A NEW MINIMAX THEOREM FOR LINEAR OPERATORS

JEAN SAINT RAYMOND

Abstract. The aim of this note is to prove the following minimax theorem which generalizes a result by B. Ricceri and extends a previous result of the author : let E be a infinite-dimensional Banach space, F be a Banach space, X be a convex subset of E whose interior is non-empty for the weak topology on bounded sets, Δ a finitedimensional convex compact subset of L(E, F), φ : F → ℝ be a continuous convex coercive map, and ψ : Δ → ℝ a convex continuous function. Assume moreover that Δ contains at most one compact operator. Then

sup_{x∈X} inf_{T ∈Δ}(φ(T x) + ψ(T)) = inf_{T ∈Δ} sup_{x∈X}(φ(T x) + ψ(T)) .

1. Introduction

For any two normed spaces E and F we will denote by L(E, F) the space of continuous linear mappings from E to F, itself normed by ∥T ∥ = sup_{∥x∥≤1} ∥T x∥.

It was shown by E. Asplund and V. Pták in [AP] that for two normed spaces E and F of dimension ≥ 2 and any two A and B in L(E, F) the following inequality holds

sup_{∥x∥≤1} inf_{λ∈ℝ} ∥Ax + λBx∥ ≤ inf_{λ∈ℝ} ∥A + λB∥ = inf_{λ∈ℝ} sup_{∥x∥≤1} ∥Ax + λBx∥

and that the equality in the above relation is attained for every pair A, B in L(E, F) if and only if both E and F are inner product spaces. In the sequel the unit ball of E is replaced by a convex set whose interior is non-empty for the weak topology σ(E, E^{∗}) or more generally the ‘weak topology on bounded sets’ and we are interested in proving such a minimax equality.

For a Banach space E, we will denote by β(E, E^{∗}) the finest locally convex topology on E which agrees with the weak topology σ(E, E^{∗}) on the bounded subsets of E. Converging sequences for β(E, E^{∗}) are all weakly converging sequences. If a subset U of E is open for β(E, E^{∗}) then every weakly converging sequence in E whose weak limit is in U has all but finitely many terms inside U. Conversely if E^{∗} is separable (e.g. if E is reflexive and separable) this characterizes the β(E, E^{∗})-open subsets of E.

The following theorem has been proved by B. Ricceri in [R] (Theorem 3).

Theorem 1.1. Let E be an infinite-dimensional reflexive Banach space, T : E → E a non-zero linear compact operator, φ : E → ℝ a convex continuous and coercive functional, J ⊂ ℝ a compact interval with 0 ∈ J, ψ : J → ℝ be a continuous convex function. Then, for each r > φ(0), one has

sup_{x∈X} inf_{λ∈J}(φ(T x − λx) + ψ(λ)) = r + ψ(0)

where X = {x ∈ E : φ(T x) ≤ r}.

2010 Mathematics Subject Classification. 49J35, 46B04, 46B50.

Key words and phrases. minimax, Banach spaces, linear operators.

Our main theorem extends this latter statement : indeed X has non-empty interior for β(E, E^{∗}) because of Theorem 4.4, and for Δ = {T − λI : λ ∈ J}, we then have sup_{x∈X} φ(T x − λx) = +∞ for all λ ∈ J \ {0} and so inf_{λ∈J} sup_{x∈X} φ(T x − λx) + ψ(λ) is clearly equal to r + ψ(0).

In the previous paper [SR], the author proved the same result with the additional hypotheses that the space E does not contain ℓ^{1} and Δ is a segment. It will be proved here in Theorem 2.8 that in this case the sets U which are β(E, E^{∗})-open are characterized by the fact that every weakly converging sequence in E whose weak limit is in U has all but finitely many terms inside U.

We will also show by providing counter-examples that the hypotheses are necessary : the minimax equality can fail if there are several compact operators in Δ or if X has empty interior for β(E, E^{∗}).

The scheme of proof in this paper is quite similar to this of [SR] but it has been of course necessary to change a large part of the proofs for getting rid of this geometrical hypothesis and generalizing to the multidimensional case of Δ. We refer to [SR] for the proofs which have not been modified.

2. Weak topologies

As said in the introduction, for a Banach space E, we will denote by β(E, E^{∗}) the finest locally convex topology on E which agrees with the weak topology σ(E, E^{∗}) on the bounded subsets of E. This topology is finer than σ(E, E^{∗}) and coarser than the norm topology : so it is compatible with the duality betweeen E and E^{∗}. We will also consider the locally convex topology τ_{s} for which the convex open sets are the convex subsets U ⊂ E such that every sequence which converges weakly to some point a ∈ U has cofinitely many terms in U. Of course such sets are norm-open, and τ_{s} is coarser than the norm-topology. Since every weakly converging sequence is bounded hence also β(E, E^{∗})-converging, it is clear that every β(E, E^{∗})-open set is τ_{s}-open. So τ_{s} is finer than β(E, E^{∗}).

If E is infinite-dimensional and has the Schur property, the weakly converging sequences are norm-converging : so in this case τ_{s} agrees with the norm-topology and is strictly finer than β(E, E^{∗}) since for this latter topology the origin is a cluster point of the unit sphere. But if E is a separable Banach space not containing ℓ^{1}, e.g. a separable reflexive space, we shall prove that τ_{s} coincides with β(E, E^{∗}).

Definition 2.1. Let E be a metric space, X be a subset of E and ε > 0. Then X will be said to be ε-distal if d(x, y) ≥ ε for any two distinct points x and y of X. And E will be said to be ε-precompact iff every ε-distal subset of E is finite (in particular E is covered by finitely many open balls of radius ε).

Conversely, in an open ball of radius ε, a 2ε-distal subset has at most one point. So if E is covered by finitely many open balls of radius ε it is 2ε-precompact.

Lemma 2.2. If A ⊂ E^{∗} is norm-compact, then its polar set

A^{o} = {x ∈ E : ∀x^{∗} ∈ A ⟨x^{∗}, x⟩ ≥ −1}

is a neighborhood of 0 for β(E, E^{∗}).

Proof. Let R > 0. Then A is covered by finitely many balls of radius (2R)^{−1} centered at x_{j}^{∗} (1 ≤ j ≤ m). And the set

W = ⋂_{j=1}^{m}\{x ∈ E : |⟨x_{j}^{∗}, x⟩| ≤ \frac{1}{2} \}

is a weak neighborhood of 0 in E which satisfies A^{o} ⊃ W ∩ B(0, R) : indeed for x ∈ W and

x^{∗} ∈ A there exists j ≤ m such that ∥x_{j}^{∗} − x^{∗}∥ ≤ \frac{1}{2R}. If moreover ∥x∥ ≤ R we get

|⟨x^{∗}, x⟩| ≤ |⟨x^{∗} − x_{j}^{∗}, x⟩| + |⟨x_{j}^{∗}, x⟩| ≤ ∥x∥ · ∥x^{∗} − x_{j}^{∗}∥ + \frac{1}{2} ≤ R.\frac{1}{2R} + \frac{1}{2} = 1

hence ⟨x^{∗}, x⟩ ≥ − |⟨x^{∗}, x⟩| ≥ −1, so x ∈ A^{o}. □

Lemma 2.3. If C ⊂ E is a neighborhood of 0 for β(E, E^{∗}), then C^{o} is norm-compact in E^{∗}.

Proof. For all n, C ∩ B(0, 2^{n}) is a weak neighborhood of 0 in B(0, 2^{n}). Thus there exists some finite subset J_{n} of E^{∗} such that

(∥x∥ ≤ 2^{n} and ∀x^{∗} ∈ J_{n} |⟨x^{∗}, x⟩| ≤ 1) ⟹ x ∈ C .

It follows that for all n :

C^{o} ⊂ \overline{conv}(B(0, 2^{−n}) ∪ \{±x^{∗} : x^{∗} ∈ J_{n}\}) ⊂ B(0, 2^{−n}) + K ,

where K is the convex norm-compact set {∑_{x^{∗}∈J_{n}} λ_{x^{∗}}x^{∗} : ∑ |λ_{x^{∗}}| ≤ 1} which is itself covered by finitely many open balls of radius 2^{−n}. We conclude that C^{o} is covered by finitely many open balls of radius 2^{1−n}, hence is 2^{2−n}-precompact. Thus C^{o} is totally bounded, and complete since norm-closed in E^{∗}. This completes the proof of compactness of C^{o}. □

Corollary 2.4. Let C be a closed convex subset of E containing 0. Then C is a neighborhood of 0 for β(E, E^{∗}) if and only if C^{o} is norm-compact in E^{∗}.

