuous retraction from the unit ball B_{X} onto S_{X}, this statement is equivalent to the following one where S_{X} is replaced by B_{X}.

Theorem. Let X be a reflexive infinite dimensional Banach space having the Kadec-Klee property. If φ : B_{X} → X^{∗} \ {0} is a compact function then there exists some x̂ ∈ S_{X} such that ⟨φ(x̂), x̂⟩ = ∥φ(x̂)∥.

In the same paper B. Ricceri asked whether this property could characterize the Kadec-Klee property among the reflexive Banach spaces, and in [6] the present author proved that it is the case.

2010 Mathematics Subject Classification. 52A40 , 46A25, 46B10 .

Key words and phrases. convex functions, Banach spaces, dual Kadec-Klee property, fixed points.

Recently A. Amini-Harandi and M. Fakhar (in the so far unpublished paper [1]) considered a variant of the previous problem. A Banach space X is said to have the dual Kadec-Klee property if the weak*-topology and the norm-topology of X^{∗} agree on the unit sphere S_{X^{∗}} of X^{∗}. When X is separable this is equivalent to the following property : If the sequence (x_{n}^{∗}) weak*-converges to x^{∗} in X^{∗} and ∥x_{n}^{∗}∥ → ∥x^{∗}∥ then ∥x_{n}^{∗} − x^{∗}∥ → 0.

Following the scheme of proof of Ricceri in [5] these two authors prove in [1] the following

Theorem. Let X be an infinite-dimensional Banach space having the dual Kadec-Klee property. If φ : S_{X^{∗}} → X \ {0} is a compact function then there exists some x̂^{∗} ∈ S_{X^{∗}} such that ⟨x̂^{∗}, φ(x̂^{∗})⟩ = ∥φ(x̂^{∗})∥.

The authors remark that this statement extends the theorem proved by Ricceri in [5] : indeed if X is a reflexive Banach space then it has the Kadec-Klee property iff X^{∗} has the dual Kadec-Klee property.

And in the same paper they conjectured that for an infinite-dimensional Banach space this property characterizes the dual Kadec-Klee property. In fact as we shall see below, one can introduce a slightly weaker property that we call here the normattaining dual Kadec-Klee property (NAKK* for short) and show in the same way that in the previous statement one can replace the dual Kadec-Klee property by the NAKK* property. The main goal of this paper is to prove the theorem 6.1 below which states that this new property is equivalent to the fixed-point property for the compact mappings from S_{X^{∗}} to X \ {0}.

In the sequel all Banach spaces are assumed to be infinite-dimensional unless explicitly said. For a Banach space X we denote by S_{X} its unit sphere, B_{X} its unit ball, X^{∗} its dual, S_{X^{∗}} and B_{X^{∗}} the unit sphere and the unit ball of X^{∗}. If x ∈ X and x^{∗} ∈ X^{∗} we denote by ⟨x^{∗}, x⟩ the value at x of the linear functional x^{∗}. We denote by NA(X) the set of all f ∈ S_{X^{∗}} which attain their norm on S_{X}, that is NA(X) = {f ∈ S_{X^{∗}} : ∃x ∈ S_{X} ⟨f, x⟩ = 1 }.

Let X be a Banach space. We will denote by κ_{X} (shortened in κ when there is no ambiguity on X) the identity mapping from S_{X^{∗}} equipped with the weak*-topology to S_{X^{∗}} equipped with the norm-topology. So X has the dual Kadec-Klee property iff κ is a homeomorphism. And we will say that X has the NAKK* property iff κ is continuous at every f ∈ NA(X). For any reflexive Banach space X we have NA(X) = S_{X^{∗}} : so in this case it is clear that the NAKK* property and the dual Kadec-Klee property are equivalent.

Let X be a separable Banach space whose dual X^{∗} is non-separable. The set C_{κ} of points of continuity of κ is separable since B_{X^{∗}} is a compact metrizable space for the weak*-topology. So κ(C_{κ}) is a separable subset of S_{X^{∗}} for the norm-topology, hence it is nowhere dense in S_{X^{∗}} since X^{∗} is non-separable. And since NA(X) is norm-dense in S_{X^{∗}} by Bishop-Phelps’ theorem, there are points of NA(X) which do not belong to \overline{κ(C_{κ})}. This shows that such a Banach space cannot have the NAKK* property.

Recall that a multivalued mapping F : X → 2^{Y} associating a non-empty subset F (x) of the topological space Y to each point x ∈ X is said to be upper semicontinuous (u.s.c. for short) if for every closed subset H of Y the set {x ∈ X : F (x)∩H ≠ ∅} is closed in X. When Y is a topological linear space, F is said to be cusco if it is u.s.c. and moreover the values of F are convex and compact subsets of Y.

In section 2 we will show that the NAKK* property is strictly weaker than the dual Kadec-Klee property and that it is a sufficient condition for getting the conclusion of the above theorem.

In section 3 we shall provide the construction of a continuous function without fixed point in a bounded weak*-closed convex subset of the dual W^{∗} of a separable Banach space W.

In section 4 we prove that if X is a separable Banach space and δ > 0 there exists a continuous function ρ : S_{X^{∗}} → S_{X} such that for all x^{∗} ∈ S_{X^{∗}} one can find some z^{∗} ∈ S_{X^{∗}} which attains its norm precisely at ρ(x^{∗}) and moreover satisfies ∥x^{∗} − z^{∗}∥ ≤ δ. So this result yields a “continuous version” of Bishop-Phelps’ theorem.

Our main theorem states that the NAKK* property is in fact a necessary condition for the validity of the conclusion of the above theorem. The section 5 is devoted to the proof of this main theorem for separable Banach spaces, and we reduce the general case to the separable case in section 6.

2. A sufficient condition

I am grateful to G. Godefroy who gave me the following example, essentially extracted from the book [2] (p. 313), which shows that the NAKK* property is strictly weaker than the dual Kadec-Klee property.

Example. There exists a separable Banach space Y which does not have the dual Kadec-Klee property but for which every norm-attaining linear functional in S_{Y^{∗}} is a point of continuity of the mapping κ_{Y}.

Proof. Let X be the space of continuous functions on the compact space Ω = [0, ω_{1}], where ω_{1} is the least uncountable ordinal. For α < ω_{1} we denote by δ_{α} the Dirac measure at α : x ↦ x(α).

It is a known fact [8] that there exists on X a Fréchet differentiable norm N which is equivalent to the uniform norm. The mapping α ↦ N^{∗}(δ_{α}) is l.s.c. from Ω to ℝ^{+}, hence constant on a club C ⊂ ω_{1}. In particular there exists an increasing sequence (α_{m})_{m∈ℕ} in C and since λ = sup_{m} α_{m} = lim_{m} α_{m} ∈ C we have N^{∗}(δ_{α_{m}}) = N^{∗}(λ) for all m ∈ ℕ. Set S = {λ} ∪ {α_{m} : m ∈ ℕ}

Choose a separable subspace Y of X such that N^{∗}(δ_{α}) = sup_{y∈y,N(y)≤1}⟨δ_{α}, y⟩ for all α ∈ S. Then the restriction N_{|Y} of N to Y is a Fréchet-differentiable norm on Y. Hence in particular the norm-attaining linear functionals on Y are points of weak*-norm continuity by Smulyan’s lemma (cf. [7]).

Nevertheless we have N_{|Y}^{∗} (δ_{λ|Y}) = N_{|Y}^{∗} (δ_{α_{m}|Y}) for all m and δ_{α_{m}|Y} → δ_{λ|Y} for the weak*-topology but inf_{m} N_{|Y}^{∗} (δ_{λ|Y} −δ_{α_{m}|Y}) > 0. So δ_{λ|Y} is not a point of weak*- norm continuity, and Y does not have the dual Kadec-Klee property although it has the NAKK* property. □

Theorem 2.1. Let X be an infinite-dimensional Banach space having the NAKK* property. If φ : S_{X^{∗}} → X \ {0} is a compact function then there exists some x̂^{∗} ∈ S_{X^{∗}} such that ⟨x̂^{∗}, φ(x̂^{∗})⟩ = ∥φ(x̂^{∗})∥.

Proof. The proof of this theorem is essentially the same as this of Amini-Harandi and Fakhar. It is also the same as Ricceris’ proof in the reflexive case by interverting X and X^{∗}.

Consider the multifunction Γ : X → 2^{S_{X^{∗}}} defined by

Γ(x) = {x^{∗} ∈ S_{X^{∗}} : ⟨x^{∗}, x⟩ = ∥x∥}

and let us show that the restriction of Γ to X′ = X \ {0} is upper semicontinuous when S_{X^{∗}} is equipped with the weak*-topology. It is enough to notice that the graph of Γ,

G = {(x, x^{∗}) ∈ X′ × B_{X^{∗}} : x^{∗} ∈ Γ(x)} = {(x, x^{∗}) : ⟨x^{∗}, x⟩ ≥ ∥x∥} ,

is closed in the product of X′ equipped with the norm-topology by B_{X^{∗}} equipped with the weak*-topology since this latter space is compact.

