Weak compactness and variational characterization of the convexity

Jean SAINT RAYMOND

Abstract. We prove that if X is a Banach space and f : X → ℝ∪{+∞} is a proper function such that f − ℓ attains its minimum for every ℓ ∈ X^{∗}, then the sublevels of f are all relatively weakly compact in X. As a consequence we show that a Banach space X where there exists a function f : X → ℝ such that f − ℓ attains its minimum for every ℓ ∈ X^{∗} is reflexive.

We also prove that if f : X → ℝ ∪ {+∞}is a weakly lower semicontinuous function on the Banach space X and if for every continuous linear functional ℓ on X the set where the function f − ℓ attains its minimum is convex and non-empty then f is convex. Mathematics Subject Classification (2010). 52A40 , 46A25, 46B10 . Keywords. convex functions, reflexive Banach spaces, weak compactness.

The first aim of this note is to answer questions asked by B. Ricceri in private communications about characterizing convex functions defined on a reflexive Banach space. The first one was the following : “Let X be a reflexive Banach space and f : X → ℝ a weakly l.s.c. function. If for all continuous linear functional ℓ on X the function f − ℓ attains its global minimum at a single point, must f be convex ?” He also asked :“If f : X → ℝ is a coercive C^{1} function such that the derivative f′ : X → X^{∗} is one-to-one, must f be convex ?”.

Positive solutions are given here to these questions (see Theorem 9 and Corollary 10). Actually the hypothesis of reflexivity can be removed from the first one since we prove that since f − ℓ attains its minimum for all ℓ ∈ X^{∗} the space X is necessarily reflexive. In either case we prove even that f is strictly convex.

More generally, we look at proper functions f : X → ℝ ∪ {+∞} such that f −ℓ attains its minimum for all ℓ ∈ X^{∗} and show that for such functions the sublevels S_{f}(α) = {x : f(x) ≤ α} are all weakly relatively compact. In this frame we obtain the following generalization of Theorem 9 : “If f : X → ℝ ∪ {+∞} is a proper weakly l.s.c. function such that the set where f − ℓ attains its minimum is convex and non-empty for all ℓ ∈ X^{∗}, then f is convex.”. Theorem 4 below solves a problem asked at least in the Erratum paper [4], in the first version of [7] as well as in [8]. Since S. Simons found the gap in Calvert and Fitzpatrick paper [3] the validity of those results have been questioned and the Erratum paper [4] seems not to be clear enough with the matter.

In order to make this paper more self-contained we will give here proofs of well-known results when they can be written in few lines.

Lemma 1. Let X be a Banach space, U be a convex open subset of X and f a convex real function on U. Then f is continuous if and only if it is weakly lower semicontinuous (l.s.c.) and if and only if it is strongly lower semicontinuous.

This well-know lemma follows from Proposition 4.1.5 in [2].

Whenever f : X → ℝ ∪ {∞} is bounded from below, we will denote by f̆ the weakly l.s.c. convex functional defined by

f̆(x) = sup{g(x) : g continuous affine , g ≤ f}

It is easily checked that f̆ ≤ f, that inf_{x} f(x) = inf_{x} f̆(x) and that the epigraph of f̆, {(x, t) ∈ X × ℝ : t ≥ f̆(x)}, is the closed convex hull of this of f. In particular, if f attains its minimum m at a point a then f̆ also attains its minimum m at a.

We now study how are linked for a convex function f the coercivity and the property that f − ℓ is bounded from below for all continuous linear functional ℓ. The following simple lemma follows from Lemma 3.2 in [1].

Lemma 2. Let X be a Banach space and f : X → ℝ∪{+∞} be convex. If the Legendre transform g of f is everywhere finite on some open neighborhood W

of 0 in X^{∗}, then necessarily lim inf_{∥x∥→∞} \frac{f(x)}{∥x∥} is positive.

For all integer n the set A_{n} = {ξ ∈ X^{∗} : g(ξ) ≤ n and g(−ξ) ≤ n} is closed convex and symmetric in X^{∗}, and we have ⋃_{n} A_{n} ⊃ W ∩ (−W). It follows from Baire’s lemma that there is some n such that A_{n} has non-empty interior, hence contains some closed ball B̃(a, ρ). By symmetry A_{n} contains

B̃(−a, ρ) too and by convexity contains B̃(0, ρ) = \frac{1}{2}(B̃(a, ρ) + B̃(−a, ρ)).

Fix x ∈ X \ {0}. Then f̆ is the Legendre transform of g, and :

f(x) ≥ f̆(x) = sup_{ξ∈X^{∗}}(⟨x, ξ⟩ − g(ξ)) ≥ sup_{∥ξ∥=ρ}(⟨x, ξ⟩ − g(ξ)) ≥ ρ ∥x∥ − n

and \frac{f(x)}{∥x∥} ≥ ρ − \frac{n}{∥x∥}, and that implies lim inf_{∥x∥→∞} \frac{f(x)}{∥x∥} ≥ ρ > 0. □

Actually in every infinite-dimensional reflexive Banach space one can

find convex functions f satisfying lim inf_{∥x∥→∞} \frac{f(x)}{∥x∥} < +∞ such that their Legendre transforms are everywhere finite (see Theorem 3.6 in [1] for example). Nevertheless this cannot happen in finite-dimensional spaces. So the non-existence of such functions characterizes the finite-dimensional spaces.