Proof. Indeed we then have C = (C^{o})^{o}. □

Lemma 2.5. Let E be a Banach space, C a bounded convex symmetric subset of E and p a continuous linear mapping from E to a finite-dimensional normed space V. Then there exists some constant M such that the inequality d(x, C ∩ ker p) ≤ M · ∥p(x)∥ holds for every x ∈ C.

Proof. Up to replacing V by its linear subspace generated by p(C), we can assume that the convex symmetric set W = p(C) is a neighborhood of 0 in V. If (e_{1}, e_{2}, …, e_{k}) is a

basis of V consisting of members of W, we let a_{0} = −\frac{1}{k} ∑_{j=1}^{k} e_{j} ∈ W and a_{j} = \frac{e_{j}}{k} for

1 ≤ j ≤ k. We then have ∑_{j=0}^{k} a_{j} = 0 and a_{j} ∈ W. Then K = conv(a_{j} : 0 ≤ j ≤ k) is a neighborhood of 0 in V, and there exists r > 0 such that B(0, r) ⊂ K ⊂ W ^{1}. There are then (ξ_{j})_{0≤j≤k} ∈ C such that p(ξ_{j}) = a_{j}.

If x ∈ C, w = p(x) and 0 < ∥p(x)∥ = ρ ≤ r, then we have −\frac{r}{ρ}w ∈ K and there exist

non-negative (λ_{j})_{0≤j≤k} with sum 1 such that −\frac{r}{ρ}w = ∑_{j} λ_{j}a_{j}. It follows that

0 = \frac{ρ}{r + ρ} · ∑_{j} λ_{j}a_{j} + \frac{r}{r + ρ} · w ,

1 In fact it is possible to prove that if dim(V) = k and W ⊃ B(0, r_{0}) there are k+1 elements a_{0}, a_{1}, … a_{k} of W such that ∑_{j} a_{j} = 0 and B(0, r) ⊂ K = conv(a_{0}, a_{1}, …, a_{k}) where r = k^{−3/2}r_{0} : take the vertices of a regular simplex and r = k^{−1}.r_{0} if V is euclidean, and use Dvoretzky-Rogers theorem ([DR]) to reduce the problem to the euclidean case.

equality which still holds if w = 0 with any (λ_{j}), hence that x̂ = \frac{ρ}{r + ρ} ∑_{j} λ_{j}ξ_{j} + \frac{r}{r + ρ} · x

satisfies p(x̂) = 0 and x̂ ∈ C. It results that

d(x, C ∩ ker p) ≤ ∥x − x̂∥ = \frac{ρ}{r + ρ} ∥x − ∑_{j} λ_{j}ξ_{j}∥ ≤ \frac{ρ}{r + ρ} · diam(C) ≤ ∥p(x)∥ · \frac{diam(C)}{r}

This inequality still holds if ∥p(x)∥ ≥ r since 0 ∈ C ∩ ker p and

d(x, C ∩ ker p) ≤ ∥x∥ ≤ diam(C) ≤ \frac{∥p(x)∥}{r} · diam(C) .

It is then enough to take M = \frac{diam(C)}{r}. □

Lemma 2.6. Let E be a Banach space, C a non-compact bounded convex symmetric closed subset of E and p a continuous linear mapping from E to a finite-dimensional normed space V. Then C ∩ ker p is not compact.

More precisely if C is not δ-precompact, then C ∩ ker p is δ′-precompact for none δ′ < \frac{δ}{2}.

Proof. If the complete set C is not compact, there exists δ > 0 and an infinite δ-distal

sequence (x_{j}) in C. Let δ′ < \frac{δ}{2}. Since W = p(C) is a bounded subset of V, \overline{W} is compact

and the sequence (p(x_{j})) has a subsequence (p(x_{j_{ℓ}})) which converges to some w ∈ \overline{W}.

If w = 0, it follows from Lemma 2.5 that there are ξ_{ℓ} ∈ C_{0} = C ∩ ker p such that ∥x_{j_{ℓ}} − ξ_{ℓ}∥ ≤ M · ∥p(x_{j_{ℓ}})∥. Thus we have for ℓ < k, ∥ξ_{ℓ} − ξ_{k}∥ ≥ δ − M(∥p(x_{j_{ℓ}}∥ + ∥p(x_{j_{k}})∥),

and in particular ∥ξ_{ℓ} − ξ_{k}∥ ≥ \frac{δ}{2} > δ′ for large ℓ, which contradicts the δ′-precompactness of C_{0}.

Conversely if w ≠ 0, let t ∈]\frac{δ′}{δ} , \frac{1}{2}[ and η = \frac{t}{1 − t} ∈]0, 1[. We have −w ∈ \overline{W} and 0 ∈ W^{o},

hence −ηw = η.(−w) + (1 − η).0 ∈ W. Thus there exists y ∈ C such that p(y) = −ηw. Letting x′_{j} = tx_{j} + (1 − t)y, we get x′_{j} ∈ C, ∥x′_{j} − x′_{k}∥ = t ∥x_{j} − x_{k}∥ ≥ tδ > δ′ and p(x′_{j_{ℓ}}) = tp(x_{j_{ℓ}}) − (1 − t)ηw = t(p(x_{j_{ℓ}}) − w) → 0. As in the previous case, we then can find ξ_{ℓ} ∈ C_{0} such that ∥ξ_{ℓ} − ξ_{k}∥ ≥ δ′ for large ℓ and k. □

Theorem 2.7. Let E be a Banach space and U ⊂ E be a symmetric convex set. Assume that U contains some ball B(0, ε) and that U is not a neighborhood of 0 for β(E, E^{∗}). Then there exists R > 0 such that for all finite-dimensional linear subspace L of E, U + L does not contain the ball B(0, R).

Proof. Consider the polar set C = U^{o} = (\overline{U})^{o} of U, which is a symmetric closed convex

subset of E^{∗}. Since U ⊃ B(0, ε), we have C ⊂ B(0, \frac{1}{ε}), which shows that C is bounded. Since U is not a neighborhood of 0 for β(E, E^{∗}), we know by Corollary 2.4 that C is not compact.

Let L be a finite-dimensional subspace of E. We have \overline{conv}(U ∪ L) = \overline{U + L}, whence (U + L)^{o} = U^{o} ∩ L^{o} = C ∩ L^{o}. Then L^{o} is a closed linear finite-codimensional subspace of E^{∗} , and the canonic mapping p : E^{∗} → E^{∗}/L^{o} has finite rank. It follows from Lemma 2.5 that there exists δ > 0 depending only on U such that C is not δ-precompact, and by

Lemma 2.6 that C ∩ ker p = C ∩ L^{o} cannot be \frac{2δ}{5} -precompact. In particular, we cannot have C ∩ L^{o} ⊂ B(0, \frac{δ}{5}). And if we had \overline{U} + L ⊃ B(0, R), we would have C ∩ L^{o} ⊂ B(0, \frac{1}{R}).

The statement follows with R = \frac{5}{δ}. □

Theorem 2.8. Let E be a separable Banach space not containing ℓ^{1} and U be a convex subset of E. If a ∈ U and if every sequence weakly converging to a has cofinitely many terms in U, then U is a neighborhood of a for β(E, E^{∗}). In particular τ_{s} coincides with β(E, E^{∗}).

Proof. Under these hypotheses, it is clear that U is a norm-neighborhood of a in E. Let U′ = (U −a)∩(a−U) ; then U′ is symmetric and contains a ball centered at 0. If U is not a neighborhood of a for β(E, E^{∗}), then U′ is not a neighborhood of 0 for β(E, E^{∗}) and there exists by previous theorem some R > 0 such that, for every finite-dimensional subspace L of E, \overline{U′} + L does not contain B(0, R).

We construct inductively a sequence (x_{p}) in B(0, R) such that x_{q} − x_{p} ∉ U′ whenever p < q. Put x_{0} = 0. If the x_{p}’s are defined for p < q, we take for L the linear space generated by (x_{p})_{p<q}, and since B(0, R) \ (\overline{U′} + L) ≠ ∅, we choose x_{q} such that ∥x_{q}∥ ≤ R and that x_{q} ∉ \overline{U′} + L. In particular, for p < q, x_{q} ∉ x_{p} + U′, it is x_{q} − x_{p} ∉ U′.

