KADEC-KLEE PROPERTY AND FIXED POINTS

JEAN SAINT RAYMOND

Abstract. Let X be a reflexive Banach space which does not have the Kadec-Klee property. Then there exists a compact mapping f from the unit ball B_{X} of X to the dual space X^{∗} such that inf_{x∈B_{X}} ∥f(x)∥ > 0 and ⟨f(x), x⟩ < ∥f(x)∥ for every x ∈ B_{X}.

A Banach space X is locally uniformly rotund (lur for short) if whenever x is in the unit sphere S_{X} and the sequence (x_{n}) in the unit ball B_{X} satisfies ∥x + x_{n}∥ → 2 then ∥x − x_{n}∥ → 0. It is well known ( cf [1]) that the norm is a C^{1} function on X \ {0} if the dual norm is lur on X^{∗}.

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 a weaker than being lur, was introduced by M. Kadec in his proof that every separable Banach space can be equipped with an equivalent lur 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 [2], B. Ricceri has shown the following :

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

In the same paper B. Ricceri asked whether this property could characterize the Kadec-Klee property among the reflexive Banach spaces. The aim of this note is to give a positive answer to this question.

Observe first that Ricceri’s theorem does not hold anymore if the hypothesis of reflexivity is removed. Indeed by James’ theorem, if X is not reflexive there exists a ξ ∈ X^{∗} of norm 1 which does not attain on B_{X} its supremum; then the constant function f on B_{X} with value ξ is compact but cannot satisfy the conclusion. So along this note we will always assume that the space X is reflexive even though some of the lemmas proved here would hold in a more general setting.

The method used by Ricceri consists to apply Fan-Kakutani’s fixed point theorem to the cusco multivalued mapping x ↦ {y ∈ B_{X} : ⟨f(x), y⟩ = ∥f(x)∥}. Hence it is not surprizing that we had to construct (Lemma 5) a fixed point free function when X is a reflexive Banach space without the Kadec-Klee property.

2000 Mathematics Subject Classification. 52A40 , 46A25, 46B10 .

Key words and phrases. convex functions, reflexive Banach spaces, weak compactness, Kadec-Klee property.

Lemma 1. Let X be a separable reflexive Banach space. Then there exists a linear compact mapping K : X → X^{∗} such that ⟨Kx, x⟩ > 0 holds for all x ≠ 0 in X.

Proof. There exists in X^{∗} a sequence (α_{k})_{k≥1} of continuous linear functionals of norm 1 which separates X. Define the linear operator T : X → ℓ_{2} by

T x = (\frac{1}{k} ⟨α_{k}, x⟩)_{k}

it is T = ∑_{k≥1} \frac{1}{k} α_{k} ⊗ e_{k}, if (e_{k}) stands for the canonical basis of ℓ_{2}.

It is easily checked that if T_{n} denotes the continuous linear operator of finite

rank defined by T_{n}x = ∑_{k≤n} \frac{1}{k} ⟨α_{k}, x⟩e_{k}, we have

∥T x − T_{n}x∥^{2} ≤ ∑_{k=n+1}^{∞} \frac{1}{k^{2}} ∥x∥^{2} ≤ \frac{∥x∥^{2}}{n}

hence ∥T − T_{n}∥ ≤ n^{−1/2} and this shows the compactness of T. Then T^{∗} : ℓ^{2} → X^{∗} is a compact operator too, and so is K = T^{∗}T : X → X^{∗}.

Moreover, for x ∈ X, we have ⟨Kx, x⟩ = ⟨T x, T x⟩ = ∑_{n=1}^{∞} n^{−2} · ⟨α_{n}, x⟩^{2} ≥ 0. Finally, if ⟨Kx, x⟩ = 0, we must have ⟨α_{n}, x⟩ = 0 for all n, thus x = 0 since the sequence (α_{n}) separates X. □

Lemma 2. Let X be a reflexive separable Banach space, C ⊂ X a bounded closed convex set of diameter R containing 0, h : C → ℝ a convex lsc non-negative function such that h(0) = 0. Then, either all minimizing sequences of h have converging subsequences, or there exist a point a ∈ h^{−1}(0), a sequence (x_{n}) in C converging weakly to a and some δ > 0 satisfying:

i) ∀n ∥x_{n} − a∥ > \frac{2δ}{3} .

ii) ∀b ∈ X ∃k ∈ ℕ ∀n ≥ k ∀t ∈ [0, 1] ∥b − (tx_{n} + (1 − t)x_{n+1})∥ > θ := \frac{δ^{2}}{2R}

iii) ∀n h(x_{n}) < 2^{−n} · min(1, \frac{θ^{2}}{8} ).