Theorem 3. Let X be a Banach space and f : X → ℝ ∪ {+∞} a proper convex l.s.c. function. Then the following three are equivalent :

i) for every ℓ ∈ X^{∗} the function f − ℓ is bounded from below.

ii) for every ℓ ∈ X^{∗} lim inf_{∥x∥→∞} \frac{f(x) − ℓ(x)}{∥x∥} > 0.

iii) for every ℓ ∈ X^{∗}, lim_{∥x∥→∞} f(x) − ℓ(x) = +∞. If moreover X is reflexive the above properties are equivalent to the stronger one:

iv) for every ℓ ∈ X^{∗} f − ℓ attains its minimum on X.

The implication i) ⇒ ii) follows from Lemma 2, and ii) ⇒ iii) is obvious. Moreover, if f satisfies iii) there exists some R such that f(x)−ℓ(x) > 0 whenever ∥x∥ > R. So the set K = {x : f(x) − ℓ(x) ≤ 0} is convex closed and bounded in X. Then it follows from Théorème 3 in [10] that the Borel function f − ℓ is bounded from below on K, hence on E.

Clearly iv) ⇒ i). And if f satisfies iii) there exists a M ∈ ℝ such that K = {x : f(x) − ℓ(x) ≤ M} is bounded and non-empty. Then K is convex weakly closed and bounded hence weakly compact if X is reflexive, and the weakly l.s.c. function f − ℓ attains its minimum on K, hence on X. □

The classical James’ theorem (cf. [6]) asserts that in a Banach space X a weakly closed bounded subset C is weakly compact if every continuous linear functional ℓ on X attains at some point of C its minimum inf_{x∈C} ℓ(x). We now prove a similar statement for unbounded subsets of X × ℝ which are epigraphs of a weakly l.s.c. function.

Theorem 4. Let X be a Banach space and f : X → ℝ∪{+∞} a weakly l.s.c. functional. If f − ℓ attains its minimum for every ℓ ∈ X^{∗} then the sublevels S_{f}(α) = {x ∈ X : f(x) ≤ α} are weakly compact.

Since, for ℓ ∈ X^{∗}, the function f − ℓ attains its minimum m at some point b, the function f̆− ℓ attains also its minimum m at b. Moreover S_{f}(α) is a weakly closed subset of Sf̆(α). Thus we can replace f by f̆ and assume that f itself is convex. By hypothesis f attains its minimum m at some point a. So up to replacing f by x ↦ f(x + a) − m + 1 we can assume that f(x) ≥ f(0) = 1 for all x ∈ X. And still f − ℓ attains its minimum for all ℓ ∈ X^{∗}. Moreover if X_{0} denotes the closed linear span of the proper domain D_{f} = {x ∈ X : f(x) < +∞} of f, every continuous linear functional φ on X_{0} extends to a continuous linear functional ℓ on X : so f − φ attains its minimum over X_{0} for every φ ∈ X_{0}^{∗} since f − ℓ is infinite on X \ X_{0}. Thus we can also assume that D_{f} is total in X.

Consider first the case where D_{f} is dense in X. Define the closed convex subset Γ of X × ℝ by Γ = {(x, t) : t ≥ f(x)}, then Γ̃ as the closed convex cone in X × ℝ × ℝ generated by Γ × {1}, it is the closure of

Γ′ = \{(x, t, s) : s > 0 and \frac{t}{s} ≥ f(\frac{x}{s})\}

and finally C ⊂ X × ℝ as \{(x, s) : (x, 1, s) ∈ Γ̃\}. Since \frac{t}{s} ≥ f(\frac{x}{s}) ≥ 1 for (x, t, s) ∈ Γ′, we have 0 ≤ s ≤ 1 whenever (x, s) ∈ C. It is easy to check that Γ′ is closed in {(x, t, s) ∈ X × ℝ × ℝ : s > 0}.

Moreover if (x, t, 0) belongs to Γ̃ there are sequences (x_{n}) converging to x, (t_{n}) converging to t and (s_{n}) converging to 0 such that (x_{n}, t_{n}, s_{n}) ∈ Γ′.

Then f(\frac{x_{n}}{s_{n}}) ≤ \frac{t_{n}}{s_{n}}. If x ≠ 0 choose ℓ ∈ X^{∗} such that ℓ(x) > 0 : we have for

positive θ

\frac{θℓ(x_{n}) − t_{n}}{s_{n}} = θℓ(\frac{x_{n}}{s_{n}}) − \frac{t_{n}}{s_{n}} ≤ θℓ(\frac{x_{n}}{s_{n}}) − f(\frac{x_{n}}{s_{n}})

≤ f^{∗}(θℓ) := sup_{x∈X} θℓ(x) − f(x) < +∞

whence we deduce that θℓ(x_{n}) ≤ t_{n} + s_{n}f^{∗}(θℓ), and passing to the limit we see that θℓ(x) ≤ t for all θ > 0, a contradiction which shows that x = 0. Thus (x, 0) ∈ C implies x = 0.

Following Lemma 2 we know that there are real numbers M and ε > 0 such that f(x) ≥ ε ∥x∥ if ∥x∥ ≥ M, hence f(x) ≥ ε(∥x∥ − M) for all x ∈ X. It follows that if (x, s) ∈ C with s ≠ 0, we have

\frac{1}{s} ≥ ε(∥\frac{x}{s}∥ − M) = \frac{ε}{s}(∥x∥ − Ms) ,

hence that ∥x∥ ≤ Ms + \frac{1}{ε} ≤ M + \frac{1}{ε}. So C is a bounded closed convex subset of X × ℝ.

We shall prove that C is weakly compact in X × ℝ. So, for α ≥ 1, the set C_{α} = {x : (x, α^{−1}) ∈ C} is weakly compact too. And

(x, α^{−1}) ∈ C ⟺ (x, 1, α^{−1}) ∈ Γ̃ ⟺ (αx, α) ∈ Γ ⟺ f(αx) ≤ α

which implies that S_{f}(α) = α^{−1}.C_{α} is weakly compact in X.