The sequence (x_{p}) has at least a cluster value x^{∗∗} in the ball of E^{∗∗} of radius R equipped with the topology σ(E^{∗∗}, E^{∗}) which is compact. And since E does not contain ℓ^{1}, it follows from Bourgain-Fremlin-Talagrand’s theorem (cf. [BFT]) that there exists a sequence (x_{p_{j}}) extracted from (x_{p}) which converges to some x^{∗∗} for σ(E^{∗∗}, E^{∗}). Then the sequence y_{j} = x_{p_{j+1}}−x_{p_{j}} converges to 0 for σ(E^{∗∗}, E^{∗}) it is for the weak topology σ(E, E^{∗}). Moreover y_{j} ∉ U′ for all j. This means that for all j either a + y_{j} ∉ U or a − y_{j} ∉ U. So we can find a sequence (ε_{j}) ∈ {−1, 1} such that a + ε_{j}y_{j} ∉ U while (a + ε_{j}y_{j}) converges weakly to a. This shows that there exists a sequence which converges to a ∈ U while none of its terms belongs to U, a contradiction.

It follows that U is a neighborhood of a for β(E, E^{∗}). And since this holds for any convex subset U, we conclude that τ_{s} = β(E, E^{∗}). □

3. A geometric lemma

Let Λ be a finite-dimensional normed space, F be a Banach space, E be an infinitedimensional normed space and Φ : Λ × E → F be a continuous bilinear functional. To each element x of E corresponds a linear operator Φ_{x} ∈ L(Λ, F) defined by Φ_{x}(λ) = Φ(λ, x) and by abuse we will denote for x ∈ E : ker(x) := ker(Φ_{x}) ⊂ Λ the kernel of the operator Φ_{x}. It is quite clear that the mapping x ↦ Φ_{x} is continuous from E to L(Λ, F).

The aim of this section is to prove the following:

Theorem 3.1. Let Λ be a finite-dimensional normed space, F be a Banach space, E be an infinite-dimensional normed space and Φ : Λ × E → F be a continuous bilinear functional. Assume that there exists some non-empty compact subset K of Λ not containing 0 such that ker(x) ∩ K ≠ ∅ for every x ∈ E. Then there exists some a ∈ K and some closed finite-codimensional subspace Ẽ of E such that Φ(a, x) = 0 for every x ∈ Ẽ.

The previous statement will be derived from the following theorem.

Theorem 3.2. Let Λ be a normed space of finite dimension ℓ, K be a compact subset of Λ not containing 0, F be a Banach space, E be a linear normed space and Φ : Λ × E → F be a continuous bilinear functional. Assume that K ∩ ker(x) ≠ ∅ for each x ∈ E, that K is minimal for this property and spans linearly the space Λ. Then there exists a subspace F_{1} of F with dimension at most ℓ(ℓ − 1) which contains Φ(a, x) for every x ∈ E and every a ∈ Λ.

We need several lemmas for proving this theorem. For each of them we assume silently the conditions of Theorem 3.2 are satisfied.

Lemma 3.3. The set U_{0} of those x ∈ E such that the rank rk(Φ_{x}) of the linear mapping Φ_{x} is maximal is a dense open subspace of E.

Proof. Since rk(Φ_{x}) ≤ ℓ, there exists some integer p ≤ ℓ such that p = max_{x∈E} rk(Φ_{x}). For every α = (a_{1}, a_{2}, …, a_{p}) ∈ Λ^{p} the set

W_{α} = {T ∈ L(Λ, F) : (T a_{1}, T a_{2}, …, T a_{p}) is linearly independent }

is an open subset of L(E, F) and U_{0} = ⋃_{α∈Λ^{p}}{x ∈ E : Φ_{x} ∈ W_{α}} ; so U_{0} is open.

Furthermore let y ∈ U_{0}, x ∈ E and ε > 0. There exists α ∈ Λ^{p} such that Φ_{y} ∈ W_{α} and for t ∈ ℝ we denote x_{t} = (1 − t)x + ty ∈ E. Of course ∥x_{t} − x∥ = |t| · ∥x − y∥ < ε for t small enough. The exterior product P (t) = ⋀_{j=1}^{p} Φ(a_{j}, x_{t}) ∈ ⋀^{p}F is polynomial in t (of degree ≤ p) and does not vanish at 1. So there are at most p values of t such that P (t) = 0, and one can find some t^{∗} such that both P (t^{∗}) ≠ 0 and |t^{∗}| < \frac{ε}{∥x − y∥}. For such a t^{∗} we

have p ≤ rk(Φ_{x_{t^{∗}}}) ≤ max_{z∈E} rk(Φ_{z}) = p, hence x_{t^{∗}} ∈ U_{0}. This shows that d(x, U_{0}) < ε. Therefore U_{0} is dense in E. □

Lemma 3.4. The set of those a ∈ K such that a ∈ ker(x) for some x ∈ U_{0} spans linearly the space E.

Proof. Assume towards a contradiction that there exists an hyperplane H of Λ containing ⋃_{x∈U_{0}} ker(x). We would have ker(x)∩K ⊂ H for all x ∈ U_{0}. Following Lemma 3.3 for every x ∈ E there exists a sequence (x_{n}) in U_{0} converging to x. By hypothesis we can find for every n some a_{n} ∈ K ∩ ker(x_{n}). Then a_{n} ∈ K ∩ H and by compactness of K ∩ H there is a subsequence (a_{n_{k}}) which converges to some â ∈ K ∩ H. So Φ(â, x) = lim_{k} Φ(a_{n_{k}}, x_{n_{k}}) = 0. This proves that ker(x) ∩ (K ∩ H) ≠ ∅ for every x ∈ E, and contradicts the minimality of K : indeed K ∩ H is compact, meets every ker(x) for x ∈ E and has linear dimension at most ℓ − 1 while the linear dimension of K is ℓ. □

Lemma 3.5. If x ∈ U_{0} and a_{0} ∈ ker(x) then Φ_{y}(a_{0}) ∈ Φ_{x}(Λ) holds for every y ∈ E.

Proof. Let T = Φ_{x} ∈ L(Λ, F) and p = rk(Φ_{x}) = max_{y∈E} rk(Φ_{y}). Then there exists α = (a_{1}, a_{2}, …, a_{p}) ∈ Λ^{p} such that T α = (T a_{1}, T a_{2}, …, T a_{p}) is a basis of the subspace F_{0} = Φ_{x}(Λ) of F. Let (b_{1}^{∗}, b_{2}^{∗}, …, b_{p}^{∗}) be the dual basis of F_{0}^{∗} (i.e. ⟨b_{i}^{∗}, T a_{j}⟩ = δ_{i}^{j}). By Hahn-Banach’s theorem we can extend each b_{i}^{∗} to some f_{i}^{∗} ∈ F^{∗}.

Let y ∈ E and consider the matrix M = (m_{i,j}) where m_{i,j} = ⟨f_{i}^{∗}, Φ_{y}(a_{j})⟩ and the vector X_{0} = (z_{1}, z_{2}, …, z_{p}) ∈ ℝ^{p} such that z_{i} = ⟨f_{i}^{∗}, Φ_{y}(a_{0})⟩. For t small enough, Φ_{x} + tΦ_{y} belongs to the open set W_{α} and since rk(Φ_{x} + tΦ_{y}) = p, the vectors Φ_{x+ty}(a_{j}) form a basis of Φ_{x+ty}(Λ). So there is a unique γ(t) ∈ ℝ^{p} with coordinates γ_{j}(t), such that

(∗)  ∑_{j=1}^{p} γ_{j}(t)Φ_{x}(a_{j}) + ∑_{j=1}^{p} tγ_{j}(t)Φ_{y}(a_{j}) = ∑_{j=1}^{p} γ_{j}(t).Φ_{x+ty}(a_{j}) = Φ_{x+ty}(a_{0})

In particular for t = 0, Φ_{x}(a_{0}) = 0, hence γ_{j}(0) = 0 for all j ≤ p. And for all i ≤ p,

∑_{j=1}^{p}(δ_{i}^{j} + tm_{i,j})γ_{j}(t) = ∑_{j=1}^{p} γ_{j}(t)⟨f_{i}^{∗}, Φ_{x+ty}(a_{j})⟩ = ⟨f_{i}^{∗}, Φ_{x+ty}(a_{0})⟩ = t⟨f_{i}^{∗}, Φ_{y}(a_{0})⟩ = tz_{i} ,

it is (I + tM)γ(t) = tX_{0}, or γ(t) = t(I + tM)^{−1}X_{0}. We conclude that the γ_{j}’s are C^{1} functions (even rational fractions) in a neighborhood of 0. Derivating (∗) with respect to t at 0 we get :

∑_{j=0}^{p}(γ′_{j}(0) · Φ_{x}(a_{j}) + γ_{j}(0) · Φ_{y}(a_{j})) = Φ_{y}(a_{0})

hence Φ_{y}(a_{0}) = Φ_{x}(λ) for λ = ∑_{j} γ′_{j}(0) · a_{j} ∈ Λ. Thus Φ_{y}(a_{0}) ∈ Φ_{x}(Λ). □

Proof of Theorem 3.2

It follows from Lemma 3.4 that there are families ā = (a_{1}, a_{2}, …, a_{ℓ}) of points of K and (x_{1}, x_{2}, …, x_{ℓ}) of U_{0} such that a_{j} ∈ ker(x_{j}) and ā spans the whole Λ. Consider the ℓ(ℓ − 1) vectors b_{i,j} = Φ_{x_{i}}a_{j} for i ≠ j in [1, ℓ] and the linear subspace F_{1} they span, which has obviously dimension at most ℓ(ℓ − 1).