Since X has the NAKK* property and Γ(x) ⊂ NA(X) for every x in X′, we see that on G the mapping Id × κ is a homeomorphism, hence that Γ is again cusco when B_{X^{∗}} is equipped with the norm-topology. Since φ is a compact mapping the set K = \overline{φ(S_{X^{∗}})} is compact in X′ and so is Γ(K) ⊂ B_{X^{∗}} by upper semicontinuity, and also its closed convex hull K_{0} = \overline{conv(Γ(K))} by Mazur’s theorem.

Since X^{∗} is infinite-dimensional there is a continuous retraction ω from B_{X^{∗}} to S_{X^{∗}} and we consider the cusco mapping F : x^{∗} ↦ Γ(φ ∘ ω(x^{∗})) defined on the compact convex set K_{0} ⊂ B_{X^{∗}} with values in K_{0}. By Fan-Kakutani’s theorem F has a fixed point x̂^{∗} ∈ K_{0}, it is x̂^{∗} ∈ F (x̂^{∗}). Since F (x̂^{∗}) = Γ(y) ⊂ S_{X^{∗}} where y = φ ∘ ω(x̂^{∗}), we have x̂^{∗} ∈ F (x̂^{∗}) ⊂ S_{X^{∗}}, hence ω(x̂^{∗}) = x̂^{∗} and y = φ(x̂^{∗}). It follows that x̂^{∗} ∈ Γ(y) what means ⟨x̂^{∗}, φ(x̂^{∗})⟩ = ⟨x̂^{∗}, y⟩ = ∥y∥ = ∥φ(x̂^{∗})∥. □

3. A fixed-point free mapping

The content of this section is almost identical to lemmas 2 to 5 in the paper [6] up to replacing the space X by a dual space W^{∗}. It is essentially provided for sake of self-containedness. Nevertheless some modifications have been brought for taking into account that W^{∗} could be non-separable.

Lemma 3.1. Let W be a separable Banach space, C ⊂ W^{∗} a bounded weak*-closed convex set of diameter R containing 0, h : C → ℝ a convex weak*-l.s.c. nonnegative function such that h(0) = 0. Then – either all minimizing sequences of h have norm-converging subsequences – or there exist a point a^{∗} ∈ h^{−1}(0), a sequence (u_{n}^{∗}) in C weak* converging to a^{∗} and some δ > 0 satisfying:

(1) lim inf_{n→∞} ∥u_{n}^{∗} − a^{∗}∥ ≥ δ.

(2) ∀b^{∗} ∈ W^{∗} lim inf_{n→∞} inf_{t∈[0,1]} ∥b^{∗} − (tu_{n}^{∗} + (1 − t)u_{n+1}^{∗})∥ ≥ θ := \frac{δ^{2}}{R}

(3) ∀n h(u_{n}^{∗}) ≤ 2^{−n}.

Proof. We define inductively an increasing sequence (V_{n}) of finite-dimensional linear subspaces of W^{∗}, a non-increasing sequence (δ_{n}) and a sequence (x_{n}^{∗}) ∈ C in such a way that V_{n} is the linear space spanned by (x_{k}^{∗})_{k<n},

δ_{n} = sup{d(x^{∗}, V_{n}) : x^{∗} ∈ C and h(x^{∗}) ≤ 2^{−n}}

h(x_{n}^{∗}) ≤ 2^{−n} and d(x_{n}^{∗}, V_{n}) > δ_{n} · (1 − 2^{−n}). One has δ_{0} ≤ R = diam(C) < +∞.

If the sequence (δ_{n}) converges to 0 every minimizing sequence (z_{k}^{∗}) for h is totally bounded, hence relatively compact and possesses a norm-converging subsequence : indeed, for ε > 0, there exist some n such that δ_{n} < \frac{ε}{2} and a finite covering of the ball of radius R + ε in V_{n} by balls (B(y_{j}^{∗}, ε/2))_{j≤J}.

Then the balls B(y_{j}^{∗}, ε) cover {z_{m}^{∗} : h(z_{m}^{∗}) ≤ 2^{−n}}, it is {z_{m}^{∗} : m > n_{0}} for some n_{0} and {z_{m}^{∗} : m ∈ ℕ} ⊂ ⋃_{j≤J} B(y_{j}^{∗}, ε) ∪ ⋃_{m≤n_{0}} B(z_{m}^{∗}, ε).

Thus assume that δ′ = inf_{n} δ_{n} > 0, and set δ = \frac{δ′}{2} and δ′_{k} = δ′(1 − 2^{−k}). Since W is separable C is compact and metrizable for the weak*-topology : the sequence (x_{n}^{∗}) has a subsequence (u_{n}^{∗}) which converges weak* to some a^{∗} ∈ C. Since h is convex and weak*-l.s.c. we necessarily have

0 ≤ inf_{x^{∗}∈C} h(x^{∗}) ≤ h(a^{∗}) ≤ lim inf_{n→∞} h(u_{n}^{∗}) ≤ 0

and this shows that a^{∗} ∈ h^{−1}(0). Denote by Y the norm-closed subspace \overline{⋃_{j∈ℕ} V_{j}} of W^{∗}. Notice that since u_{n}^{∗} = x_{m}^{∗} for some m ≥ n we have h(u_{n}^{∗}) ≤ 2^{−m} ≤ 2^{−n}.

For all b^{∗} ∈ Y and all ε > 0, there is some j ∈ ℕ and some u^{∗} ∈ V_{j} such that ∥b^{∗} − u^{∗}∥ ≤ ε. Then, for k ≥ j, ∥x_{k}^{∗} − u^{∗}∥ ≥ d(x_{k}^{∗}, V_{j}) ≥ d(x_{k}^{∗}, V_{k}) ≥ δ′_{k} and ∥x_{k}^{∗} − b^{∗}∥ ≥ ∥x_{k}^{∗} − u^{∗}∥ − ∥b^{∗} − u^{∗}∥ ≥ δ′_{k} − ε. Thus for all b^{∗} ∈ Y we have lim inf_{n→∞} ∥x_{n}^{∗} − b^{∗}∥ ≥ lim inf_{k} δ′_{k} = δ′ = 2δ.

If b^{∗} ∈ W^{∗} and d(b^{∗}, Y) < δ, there is some y^{∗} ∈ Y such that ∥y^{∗} − b^{∗}∥ < δ. Then ∥x_{k}^{∗} − b^{∗}∥ ≥ ∥x_{k}^{∗} − y^{∗}∥ − ∥y^{∗} − b^{∗}∥ > ∥x_{k}^{∗} − y^{∗}∥ − δ. So lim inf_{n→∞} ∥x_{n}^{∗} − b^{∗}∥ ≥ δ. And if b^{∗} ∈ W^{∗} and d(b^{∗}, Y) ≥ δ we have ∥x_{k}^{∗} − b^{∗}∥ ≥ d(b^{∗}, Y) ≥ δ for all k, hence lim inf_{n→∞} ∥x_{n}^{∗} − b^{∗}∥ ≥ δ. And in particular lim inf_{n→∞} ∥x_{n}^{∗} − a^{∗}∥ ≥ δ.

If b^{∗} ∈ Y and ε > 0 there exists j ∈ ℕ and u^{∗} ∈ V_{j} such that ∥b^{∗} − u^{∗}∥ < ε. Then if ℓ > k > j, t ∈ [0, 1] and y^{∗} = (1 − t)x_{k}^{∗} + tx_{ℓ}^{∗}, we have on one hand

∥b^{∗} − x_{k}^{∗}∥ ≥ d(x_{k}^{∗}, V_{j}) − d(b^{∗}, V_{j}) ≥ d(x_{k}^{∗}, V_{k}) − d(b^{∗}, V_{j}) ≥ δ′_{k} − ε

so

∥b^{∗} − y^{∗}∥ ≥ ∥b^{∗} − x_{k}^{∗}∥ − t. ∥x_{k}^{∗} − x_{ℓ}^{∗}∥ ≥ δ′_{k} − ε − t.R

and on the other hand since u^{∗} − (1 − t)x_{k}^{∗} ∈ V_{k+1} ⊂ V_{ℓ},

∥b^{∗} − y^{∗}∥ ≥ ∥u^{∗} − y^{∗}∥−∥u^{∗} − b^{∗}∥ = t·∥x_{ℓ}^{∗} − \frac{u^{∗} − (1 − t)x_{k}^{∗}}{t} ∥−∥u^{∗} − b^{∗}∥ ≥ t.δ′_{ℓ} −ε

It is easily checked that inf_{t∈[0,1]} max(δ′_{k} − tR, tδ′_{ℓ}) = \frac{δ′_{k} · δ′_{ℓ}}{R + δ′_{ℓ}} ≥ \frac{δ′_{k} · δ′_{ℓ}}{2R}, so

inf_{t∈[0,1]} ∥b^{∗} − ((1 − t)x_{k}^{∗} + tx_{ℓ}^{∗})∥ ≥ \frac{δ′_{k} · δ′_{ℓ}}{2R} − ε

and finally lim inf_{{k<ℓ \atop k→∞}} inf_{t∈[0,1]} ∥b^{∗} − ((1 − t)x_{k}^{∗} + tx_{ℓ}^{∗})∥ ≥ \frac{δ^{′2}}{2R} = \frac{2δ^{2}}{R}.