By James’ theorem (cf. [6]) it is enough to prove that every continuous linear functional on X × ℝ attains its maximum on C. So let ℓ ∈ X^{∗} and β ∈ ℝ. We want to prove that the functional ψ : (y, s) ↦ ℓ(y) − βs attains its maximum on C.

We have 0 ≤ s ≤ 1 for every (x, s) ∈ C. Since (0, n) ∈ Γ for all integer

n ≥ 1, we have (0, 1, \frac{1}{n}) ∈ Γ̃, hence (0, 1, 0) ∈ Γ̃ and (0, 0) ∈ C. So for ℓ = 0,

ψ attains its maximum 0 at (0, 0) if β ≥ 0. Moreover (0, 1, 1) ∈ Γ̃, hence (0, 1) ∈ C and ψ attains its maximum −β at (0, 1) if ℓ = 0 and β < 0.

Assume now that ℓ ≠ 0, and define for α ≥ 1 :

ρ̃(α) = sup{ℓ(x) : f(x) ≤ α}

Since f(x) ≥ ε(∥x∥ − M), we have ρ̃(α) ≤ ∥ℓ∥ .(M + \frac{α}{ε}), and this shows that ρ̃(α) < +∞. And since f(0) = 1 ≤ α, we have ρ̃(α) ≥ 0.

Lemma 5. The function ρ̃ is concave and non-decreasing on [1, +∞[, satisfies

lim_{α→∞} ρ̃(α) = +∞ and lim_{α→∞} \frac{ρ̃(α)}{α} = 0.

It is clear from its definition that ρ̃ is non-decreasing. Let α, α′ be ≥ 1, 0 ≤ θ ≤ 1 and η > 0. There are y and y′ ∈ X such that ℓ(y) > ρ̃(α) − η, ℓ(y′) > ρ̃(α′) − η, f(y) ≤ α and f(y′) ≤ α′. Then, for y′′ = θy + (1 − θ)y′,

f(y′′) = f(θ.y + (1 − θ)y′) ≤ θf(y) + (1 − θ)f(y′) ≤ α′′ := θα + (1 − θ)α′

and

ρ̃(α′′) ≥ ℓ(y′′) = θℓ(y) + (1 − θ)ℓ(y′) > θρ̃(α) + (1 − θ)ρ̃(α′) − η

which proves the concavity of ρ̃ since η is arbitrary.

If A is any positive real number the dense set D_{f} intersects the nonempty open set {y : ℓ(y) > A}, and there is a point y_{0} ∈ D_{f} such that ℓ(y_{0}) > A. Then, for α ≥ f(y_{0}) we have ρ̃(α) ≥ ℓ(y_{0}) ≥ A. And this shows that lim_{α→∞} ρ̃(α) = +∞.

Finally, let n ∈ ℕ. Since f − nℓ attains on X a finite minimum, we have f^{∗}(nℓ) := sup_{x∈X} nℓ(x) − f(x) < +∞. So for all x such that f(x) ≤ α :

nℓ(x) ≤ f(x) + f^{∗}(nℓ) ≤ α + f^{∗}(nℓ)

hence 0 ≤ ρ̃(α) ≤ \frac{α}{n} + \frac{1}{n}f^{∗}(nℓ) and this shows that lim sup_{α→∞} \frac{ρ̃(α)}{α} ≤ \frac{1}{n} for all

n, hence that ρ̃(α) = o(α). □

The concave and non-decreasing function ρ̃ is continuous on ]1, +∞[. Since it might be discontinuous at 1, we replace it by the continuous function ρ which agrees with ρ̃ on ]1, +∞[. So ρ(1) = inf_{α>1} ρ̃(α). Then still ρ is concave on [1, +∞[ and satisfies the conclusions of Lemma 5.

It follows from Lemma 5 that the function α ↦ \frac{ρ(α) − β}{α} which takes positive values for α large enough and goes to 0 at the infinity attains its

maximum μ > 0 over [1, +∞[ at some α_{0}. By hypothesis, the function f − \frac{1}{μ}ℓ

attains its minimum over X at some x^{∗}. Put α^{∗} = f(x^{∗}). So for all x ∈ X

and all t ≥ f(x) we have t − \frac{1}{μ}ℓ(x) ≥ α^{∗} − \frac{1}{μ}ℓ(x^{∗}). Thus

ℓ(x) − β − μ.t ≤ ℓ(x^{∗}) − β − μ.α^{∗} ,

and for any η > 0 :

ρ̃(α_{0} +η)−β −μ.(α_{0} +η) = sup_{f(x)≤α_{0}+η} ℓ(x)−β −μ.(α_{0} +η) ≤ ℓ(x^{∗})−β −μ.α^{∗}

whence we get

ℓ(x^{∗}) − β − μ.α^{∗} ≥ lim_{η→0} ρ̃(α_{0} + η) − β − μ.(α_{0} + η) = ρ(α_{0}) − β − μ.α_{0} = 0

hence ℓ(x^{∗}) − β ≥ μ.α^{∗}, and ψ(\frac{x^{∗}}{α^{∗}} , \frac{1}{α^{∗}}) = \frac{ℓ(x^{∗}) − β}{α^{∗}} ≥ μ.