If x ∈ E and a ∈ Λ there are real coefficients (β_{i})_{i≤ℓ} such that a = ∑_{i} β_{i}a_{i}. So Φ(a, x) = ∑_{i} β_{i}Φ_{x}(a_{i}), and by Lemma 3.5 Φ_{x}(a_{i}) ∈ Φ_{x_{i}}(Λ). Therefore there are real numbers μ_{i,j} such that Φ_{x}(a_{i}) = ∑_{j} μ_{i,j}Φ_{x_{i}}(a_{j}) = ∑_{j≠i} μ_{i,j}b_{i,j} since Φ_{x_{i}}(a_{i}) = 0. Thus Φ(a, x) = Φ_{x}(a) = ∑_{i≠j} λ_{i}μ_{i,j}b_{i,j} ∈ F_{1}. □

Proof of Theorem 3.1

We now intend to derive Theorem 3.1 from Theorem 3.2. Clearly there exists a minimal compact subset K_{0} of K which meets ker(x) for every x ∈ E. Define Λ_{0} = span(K_{0}) ⊂ Λ and ℓ_{0} ≤ ℓ the dimension of Λ_{0}. Since 0 ∉ K_{0} and K_{0} ≠ ∅, we have ℓ_{0} ≥ 1. Replacing Λ by Λ_{0} and K by K_{0} we can apply Theorem 3.2 and find a subspace F_{1} of F of dimension N ≤ ℓ_{0}(ℓ_{0} − 1) ≤ ℓ(ℓ − 1) containing Φ(λ, x) for every x ∈ E and every λ ∈ Λ_{0}, hence Φ(a, x) for x ∈ E and a ∈ K_{0}.

Choose a basis (b_{1}, b_{2}, …, b_{N}) of F_{1} and denote (b_{1}^{∗}, b_{2}^{∗}, …, b_{N}^{∗}) the dual basis in F_{1}^{∗}. Extend by Hahn-Banach’s theorem each b_{j}^{∗} into a φ_{j} ∈ F^{∗}.

So we get a finite family (φ_{j})_{1≤j≤N} of continuous linear functionals on F such that F_{1} ∩ ⋂_{j≤N} ker(φ_{j}) = ⋂_{j} ker(b_{j}^{∗}) = {0}. Choose any a ∈ K_{0}. For 1 ≤ j ≤ N consider the continuous linear functional on E defined by τ_{j} : x ↦ ⟨φ_{j}, Φ(a, x)⟩ and

Ẽ = {x ∈ E : ∀j ∈ [1, N] τ_{j}(x) = 0 } ,

which is a closed finite-codimensional subspace of E. Then for every x ∈ Ẽ and b = Φ_{x}(a), we have ⟨φ_{j}, b⟩ = τ_{j}(x) = 0 and b ∈ F_{1}, hence b = 0. □

Notice that in order to get Theorem 3.1 we have to restrict the conclusion to some subspace Ẽ of E. In fact we can construct the following example :

Example 3.6. There exists a compact subset Δ of ℝ^{3} not containing 0 and a 2-dimensional subspace E of L(ℝ^{3}, ℝ^{2}) such that Δ∩ker U ≠ ∅ for every U ∈ E but Δ∩⋂_{U∈E} ker(U) = ∅.

Proof. Take Δ = \{(x, y, z) ∈ ℝ^{3} : z = 1 and x^{2} + y^{2} ≤ \frac{1}{4} \} which is a compact disc in the

affine hyperplane H = {(x, y, z) : z = 1}. Define the linear mappings S, T in L(ℝ^{3}, ℝ^{2}) having respective matrices

(\begin{matrix}0 & 1 & −1/2 \\ −1 & 0 & 0 \end{matrix}) , (\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 1/2 \end{matrix})

and let E be the linear space spanned by S and T. For λ ∈ ℝ and X = (x, y, z), we have X ∈ H ∩ ker(S + λT) iff

\{ \begin{matrix}λx + (y + z/2) & = z \\ −x + λ(y + z/2) & = 0 \\ z & = 1 \end{matrix}

hence x = λ(y + 1/2) and (y + 1/2)(1 + λ^{2}) = 1, thus x = \frac{λ}{1 + λ^{2}} , y = \frac{1 − λ^{2}}{2(1 + λ^{2})}.

So H ∩ ker(S + λT) = \{(\frac{λ}{1 + λ^{2}} , \frac{1 − λ^{2}}{2(1 + λ^{2})}, 1)\} and H ∩ ker(T) = \{(0, −1/2, 1)\}.

For every non-zero V ∈ E there is a unique point X_{V} in H∩ker(V) satisfying X_{V} ∈ Δ and

x^{2} + y^{2} = \frac{1}{4}. It appears that this unique point X_{V} depends on V, so that Δ ∩ ⋂_{V ∈E} ker(V) is empty. □

Corollary 3.7. Let E and F be infinite-dimensional Banach spaces and (S_{j})_{1≤j≤ℓ} a linearly independent finite family of continuous linear operators from E to F. Then either there exists some a ∈ E such that the family (S_{j}a)_{j≤ℓ} is independent, or there is some non-zero μ = (μ_{j}) ∈ ℝ^{ℓ} such that ∑_{j} μ_{j}S_{j} has finite rank.

Proof. Suppose that the family (S_{j}a)_{j≤ℓ} is independent for none a ∈ E and consider the bilinear mapping Φ : ℝ^{ℓ} ×E → F defined by Φ(λ, x) = ∑_{j=1}^{ℓ} λ_{j}S_{j}x for λ = (λ_{1}, λ_{2}, …, λ_{ℓ}). Let K = {λ = (λ_{1}, λ_{2}, …, λ_{ℓ}) ∈ ℝ^{ℓ} : ∑_{j} λ_{j}^{2} = 1}. Denote S_{λ} = ∑_{j} λ_{j}S_{j} for λ ∈ ℝ^{ℓ}. Then for every x ∈ E there is λ ∈ K such that Φ(λ, x) = S_{λ}(x) = 0, and it follows from Theorem 3.1 that there are Ẽ ⊂ E finite-codimensional and μ ∈ K such that Φ(μ, x) = 0 for all x ∈ Ẽ. Thus rk(S_{μ}) ≤ codim(Ẽ). □

4. E′-convergence and compact operators

Definition 4.1. Let E be a Banach space, E^{∗} its dual and E′ a linear subspace of E^{∗}. We will say that the sequence (x_{n}) E′-converges to a ∈ E iff it is bounded in E and converges to a for the weak topology σ(E, E′), it is lim_{n}⟨θ, x_{n} − a⟩ = 0 for every θ ∈ E′. We also will say that a sequence (x_{n}) is E′-null if it E′-converges to 0.

We will denote by Z the linear space of E′-null sequences.

Of course if E′ = E^{∗}, the sequence E^{∗}-converges to a iff it converges weakly to a.

Lemma 4.2. Assume that E′ is a separable linear subspace of E^{∗} and that (x_{n}) is a bounded sequence in E. Then there exists a subsequence (x_{n_{k}}) such that lim_{k→∞}⟨θ, x_{n_{k}}⟩ exists in ℝ for all θ ∈ E′.

Proof. If (θ_{j})_{j∈ℕ} is a norm-dense sequence in E′, each sequence (⟨θ_{j}, x_{n}⟩)_{n} is bounded in ℝ. Thus there exists a subsequence (x_{n_{k}}) such that : lim_{k→∞}⟨θ_{j}, x_{n_{k}}⟩ exists for all j. And it is easy to see that the sequence (⟨θ, x_{n_{k}}⟩)_{k} is then Cauchy in ℝ for every θ ∈ E′. □

Lemma 4.3. If T is not compact, there exists an ε > 0 and a sequence (a_{n}) in the unit ball B of E such that : d(T a_{n}, span({T a_{p} : p < n})) ≥ ε holds for all n ∈ ℕ.