Proof. Let (u_{j})_{j∈ℕ} be a dense sequence in X. We define inductively a sequence (V_{n}) of finite-dimensional linear subspaces of X, 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 (u_{j})_{j<n} and (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}/2. One has δ_{0} ≤ R = diam(C) < +∞.

If the sequence (δ_{n}) converges to 0 every minimizing sequence (z_{k}) of h is totally bounded, hence relatively compact and possesses a 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} ≤ \frac{R}{2}. By weak compactness of C and Eberlein theorem, we can pass to a subsequence and assume that both

the sequence (x_{n}) converges weakly to some a ∈ C and h(x_{n}) < 2^{−n} · \frac{θ^{2}}{8} . Since h is convex and lsc, hence weakly lsc, we necessarily have

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

and this shows that a ∈ h^{−1}(0).

For all b ∈ X there is some j such that ∥b − u_{j}∥ < \frac{δ′}{6} .Then for k > j, we have

u_{j} ∈ V_{k}, hence d(x_{k}, u_{j}) ≥ d(x_{k}, V_{k}) > \frac{1}{2}δ_{k} ≥ \frac{δ′}{2} , and

∥b − x_{k}∥ ≥ \frac{δ′}{2} − ∥b − u_{j}∥ > \frac{δ′}{3} = \frac{2δ}{3}

In particular, up to removing finitely many members of the sequence (x_{n}) we can

assume that for each n ∥x_{n} − a∥ > \frac{2δ}{3} and h(x_{n}) < 2^{−n} · \frac{θ^{2}}{8} .

More generally, if b ∈ X and ε > 0, there exists a j such that ∥b − u_{j}∥ < ε. Then if ℓ > k > j, t ∈ [0, 1] and y = (1 − t)x_{k} + tx_{ℓ}, we have on the one hand ∥u_{j} − x_{k}∥ ≥ d(x_{k}, V_{k}) and

∥b − y∥ ≥ ∥b − x_{k}∥ − t · ∥x_{k} − x_{ℓ}∥ ≥ ∥u_{j} − x_{k}∥ − ∥b − u_{j}∥ − t · R ≥ δ − ε − t · R

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

∥b − y∥ ≥ ∥u_{j} − y∥ − ∥u_{j} − b∥ = t · ∥x_{ℓ} − \frac{u_{j} − (1 − t)x_{k}}{t} ∥ − ∥u_{j} − b∥ ≥ t · δ − ε

It follows that taking ε < \frac{δ^{2}}{6R}, we get for all t ∈ [0, 1],

∥b − y∥ ≥ \frac{δ}{R + δ} (δ − ε − t · R) + \frac{R}{R + δ} (t · δ − ε) = \frac{δ^{2}}{R + δ} − ε > \frac{2δ^{2}}{3R} − \frac{δ^{2}}{6R} = \frac{δ^{2}}{2R} .

And it follows from what precedes that all conditions of the lemma are satisfied by the sequence we constructed this way. □

If there exists a minimizing sequence without converging subsequence we choose a point a , a δ > 0 and a sequence (x_{n}) satisfying conditions i), ii) and iii) of Lemma 2, then define for t ∈ ℝ^{+},

γ(t) = (t − k)x_{k+1} + (k + 1 − t)x_{k} ∈ C if k ≤ t ≤ k + 1

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

Lemma 3. h ∘ γ(t) ≤ 2^{−n} · \frac{θ^{2}}{8} holds whenever t ≥ n.