If b^{∗} ∈ W^{∗} and d(b^{∗}, Y) < θ := \frac{δ^{2}}{R} there is some u^{∗} ∈ Y such that ∥u^{∗} − b^{∗}∥ < θ. So

lim inf_{{k<ℓ \atop k→∞}} inf_{t∈[0,1]} ∥b^{∗} − ((1 − t)x_{k}^{∗} + tx_{ℓ}^{∗})∥ ≥ lim inf_{{k<ℓ \atop k→∞}} inf_{t∈[0,1]} ∥u^{∗} − ((1 − t)x_{k}^{∗} + tx_{ℓ}^{∗})∥ − θ

≥ \frac{2δ^{2}}{R} − θ = θ

And if b^{∗} ∈ W^{∗} with d(b^{∗}, Y) ≥ θ we have ∥b^{∗} − ((1 − t)x_{k}^{∗} + tx_{ℓ}^{∗})∥ ≥ θ for all k, ℓ and t, hence lim inf_{{k<ℓ \atop k→∞}} inf_{t∈[0,1]} ∥b^{∗} − ((1 − t)x_{k}^{∗} + tx_{ℓ}^{∗})∥ ≥ θ.

It follows from what precedes that all clauses of the lemma are satisfied by the sequence (u_{n}^{∗}) we constructed this way. □

If there exists θ > 0 and a sequence (u_{n}^{∗}) satisfying conditions (2) and (3) of lemma 3.1, we define for t ∈ ℝ^{+},

γ(t) = (t − k)u_{k+1}^{∗} + (k + 1 − t)u_{k}^{∗} ∈ C if k ≤ t ≤ k + 1

and γ is a continuous R-Lipschitz locally piecewise-affine function from ℝ^{+} to C such that γ(k) = u_{k}^{∗}. For s ∈ ℝ^{+} we set also G_{s} = {γ(t) : t ≥ s}.

Lemma 3.2. h ∘ γ(t) ≤ 2^{−n} holds whenever t ≥ n.

Proof. Indeed if t ∈ [k, k + 1] and k ≥ n, we have by convexity

h(γ(t)) = h((t − k)u_{k+1}^{∗} + (k − 1 − t)u_{k}^{∗}) ≤ (t − k)h(u_{k+1}^{∗}) + (k + 1 − t)h(u_{k}^{∗})
≤ max(h(u_{k}^{∗}), h(u_{k+1}^{∗})) ≤ 2^{−k} ≤ 2^{−n}

from what the result follows. □

Lemma 3.3. There exists a continuous function ψ : W^{∗} → ℝ^{+} such that for all

s ≥ 0 and all x^{∗} ∈ W^{∗}, the implication ∥x^{∗} − γ(s)∥ < θ · \frac{s + 2}{2s + 2} ⟹ ψ(x^{∗}) > s

holds.

Proof. The function (x^{∗}, s) ↦ d(x^{∗}, G_{s}), which is non-decreasing with respect to s for fixed x^{∗}, is continuous on W^{∗} × ℝ^{+} : indeed for s ≤ t,

d(y^{∗}, G_{s}) ≤ d(y^{∗}, G_{t}) ≤ inf_{z^{∗}∈G_{s}} d(y^{∗}, z^{∗}) + sup_{z^{∗}∈G_{s}} d(z^{∗}, G_{t})

= d(y^{∗}, G_{s}) + sup_{s′≥s} d(γ(s′), G_{t})

thus

|d(x^{∗}, G_{s}) − d(y^{∗}, G_{t})| ≤ |d(x^{∗}, G_{s}) − d(y^{∗}, G_{s})| + |d(y^{∗}, G_{s}) − d(y^{∗}, G_{t})|

≤ ∥x^{∗} − y^{∗}∥ + sup_{s′≥s} d(γ(s′), G_{t})

≤ ∥x^{∗} − y^{∗}∥ + sup_{s≤s′≤t} ∥γ(s′) − γ(t)∥

≤ ∥x^{∗} − y^{∗}∥ + R · |t − s|

Moreover following lemma 3.1, one has sup_{s∈ℝ^{+}} d(x^{∗}, G_{s}) ≥ lim inf_{t→∞} ∥x^{∗} − γ(t)∥ ≥ θ for all x^{∗} ∈ W.

The function s ↦ θ · \frac{s + 2}{2s + 2} is decreasing on ℝ^{+} from θ to \frac{θ}{2}. Then set for

x^{∗} ∈ W^{∗}: χ(x^{∗}, s) = min(d(x^{∗}, G_{s}), θ)−θ· \frac{s + 2}{2s + 2}. The function χ(x^{∗}, .) is increasing

on \overline{ℝ}_{+} = [0, +∞] and satisfies χ(x^{∗}, 0) ≤ 0 and χ(x^{∗}, +∞) = \frac{θ}{2}. There exists then

a unique s^{∗} = ψ(x^{∗}) ∈ \overline{ℝ}_{+} = [0, +∞] such that χ(x^{∗}, s^{∗}) = 0. It is immediately checked that the graph of ψ is closed in W^{∗} × \overline{ℝ}_{+}, hence that ψ is continuous since

\overline{ℝ}_{+} is compact. Since χ(x^{∗}, +∞) = \frac{θ}{2} > 0 we have ψ(x^{∗}) < +∞ for every x^{∗} ∈ W^{∗}.

Finally if ∥x^{∗} − γ(s)∥ < θ· \frac{s + 2}{2s + 2}, we get d(x^{∗}, G_{s}) ≤ ∥x^{∗} − γ(s)∥ < θ· \frac{s + 2}{2s + 2} hence

χ(x^{∗}, s) < 0 = χ(x^{∗}, s^{∗}), thus s < s^{∗} = ψ(x^{∗}), and this achieves the proof. □

Lemma 3.4. The continuous function F : W^{∗} → C defined by F = γ ∘ ψ satisfies

∥x^{∗} − F (x^{∗})∥ > \frac{θ}{2} for all x^{∗} ∈ W.

Proof. If there was some x^{∗} ∈ W^{∗} such that ∥x^{∗} − F (x^{∗})∥ ≤ \frac{θ}{2} we would have

F (x^{∗}) = γ(s) for s = ψ(x^{∗}), hence ∥x^{∗} − γ(s)∥ ≤ \frac{θ}{2} < θ · \frac{s + 2}{2s + 2}, from what

ψ(x^{∗}) > s = ψ(x^{∗}) would follow by lemma 3.3, a contradiction. □

Lemma 3.5. Let X be a separable Banach space, a be a point in the unit sphere S_{X} of X, a^{∗} in the unit sphere S_{X^{∗}} of the dual X^{∗} attaining its norm at a. Assume that the mapping κ is discontinuous at a^{∗}. Then there exists a continuous function ψ̃ from the unit ball B_{X^{∗}} of X^{∗} into ℝ^{+}, a continuous function γ̃ : ℝ^{+} → B_{X^{∗}} and δ ∈]0, 1] such that

(1) for every x^{∗} ∈ B_{X^{∗}} ∥x^{∗} − γ̃ ∘ ψ̃(x^{∗})∥ > δ,
(2) lim_{t→∞}⟨γ̃(t), a⟩ = ⟨a^{∗}, a⟩ = 1.

Proof. By hypothesis there is a sequence (x_{n}^{∗}) in the unit ball B_{X^{∗}} such that (x_{n}^{∗}) converges weak* to a^{∗} but does not converge in norm. Denote by W the quotient space X/ℝ·a. Then the dual space W^{∗} is isomorphic to the weak*-closed hyperplane Z = {z^{∗} ∈ X^{∗} : ⟨z^{∗}, a⟩ = 0}. Let π : X^{∗} → Z be the projection defined by π(x^{∗}) = x^{∗} − ⟨x^{∗}, a⟩ · a^{∗} whose kernel is ℝ · a^{∗}. Clearly ∥π(x^{∗})∥ ≤ 2 ∥x^{∗}∥, hence ∥π∥ ≤ 2. The mapping x^{∗} ↦ (π(x^{∗}), ⟨x^{∗}, a⟩) is then an isomorphism from X^{∗} onto Z ×ℝ. And C = π(B_{X^{∗}}) is a bounded convex weak*-closed neighborhood of 0 in Z. For z^{∗} ∈ C there is some s ∈ ℝ such that z^{∗}+s·a^{∗} ∈ B_{X^{∗}} and we denote by c(z^{∗}) the largest such s. Moreover one has 1 ≥ ∥z^{∗} + c(z^{∗}) · a^{∗}∥ ≥ ⟨z^{∗} + c(z^{∗}) · a^{∗}, a⟩ = c(z^{∗}), thus c(z^{∗}) ≤ 1.