On the other hand, if (y, s) ∈ C and 0 < s ≤ 1, we have (y, 1, s) ∈ Γ̃,

hence (\frac{y}{s} , \frac{1}{s}) ∈ Γ, which means f(\frac{y}{s}) ≤ \frac{1}{s}, or f(x) ≤ α with α = \frac{1}{s} ≥ 1 and x = αy. This implies

ψ(y, s) = ℓ(y) − βs = \frac{ℓ(x) − β}{α} ≤ \frac{ρ̃(α) − β}{α} ≤ \frac{ρ(α) − β}{α} ≤ μ

whenever (y, s) ∈ C and s > 0.

Thus sup_{(y,s)∈C} ψ(y, s) ≤ μ. Since (\frac{x^{∗}}{α^{∗}} , \frac{1}{α^{∗}}) ∈ C, we conclude that

sup_{(y,s)∈C} ψ(y, s) ≤ μ ≤ ψ(\frac{x^{∗}}{α^{∗}} , \frac{1}{α^{∗}}) ≤ sup_{(y,s)∈C} ψ(y, s) ,

which shows that ψ attains its maximum μ over C at (\frac{x^{∗}}{α^{∗}} , \frac{1}{α^{∗}}) and completes the proof in the case where D_{f} is dense in X.

In the general case define for n ∈ ℤ^{∗} = ℤ \ \{0\} : h_{n}(x) = n^{2}f(\frac{x}{n}),
H_{n} = {(x, t) ∈ X × ℝ : t ≥ h_{n}(x)}

and H as the closed convex hull of ⋃_{n∈ℤ^{∗}} H_{n}. Since h_{n}(x) ≥ n^{2} ≥ 1 we have for all n ∈ ℤ^{∗} : H_{n} ⊂ X × [1, +∞[. It follows that H_{1} ⊂ H ⊂ X × [1, +∞[. Then the function h defined by h(x) = inf{t : (x, t) ∈ H} is convex l.s.c. and satisfies

1 ≤ h(x) ≤ h_{1}(x) = f(x) .

Since h(x) ≤ h_{n}(x) = n^{2}f(\frac{x}{n}) < +∞ whenever x ∈ nD_{f}, it is clear that n.D_{f} = D_{h_{n}} ⊂ D_{h}. And since D_{f} generates X, we have sup_{y∈D_{f}} |ℓ(y)| > 0 , hence sup_{y∈D_{h}} ℓ(y) = +∞, for all non-zero ℓ ∈ X^{∗}. Thus the convex set D_{h} is dense in X.

We shall prove that h − ℓ attains its minimum for every ℓ ∈ X^{∗}. It will follow from the first case that the sublevels of h are weakly compact. And since S_{f}(α) is a closed subset of the weakly compact set S_{h}(α) we are done.

Fix ℓ ∈ X^{∗}. Then for β ∈ ℝ,

β ≤ inf_{x∈X} h(x) − ℓ(x) ⟺ ⋃_{n∈ℤ^{∗}} H_{n} ⊂ {(x, t) : t ≥ ℓ(x) + β}

⟺ ∀n ∈ ℤ^{∗} ∀(x, t) ∈ H_{n} t ≥ ℓ(x) + β

⟺ ∀n ∈ ℤ^{∗} ∀x ∈ X n^{2}f(\frac{x}{n}) ≥ ℓ(x) + β

⟺ ∀n ∈ ℤ^{∗} ∀y ∈ X f(y) ≥ \frac{1}{n}ℓ(y) + \frac{β}{n^{2}}

⟺ ∀n ∈ ℤ^{∗} β ≤ n^{2} inf_{y} f(y) − \frac{ℓ}{n}(y) = −n^{2}f^{∗}(\frac{ℓ}{n})

thus inf_{x∈X} h(x) − ℓ(x) = inf_{n∈ℤ^{∗}} −n^{2}f^{∗}(\frac{ℓ}{n}).

The function g_{ℓ} : θ ↦ f^{∗}(θ.ℓ) is convex on ℝ hence continuous at 0. It

follows that g_{ℓ}(\frac{1}{n}) → g_{ℓ}(0) = f^{∗}(0) = −1 as |n| → ∞, and −n^{2}f^{∗}(\frac{ℓ}{n}) → +∞.

So there exists a m ∈ ℤ^{∗} such that for all n ∈ ℤ^{∗} : −m^{2}f^{∗}(\frac{ℓ}{m}) ≤ −n^{2}f^{∗}(\frac{ℓ}{n}).

And by hypothesis there is some a ∈ X such that f^{∗}(\frac{ℓ}{m}) = \frac{ℓ(a)}{m} − f(a). Then we conclude that

inf_{x} h(x) − ℓ(x) = inf_{n∈ℤ^{∗}} −n^{2}f^{∗}(\frac{ℓ}{n}) = −m^{2}f^{∗}(\frac{ℓ}{m}) = m^{2}f(a) − ℓ(ma)
≥ h(ma) − ℓ(ma)

which means that h − ℓ attains its minimum at ma. □

It should be noticed that there is only one place where we used the fact that f is l.s.c. : we said that the sublevel S_{f}(α) is a weakly closed subset of the weakly compact set Sf̆(α). So without this assumption we can only conclude that S_{f}(α) is a subset of the weakly compact set Sf̆(α), hence that the sublevels of f are relatively weakly compact.

Corollary 6. Let X be a Banach space. Suppose that there exists some functional f : X → ℝ ∪ {+∞} such that the proper domain D_{f} has non-empty interior and that f − ℓ attains its minimum for every ℓ ∈ X^{∗}. Then X is reflexive.