Proof. Indeed if n ≤ k ≤ t ≤ k + 1, we have

h(γ(t)) = h((t − k)x_{k+1} + (k − 1 − t)x_{k}) ≤ (t − k)h(x_{k+1}) + (k + 1 − t)h(x_{k})

≤ max(h(x_{k}), h(x_{k+1})) ≤ 2^{−k} · \frac{θ^{2}}{8} ≤ 2^{−n} · \frac{θ^{2}}{8}

from what the result follows. □

Lemma 4. There exists a continuous function ψ : X → ℝ^{+} such that for all s ≥ 0

and all x ∈ X, the implication d(x, G_{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 X × 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 2 ii), one has sup_{s∈ℝ^{+}} d(x, G_{s}) ≥ θ for all x ∈ X.

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

x ∈ X, 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 X × \overline{ℝ}_{+}, hence that ψ is continuous since

\overline{ℝ}_{+} is compact. Finally if d(x, G_{s}) < θ · \frac{s + 2}{2s + 2}, we get χ(x, s) < 0, hence ψ(x) > s,

and this achieves the proof. □

Lemma 5. The continuous function F : X → C defined by F = γ ∘ ψ satisfies

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

Proof. If there was x ∈ X such that ∥x − F (x)∥ ≤ \frac{θ}{2} and if s = ψ(x), we would have

F (x) = γ(s) ∈ G_{s} hence d(x, G_{s}) ≤ \frac{θ}{2} < θ · \frac{s + 2}{2s + 2}, from what ψ(x) > s = ψ(x)

would follow, a contradiction. □

Since X is a separable reflexive Banach space, there exists on X an equivalent norm ν which is lur like the dual norm ν^{∗} on the dual space X^{∗} ([1], Corollary 4.3 (ii)). Then ν is rotund and of class C^{1}. And since ν is lur, it has the Kadec-Klee property. From now on we will assume that the norm ∥.∥ of X possesses these two

properties and set q(x) = \frac{1}{2} ∥x∥^{2} for x ∈ X. Then the function q is C^{1} on X and we have q′(0) = 0.

It is well known that for x ≠ 0, ν′(x) has norm 1 in X^{∗} and that ⟨ν′(x), x⟩ = ∥x∥. Thus q′(x) = ∥x∥ · ν′(x) and ∥q′(x)∥ = ∥x∥.

In the whole sequel h will denote a convex lower semi-continuous (lsc for short) function X → \overline{ℝ}_{+} vanishing at 0 and we set for every u ∈ X :

g(u) = inf {h(y) + q(z) : y + z = u}

The epigraph Epi(g) of g is the sum in X × ℝ of the epigraphs Epi(h) and Epi(q), hence is convex, and this shows that g is convex.

Clearly 0 ≤ g(u) ≤ h(u) + q(0) = h(u) and g(u) ≤ h(0) + q(u) = q(u) < +∞. In particular, the convex function g is, like q, locally bounded on X, hence continuous.

Lemma 6. For all u ∈ X there exists a unique pair (y, z) = (σ(u), ζ(u)) ∈ X × X such that y + z = u and g(u) = h(y) + q(z). In particular we have h^{−1}(0) = g^{−1}(0).

Proof. If there were two such pairs (y, z) and (y′, z′) we would also have u =

\frac{y + y′}{2} + \frac{z + z′}{2} . Since h is convex we would have h(\frac{y + y′}{2} ) ≤ \frac{1}{2}(h(y) + h(y′)) and

since X is rotund the function q is strictly convex. So q(\frac{z + z′}{2} ) < \frac{1}{2}(q(z) + q(z′)) if z ≠ z′ and finally

g(u) ≤ h(\frac{y + y′}{2} )+q(\frac{z + z′}{2} ) < \frac{1}{2}(h(y)+q(z)+h(y′)+q(z′)) = \frac{g(u) + g(u)}{2} = g(u)

and this contradiction completes the proof of uniqueness.

If (y_{n}) and (z_{n}) are sequences such that y_{n} + z_{n} = u and h(y_{n}) + q(z_{n}) → g(u), we have lim sup_{n} q(z_{n}) ≤ g(u), hence sup_{n} q(z_{n}) < +∞, and the sequence (z_{n}) is bounded in X hence weakly relatively compact.