And it is easily checked that the function z^{∗} ↦ c(z^{∗}), which satisfies c(0) = 1, is concave and weak*-u.s.c. since

c(z^{∗}) = inf_{w∈B_{X},⟨a^{∗},w⟩>0} \frac{1 − ⟨z^{∗}, w⟩}{⟨a^{∗}, w⟩} .

Indeed if w ∈ B_{X} and ⟨a^{∗}, w⟩ > 0, one has

c(z^{∗}) · ⟨a^{∗}, w⟩ + ⟨z^{∗}, w⟩ = ⟨z^{∗} + c(z^{∗}) · a^{∗}, w⟩ ≤ ∥w∥ · ∥z^{∗} + c(z^{∗}) · a^{∗}∥ ≤ 1 .

Conversely if s = c(z^{∗}) and ε > 0 we have ∥z^{∗} + (s + ε) · a^{∗}∥ > 1. Thus there is a w ∈ B_{X} such that ⟨z^{∗} + (s + ε) · a^{∗}, w⟩ > 1. Then

1 < ⟨z^{∗} + (s + ε) · a^{∗}, w⟩ = ⟨z^{∗} + s · a^{∗}, w⟩ + ε⟨a^{∗}, w⟩ ≤ ∥w∥ · ∥z^{∗} + s · a^{∗}∥ + ε⟨a^{∗}, w⟩
≤ 1 + ε⟨a^{∗}, w⟩

hence ⟨a^{∗}, w⟩ > 0 and (s + ε)⟨a^{∗}, w⟩ + ⟨z^{∗}, w⟩ > 1, thus c(z^{∗}) = s > \frac{1 − ⟨z^{∗}, w⟩}{⟨a^{∗}, w⟩} − ε.

Then consider the function h : C → ℝ defined by h(z^{∗}) = 1 − c(z^{∗}), which is weak*-l.s.c., convex, non-negative and vanishes at 0. Since the sequence (x_{n}^{∗}) converges weakly to a^{∗} but does not converge in norm, one has ⟨x_{n}^{∗}, a⟩ → ⟨a^{∗}, a⟩ = 1, hence z_{n}^{∗} = π(x_{n}^{∗}) = x_{n}^{∗} − ⟨x_{n}^{∗}, a⟩ · a^{∗} → 0 for the weak*-topology, but

lim sup_{n} ∥z_{n}^{∗}∥ = lim sup_{n} ∥x_{n}^{∗} − ⟨x_{n}^{∗}, a⟩a^{∗}∥ = lim sup_{n} ∥x_{n}^{∗} − a^{∗}∥ > 0 .

Since z_{n}^{∗} + ⟨x_{n}^{∗}, a⟩ · a^{∗} = x_{n}^{∗} ∈ B_{X^{∗}}, one sees that c(z_{n}^{∗}) ≥ ⟨x_{n}^{∗}, a⟩, hence that h(z_{n}^{∗}) ≤ 1−⟨x_{n}^{∗}, a⟩ → 0. Thus (z_{n}^{∗}) is a minimizing non relatively compact sequence for h. Then it follows from lemmas 3.2 and 3.3 applied to W and to the sequence (z_{n}^{∗}) in Z ≃ Y^{∗} that there exist continuous functions γ : ℝ^{+} → C ⊂ Y^{∗} and ψ : Y^{∗} → ℝ^{+} and θ > 0 satisfying

• for every k ∈ ℕ, γ is affine on the segment [k, k + 1],

• lim inf_{t→∞} ∥b^{∗} − γ(t)∥ ≥ θ for all b^{∗} ∈ W^{∗},

• lim_{t→∞} h ∘ γ(t) = 0,

• ∀s ≥ 0 ∀x^{∗} ∈ W^{∗} ∥x^{∗} − γ(s)∥ < θ · \frac{s + 2}{2s + 2} ⟹ ψ(x^{∗}) > s.

Then define γ̃(t) = γ(t) + c ∘ γ(t) · a^{∗} and ψ̃ = ψ ∘ π. So π ∘ γ̃(t) = γ(t) and γ(t) ∈ S_{X^{∗}}. Since h is convex and l.s.c., h ∘ γ is convex and l.s.c. on each interval [k, k + 1] hence continuous on [k, k + 1]. Thus c ∘ γ = 1 − h ∘ γ is continuous on ℝ^{+} and so is γ̃.

Define δ = min(1, \frac{θ}{4}). Assume ∥x^{∗} − γ̃ ∘ ψ̃(x^{∗})∥ ≤ δ for some x^{∗} ∈ B_{X^{∗}}, put z^{∗} = π(x^{∗}) and y^{∗} = x^{∗} − γ̃ ∘ ψ ∘ π(x^{∗}). We have π(y^{∗}) = z^{∗} − γ ∘ ψ(z^{∗}) and

∥y^{∗}∥ ≤ δ, hence ∥π(y^{∗})∥ ≤ ∥π∥ · ∥y^{∗}∥ ≤ 2δ ≤ \frac{θ}{2}. Then lemma 3.4 applied to z^{∗} brings a contradiction which proves (1).

Moreover since γ(t) ∈ Z, we have ⟨γ(t), a⟩ = 0, hence

lim_{t→∞}⟨γ̃(t), a⟩ = lim_{t→∞}(⟨γ(t), a⟩ + ⟨(1 − h) ∘ γ(t)⟨a^{∗}, a⟩) = 1 − lim_{t→∞} h ∘ γ(t) = 1

which proves (2). □

4. A parametrized Bishop-Phelps theorem

The theorem 4.1 we prove in this section is very close to (a particular case of) this proved in a very general frame in [3]. And the methods we use in its proof are also very close. Nevertheless we need an extra condition (the limit of ρ(f) as ⟨f, a⟩ converges to 1) which does not seem to easily follow from the result of [3]. It is why we give the below proof which follows the standard proof of Bishop-Phelps’ theorem and should be easy to understand in this concrete situation.

Theorem 4.1. Let X be a separable Banach space, S be its unit sphere, X^{∗} be the dual space of X, S^{∗} be the unit sphere of X^{∗}, a ∈ S and δ > 0. Then there exist a continuous function ρ : S^{∗} → S and a function σ : S^{∗} → X^{∗} such that for every f ∈ S^{∗} ∥σ(f) − f∥ ≤ δ and σ(f) attains its norm at ρ(f), and moreover ∥ρ(f) − a∥ → 0 when ⟨f, a⟩ converges to 1.

Proof. We need some lemmas the first two of them are classical ingredients of the standard proof of Bishop-Phelps’ theorem.

Lemma 4.2. Let f ∈ X^{∗} with ∥f∥ > δ, Γ be the cone {z ∈ X : ⟨f, z⟩ ≥ δ ∥z∥ } and g ≠ 0 in X^{∗}. If g is non-negative on Γ then there exists some λ > 0 such that ∥f − λg∥ ≤ δ.

Proof. Since ∥f∥ > δ one can find some x_{0} ∈ X such that ⟨f, x_{0}⟩ > δ ∥x_{0}∥. Then x_{1} = \frac{x_{0}}{⟨f, x_{0}⟩ − δ ∥x_{0}∥} satisfies ⟨f, x_{1}⟩ = 1 + δ ∥x_{1}∥. Let H be the polar set of the

closed ball B(f, δ), it is

H = B(f, δ)^{o} = {x ∈ X : ∀g ∈ B(f, δ) ⟨g, x⟩ ≥ −1 } = {x : inf_{∥g−f∥≤δ}⟨g, x⟩ ≥ −1 }

And since for fixed x, inf_{∥g−f∥≤δ}⟨g, x⟩ = ⟨f, x⟩ − δ ∥x∥, we have

H = {x ∈ X : ⟨f, x⟩ ≥ −1 + δ ∥x∥ } .

Thus if x ∈ H we get ⟨f, x + x_{1}⟩ ≥ −1 + δ ∥x∥ + δ ∥x_{1}∥ + 1 ≥ δ ∥x + x_{1}∥, hence x + x_{1} ∈ Γ.