If the interior U of D_{f} is non-empty, it follows from Lemma 1 that the convex function f̆ is continuous on U. So for a ∈ U and α > f(a) the sublevel Sf̆(α) which is weakly compact by Theorem 4 contains a closed ball B̃(a, r). And since B̃(a, r) is weakly closed in the weakly compact set Sf̆(α), we can conclude that this ball is weakly compact and that X is reflexive. This completes the proof of corollary 6. □

If X is a reflexive Banach space every convex continuous function φ such that φ − ℓ is bounded from below for every ℓ ∈ X^{∗} satisfies ∂φ(X) = X^{∗} since φ − ℓ attains its minimum for every ℓ ∈ X^{∗}, and ℓ ∈ ∂φ(a) when φ − ℓ attains its minimum at a.

Conversely one deduces immediately from Corollary 6 that if X is nonreflexive and if f is a convex l.s.c. function X → ℝ ∪ {+∞} such that D_{f} has non-empty interior, there exists some ℓ ∈ X^{∗} such that f − ℓ does not attain its minimum, hence ℓ is not in the range of the subdifferential ∂f. So ∂f(X) ≠ X^{∗}.

And this yields a simple proof of the result announced by S. Fitzpatrick and B. Calvert in [3] and [4], characterizing the reflexivity of a Banach space in terms of the surjectivity of the subdifferential of a convex function.

As noticed in the introduction Theorem 4 gives in a full generality a positive answer to a conjecture proposed by J. Orihuela and M. Ruiz-Galán and solved by them in particular cases, either assuming the coercivity of the function f or the sequential w^{∗}-compactness of the unit ball of the dual space X^{∗} (see [8] and [9]). It is worth noting that their method of proof consisted to adapt the arguments of James’ theorem for covering the unbounded case, hence reproving James’ theorem. On the other hand the method used here consists to reduce the proof to an application of James’ theorem via a conical projection transforming the epigraph of f into a bounded convex set.

Notice also that this Theorem 4 has consequences for the study of the socalled Jouini-Schachermayer-Touzi theorem for convex risk in financial mathematics (see [7]) and that in this frame F. Delbaen gave a proof of Theorem 4 for X = L^{1} (even in the non-separable case) using also a reduction to the classical James’ theorem via an homogenisation trick (see [5]).

The sequel of this paper is devoted to characterizing the convexity of weakly l.s.c. functions by variational conditions.

Lemma 7. Let X be a real Banach space and f : X → ℝ^{+} ∪ {∞} be a

weakly l.s.c. functional on X satisfying f(0) = 0 and lim inf_{∥x∥→∞} \frac{f(x)}{∥x∥} > 0.

Assume that for some positive ε the sublevel {x : f(x) ≤ ε} is weakly compact. If moreover f^{−1}(0) is contained in a closed bounded convex subset K of X, then f̆^{−1}(0) ⊂ K.

Let φ be a continuous linear functional on X and m > sup_{x∈K} φ(x). We will show that φ(a) ≤ m for all a ∈ f̆^{−1}(0), and this will prove that f̆^{−1}(0) ⊂ K.

Since for every s ∈]0, ε[ the set {x : f(x) ≤ s} is weakly closed, hence weakly compact, the weakly compact sets

H(s) = {x : f(x) ≤ s} ∩ {x : φ(x) ≥ m}

have an empty intersection for s ∈]0, ε[. Thus there is some positive λ such that H(λ) = ∅, hence {x : f(x) ≤ λ} ⊂ U = {x : φ(x) < m}.

The closed convex hull L of the weakly compact set {f(x) ≤ λ} is weakly compact too, and contained in Ū : we have sup_{x∈L} φ(x) ≤ sup_{x∈Ū} φ(x) ≤ m.

Since lim inf_{∥x∥→∞} \frac{f(x)}{∥x∥} > 0, there are real numbers M and η > 0 such

that f(x) ≥ η ∥x∥ whenever ∥x∥ ≥ M. It follows that the epigraph of f, Γ = {(x, t) : t ≥ f(x)} is contained in the union of the three sets L × [0, λ], {x : ∥x∥ ≤ M} × [λ, +∞[ and C_{η} = {(x, t) : t ≥ η ∥x∥}.

The second one itself is contained in \{(x, t) : t ≥ \frac{λ}{M} ∥x∥\}. So replacing η

by δ = min(η, \frac{λ}{M}), we get Γ ⊂ (L×[0, λ])∪C_{δ} ⊂ Γ_{0} = \{(x, t) : t ≥ δ.d(x, L)\},

and the epigraph of f̆ is contained in the closed convex set Γ_{0}.

Then if f̆(a) = 0, one has (a, 0) ∈ Γ_{0}, hence d(a, L) = 0. It follows that a ∈ L, hence that φ(a) ≤ m. □

From now on and in all the sequel we will denote by M_{g}, for any function g : X → ℝ ∪ {+∞}, the set of points where g attains its global minimum, it is

M_{g} = {x ∈ X : ∀y ∈ X g(y) ≥ g(x)} .

Lemma 8. Let X be a real Banach space and f : X → ℝ ∪ {+∞} a weakly l.s.c. function. Assume that f − φ attains its minimum for all φ ∈ X^{∗}. If, for some ℓ ∈ X^{∗} and some a ∈ X, the function f̆− ℓ attains its minimum at a and if M_{f−ℓ} is convex then f(a) = f̆(a).

Notice first that by Theorem 4 the sublevels of f − ℓ are all weakly compact. Let b ∈ M_{f−ℓ}. If m = f(b)−ℓ(b) the affine function h : x ↦ ℓ(x)+m satisfies h ≤ f, hence h ≤ f̆. So h(a) ≤ f̆(a), which implies ℓ(a) + m ≤ f̆(a). Moreover since f̆− ℓ attains its minimum at a we alse have

m ≤ f̆(a) − ℓ(a) ≤ f̆(b) − ℓ(b) ≤ f(b) − ℓ(b) = m

whence we get f̆(b) = f(b) and f̆(b) − ℓ(b) = f̆(a) − ℓ(a).