Un to extracting a subsequence, we can and do assume that (z_{n}) converges weakly to z ∈ X. Then y_{n} = u − z_{n} ⇀ y = u − z. Since q and h are weakly lsc we have h(y) ≤ lim inf_{n} h(y_{n}) and q(z) ≤ lim inf_{n} q(z_{n}) hence

g(u) ≤ h(y) + q(z) ≤ lim inf(h(y_{n}) + lim inf q(z_{n})) ≤ lim inf(h(y_{n}) + q(z_{n})) = g(u)

These inequalities show that g(u) = h(y) + q(z) and prove the existence of the pair (y, z). Then define (σ(u), ζ(u)) = (y, z).

Since g ≤ h, we have h^{−1}(0) ⊂ g^{−1}(0). Conversely if (y, z) = (σ(u), ζ(u)) and g(u) = 0, we get h(y) = q(z) = 0, whence z = 0 and y = u, so h(u) = g(u) = 0. □

Lemma 7. We have ∥u − σ(u)∥ ≤ \sqrt{2 · h(u)}.

Proof. Since u − σ(u) = ζ(u) and

h(u) ≥ g(u) = q ∘ ζ(u) + h ∘ σ(u) ≥ q ∘ ζ(u) = \frac{1}{2} ∥ζ(u)∥^{2}

we get the announced statement. □

Lemma 8. The functions σ and ζ are continuous from X to X.

Proof. Since σ(u) = u − ζ(u), it is enough to prove the continuity of ζ.

So assume that u_{n} → u, then set z_{n} = ζ(u_{n}) and y_{n} = u_{n} − z_{n}. Since g is continuous, we have M = sup_{n} g(u_{n}) < +∞, and q(z_{n}) ≤ g(u_{n}) ≤ M. Thus (z_{n}) is bounded and passing to a subsequence we can assume that (z_{n}) converges weakly to some z. Set y = u − z. By weak semi-continuity, we have h(y) ≤ lim inf_{n} h(y_{n}) and q(z) ≤ lim inf q(z_{n}), hence

g(u) ≤ h(y) + q(z) ≤ lim inf h(y_{n}) + lim inf q(z_{n})
≤ lim inf(h(y_{n}) + q(z_{n})) = lim g(u_{n}) = g(u)

from what follows that z = ζ(u) and y = σ(u). Moreover we have

lim sup q(z_{n}) ≤ lim g(u_{n}) − lim inf h(y_{n}) ≤ g(u) − h(y) = q(z) ≤ lim inf q(z_{n})

and this shows that q(z_{n}) → q(z), hence ∥z_{n}∥ → ∥z∥. Since z_{n} ⇀ z and X has the Kadec-Klee property we conclude that ∥z_{n} − z∥ → 0, it is ζ(u_{n}) → ζ(u), and this completes the proof of the continuity of ζ. □

Lemma 9. The function g is C^{1} and we have g′(u) = q′ ∘ ζ(u) and ∥g′(u)∥ ≤ \sqrt{2 · h(u)}. Moreover, for all u ∈ X, the function v ↦ h(v) − ⟨g′(u), v⟩ attains at σ(u) its minimum over X.

Proof. Let y = σ(u) and z = ζ(u). Since the norm of X is Fréchet-differentiable, so is q. Then if w = q′(z) we have q(z + k) = q(z) + ⟨w, k⟩ + o(k), hence

g(u + k) ≤ h(y) + q(z + k) = h(y) + q(z) + ⟨w, k⟩ + o(k) = g(u) + ⟨w, k⟩ + o(k)

and by convexity of g : g(u + k) ≥ 2g(u) − g(u − k) ≥ g(u) − ⟨w, −k⟩ + o(k), so finally g(u + k) = g(u) + ⟨w, k⟩ + o(k), what means that g is Fréchet-differentiable and that g′(u) = w = q′ ∘ ζ(u). Since q′ and ζ are continuous, we see that g is C^{1}. By convexity we have even g(u + k) ≥ g(u) + ⟨g′(u), k⟩ for all k ∈ X. And since g′(u) = q′(ζ(u)), we get

∥g′(u)∥ = ∥q′(ζ(u)∥ = ∥ζ(u)∥ ≤ \sqrt{2 · g(u)} ≤ \sqrt{2 · h(u)}

We also have for all k ∈ X,

h(y) + q(z) + ⟨g′(u), k⟩ = g(u) + ⟨g′(u), k⟩ ≤ g(u + k) ≤ h(y + k) + q(z)

hence h(y + k) ≥ h(y) + ⟨g′(u), k⟩.