If g ≥ 0 on Γ and x ∈ H we have ⟨g, x + x_{1}⟩ ≥ 0 since x + x_{1} ∈ Γ, and

g(x) = g(x + x_{1}) − g(x_{1}) ≥ −g(x_{1}) ≥ − ∥g∥ · ∥x_{1}∥

It follows that g̃ = \frac{g}{∥g∥ · ∥x_{1}∥} satisfies g̃ ≠ 0 and ⟨g̃, x⟩ ≥ −1 for all x ∈ H, hence

g̃ ∈ H^{o} = B(f, δ)^{oo} = \overline{conv(B(f, δ) ∪ \{0\})} = conv(B(f, δ) ∪ \{0\}) = ⋃_{0≤t≤1} t · B(f, δ)

So, since g̃ ≠ 0, there exists t ∈]0, 1] such that g̃ ∈ t·B(f, δ) and we get ∥f − λg∥ ≤ δ

for λ = \frac{1}{t · ∥g∥ · ∥x_{1}∥}. □

Lemma 4.3. Let f ∈ S^{∗}, x̃ ∈ B_{X}, δ > 0 with δ < 1. Assume that ∥x̃ + z∥ ≥ 1 for every z ∈ X such that ⟨f, z⟩ ≤ δ · ∥z∥. Then there exists g ∈ X^{∗} which attains its norm at x̃ and satisfies ∥f − g∥ ≤ δ.

Proof. Denote by U the open unit ball of X. The cone Γ = {z ∈ X : ⟨f, z⟩ ≥ δ ∥z∥ } is convex closed and non-empty since sup_{z∈S}⟨f, z⟩ = ∥f∥ = 1 > δ. By hypothesis the convex sets x̃ + Γ and U are disjoint. Because U is open, it follows from Hahn-Banach’s theorem that there exists a non-zero g_{1} ∈ X^{∗} such that

∥g_{1}∥ = sup_{z∈U}⟨g_{1}, z⟩ ≤ inf_{y∈x̃+Γ}⟨g_{1}, y⟩ = inf_{w∈Γ}⟨g_{1}, x̃ + w⟩ = ⟨g_{1}, x̃⟩ + inf_{w∈Γ}⟨g_{1}, w⟩

Since Γ is a cone we have either inf_{w∈Γ}⟨g_{1}, w⟩ = 0 or inf_{w∈Γ}⟨g_{1}, w⟩ = −∞. The second case cannot happen since inf_{w∈Γ}⟨g_{1}, w⟩ ≥ ∥g_{1}∥ − ⟨g_{1}, x̃⟩. So ⟨g_{1}, x̃⟩ ≥ ∥g_{1}∥ and we conclude that x̃ ∈ S, that g_{1} attains its norm at x̃ and that g_{1} ≥ 0 on Γ. Then it follows from lemma 4.2 that there is λ > 0 such that ∥f − λg_{1}∥ ≤ δ. And since g = λg_{1} attains its norm at x̃ too, this completes the proof. □

Fix δ ∈]0, 1[. For f ∈ S^{∗}, x ∈ U, η > 0 we put

Γ_{f} = {y ∈ X : ⟨f, y⟩ > δ ∥y∥}

M(f, x) = sup_{z∈U∩(x+Γ_{f})}⟨f, z − x⟩

C_{f,x}^{η} = {y ∈ U : y ∈ x + Γ_{f} and ⟨f, y − x⟩ > M(f, x) − η }

Ω_{η} = {(f, x, y) ∈ S^{∗} × U × U : ⟨f, y − x⟩ > max (δ ∥y − x∥ , M(f, x) − η) }

It is clear that for all f ∈ S^{∗} and all x ∈ U there exists some y ∈ X such that y ∈ Γ_{f}, hence some t > 0 such that z = x + ty ∈ U and z ∈ x + Γ_{f}; in particular M(f, x) is well defined and 0 < M(f, x) ≤ sup_{z∈U}⟨f, z − x⟩ = 1 − ⟨f, x⟩. Moreover the set C_{f,x}^{η} is a non-empty open convex subset of X and C_{f,x}^{η} = {y : (f, x, y) ∈ Ω_{η}}.

Lemma 4.4. Let V be an bounded open convex neighborhood of 0, r > 0 such that r · B_{X} ⊂ V, R = sup_{y∈V} ∥y∥. Then the mapping f ↦ V ∩ Γ_{f} is locally μ-Lipschitz from S^{∗} into the set BC(X) of bounded convex subsets of X equipped with

the Hausdorff distance, where μ = \frac{R · (R + r)}{r(1 − δ)}.

Proof. Let θ ∈]δ, 1[. If f ∈ S^{∗} there is some w ∈ X such that ∥w∥ = r and ⟨f, w⟩ > θr. Then if y ∈ V ∩ Γ_{g} and t ∈]0, 1[, the point z = t · w + (1 − t) · y belongs to V. Moreover the function z ↦ ⟨f, z⟩ − δ ∥z∥ is concave, whence

⟨f, z⟩ − δ ∥z∥ ≥ t(⟨f, w⟩ − δ ∥w∥) + (1 − t)(⟨f, y⟩ − δ ∥y∥)
> t(θ − δ)r + (1 − t)(⟨g, y⟩ − δ ∥y∥ − ∥f − g∥ · ∥y∥)
> t(θ − δ)r − R · ∥f − g∥

If follows that z ∈ V ∩ Γ_{f} if t = \frac{R · ∥f − g∥}{r(θ − δ)} provided that t ≤ 1, what means

∥f − g∥ ≤ α′ = \frac{r(θ − δ)}{R}. And since ∥y − z∥ = t ∥w − y∥ ≤ t(R + r), we get

sup_{y∈V ∩Γ_{g}} d(y, V ∩ Γ_{f}) ≤ ∥f − g∥ · \frac{R · (R + r)}{r(θ − δ)}, it is

∥f − g∥ ≤ α′ ⟹ d_{H}(V ∩ Γ_{g}, V ∩ Γ_{f}) ≤ \frac{R · (R + r)}{r(θ − δ)} · ∥f − g∥ .

Since this holds for any θ ∈]δ, 1[ we get

∥f − g∥ < α = \frac{r(1 − δ)}{R} ⟹ d_{H}(V ∩ Γ_{g}, V ∩ Γ_{f}) ≤ \frac{R · (R + r)}{r(1 − δ)} · ∥f − g∥ ,

and this shows that the function f ↦ V ∩ Γ_{f} is locally μ-Lipschitz. □

Lemma 4.5. The function M is continuous on S^{∗} × U.

Proof. For fixed x ∈ U and r ≤ 1 − ∥x∥ we have M(f, x) = sup_{y∈(U−x)∩Γ_{f}} ⟨f, y⟩. If

g ∈ S^{∗} with ∥g − f∥ < α = \frac{r(1 − δ)}{R}, and z ∈ (U − x) ∩ Γ_{g}, we have ∥z∥ ≤ 2 and

⟨g, z⟩ ≤ ⟨f, z⟩ + ∥f − g∥ · ∥z∥ ≤ ⟨f, z⟩ + 2 ∥f − g∥ .

For all y ∈ (U − x) ∩ Γ_{f}, we also have

⟨f, z⟩ ≤ ⟨f, y⟩ + ∥f∥ · ∥z − y∥ ≤ M(f, x) + ∥y − z∥ ,

thus

⟨g, z⟩ ≤ ⟨f, z⟩ + 2 ∥f − g∥ ≤ M(f, x) + d_{H}((U − x) ∩ Γ_{f}, (U − x) ∩ Γ_{g}) + 2 ∥f − g∥
≤ M(f, x) + λ · ∥f − g∥

where λ = 2 + μ. It follows that M(g, x) ≤ M(f, x) + λ ∥f − g∥, and by symmetry |M(f, x) − M(g, x)| ≤ λ ∥f − g∥ as soon as ∥f − g∥ < α.

For fixed f the function M_{f} : x ↦ M(f, x) takes values in [0, 2] and is concave : indeed let x and x′ in U, t ∈ [0, 1] and x′′ = t · x + (1 − t) · x′ ; for positive ε one can find z and z′ in U such that ⟨f, z⟩ > M(f, x) − ε, ⟨f, z′⟩ > M(f, x′) − ε, ⟨f, z−x⟩ > δ ∥z − x∥ and ⟨f, z′−x′⟩ > δ ∥z′ − x′∥. Then the point z′′ = t·z+(1−t)·z′ belongs to U, satisfies ⟨f, z′′ − x′′⟩ > δ ∥z′′ − x′′∥ and we have

M(f, x′′) ≥ ⟨f, z′′−x′′⟩ = t⟨f, z−x⟩+(1−t)⟨f, z′−x′⟩ > t·M(f, x)+(1−t)M(f, x′)−ε

from what we conclude that M_{f} is concave and a fortiori continuous on U. Finally if ∥x∥ < 1 − r and ∥y∥ < 1 − r, we have for f and g ∈ S^{∗} such that ∥f − g∥ < α :

|M(g, y) − M(f, x)| ≤ |M(g, y) − M(f, y)| + |M(f, y) − M(f, x)|
≤ λ ∥g − f∥ + |M_{f}(y) − M_{f}(x)|

and we conclude that M is continuous at (f, x). □

It follows from previous lemma that the set Ω_{η} is open in S^{∗} × U × U.