Proof. Let B be the unit ball of E. Define inductively ε_{n} ≥ 0 and a_{n} ∈ B such that we have

ε_{n} = sup_{a∈B} d(T a, L_{n}) and d(T a_{n}, L_{n}) ≥ \frac{ε_{n}}{2}

where L_{n} = span(T a_{1}, T a_{2}, …, T a_{n−1}). Of course L_{0} = {0} and ε_{0} = ∥T ∥.

It is clear that the sequence (ε_{n}) is non-increasing and does not vanish if rk(T) = +∞. Of course T is compact if it has finite rank.

Suppose T has infinite rank. First show that if ε = inf_{n} \frac{ε_{n}}{2} = 0 then T is compact. Indeed

let η > 0 ; there exists n such that ε_{n} < η′ = \frac{η}{2} . Then C = \{y ∈ L_{n} : ∥y∥ ≤ ∥T ∥ + η′\} is compact, hence covered by finitely many balls (B(y_{j}, η′))_{j≤N}. For all a ∈ B, we have by definition d(T a, L_{n}) ≤ ε_{n} < η′. Then there exists y ∈ L_{n} such that ∥T a − y∥ < η′, thus ∥y∥ ≤ ∥T a∥ + ∥y − T a∥ < ∥T ∥ + η′ and there exists j ≤ N such that ∥y − y_{j}∥ < η′, hence ∥T a − y_{j}∥ < 2η′ = η, what shows that T (B) ⊂ ⋃_{j≤N} B(y_{j}, η). It follows that T (B) is totally bounded, hence that T is a compact operator.

Conversely if ε = inf_{n} \frac{ε_{n}}{2} > 0 we have for all n : d(T a_{n}, L_{n}) ≥ \frac{ε_{n}}{2} ≥ ε. □

Theorem 4.4. Let E be a separable Banach space, V be a Banach space and T ∈ L(E, V). Then the operator T is compact if and only if there exists a separable subspace E′ of E^{∗} such that for every E′-null sequence (x_{n}) the sequence (T x_{n}) converges in norm to 0 in V.

More precisely, if T is not compact, E′ is a separable subspace of E^{∗} and m is any integer, then there is a linear subspace Y of Z of dimension m such that for every non-zero v = (v_{n}) ∈ Y the inequality inf_{n} ∥T v_{n}∥ > 0 holds.

Proof. Suppose T is compact. Then T^{∗} : V^{∗} → E^{∗} is compact too, so T^{∗}(B^{∗}) is relatively compact in E^{∗} hence norm-separable. Let (φ_{k}) be a sequence in B^{∗} such that (ψ_{k}) = (T^{∗}φ_{k}) is norm-dense in T^{∗}(B^{∗}) and denote by E′ the linear span of the ψ_{k}’s. Suppose that (x_{n}) is a bounded sequence in E and that lim_{n}⟨ψ_{k}, x_{n}⟩ = 0 for all k. Let M = sup_{n} ∥x_{n}∥. Then the sequence (T x_{n}) has at least one cluster point b (for the norm-topology), and if this cluster point is unique, then (T x_{n}) converges to b. Assume b ≠ 0 ; there is some φ ∈ B^{∗}

such that ⟨φ, b⟩ > δ = \frac{∥b∥}{2}, then some k such that ∥ψ_{k} − T^{∗}φ∥ < \frac{δ}{2M}. There exists an increasing sequence (n_{j}) such that ⟨φ, T x_{n_{j}}⟩ > δ for all j. So

⟨ψ_{k}, x_{n_{j}}⟩ ≥ ⟨T^{∗}φ, x_{n_{j}}⟩ − ∥ψ_{k} − T^{∗}φ∥ · ∥x_{n_{j}} ∥ = ⟨φ, T x_{n_{j}}⟩ − ∥ψ_{k} − T^{∗}φ∥ · ∥x_{n_{j}} ∥

> δ − M · \frac{δ}{2M} = \frac{δ}{2}

contradicting the hypothesis ⟨ψ_{k}, x_{n}⟩ → 0.

Conversely assume that E′ is a separable subspace of E^{∗} and T is not compact. Choose first following Lemma 4.3 some ε > 0 and some sequence (a_{n}) in the unit ball B of E such that d(T a_{n}, L_{n}) ≥ ε for all n where L_{n} = span(T a_{1}, T a_{2}, …, T a_{n−1}). By Lemma 4.2 we can and do assume that λ(θ) := lim_{n→∞}⟨θ, a_{n}⟩ exists for all θ ∈ E′. Then put for 1 ≤ j ≤ m: b_{j} = (b_{n}^{(j)}), where b_{n}^{(j)} = a_{n+j} − a_{n}, and let Y be the linear space of sequences spanned by the b_{j}’s. We clearly have sup_{n} ∥b_{n}^{(j)}∥ ≤ 2, what shows that the sequences b_{j} are bounded and so are all members of Y. Moreover for θ ∈ E′ we have

lim_{n→∞}⟨θ, b_{n}^{(j)}⟩ = lim_{n→∞}⟨θ, a_{n+j}⟩ − lim_{n→∞}⟨θ, a_{n}⟩ = λ(θ) − λ(θ) = 0 .

It follows that Y ⊂ Z. Finally if v = (v_{n}) ∈ Y, there are real coefficients (α_{j})_{1≤j≤m} such that v = ∑_{j=1}^{m} α_{j}b_{j}. If these coefficients are not all zero there exists p ≤ m such that α_{p} ≠ 0 and α_{j} = 0 for p < j ≤ m. We then have

T v_{n} = ∑_{j=1}^{m} α_{j}(T a_{n+j} − T a_{n}) = ∑_{j=1}^{p} α_{j}(T a_{n+j} − T a_{n}) = α_{p}.T a_{n+p} + y

where y ∈ L_{p+n}, hence ∥v_{n}∥ ≥ |α_{p}| .ε, and inf_{n} ∥T v_{n}∥ ≥ |α_{p}| .ε > 0. In particular this shows that v ≠ 0, thus that the (b_{j})_{j≤m} form a basis of Y and that dim(Y) = m.

Applying what precedes for m = 1, we get a E′-null sequence (x_{n}) in E such that inf_{n} ∥T x_{n}∥ > 0, hence (T x_{n}) does not converge to 0 in F. □

Lemma 4.5. Let E and V be Banach spaces, π ∈ L(E, V) be a compact operator and Y be a convex non-empty open subset of V. Then π^{−1}(Y) is open for β(E, E^{∗}).

Proof. Up to a translation in E we can assume that 0 ∈ Y hence that Y ⊃ B(0, 2^{−m}) for some m ∈ ℕ. Then the polar set Y^{o} is contained in 2^{m} · B^{∗} (where B^{∗} denotes the unit ball of E^{∗}) and

π^{−1}(Y)^{o} = {φ ∈ E^{∗} : ∀x ∈ π^{−1}(Y) ⟨φ, x⟩ ≥ −1}

Since εY is a convex neighborhood of 0 we have for every ε > 0 :

\overline{Y} ⊂ Y + εY = (1 + ε)Y

hence π^{−1}(\overline{Y}) ⊂ (1 + ε)π^{−1}(Y) and π^{−1}(Y)^{o} ⊃ π^{−1}(\overline{Y})^{o} ⊃ \frac{1}{(1 + ε)}π^{−1}(Y)^{o}. Therefore

π^{−1}(Y)^{o} = π^{−1}(\overline{Y})^{o}.

For z ∈ (π^{∗}(Y^{o}))^{o} and any ξ ∈ Y^{o} we have ⟨ξ, πz⟩ = ⟨π^{∗}ξ, z⟩ ≥ −1, that means πz ∈ (Y^{o})^{o} = \overline{Y}, whence z ∈ π^{−1}(\overline{Y}). Thus (π^{∗}(Y^{o}))^{o} ⊂ π^{−1}(\overline{Y})^{o} = (π^{−1}(Y))^{o} and it follows that 2^{m} · \overline{π^{∗}(B^{∗})} ⊃ \overline{π^{∗}(Y^{o})} ⊃ π^{−1}(Y)^{o}.

Since π is compact so is π^{∗}, and we conclude that the convex weakly closed set π^{−1}(Y)^{o} is norm-compact thus that π^{−1}(Y) is a neighborhood of 0 in E for β(E, E^{∗}). □

Lemma 4.6. Let E be a infinite-dimensional Banach space and X be a convex subset of E whose interior for β(E, E^{∗}) is non-empty. Then there exists a Banach space V, a compact linear mapping π : E → V with dense range and a convex open subset Y of V such that π^{−1}(Y) ⊂ X ⊂ \overline{π^{−1}(Y)}.