And with k = v − y we get h(v) ≥ h(y) + ⟨g′(u), v − y⟩, it is

∀v ∈ X h(v) − ⟨g′(u), v⟩ ≥ h ∘ σ(u) − ⟨g′(u), σ(u)⟩

and that means that the function v ↦ h(v)−⟨g′(u), v⟩ attains at σ(u) its minimum over X. □

Theorem 10. If h is a non-negative lsc convex function vanishing at 0 defined on a bounded closed convex subset C of X and if the minimizing sequences of h are not all relatively compact, then there exists a compact function f : X → X^{∗} such that

h(x) − ⟨f(x), x⟩ > inf_{u∈C}(h(u) − ⟨f(x), u⟩)

for all x ∈ C.

Proof. Extend h to a convex lsc function from X to \overline{ℝ}_{+} by setting h(x) = +∞ if x ∉ C. Then every minimizing sequence for h is minimizing for the function g : u ↦ inf {h(y) + q(z) : y + z = u} since 0 ≤ g ≤ h. It follows that the minimizing sequences for g_{|C} are not all relatively compact. Applying to g Lemmas 2 to 5, one finds a continuous function γ : ℝ^{+} → C, a continuous function ψ : X → ℝ^{+} and

θ > 0 such that F = γ ∘ ψ : X → C satisfies ∥x − F (x)∥ > \frac{θ}{2} for all x ∈ X and

that g ∘ γ(t) ≤ 2^{−n} · \frac{θ^{2}}{8} if t ≥ n.

Then set u = F (x), w = g′(u) and γ̃(t) = σ ∘ γ(t). Since h(γ̃(t)) ≤ g(γ(t)) < +∞ one sees that following Lemma 9, h(x) − ⟨w, x⟩ ≥ h ∘ γ̃(t) − ⟨w, γ̃(t)⟩. The function x ↦ g′(F (x)) = q′ ∘ ζ ∘ F (x) is continuous. Moreover since t ↦ g′(γ(t)) is continuous on ℝ^{+} and tends to 0 as t → +∞, the set {g′(γ(t)) : t ∈ ℝ^{+}} is relatively compact in X^{∗}. It follows that the function g′ ∘ F is compact from X to X^{∗}.

It follows from Lemma 7 that ∥F (x) − γ̃(ψ(x))∥ ≤ \sqrt{2 · \frac{θ^{2}}{8}} = \frac{θ}{2} < ∥F (x) − x∥

whence we get that x ≠ γ̃(ψ(x)). Then put, with the compact operator K defined in Lemma 1,

f(x) = g′(F (x)) + K(γ̃ ∘ ψ(x) − x)

The mapping x ↦ K(γ̃ ∘ ψ(x) − x) is compact from C to X^{∗} and one has with y = γ̃ ∘ ψ(x) and w = g′(F (x)),

h(x) − ⟨f(x), x⟩ = h(x) − ⟨w, x⟩ − ⟨K(y − x), x⟩

≥ h(y) − ⟨w, y⟩ − ⟨K(y − x), x⟩

= h(y) − ⟨f(x), y⟩ + ⟨K(y − x), y⟩ − ⟨K(y − x), x⟩

= h(y) − ⟨f(x), y⟩ + ⟨K(y − x), y − x⟩

> h(y) − ⟨f(x), y⟩ ≥ inf_{u∈C}(h(u) − ⟨f(x), u⟩)

which completes the proof since f is compact. □

Theorem 11. Let X be a separable reflexive Banach space without the Kadec-Klee property and B_{X} its unit ball. Then there exists a compact function f : B_{X} → X^{∗} such that ∥f(x)∥ ≥ 1 and ⟨f(x), x⟩ < ∥f(x)∥ for all x ∈ B_{X}.