Consider the function g : x ↦ f(x+b)−ℓ(x)−f(b) which is non-negative weakly l.s.c. and vanishes at 0. All sublevels of g are weakly compact. Thus g satisfies the hypotheses of Lemma 7 with K = M_{f−ℓ} − b.

Moreover ğ(x) = f̆(x + b) − ℓ(x) − f(b), so

ğ(a − b) = f̆(a) − ℓ(a − b) − f(b) = f̆(a) − ℓ(a) − (f(b) − ℓ(b)) = 0 .

It follows from Lemma 7 that a − b ∈ ğ^{−1}(0) ⊂ K = M_{f−ℓ} − b, hence that a ∈ M_{f−ℓ}, it is f(a) − ℓ(a) = m = f̆(a) − ℓ(a). Finally we conclude that f(a) = f̆(a). □

We now give a criterion of strict convexity for weakly l.s.c. functions, which answers a question of B. Ricceri mentioned in the introduction.

Theorem 9. Let X be a real Banach space and f be a real weakly l.s.c. functional on X such that f − ℓ has a unique global minimum for every continuous linear functional ℓ on X. Then f is strictly convex.

Let f satisfy the hypotheses of the theorem. Then the function f̆ is convex continuous and weakly l.s.c. and the set W = {(x, t) : t > f̆(x)} is an open convex subset of X × ℝ. Fix a ∈ X ; then (a, f̆(a)) ∉ W. By Hahn-Banach’s theorem there exists a continuous linear functional on X × ℝ which separates (a, f̆(a)) from W, hence a continuous linear functional ℓ on X such that

t − ℓ(x) ≥ f̆(a) − ℓ(a)

for all x ∈ X and all t > f̆(x). In particular we have f̆(x)−ℓ(x) ≥ f̃(a)−ℓ(a) for all x ∈ X. This means that f̆− ℓ attains its minimum at a. Since M_{f−ℓ} is a singleton for all ℓ ∈ X^{∗}, it follows then from Lemma 8 that f(a) = f̆(a). And since a is arbitrary, we get that f = f̆ is convex.

Finally, if f was not strictly convex, there would exist two distinct points a and b such that f is affine on the segment [a, b]. One sees as above that it would exist a continuous linear functional ℓ such that f − ℓ attains its

minimum at \frac{a + b}{2} ; so

g(x) = f(x) − f(\frac{a + b}{2}) − ℓ(x − \frac{a + b}{2}) ≥ 0

for all x. We would then have

g(a) + g(b) = f(a) + f(b) − 2f(\frac{a + b}{2}) − ℓ(\frac{a − b}{2}) − ℓ(\frac{b − a}{2}) = 0 ,

hence g(a) = g(b) = 0 and f − ℓ would attain its minimum both at a and at b, contradicting the hypothesis. □

We deduce from the previous theorem the following corollary which also answers a question raised by B. Ricceri.

Corollary 10. Let X be a reflexive real Banach space, f a real coercive weakly l.s.c. functional of class C^{1} on X. Suppose that the derivative f′ : X → X^{∗} is one-to-one. Then f is strictly convex.

Since X is reflexive and f is assumed to be coercive, the function f − ℓ attains its minimum for all ℓ ∈ X^{∗}. Following the above theorem, if f was not strictly convex it would exist a continuous linear functional ℓ and two distinct points a and b such that f − ℓ attains its minimum both at a and at b. We would then have

f′(a) − ℓ = (f − ℓ)′(a) = 0 = (f − ℓ)′(b) = f′(b) − ℓ

hence f′(a) = f′(b), contradicting the hypothesis if a ≠ b. □

It should be noticed that in Corollary 10 above the coercivity of f is only used to guarantee that f − ℓ is bounded from below for every ℓ ∈ X^{∗}.

Theorem 11. Let X be a reflexive Banach space, f : X → ℝ ∪ {+∞} be a functional such that f − ℓ is bounded from below for all continuous linear functional ℓ. Then f is weakly l.s.c. if and only if it is sequentially weakly l.s.c.

If f −ℓ is bounded from below so is f̆−ℓ. It follows then from Theorem 3 that lim_{∥x∥→∞} f̆(x) = +∞ and a fortiori lim_{∥x∥→∞} f(x) = +∞. Hence the sublevels S_{f}(α) of f are all bounded. Since a bounded set is weakly compact if and only it is sequentially weakly closed, the sets S_{f}(α) are weakly closed if and only they are sequentially weakly closed. So f is weakly l.s.c. if and only if it is sequentially weakly l.s.c. □

We now give a generalization of Theorem 9 : we no longer assume that the proper domain of f is the whole space X and that f − ℓ attains its minimum at a single point.

Theorem 12. Let X be a real Banach space, f : X → ℝ ∪ {+∞} be a proper weakly l.s.c. functional on X. Assume that for all continuous linear functional ℓ on X the set M_{f−ℓ} where f − ℓ attains its global minimum is convex and non-empty. Then f is convex.

Notice first that the convexity of this set is clearly necessary : if f is convex and m = inf_{x} f(x) − ℓ(x), then the set M_{f−ℓ} = {x : f(x) − ℓ(x) ≤ m} is convex.