Proof. Let X_{0} be the interior of X for β(E, E^{∗}) and a ∈ X_{0}. Put W = (X_{0} − a) ∩ (a − X_{0}) which is a symmetric convex subset of E. Then W is open for β(E, E^{∗}) and contains 0. Since β(E, E^{∗}) is coarser than the norm-topology, W is absorbing and the Minkowski functional p_{W} : x ↦ inf{r > 0 : r^{−1}x ∈ W } is a semi-norm on E such that W = {x : p_{W} (x) < 1}. Denote by V the separated completion of (E, p_{W}) and π the canonical mapping from E to V. By definition π(E) is dense in the Banach space V, and π(W) = π(E) ∩ B(0, 1). We will show that π(X_{0}) is open in π(E) and that if Y denotes the interior of \overline{π(X_{0})} = \overline{π(X)} then we have X_{0} = π^{−1}(Y). Then since X_{0} ≠ ∅ we get π^{−1}(Y) = X_{0} ⊂ X ⊂ \overline{X_{0}} = \overline{π^{−1}(Y)}.

Indeed let b ∈ π(X_{0}) and u ∈ X_{0} such that πu = b. The set {t ∈ ℝ : u − t(a − u) ∈ X_{0}} is open and contains 0. Hence it contains also some ε > 0 and we let v = u − ε(a − u). The homothety with center v ∈ X_{0} and ratio η = \frac{ε}{1 + ε} < 1 transforms X_{0} into itself and

in particular a into u, hence a + W ⊂ X_{0} into u + η · W. Thus π(X_{0}) ⊃ π(E) ∩ B(b, η) and π(X_{0}) is a neighborhood of b in π(E). Finally if Y is the interior of the convex set \overline{π(X_{0})} then Y ∩ π(E) is a convex open subset of π(E) contained in π(E) ∩ \overline{π(X_{0})} and containing π(X_{0}) : indeed if u ∈ π(X_{0}) we have shown the existence of a ball B(u, r) such that π(E) ∩ B(u, r) ⊂ π(X_{0}), hence that \overline{π(E) ∩ B(u, r)} ⊂ B(u, r) ⊂ \overline{π(X_{0})} and u ∈ Y.

It remains to show the compactness of π. By definition of the topology β(E, E^{∗}), for each q ∈ ℕ there is a finite set J_{q} ⊂ E^{∗} such that

∥x∥ ≤ 2^{q} and (∀φ ∈ J_{q} |⟨φ, x⟩| < 1) ⟹ x ∈ W .

Let E′ be the linear span of ⋃_{q} J_{q}. We show that any sequence (x_{n}) in B which E′-converges to 0 satisfies p_{W} (x_{n}) → 0, hence πx_{n} → 0. Indeed for any integer q there is some N such that for all ℓ ∈ J_{q} and all n ≥ N we have |⟨ℓ, x_{n}⟩| < 2^{−q}. So for n ≥ N,

∥2^{q}x_{n}∥ ≤ 2^{q} and (∀φ ∈ J_{q} |⟨φ, 2^{q}x_{n}⟩| < 1)

what implies 2^{q}x_{n} ∈ W, it is p_{W} (x_{n}) ≤ 2^{−q}. It follows that πx_{n} → 0 in V. And by Theorem 4.4 this shows the compactness of π. □

Theorem 4.7. Let E and V be Banach spaces, U a bounded convex open set in V and T ∈ L(E, V). Then T^{−1}(U) has non-empty interior for β(E, E^{∗}) iff T is compact.

Proof. It follows from Lemma 4.5 that T^{−1}(U) has non-empty interior for β(E, E^{∗}) if T is compact.

Conversely, denote M = sup_{v∈U} ∥v∥. If T^{−1}(U) has non-empty interior it follows from Lemma 4.6 that there exists a Banach space W, a compact operator S ∈ L(E, W) with dense range and an open set Y ⊂ W such that T^{−1}(U) ⊃ S^{−1}(Y). Then there is x_{0} ∈ E and ε > 0 such that Y ⊃ B(Sx_{0}, ε). We claim that T factorizes through S : there exists some linear continuous operator R : W → V such that T = R ∘ S, and this will imply that T is compact.

Indeed let x and z be points in E such that Sx = Sz. Then for all integer n we have S(n(x−z)) = 0 ∈ B(0, ε) hence S(x_{0}+n(x−z)) ∈ Y and x_{0}+n(x−z) ∈ S^{−1}(Y) ⊂ T^{−1}(U). So we get T x_{0} + n · T (x − z) ∈ U and

sup_{n}(n · ∥T (x − z)∥) ≤ M + ∥T x_{0}∥ < +∞ .

Thus T x − T z = T (x − z) = 0. It follows that there exists a mapping r : S(E) → V such that T x = r(Sx) for every x ∈ E. One readily checks that r is linear. Moreover, if w ∈ S(E) and ∥w∥ < 1, we have w = Sx for some x ∈ E and S(x_{0}+εx) = Sx_{0}+εSx ∈ B(Sx_{0}, ε) ⊂ Y hence x_{0} + εx ∈ S^{−1}(Y) ⊂ T^{−1}(U) and T x_{0} + εT x ∈ U. So

∥r(w)∥ = ∥r(Sx)∥ = ∥T x∥ ≤ M_{1} = \frac{1}{ε}(M + ∥T x_{0}∥)

what implies that r is continuous on S(E) with norm ≤ M_{1}. So r extends to a continuous linear operator R from \overline{S(E)} = W to V and we get T = S ∘ R. □

Corollary 4.8. Let E be a Banach space, X be a non-empty convex symmetric open subset of E and p_{X} : E → ℝ^{+} be the Minkowski functional : x ↦ inf{t > 0 : x ∈ t · X}. Then p_{X} is a semi-norm on E, the separated completion V of (E, p_{X}) is a Banach space, and the canonical mapping π : E → V is compact iff X has non-empty interior for β(E, E^{∗}).

Proof. Indeed π^{−1}(B_{V}) = X, where B_{V} denotes the open unit ball of V. □

The following lemma will not be used in the sequel but it is interesting by itself.

Lemma 4.9. Let E be a separable Banach space, N ⊂ E be a closed subspace, q : E → Q = E/N be the canonical projection, X be a convex subset of E with non-empty interior for the topology β(E, E^{∗}). Then q(X) is a convex subset of Q with non-empty interior for β(Q, Q^{∗}).

Proof. Following lemma 4.6, there are a Banach space V, a compact linear mapping π : E → V, and an open convex subset Y ⊂ V such that X_{0} = π^{−1}(Y) ⊂ X ⊂ \overline{π^{−1}(Y)}. Then M = \overline{π(N)} is a closed linear subspace of V. Consider the canonical projection q̃: V → Q̃ = V/M, which is an open mapping. Then Ỹ = q̃(Y) is open in Q̃. Since q̃ ∘ π is 0 on N = ker q, there is a linear mapping π̃ : Q = E/N → Q̃ such that π̃ ∘ q = q̃ ∘ π. Moreover if U is any open subset of Q̃ then π̃^{−1}(U) = q((q̃ ∘ π)^{−1}(U)), and this set is open since q̃ ∘ π is continuous and q open. So π̃ is continuous. We now show that π̃ is compact. Let (z_{n}) be a bounded sequence in Q, and suppose sup_{n} ∥z_{n}∥ < M. Then for each n we can find x_{n} ∈ E such that ∥x_{n}∥ < M and qx_{n} = z_{n}. Since q̃ ∘ π is compact there is a convergent subsequence (q̃ ∘ πx_{n_{k}}) of the sequence (q̃ ∘ πx_{n}). Finally (π̃z_{n_{k}}) = (π̃ ∘ qx_{n_{k}}) = (q̃ ∘ πx_{n_{k}}) is a convergent subsequence of (π̃z_{n}).

We now prove that the convex set q(X) ⊂ Q has non-empty interior for the topology β(Q, Q^{∗}). Indeed Ỹ is a non-empty convex subset of Q̃ and we claim that

q(X) ⊃ q(X_{0}) = π̃^{−1}(Ỹ) .

Since T̃ is compact, this will show that q(X_{0}) is open for β(Q̃, Q̃^{∗}). In fact

z ∈ q(X_{0}) ⟺ ∃x (q(x) = z and π(x) ∈ Y )

⟹ ∃x ∈ π^{−1}(Y) π̃(z) = q̃ ∘ π(x) ⟹ π̃(z) ∈ q̃(Y) = Ỹ .