Proof. By hypothesis there is a ∈ X of norm 1 and a sequence (x_{n}) in the unit ball B_{X} such that (x_{n}) converges weakly to a but does not converge in norm. There is then a^{∗} ∈ X^{∗} such that ∥a^{∗}∥ = 1 and ⟨a^{∗}, a⟩ = 1. Consider the closed hyperplane Y = ker(a^{∗}) of X and the projection π : X → Y defined by π(x) = x − ⟨a^{∗}, x⟩ · a whose kernel is ℝ·a. The mapping x ↦ (π(x), ⟨a^{∗}, x⟩) is then an isomorphism from X onto Y × ℝ. And C = π(B_{X}) is a bounded convex closed neighborhood of 0 in Y.

For y ∈ C there is some s ∈ ℝ such that y + s.a ∈ B_{X} and we denote by s(y) the largest such s. Moreover one has 1 ≥ ∥y + s(y) · a∥ ≥ ⟨a^{∗}, y + s(y) · a⟩ = s(y), thus s(y) ≤ 1.

And it is easily checked that the function y ↦ s(y), which satisfies s(0) = 1, is

concave and upper semi-continuous since s(y) = inf_{w∈X^{∗},∥w∥≤1,⟨w,a⟩>0} \frac{1 − ⟨w, y⟩}{⟨w, a⟩}.

Indeed if ∥w∥ ≤ 1 and ⟨w, a⟩ > 0, one has

s(y).⟨w, a⟩ + ⟨w, y⟩ = ⟨w, y + s(y).a⟩ ≤ ∥w∥ . ∥y + s(y).a∥ = 1

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

1 < ⟨w, y+(s+ε).a⟩ = ⟨w, y+s.a⟩+ε⟨w, a⟩ ≤ ∥w∥ . ∥y + s · a∥+ε⟨w, a⟩ = 1+ε⟨w, a⟩

hence ⟨w, a⟩ > 0 and (s + ε)⟨w, a⟩ + ⟨w, y⟩ > 1, thus s > \frac{1 − ⟨w, y⟩}{⟨w, a⟩} − ε.

Consider then the function h : C → ℝ defined by h(y) = 1 − s(y), which is lsc, convex, non-negative and vanishes at 0. Since the sequence (x_{n}) converges weakly to a but does not converge strongly, one has ⟨a^{∗}, x_{n}⟩ → ⟨a^{∗}, a⟩ = 1, hence y_{n} = π(x_{n}) ⇀ 0, but

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

Since y_{n} + ⟨a^{∗}, x_{n}⟩ · a = x_{n} ∈ B_{X}, one sees that s(y_{n}) ≥ ⟨a^{∗}, x_{n}⟩, hence that h(y_{n}) ≤ 1 − ⟨a^{∗}, x_{n}⟩ → 0. Thus (y_{n}) is a minimizing non relatively compact sequence for h. Then it follows from Theorem 10 that one can find a compact function f_{0} : C → Y^{∗} ⊂ X^{∗} such that

h(y) − ⟨f_{0}(y), y⟩ > inf_{u∈C}(h(u) − ⟨f_{0}(y), u⟩)

for all y ∈ C, and put f(x) = f_{0} ∘ π(x) + a^{∗} for x ∈ B_{X}. It is obvious that ∥f(x)∥ ≥ ⟨f(x), a⟩ = 1 and that f is compact from B_{X} to X^{∗}. Then, for x ∈ B_{X} and y = π(x), one has ⟨f_{0}(y), a⟩ = 0, hence ⟨f_{0}(y), x⟩ = ⟨f_{0}(y), y⟩ and

⟨f(x), x⟩ = ⟨f_{0}(y), x⟩ + ⟨a^{∗}, x⟩ = ⟨f_{0}(y), y⟩ + ⟨a^{∗}, x⟩
≤ ⟨f_{0}(y), y⟩ + s(y) = ⟨f_{0}(y), y⟩ + 1 − h(y)

And there exists u ∈ C such that h(y) − ⟨f_{0}(y), y⟩ > h(u) − ⟨f_{0}(y), u⟩. Since v = u + (1 − h(u)) · a ∈ B_{X} by the definition of h(u), one gets

h(u) − ⟨f_{0}(y, u⟩ = 1 − ⟨a^{∗}, v) − ⟨f_{0} ∘ π(x), u⟩ = 1 − ⟨a^{∗}, v) − ⟨f_{0} ∘ π(x), v⟩
= 1 − ⟨f(x), v⟩

it is

∥f(x)∥ ≥ ⟨f(x), v⟩ = 1 − h(u) + ⟨f_{0}(y), u⟩ > 1 − h(y) + ⟨f_{0}(y), y⟩ ≥ ⟨f(x), x⟩

which is the desired inequality. □

Theorem 12. Let X be a reflexive Banach space without the Kadec-Klee property and B_{X} its unit ball. Then there exists a compact function f : B_{X} → X^{∗} such that ∥f(x)∥ ≥ 1 and ⟨f(x), x⟩ < ∥f(x)∥ for all x ∈ B_{X}.