Lemma 4.6. Let η ∈]0, 1 − δ[ and f ∈ S^{∗}. If 0 ≤ λ ≤ 1 − η < 1 − \frac{η}{2} ≤ μ < 1 and

if ⟨f, a⟩ ≥ 1 − \frac{η}{2} , we have (f, λ · a, μ · a) ∈ Ω_{η}.

Proof. For x = λ·a and y = μ·a, we have ⟨f, y −x⟩ = (μ−λ)⟨f, a⟩ ≥ (μ−λ)(1− \frac{η}{2} )

and δ ∥y − x∥ = δ(μ − λ), hence y − x ∈ Γ_{f} since μ ≠ λ and 1 − \frac{η}{2} > δ.

We have shown that M(f, x) ≤ 1 − ⟨f, x⟩ whenever x ∈ U. For x = λ · a and y = μ · a, one has

⟨f, y − x⟩ − (M(f, x) − η) ≥ (μ − λ) · ⟨f, a⟩ − 1 + λ · ⟨f, a⟩ + η = μ · ⟨f, a⟩ − 1 + η

≥ (1 − \frac{η}{2} )(1 − \frac{η}{2} ) − 1 + η = \frac{η^{2}}{4} > 0 ,

what completes the proof of the lemma. □

Lemma 4.7. For positive fixed δ and η < 1 − δ there exists a continuous function h_{η} from T = S^{∗} × U to U such that h_{η}(f, x) ∈ C_{f,x}^{η} for all (f, x) ∈ T and moreover

h_{η}(f, λ · a) = (1 − \frac{η}{2} ) · a for all λ ∈ [0, 1 − η] whenever ⟨f, a⟩ ≥ 1 − \frac{η}{2} .

Proof. Since Ω_{η} is open, the multivalued mapping Q_{η} : (f, x) ↦ C_{f,x}^{η} is l.s.c. with convex open values. Moreover the set

F = {(f, λ · a) : 0 ≤ λ ≤ 1 − η and ⟨f, a⟩ ≥ 1 − η/2 }

is closed in T and the function h̃_{η} : (f, λ · a) ↦ (1 − \frac{η}{2} ) · a is a continuous selection of Q_{η} on F by lemma 4.6. By Michael’s theorem ([4], 3.1′′′), Q_{η} has a continuous selection h_{η} which extends h̃_{η} since X is separable and T metrizable. So we can choose h_{η} such that h_{η}(f, λ·a) = (1− \frac{η}{2} )·a when ⟨f, a⟩ ≥ 1−η/2 and λ ≤ 1−η. □

Then we define inductively a sequence (φ_{n})_{n∈ℕ} of continuous functions from S^{∗} to U by letting φ_{0}(f) = 0 and φ_{n+1}(f) = h_{η·2^{−n}}(f, φ_{n}(f)). For fixed f ∈ S^{∗} the sequence x_{n} = φ_{n}(f) satisfies then the properties :

i) ⟨f, x_{n+1} − x_{n}⟩ > δ ∥x_{n+1} − x_{n}∥

ii) ⟨f, x_{n+1}⟩ > M(f, x_{n}) − η · 2^{−n}

Since

⟨f, x_{n+2} − x_{n}⟩ = ⟨f, x_{n+2} − x_{n+1}⟩ + ⟨f, x_{n+1} − x_{n}⟩
> δ ∥x_{n+2} − x_{n+1}∥ + δ ∥x_{n+1} − x_{n}∥ ≥ δ ∥x_{n+2} − x_{n}∥

we see that x_{n+2} − x_{n} ∈ Γ_{f}, hence⟨f, x_{n+2} − x_{n}⟩ ≤ M(f, x_{n}), and that

δ · ∥x_{n+2} − x_{n+1}∥ < ⟨f, x_{n+2} − x_{n+1}⟩ = ⟨f, x_{n+2} − x_{n}⟩ − ⟨f, x_{n+1} − x_{n}⟩
< M(f, x_{n}) − (M(f, x_{n}) − η · 2^{−n}) = η · 2^{−n}

We then have ∥x_{n+2} − x_{n+1}∥ ≤ ηδ^{−1} · 2^{−n}, what shows that the sequence (x_{n}) converges in X to some x̃. Thus the sequence (φ_{n}) of continuous functions satisfies ∥φ_{n+2}(f) − φ_{n+1}(f)∥ ≤ ηδ^{−1}·2^{−n} and converges uniformly to a continuous function ρ : S^{∗} → B_{X}.

For n < p one has

⟨f, x_{p} − x_{n}⟩ = ∑_{j=n}^{p−1}⟨f, x_{j+1} − x_{j}⟩ > δ ∑_{j=n}^{p−1} ∥x_{j+1} − x_{j}∥ ≥ δ ∥x_{p} − x_{n}∥

and passing to the limit we get ⟨f, x̃⟩ − ⟨f, x_{n}⟩ ≥ δ ∥x̃ − x_{n}∥. If there was some z ∈ U such that z − x̃ ∈ Γ_{f} we would also have ⟨f, z⟩ − ⟨f, x_{n}⟩ > δ ∥z − x_{n}∥, hence

M(f, x_{n}) ≥ ⟨f, z − x_{n}⟩ ≥ ⟨f, z − x̃⟩ + ⟨f, x_{n+1} − x_{n}⟩ ≥ ⟨f, z − x̃⟩ + M(f, x_{n}) − η · 2^{−n}

and δ ∥z − x̃∥ ≤ ⟨f, z−x̃⟩ ≤ η·2^{−n} for all n, what would imply x̃ = z in contradiction with z − x̃ ∈ Γ_{f}. But if x̃ ∈ U, we could find some w ∈ Γ_{f} then some t > 0 such that z = x̃ + t · w ∈ U. So x̃ ∈ B \ U = S. Hence ρ(f) = x̃ ∈ S.

It follows that \left∥{ \atop x̃}\right∥ = 1 and that the open set U is disjoint from x̃+Γ_{f}, hence also disjoint from x̃ + \overline{Γ_{f}} = \{z : ⟨f, z − x̃⟩ ≥ δ ∥z − x̃∥\}. Then it follows from lemma 4.3 that there exists g ∈ X^{∗} which attains its norm at x̃ and such that ∥f − g∥ ≤ δ. We conclude that for all f ∈ S^{∗} there exists some σ(f) ∈ X^{∗} which attains its norm at ρ(f) and satisfies ∥f − σ(f)∥ ≤ δ.

Moreover, if ⟨f, a⟩ > 1−η·2^{−p−1} for some integer p one checks easily by induction on n < p that x_{n+1} = φ_{n+1}(f) = (1 − η · 2^{−n}) · a, hence that x_{p} = (1 − η · 2^{1−p}) · a. And since

δ · ∥x̃ − x_{p}∥ ≤ ⟨f, x̃ − x_{p}⟩ ≤ \left∥{ \atop x̃}\right∥ − ⟨f, x_{p}⟩ ≤ 1 − (1 − η · 2^{1−p})⟨f, a⟩
≤ 1 − (1 − η · 2^{1−p})(1 − η · 2^{−p}) ≤ η · 2^{1−p} + η · 2^{−p} = 3η · 2^{−p}

one finds

∥ρ(f) − a∥ = ∥x̃ − a∥ ≤ ∥x̃ − x_{p}∥ + ∥x_{p} − a∥ ≤ 3\frac{η}{δ} · 2^{−p} + η · 2^{1−p} ≤ (\frac{3}{δ} + 2) · η · 2^{−p}

what shows that ρ(f) tends to a as ⟨f, a⟩ tends to 1.

More precisely it follows from what precedes that ρ(f) = a when ⟨f, a⟩ = 1 and that if η · 2^{−p−2} ≤ 1 − ⟨f, a⟩ < η · 2^{−p−1} then

∥ρ(f) − a∥ ≤ (\frac{3}{δ} + 2) · η · 2^{−p} ≤ 4(\frac{3}{δ} + 2) · (1 − ⟨f, a⟩) .

Moreover, if 1 − ⟨f, a⟩ ≥ \frac{η}{2} one has ∥ρ(f) − a∥ ≤ 2 ≤ \frac{4}{η} (1 − ⟨f, a⟩).