One can also notice that if the set M_{f−ℓ} is only assumed to be connected and non-empty for all ℓ ∈ X^{∗}, the conclusion fails. Indeed the function f : (x, y) ↦ (x^{2} + y^{2})^{2} − 2(x^{2} + y^{2}) on ℝ^{2} is not convex since

f(1, 0) = f(−1, 0) = −1 < 0 = f(0, 0) .

Nevertheless an easy calculation shows that if ℓ = 0 the function f − ℓ = f attains its minimum −1 at every point of the unit circle {(x, y) : x^{2} +y^{2} = 1} which is connected, but that for ℓ ≠ 0, the function f −ℓ attains its minimum at a unique point.

Fix a ∈ Df̆. We want to prove that f(a) = f̆(a). If D_{f} has empty interior it might happen that for no ℓ ∈ X^{∗} the function f − ℓ attains its minimum at a. But using the arguments of Bishop-Phelps’ Theorem we will show that this happens for points close to a.

Let η > 0 and Γ_{η} be the closed convex cone {(x, t) ∈ X ×ℝ : ∥x∥ ≤ η.t}. Since the function f attains its minimum it is bounded from below, and the order relation R_{η} defined on the epigraph E = {(x, t) ∈ X × ℝ : t ≥ f̆(x)} of f̆ by

(x, t) R_{η} (y, s) ⟺ (x − y, t − s) ∈ Γ_{η}

is easily seen to be inductive : if (x_{j}, t_{j})_{j∈J} is a chain for R_{η} one has t_{j} ≤ t_{k} if j ≥ k and inf_{j} t_{j} ≥ μ = inf_{x} f(x) > −∞. Thus there exists an increasing sequence (j_{n}) ∈ J such that t_{j_{n}} → θ = inf_{j} t_{j}. So ∥x_{j_{n}} − x_{j_{p}} ∥ ≤ η. |t_{j_{n}} − t_{j_{p}} | and the Cauchy sequence (x_{j_{n}}) converges to some z ∈ X. Then (x_{j_{n}}, t_{j_{n}}) converges to (z, θ) ∈ E and clearly one has (x_{j}, t_{j}) R_{η} (z, θ) for all j ∈ J.

It follows then from Zorn’s lemma that there is some R_{η}-maximal (b, s) in E such that (a, f̆(a)) R_{η} (b, s), and in particular s ≤ f̆(a). Moreover since (b, s) R_{η} (b, f̆(b)), one has s = f̆(b). We conclude that f̆(b) ≤ f̆(a) and that ∥b − a∥ ≤ η.(f̆(a) − f̆(b)) ≤ η.(f̆(a) − μ).

By maximality of (b, f̆(b)) the convex subset E of X ×ℝ is disjoint from the convex open set

U = {(x, t) : f̆(b) − t > η. ∥x − b∥}

and by Hahn-Banach’s theorem we can separate them, by a continuous linear functional (x, t) ↦ t − ℓ(x) for some ℓ ∈ X^{∗}:

f̆(b) − ℓ(b) ≤ sup_{(x,t)∈U} t − ℓ(x) ≤ inf_{(y,s)∈E} s − ℓ(y) = inf_{y∈E} f̆(y) − ℓ(y) ≤ f̆(b) − ℓ(b)

This implies that f̆−ℓ attains its minimum at b. It follows then from Lemma 8 that f(b) = f̆(b), hence we get f(b) = f̆(b) ≤ f̆(a) for some b such that ∥b − a∥ ≤ η.(f̆(a) − μ). Then since f is l.s.c. and η > 0 is arbitrary we conclude that

f(a) ≤ lim inf_{x→a} f(x) ≤ f̆(a) ≤ f(a) .

Finally since a is arbitrary in Df̆, we get that f = f̆, and this shows that f is convex. □

In Theorem 9, we were able to deduce the strict convexity of the function f from the hypothesis that M_{f−ℓ} is a singleton for all ℓ ∈ X^{∗}. We now show that this is no longer true if we remove the hypothesis that the proper domain of f is the whole space X (or at least has non-empty interior).

Theorem 13. There exists a reflexive Banach space X, a proper l.s.c. convex not strictly convex function f : X → ℝ ∪ {+∞} such that for every ℓ ∈ X^{∗} the set M_{f−ℓ} is a singleton.

Let p and q be real numbers such that 1 < p < 2 < q, q′ = \frac{q}{q − 1}, X

be the space ℓ^{q}, K be the convex subset {x = (x_{n}) ∈ ℓ^{q} : ∑_{n} |x_{n}|^{p} ≤ 1} of X and D = {x = (x_{n}) ∈ ℓ^{q} : ∑_{n} |x_{n}|^{2} ≤ 2}. Notice that K ⊂ D. Define f_{0}(x) = d_{2}(x, K) = inf_{y∈K} ∥x − y∥_{2} for x ∈ D, then

f(x) = f_{0}(x)^{2} + ⟨a, x⟩

where a = (\frac{1}{(n + 1)^{1/q′}}) and ⟨a, x⟩ = ∑_{n} a_{n}.x_{n} if x ∈ D, and f(x) = +∞

whenever x ∈ X \ D.

Clearly, if u is any non-zero vector in K we have f(0) = \frac{f(u) + f(−u)}{2}; so f is not strictly convex. Nevertheless it is easily checked that f_{0} is convex on X, and so is f. Moreover, for α ≥ 0 we have f_{0}^{−1}([0, α]) = D∩(K+\sqrt{α}.B_{2}), where B_{2} is the convex weakly compact set {x : ∑_{n} |x_{n}|^{2} ≤ 1}. It follows that f_{0}^{−1}([0, α]) is weakly compact in X and that f_{0} is weakly l.s.c. ; then f_{0}^{2} is weakly l.s.c. too. On D the weak topology of X coincides with the weak topology of ℓ^{2}, and the function x ↦ ⟨a, x⟩ is weakly continuous on D. And finally f is a proper convex l.s.c. function on X.