Proof. By hypothesis there exist some a ∈ X of norm 1 and a sequence (x_{n})_{n≥1} in B_{X} which converges weakly to a with η = inf_{n} ∥x_{n} − a∥ > 0. Then one can find a sequence (ℓ_{n}) in X^{∗} such that ∥ℓ_{n}∥ = 1 for all n, ⟨ℓ_{0}, a⟩ = 1 and ⟨ℓ_{n}, x_{n} − a⟩ ≥ η for n ≥ 1.

Consider then the closed subspace X_{0} = ⋂_{n} ker(ℓ_{n}) of X and the quotient space Y = X/X_{0}. The canonical projection π : X → Y has norm 1, and π^{∗} is an isometric injection from Y^{∗} into X^{∗}. The space Y is reflexive and separated by the linear functionals ℓ̃_{n} = π^{∗}(ℓ_{n}), hence separable, and so is its dual Y^{∗}. Then we have

ẋ_{n} = π(x_{n}) ⇀ ȧ = π(a), \left∥{ \atop ẋ_{n}}\right∥ = ∥π(x_{n})∥ ≤ 1,

⟨ℓ̃_{0}, ȧ⟩ = ⟨ℓ_{0}, a⟩ = 1 ≤ ∥ℓ̃_{0}∥ · \left∥{ \atop ȧ}\right∥ = ∥ℓ_{0}∥ · \left∥{ \atop ȧ}\right∥ = \left∥{ \atop ȧ}\right∥ ≤ ∥a∥ = 1

hence \left∥{ \atop ȧ}\right∥ = 1, and similarly

η ≤ ⟨ℓ̃_{n}, ẋ_{n} − ȧ⟩ = ⟨ℓ_{n}, x_{n} − a)⟩ ≤ ∥ℓ̃_{n}∥ · ∥ẋ_{n} − ȧ∥ = ∥ℓ_{n}∥ · ∥ẋ_{n} − ȧ∥ = ∥ẋ_{n} − ȧ∥

which shows that Y does not have the Kadec-Klee property. It follows then from Theorem 11 that there exists a compact function f̃ on the ball B_{Y} with values in Y^{∗} such that ∥f̃(y)∥ ≥ 1 and ⟨f̃(y), y⟩ < ∥f̃(y)∥ for all y ∈ B_{Y}. Consider then f = π^{∗} ∘ f̃ ∘ π : B → X^{∗} which is compact since π and π^{∗} are continuous and f̃ is compact.

For x ∈ B_{X} one has : ∥f(x)∥ = ∥π^{∗}(f̃ ∘ π(x))∥ = ∥f̃(π(x))∥ ≥ 1 and

⟨f(x), x⟩ = ⟨f̃ ∘ π(x), π(x)⟩ < ∥f̃ ∘ π(x)∥ = ∥π^{∗}(f̃ ∘ π(x))∥ = ∥f(x)∥ ,

and this shows that f possesses the required properties. □

References

1. R. Deville, G. Godefroy, V. Zizler, Smoothness and Renormings in Banach Spaces, Pitman Monographs and Surveys in Pure and Applied Mathematics, 64. Longman Scientific Technical, Harlow; copublished in the United States with John Wiley & Sons, Inc., New York, 1993

2. B. Ricceri, Existence of fixed points for a particular multifunction., Fixed Point Theory 11 (2010), no. 1, 125-128

Analyse Fonctionnelle Institut de Mathématique de Jussieu Université Pierre and Marie Curie (Paris 6) Boîte 186 - 4 place Jussieu F- 75252 Paris Cedex 05 FRANCE

   E-mail address: raymond@math.jussieu.fr