So ∥ρ(f) − a∥ ≤ τ · (1 − ⟨f, a⟩) with τ = 4 · max(\frac{1}{η} , (\frac{3}{δ} + 2)). This completes the

proof of Theorem 4.1. □

Notice that the function σ above is defined pointwise after the construction of the continuous function ρ. So nothing proves that it could be continuous. The best we can say is that σ has to be a selection of the multivalued mapping Ψ : S^{∗} → X^{∗}

g ↦ Ψ(g) = {z^{∗} : ⟨z^{∗}, ρ(g)⟩ ≥ ∥z^{∗}∥ } ∩ {z^{∗} : ∥z^{∗} − g∥ ≤ δ }

which is cusco when S_{X^{∗}} is equipped with the norm-topology and X^{∗} with the weak*-topology since its graph is closed, and thus that σ could be chosen normweak* Baire-1. It follows that σ is necessarily norm-weak* continuous at each point f such that Ψ(f) is a singleton : in particular if the norm of X is Gâteauxdifférentiable this is the case at each point f ∈ S^{∗}.

Corollary 4.8. Let X be a separable Banach space, S be its unit sphere, X^{∗} be the dual space of X, S^{∗} be the unit sphere of X^{∗}, a ∈ S and δ > 0. Then there exist a continuous function ρ : S^{∗} → S and a function σ : S^{∗} → S^{∗} such that for every f ∈ S^{∗} ∥σ(f) − f∥ ≤ δ and σ(f) attains its norm at ρ(f), and moreover ρ(f) converges in norm to a when ⟨f, a⟩ converges to 1.

Proof. Apply theorem 4.1 with δ′ = δ/2. We get a continuous function ρ : S^{∗} → S and a function σ′ : S^{∗} → X^{∗} such that ∥σ′(f) − f∥ ≤ δ′ and σ′(f) attains its norm

at ρ(f). Put σ(f) = \frac{σ′(f)}{∥σ′(f)∥}. We have σ(f) ∈ S^{∗} and σ(f) attains its norm at

ρ(f). It remains to prove that ∥σ(f) − f∥ ≤ δ.

Put s = ∥σ′(f)∥. We have |1 − s| = |∥f∥ − ∥σ′(f)∥| ≤ ∥f − σ′(f)∥ ≤ δ′ and

∥σ(f) − f∥ ≤ ∥σ′(f) − f∥ + ∥σ′(f) − σ(f)∥ = ∥σ′(f) − f∥ + |1 − s^{−1}| · ∥σ′(f)∥
= ∥σ′(f) − f∥ + |1 − s^{−1}| · s ≤ δ′ + |s − 1| ≤ 2δ′ = δ .

And this completes the proof since lim_{⟨f,a⟩→1} ∥ρ(f) − a∥ = 0. □

5. Proof of the main theorem

Lemma 5.1. Let X be a separable infinite-dimensional Banach space and δ > 0. Then there exists a compact mapping K : X^{∗} → X such that ∥K(x^{∗})∥ ≤ 3^{−1} · ∥x^{∗}∥ and ⟨y^{∗}, K(x^{∗})⟩ > 0 holds for all x^{∗} and y^{∗} in X^{∗} such that ∥x^{∗} − y^{∗}∥ ≤ δ < ∥x^{∗}∥.

Proof. There exists in X a sequence (a_{k})_{k≥0} of vectors of norm 1 which spans X. Define the linear operator T : ℓ_{2} → X by T e_{k} = 2^{−k−1} · a_{k} where (e_{k})_{k≥0} is the canonical basis of ℓ^{2}, it is

T u = ∑_{k=0}^{∞} 2^{−k−1}u_{k} · a_{k} for u = (u_{k})_{k≥0} ∈ ℓ^{2}

We have

∥T u∥ ≤ ∑_{k=0}^{∞} 2^{−k−1} |u_{k}| · ∥a_{k}∥ = ∑_{k=0}^{∞} 2^{−k−1} |u_{k}| ≤ (∑_{k=0}^{∞} 2^{−2k−2})^{1/2}.(∑_{k=0}^{∞} u_{k}^{2})^{1/2}

= \frac{1}{\sqrt{3}} · ∥u∥

whence we conclude that T is continuous and that ∥T ∥ ≤ 3^{−1/2}. Moreover it is easily checked that denoting by T_{n} the continuous linear operator of finite rank : u ↦ T_{n}u = ∑_{k=0}^{n} 2^{−k−1}u_{k} · a_{k}, we have

∥T u − T_{n}u∥^{2} ≤ ( ∑_{k=n+1}^{∞} 2^{−k} |u_{k}| · ∥a_{k}∥)^{2} ≤ (∑_{n+1}^{∞} 2^{−2k}) · (∑_{n+1}^{∞} u_{k}^{2}) = \frac{2^{−2n}}{3} · ∥u∥^{2}

hence ∥T − T_{n}∥ ≤ 3^{−1/2} · 2^{−n}, and this shows the compactness of T.

Then T^{∗} : X^{∗} → ℓ^{2} is compact too and weak* continuous from X^{∗} to ℓ^{2}. Since B_{X^{∗}} is weak*-compact, so is L = T^{∗}(δ ·B_{X^{∗}}). It follows that L is a compact convex subset of ℓ^{2}, and in particular closed and convex. Notice that T^{∗} is one-to-one : indeed, if x^{∗} ∈ ker(T^{∗}) we have for all k ≥ 0 : 0 = ⟨T^{∗}x^{∗}, e_{k}⟩ = ⟨x^{∗}, T e_{k}⟩ = ⟨x^{∗}, a_{k}⟩, and since the sequence (a_{k}) spans X, this shows that x^{∗} = 0.

If p denotes the orthogonal projection on L, then the mapping p̃: u ↦ u − p(u) is continuous from ℓ^{2} to itself and the mapping

K = T ∘ p̃ ∘ T^{∗} : x^{∗} ↦ T (T^{∗}x^{∗} − p(T^{∗}x^{∗}))

is compact from X^{∗} to X.

Moreover, if u ∈ ℓ^{2} \ L we have for all v ∈ L

⟨u − v, u − p(u)⟩ ≥ ⟨u − p(u), u − p(u)⟩ = ∥u − p(u)∥^{2} = d(u, L)^{2} > 0 .

Then if ∥x^{∗}∥ > δ and ∥y^{∗} − x^{∗}∥ ≤ δ we put u = T^{∗}x^{∗} and v = T^{∗}(x^{∗} − y^{∗}) ∈ L. Then u ∉ L since T^{∗} is one-to-one and we get

⟨y^{∗}, K(x^{∗})⟩ = ⟨y^{∗}, T (T^{∗}x^{∗} − p(T^{∗}x^{∗}))⟩ = ⟨T^{∗}y^{∗}, T^{∗}x^{∗} − p(T^{∗}x^{∗})⟩
= ⟨u − v, u − p(u)⟩ > 0 .

Finally ∥K(x^{∗})∥ = ∥T (p̃ ∘ T^{∗}x^{∗})∥ ≤ ∥T ∥ · ∥T^{∗}x^{∗} − p(T^{∗}x^{∗})∥. Since 0 ∈ L, ∥p̃(u)∥ = d(u, L) ≤ ∥u∥. So ∥p̃(T^{∗}x^{∗})∥ ≤ ∥T^{∗}x^{∗}∥ ≤ ∥T^{∗}∥ · ∥x^{∗}∥ = ∥T ∥ · ∥x^{∗}∥ and

∥K(x^{∗})∥ ≤ ∥T ∥^{2} · ∥x^{∗}∥ ≤ \frac{∥x^{∗}∥}{3}. □

Theorem 5.2. Let X be a separable infinite-dimensional Banach space which does not have the NAKK* property . Then there exists a compact mapping φ : B_{X^{∗}} → X \ {0} such that ⟨f, φ(f)⟩ < ∥φ(f)∥ for every f ∈ B_{X^{∗}}.

Proof. Since X does not have the NAKK* property there are a ∈ S_{X} and a^{∗} ∈ S_{X^{∗}} such that a^{∗} attains its norm at a, it is ⟨a^{∗}, a⟩ = 1, and the map κ is discontinuous at a^{∗}.

By lemma 3.5 we can find continuous functions ψ̃ from the unit ball B_{X^{∗}} of X^{∗} into ℝ^{+} and γ̃ : ℝ^{+} → B_{X^{∗}}, and δ ∈]0, 1] such that

(1) for every f ∈ B_{X^{∗}} ∥f − γ̃ ∘ ψ̃(f)∥ > δ,
(2) lim_{t→∞}⟨γ̃(t), a⟩ = ⟨a^{∗}, a⟩ = 1.

By corollary 4.8 there are a continuous function ρ : S_{X^{∗}} → S_{X} and a function σ : S_{X^{∗}} → S_{X^{∗}} such that for every f ∈ S_{X^{∗}} ∥σ(f) − f∥ ≤ δ and ⟨σ(f), ρ(f)⟩ = 1, and moreover ρ(f) converges in norm to a when ⟨f, a⟩ converges to 1.