Conversely, if π̃(z) ∈ Ỹ, there are x_{0} ∈ E such that z = qx_{0} and y ∈ Y such that π̃(z) = q̃(y). For each x ∈ E such that π(x) = y we have π̃(z) = q̃ ∘ π(x) = π̃ ∘ q(x), so

0 = π̃ ∘ q(x − x_{0}) = q̃ ∘ π(x − x_{0})
⟹ y − π(x_{0}) = π(x − x_{0}) ∈ M = \overline{π(N)} ⟹ y ∈ \overline{π(x_{0}) + π(N)}

And we deduce that \overline{π(x_{0} + N)} ∩ Y ≠ ∅ hence that π(x_{0} + N) ∩ Y ≠ ∅ since Y is open. It follows that (x_{0} + N) ∩ X_{0} ≠ ∅, and finally that z = q(x_{0}) ∈ q(X_{0}). □

5. Unboundedness theorems

Definition 5.1. Let E be a Banach space and E^{∗} be its dual. A subspace E′ of E^{∗} will be called norming if the equality ∥x∥ = sup{⟨φ, x⟩ : φ ∈ E′ and ∥φ∥ ≤ 1} holds for all x ∈ E.

Of course, by Hahn-Banach’s theorem, E^{∗} itself is norming.

Lemma 5.2. Let F be a separable Banach space. Then there exists a norming separable subspace F′ of F^{∗}.

Proof. For every φ in the unit sphere Σ^{∗} of F^{∗}, the set

U_{k,φ} = {y ∈ F : ∥y∥ = 1 and ⟨φ, y⟩ > 2^{−1/k}}

is an open subset of the unit sphere S of F. By Hahn-Banach’s theorem S = ⋃_{φ∈Σ^{∗}} U_{k,φ} for every k. Since S is separable there is a countable set A_{k} ⊂ Σ^{∗} such that S = ⋃_{φ∈A_{k}} U_{k,φ}. So the subspace F′ spanned by ⋃_{k} A_{k} is norming. □

Denote by 𝔘 some free ultrafilter on ℕ, Z the closed subspace of ℓ^{∞}(E) consisting of all E′-null sequences in E, c_{0}^{𝔘}(F) ⊂ ℓ^{∞}(F) the space of bounded sequences (z_{n}) in F such that lim_{n,𝔘} z_{n} = 0 and by F the quotient space ℓ^{∞}(F)/c_{0}^{𝔘}(F), which is an ultrapower of F.

Theorem 5.3. Let E be a separable infinite-dimensional Banach space, E′ a separable subspace of E^{∗}, F be a separable Banach space, Δ a finite-dimensional compact subset of L(E, F) not containing 0.

If there is no compact operator T ∈ Δ then there exists a sequence (w_{n}) in Z and ε > 0 such that ∥T w_{n}∥ ≥ ε for all T ∈ Δ and all n ∈ ℕ.

For every ŵ = (w_{n}) ∈ Z and T ∈ Δ, the sequence (T w_{n}) is bounded in F and we define T̂ŵ ∈ F as the class of this sequence modulo c_{0}^{𝔘}(F). So T̂ is a continuous linear operator from Z to F : clearly ∥T̂∥ ≤ ∥T ∥.

Lemma 5.4. Let ŵ = (w_{n}) ∈ Z. If lim_{n→∞} inf_{T ∈Δ} ∥T w_{n}∥ = 0 then there exists an S ∈ Δ such that Ŝŵ = 0.

Proof. Let ε_{n} = inf_{T ∈Δ} ∥T w_{n}∥. By compactness of Δ there is some T_{n} ∈ Δ such that ∥T_{n}w_{n}∥ = ε_{n}. Define S = lim_{n,𝔘} T_{n} ∈ Δ. Since

∥Sw_{n}∥ ≤ ∥Sw_{n} − T_{n}w_{n}∥ + ∥T_{n}w_{n}∥ ≤ ∥T_{n} − S∥ . sup_{n∈ℕ} ∥w_{n}∥ + ε_{n}

we get (Sw_{n}) ∈ c_{0}^{𝔘}, hence Ŝŵ = 0 ∈ F. □

Proof of Theorem 5.3.

If there are ε > 0 and ŵ = (w_{n}) ∈ Z such that lim sup_{n} inf_{T ∈Δ} ∥T w_{n}∥ ≥ 2ε then the set D = {n ∈ ℕ : ∀T ∈ Δ ∥T w_{n}∥ ≥ ε} is infinite. Enumerate D into an increasing sequence (n_{k}) and put w′_{k} = w_{n_{k}}. Then ŵ′ = (w′_{k}) ∈ Z and we get ∥T w′_{k}∥ ≥ ε for all k and all T ∈ Δ as wanted.

Assume towards a contradiction that there is no ε > 0 and ŵ = (w_{n}) ∈ Z such that lim sup_{n} ∥T w_{n}∥ ≥ 2ε for all T ∈ Δ. Then for every ŵ = (w_{n}) ∈ Z and ε > 0 we have lim sup_{n→∞} inf_{T ∈Δ} ∥T w_{n}∥ < 2ε, it is lim_{n→∞} inf_{T ∈Δ} ∥T w_{n}∥ = 0 hence by Lemma 5.4: Ŝŵ = 0 for some S ∈ Δ. It follows that for each ŵ ∈ Z we can find S ∈ Δ such that Ŝŵ = 0.

Consider the finite-dimensional subspace Λ of L(E, F) spanned by Δ and the continuous bilinear functional Φ : Λ × Z → F defined by Φ(T, ŵ) = T̂ŵ. If follows from what precedes that for every ŵ ∈ Z there exists T ∈ Δ such that Φ(T, ŵ) = 0. Apply Theorem 3.1 to Λ, F and Δ : we can find some Θ ∈ Δ and some subspace Z̃ of Z of finite codimension m such that Φ(Θ, w̃) = 0 for every w̃ ∈ Z̃.

If Θ was not compact we could find by Theorem 4.4 a subspace Y of dimension m + 1 of Z such that inf_{n} ∥Θv_{n}∥ > 0 for every non-zero element v = (v_{n}) of Y. Thus W̃ ∩ Y ≠ {0}, and there exists a non-zero v̂ ∈ W̃ ∩ Y. So we would get

0 = ∥Φ(Θ, v̂)∥ = lim_{n,𝔘} ∥Θv_{n}∥ ≥ inf_{n} ∥Θv_{n}∥ > 0 ,

a contradiction.

Notice that this necessary condition is also sufficient if it holds for all separable subspaces E′ of E^{∗} : indeed if some S ∈ Δ was compact it would exist by Theorem 4.4 some such E′ such that Sw_{n} → 0 for all w = (w_{n}) ∈ Z ; then for all sequence (w_{n}) ∈ Z we should have

lim sup_{n→∞} ( inf_{T ∈Δ} ∥T w_{n}∥) ≤ lim_{n→∞} ∥Sw_{n}∥ = 0 .  □

Lemma 5.5. Let E be a separable infinite-dimensional Banach space, F a separable Banach space, Δ be a convex compact finite-dimensional subset of L(E, F) containing no compact operator. Assume that Y is a non-empty open subset of the normed space V, π ∈ L(E, V) is a compact operator with dense range and X = π^{−1}(Y).

Then sup_{x∈X} inf_{T ∈Δ} ∥T x∥ = +∞.

Proof. Let y_{0} ∈ Y ∩ π(E), x_{0} ∈ π^{−1}(y_{0}) and r > 0 such that the closed ball B̃(y_{0}, r) is contained in Y. Choose following Theorem 4.4 some separable subspace E′ ⊂ E^{∗} such that every E′-null sequence (x_{n}) satisfies ∥πx_{n}∥ → 0.

It follows from Theorem 5.3 (notice that Δ does not contain the compact operator 0) that one can find in B_{E} an E′-null sequence (w_{n}) and ε > 0 such that for all n ∈ ℕ and all T ∈ Δ the inequality ∥T w_{n}∥ ≥ ε holds. Then the sequence (πw_{n}) converges to 0 in V and we put for n ∈ ℕ:

x_{n} = x_{0} + \frac{r}{2^{−n} + ∥πw_{n}∥} · w_{n} .