For every ℓ ∈ X^{∗}, f − ℓ is weakly l.s.c. on the weakly compact set D thus attains its minimum. We now prove that this minimum is attained at a single point. Indeed assume that f − ℓ attains its minimum both at x + h and x − h (h ≠ 0). We then have 2f_{0}(x)^{2} = f_{0}(x + h)^{2} + f_{0}(x − h)^{2} hence

(f_{0}(x + h) + f_{0}(x − h))^{2} + (f_{0}(x + h) − f_{0}(x − h))^{2}
= 2(f_{0}(x + h)^{2} + f_{0}(x − h)^{2}) = 4f_{0}(x)^{2} ≤ (f_{0}(x + h) + f_{0}(x − h))^{2}

and this implies that f_{0}(x + h) = f_{0}(x − h) = f_{0}(x). Since K is convex and closed there are points y′ and y′′ in K such that f_{0}(x + h) = ∥x + h − y′∥_{2}

and f_{0}(x − h) = ∥x − h − y′′∥_{2}. Let y = \frac{y′ + y′′}{2} ∈ K and r = f_{0}(x).

We have ∥x − y∥_{2} ≤ \frac{1}{2}(∥x + h − y′∥_{2} + ∥x − h − y′′∥_{2}) = r. If y′ = y′′

we must have ∥x + h − y∥_{2} = ∥x − h − y∥_{2} = ∥x − y∥_{2} = r hence h = 0 since ℓ^{2} is uniformly convex, in contradiction with the hypothesis h ≠ 0.

Now if y′ ≠ y′′ and r > 0 we have ∥y∥_{p} < 1 since ℓ^{p} is uniformly convex too. Define w = (w_{n}) by w_{n} = 2^{−n}(x_{n} − y_{n}). So w ∈ ℓ_{p} ⊂ ℓ_{2}, and for t > 0 small enough we have ∥y + t.w∥_{p} ≤ ∥y∥_{p} + t ∥w∥_{p} < 1. We also have

∥x − (y + t.w)∥_{2}^{2} = ∥x − y∥_{2}^{2} + t^{2} ∥w∥_{2}^{2} − 2t⟨x − y, w⟩

= r^{2} + t^{2} ∥w∥_{2}^{2} − 2t ∑_{n} 2^{−n} |x_{n} − y_{n}|^{2} < r^{2}

which asserts that f_{0}(x) = d_{2}(x, K) < r since y + tw ∈ K. And this last contradiction shows that r = 0, it is x, x + h and x − h belong all three to K. As before we see by uniform convexity that ∥x∥_{p} < 1. Since ℓ is continuous on X there is b = (b_{n}) ∈ ℓ^{q′} such that ℓ(x) = ⟨b, x⟩. Then

M_{f−ℓ} = f(x + h) − ℓ(x + h) = f_{0}(x + h)^{2} + ⟨a − b, x + h⟩ = ⟨a − b, x + h⟩

= f(x − h) − ℓ(x − h) = f_{0}(x − h)^{2} + ⟨a − b, x − h⟩ = ⟨a − b, x − h⟩

Observe that a − b ≠ 0 since b ∈ ℓ^{q′} and a ∉ ℓ^{q′}. and define v = (v_{n}) by v_{n} = 2^{−n}(a_{n} − b_{n}). Then as above for t > 0 small enough x − tw ∈ K and ⟨a − b, x − tw⟩ = ⟨a − b, x⟩ − t ∑_{n} 2^{−n}(a_{n} − b_{n})^{2} < M_{f−ℓ}. But

(f − ℓ)(x − tw) = f_{0}(x − tw)^{2} + ⟨a − b, x − tw⟩ = ⟨a − b, x − tw⟩ < M_{f−ℓ}

and this contradicts the definition of M_{f−ℓ} and completes the proof. □

We can obtain as a corollary of Theorem 12 a simple proof of the wellknown following version of Bunt-Motzkin’s theorem.

Corollary 14. Let X be a real Hilbert space and A be a non-empty sequentially weakly closed subset of X. Assume that for every x in X there is a unique point a ∈ A such that ∥x − a∥ = d(x, A). Then A is convex.

Define the function f on X by

f(a) = \begin{cases}∥a∥^{2} & if a ∈ A \\ +∞ & if a ∉ A \end{cases}

Since S_{f}(α) = A ∩ B̃(0, \sqrt{α}) is sequentially weakly closed for every α ≥ 0, the function f is sequentially weakly l.s.c. Moreover, for y ∈ X and a ∈ A,

f(a) − ⟨2y, a⟩ = ∥a − y∥^{2} − ∥y∥^{2}

It follows that the function a ↦ ∥a − y∥ attains its minimum on C at x if and only if f − 2⟨y, .⟩ attains its minimum on X at x. By hypothesis this happens at exactly one point x for all y ∈ X. By theorem 11, f is weakly l.s.c. Thus for every ℓ ∈ X^{∗} the function f − ℓ attains its minimum on a convex set, and f is convex. And we conclude that A = D_{f} is convex. □

14 Jean SAINT RAYMOND

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Jean SAINT RAYMOND Analyse Fonctionnelle Institut de Mathématique de Jussieu Université Pierre et Marie Curie Boîte 186 - 4 place Jussieu F- 75252 Paris Cedex 05 FRANCE

e-mail: raymond@math.jussieu.fr