Consider the continuous function φ_{1} = ρ ∘ γ̃ ∘ ψ̃ : B_{X^{∗}} → S_{X}. It is clear that φ_{1}(B_{X^{∗}}) ⊂ {ρ ∘ γ̃(t) : t ∈ ℝ^{+}} and that this latter set is relatively compact in S_{X} since lim_{t→∞} ρ ∘ γ̃(t) = a. So φ_{1} is a compact mapping. Now take

φ(f) = φ_{1}(f) + K(γ̃ ∘ ψ̃(f) − f)

where K is the compact mapping defined in lemma 5.1. Then φ is a compact mapping from B_{X^{∗}} to X.

Denote g = γ̃ ∘ ψ̃(f). So ∥g − f∥ ≤ ∥g∥ + ∥f∥ ≤ 2, and ∥K(g − f)∥ ≤ \frac{2}{3}. For

f ∈ B_{X^{∗}} we have ∥φ(f)∥ ≥ ∥φ_{1}(f)∥ − \frac{2}{3} = \frac{1}{3}. So we get inf_{f∈B_{X^{∗}}} ∥φ(f)∥ > 0 and

φ is a compact mapping from B_{X^{∗}} to X \ {0}.

Finally, we have ∥σ(g) − g∥ ≤ δ < ∥g − f∥. Hence ⟨σ(g) − f, K(g − f)⟩ > 0 (put x^{∗} = g − f and y^{∗} = σ(g) − f ). Thus φ(f) = ρ(g) + K(g − f) and

⟨σ(g), φ(f)⟩ = ⟨σ(g), ρ(g)⟩ + ⟨σ(g), K(g − f)⟩
= 1 + ⟨f, K(g − f)⟩ + ⟨σ(g) − f, K(g − f)⟩
> 1 + ⟨f, φ(f) − ρ(g)⟩ = 1 + ⟨f, φ(f)⟩ − ⟨f, ρ(g)⟩
≥ ⟨f, φ(f)⟩ + 1 − ∥f∥ · ∥ρ(g)∥ = ⟨f, φ(f)⟩

hence

∥φ(f)∥ = ∥φ(f)∥ · ∥σ(g)∥ ≥ ⟨σ(g), φ(f)⟩ > ⟨f, φ(f)⟩

what shows that φ satisfies the required conditions. □

6. The non-separable case

Theorem 6.1. Let X be a Banach space which does not have the NAKK* property. Then there exists a compact mapping φ : B_{X^{∗}} → X \ {0} such that ⟨f, φ(f)⟩ < ∥φ(f)∥ for every f ∈ B_{X^{∗}}.

Proof. Suppose X does not have the NAKK* property. Then there exists a ∈ S_{X} and a^{∗} ∈ S_{X^{∗}} such that a^{∗} attains its norm at a and that the mapping κ_{X} is discontinuous at a^{∗}. So there is some δ > 0 such that S_{X^{∗}} ∩ B̃(a^{∗}, δ) is not a neighborhood of a^{∗} in S_{X^{∗}} for the weak*-topology.

We construct inductively a sequence (x_{n}^{∗}) ∈ S_{X^{∗}} and a sequence (x_{n}) in S_{X} in such a way that

(1) x_{0} = a

(2) ∥x_{n}^{∗} − a^{∗}∥ > δ
(3) |⟨x_{n}^{∗} − a^{∗}, x_{k}⟩| < 2^{−n} whenever 0 ≤ k < 3n
(4) ⟨a^{∗}, x_{3n}⟩ > 1 − 2^{−n}
(5) ⟨x_{n}^{∗} − a^{∗}, x_{3n+1}⟩ > δ
(6) ⟨x_{n}^{∗}, x_{3n+2}⟩ > 1 − 2^{−n}

The choice of x_{n}^{∗} is possible since {z^{∗} ∈ S_{X^{∗}} : ∀k < 3n |⟨z^{∗} − a^{∗}, x_{k}⟩| < 2^{−n}} is a weak* neighborhood of a^{∗} in S_{X^{∗}}, hence is not contained in the closed ball B̃(a^{∗}, δ) of S_{X^{∗}}. The choice of x_{3n} is possible since sup_{y∈S_{X}} ⟨a^{∗}, y⟩ = ∥a^{∗}∥ > 1 − 2^{−n}, and same for x_{3n+1} since ∥x_{n}^{∗} − a^{∗}∥ > δ and x_{3n+2} since ∥x_{n}^{∗}∥ > 1 − 2^{−n}.

Consider the closed linear subspace Y of X spanned by the x_{n}’s. Then Y is a separable Banach space and its dual Y^{∗} is isometric to the quotient of X^{∗} by the closed subspace Y^{⊥} = {z^{∗} ∈ X^{∗} : ∀n ⟨z^{∗}, x_{n}⟩ = 0} of X^{∗}.

Denote by j the canonical injection of Y into X ; then j^{∗} is the canonical projection : X^{∗} → X^{∗}/Y^{⊥}. We then have ∥j^{∗}(x_{n}^{∗})∥ ≤ ∥x_{n}^{∗}∥ = 1 and

∥j^{∗}(x_{n}^{∗})∥ = ∥j^{∗}(x_{n}^{∗})∥ · ∥x_{3n+2}∥ ≥ ⟨j^{∗}(x_{n}^{∗}), x_{3n+2}⟩ = ⟨x_{n}^{∗}, j(x_{3n+2})⟩ > 1 − 2^{−n}

Hence ∥j^{∗}(x_{n}^{∗})∥ → 1. Analogously we get 1 ≥ ∥j^{∗}(a^{∗})∥ ≥ sup_{n}⟨a^{∗}, x_{3n}⟩ = 1 and

lim inf_{n→∞} ∥j^{∗}(x_{n}^{∗}) − j^{∗}(a^{∗})∥ ≥ lim inf_{n→∞}⟨x_{n}^{∗} − a^{∗}, x_{3n+1}⟩ ≥ δ .

Since |⟨j^{∗}(x_{n}^{∗}) − j^{∗}(a^{∗}), x_{k}⟩| < 2^{−n} for n > k/3 and the sequence (j^{∗}(x_{n}^{∗})−j^{∗}(a^{∗})) is bounded, we have lim_{n→∞}⟨j^{∗}(x_{n}^{∗}) − j^{∗}(a^{∗}), y⟩ = 0 for all y ∈ Y. Thus the sequence

ξ_{n} = \frac{j^{∗}(x_{n}^{∗})}{∥j^{∗}(x_{n}^{∗})∥} converges to j^{∗}(a^{∗}) in S_{Y^{∗}} for the weak*-topology but not for the

norm-topology. So κ_{Y} is discontinuous at j^{∗}(a^{∗}). Moreover j^{∗}(a^{∗}) attains its norm at x_{0} = a. It follows that Y is a separable Banach space not having the NAKK* property .

By theorem 5.2 one can find a compact mapping ψ : B_{Y^{∗}} → Y \ {0} such that ⟨f, ψ(f)⟩ < ∥ψ(f)∥ for every f ∈ B_{Y^{∗}}. Then φ = j ∘ ψ ∘ j^{∗} : B_{X^{∗}} → X \ {0} is compact and for every g ∈ B_{X^{∗}} we have f = j^{∗}(g) ∈ B_{Y^{∗}} hence φ(g) = j ∘ ψ(f) and ∥φ(g)∥ = ∥j(ψ(f))∥ = ∥ψ(f)∥ whereas

⟨g, φ(g)⟩ = ⟨g, j(ψ(f)⟩ = ⟨j^{∗}(g), ψ(f)⟩ = ⟨f, ψ(f)⟩ < ∥ψ(f)∥ = ∥φ(g)∥ .

This completes the proof. □

It should be noticed that theorem 6.1 contains the result of [6] : indeed if X is a reflexive Banach space without the Kadec-Klee property then X^{∗} is a reflexive Banach space without the dual Kadec-Klee property, which is here equivalent to the NAKK* property since NA(X^{∗}) = S_{X^{∗∗}} = S_{X}. So it follows from theorem 6.1 that one can find some compact mapping φ from B_{X^{∗∗}} = B_{X} to X^{∗} \ {0} such that ⟨φ(x), x⟩ < ∥φ(x)∥ for all x ∈ B_{X}.

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Sorbonne Universités, Université Pierre et Marie Curie (Univ Paris 06), CNRS Institut de Mathématique de Jussieu - Paris Rive Gauche Université Pierre et Marie Curie (Paris 6) Boîte 186 - 4 place Jussieu, F- 75252 Paris Cedex 05, FRANCE

  E-mail address: jean.saint-raymond@imj-prg.fr