We then have

∥πx_{n} − y_{0}∥ = ∥πx_{n} − πx_{0}∥ = \frac{r. ∥πw_{n}∥}{2^{−n} + ∥πw_{n}∥} < r

hence πx_{n} ∈ B(y_{0}, r) ⊂ Y, it is x_{n} ∈ X, and for T ∈ Δ :

∥T x_{n}∥ ≥ \frac{r}{2^{−n} + ∥πw_{n}∥} · ∥T w_{n}∥ − ∥T x_{0}∥ ≥ \frac{r.ε}{2^{−n} + ∥πw_{n}∥} − ∥T ∥ · ∥x_{0}∥ ,

hence inf_{T ∈Δ} ∥T x_{n}∥ ≥ \frac{r.ε}{2^{−n} + ∥πw_{n}∥} − sup_{T ∈Δ} ∥T ∥ · ∥x_{0}∥ and

sup_{x∈X} inf_{T ∈Δ} ∥T x∥ ≥ sup_{n} \frac{r.ε}{2^{−n} + ∥πw_{n}∥} − sup_{T ∈Δ} ∥T ∥ · ∥x_{0}∥ = +∞ ,

what completes the proof of the lemma. □

Corollary 5.6. Let E be a separable infinite-dimensional Banach space, F a separable Banach space, Δ be a convex compact finite-dimensional subset of L(E, F) containing no compact operator and φ : F → ℝ a continuous and coercive function. Assume Y is a nonempty convex open subset of the normed space V, π ∈ L(E, V) a compact operator with dense range and X = π^{−1}(Y).

Then sup_{x∈X} inf_{T ∈Δ} φ(T x) = +∞.

Proof. Since φ is coercive, for each M ∈ ℝ^{+}, there is some real R > 0 such that for all u ∈ F φ(u) < M ⟹ ∥u∥ < R. Following lemma 5.5, there exists x ∈ X such that ∥T x∥ ≥ R for all T ∈ Δ ; then we have for this vector x : inf_{T ∈Δ} φ(T x) ≥ M. □

Lemma 5.7. Let E be a separable infinite-dimensional Banach space, F and V be Banach spaces, Δ ⊂ L(E, F) be a compact convex set containing exactly one compact operator H , Y be a non-empty convex open subset of V, π ∈ L(E, V) be a compact operator with dense range and X = π^{−1}(Y). Then for T ≠ H in Δ, sup_{x∈X} ∥T x∥ = +∞.

Proof. By hypothesis, for T ∈ Δ distinct from H, {T } is a convex compact subset of L(E, F) and contains no compact operator : so it follows from Lemma 5.5 that

sup_{x∈X} ∥T x∥ = sup_{x∈X} inf_{S∈{T }} ∥Sx∥ = +∞  □

Corollary 5.8. Let E be a separable infinite-dimensional Banach space, F and V be Banach spaces, Δ ⊂ L(E, F) be a compact convex set containing exactly one compact operator H, Y be a non-empty convex open subset of V, π ∈ L(E, V) be a compact operator with dense range and X = π^{−1}(Y) and φ : F → ℝ be a convex continuous and coercive function.

Then for T ≠ H in Δ, sup_{x∈X} φ(T x) = +∞.

Proof. Since φ is coercive, for each M ∈ ℝ^{+}, there is R > 0 such that for all u ∈ F φ(u) < M ⟹ ∥u∥ < R. Following lemma 5.7, there exists x ∈ X such that ∥T x∥ ≥ R ; then we have for this vector x : φ(T x) ≥ M. □

6. Approximation theorems

The first lemma here is a variation of Lemma 10 in [SR].

Lemma 6.1. Let φ be a convex continuous and coercive function on the separable Banach space F. Then the conjugate function φ^{∗} has a proper domain D which is a convex neighborhood of 0 in F^{∗}. Moreover there exists a separable subspace F′ of F^{∗} such that for any dense subset Z of F′, we have for all x ∈ F : φ(x) = sup_{ξ∈Z}⟨ξ, x⟩ − φ^{∗}(ξ).

Proof. Since φ^{∗} is convex and the proper domain of φ^{∗} is

D = {ξ : φ^{∗}(ξ) < +∞} = ⋃_{n}{ξ : φ^{∗}(ξ) ≤ n}

it is clear that D is convex. Since φ is coercive the set {x : φ(x) ≤ 1 + φ(0) } is bounded in F, and there exists R > 0 such that φ(x) < 1 + φ(0) ⟹ ∥x∥ < R. By convexity we

deduce that φ(x) ≥ φ(0) + \frac{∥x∥}{R} if ∥x∥ ≥ R hence that φ(x) − ⟨ξ, x⟩ is bounded from below

outside the ball of radius R if ξ ∈ F^{∗} and ∥ξ∥ < \frac{1}{R}.

The convex continuous function x ↦ φ(x) − ⟨ξ, x⟩ is necessarily bounded from below

on the bounded convex complete set B̃(0, R) ; it follows that D ⊃ B(0, \frac{1}{R}) hence that the interior D^{o} of D is non-empty. The convex l.s.c. function φ^{∗} is finite on D. Hence, if L_{m} denotes the closed subset of D^{o} defined by {ξ ∈ D^{o} : φ^{∗}(ξ) ≤ m} we have D^{o} = ⋃_{m∈ℕ} L_{m} and some L_{m} has non-empty interior by Baire’s Category Theorem. So φ^{∗} is bounded from above on a neighborhood of some point of D^{o} , hence is continuous on D^{o}. Since ⟨ξ, x⟩ ≤ φ(x) + φ^{∗}(ξ) for all x ∈ F and all ξ ∈ F^{∗}, we have φ(x) ≥ sup_{ξ∈Z} [⟨ξ, x⟩ − φ^{∗}(ξ)] for any Z ⊂ F^{∗}.

Let (B_{n})_{n∈ℕ} be a sequence of open balls in F which constitutes a basis for the topology. Assume F′ ⊂ F^{∗} and Z is dense in F′. Since D^{o} is open, D^{o} ∩ Z is dense in D^{o} ∩ F′. By continuity of φ^{∗} on D^{o} we necessarily have

φ(x) ≥ sup_{ξ∈D^{o}∩Z} [⟨ξ, x⟩ − φ^{∗}(ξ)] = sup_{ξ∈D^{o}∩F′} [⟨ξ, x⟩ − φ^{∗}(ξ)] .

If α < φ(x), the point (x, α) does not belong to the epigraph G of φ, which is the closed convex subset G = {(u, t) : t ≥ φ(u)} of F × ℝ. And there exists (q, n) ∈ ℚ × ℕ such that α < q, x ∈ B_{n} and inf_{y∈B_{n}} φ(y) > q.

Denote by A the countable set {(q, n) ∈ ℚ × ℕ : inf_{y∈B_{n}} φ(y) > q}. Then it follows from Hahn-Banach’s theorem that for (q, n) ∈ A there exists ξ_{q,n} ∈ F^{∗} such that

⟨ξ_{q,n}, x⟩ − q > sup_{(u,s)∈G} [⟨ξ_{q,n}, u⟩ − s] = sup_{u∈F} [⟨ξ_{q,n}, u⟩ − φ(u)] = φ^{∗}(ξ_{q,n})

what implies ξ_{q,n} ∈ D and ⟨ξ_{q,n}, x⟩ − φ^{∗}(ξ_{q,n}) > q. Let F′ be the linear space spanned by the linear functionals (ξ_{q,n})_{(q,n)∈A}. Thus φ(x) = sup_{ξ∈D∩F′} [⟨ξ, x⟩ − φ^{∗}(ξ)]. For ξ ∈ D∩F′, we have tξ ∈ D^{o} ∩ F′ for all t ∈ [0, 1[ since 0 ∈ D^{o}. The function t ↦ φ^{∗}(tξ) is convex and l.s.c. on [0, 1] , hence continuous at 1, and it follows that

⟨ξ, x⟩ − φ^{∗}(ξ) = lim_{t→1,t<1}⟨tξ, x⟩ − φ^{∗}(tξ) ≤ sup_{ξ∈D^{o}∩F′} [⟨ξ, x⟩ − φ^{∗}(ξ)] ,

hence that

φ(x) = sup_{ξ∈D∩F′} [⟨ξ, x⟩ − φ^{∗}(ξ)] ≤ sup_{ξ∈D^{o}∩F′} [⟨ξ, x⟩ − φ^{∗}(ξ)] ≤ sup_{ξ∈Z} [⟨ξ, x⟩ − φ^{∗}(ξ)] ≤ φ(x) .

what is the wanted conclusion. □

From now on we are interested in the following frame : Δ is a finite-dimensional convex compact subset of L(E, F) containing only one compact operator H and D is the linear subspace of L(E, F) generated by Δ − H = {T − H : T ∈ Δ}. We denote by K = K(E, F) the set of compact operators from E to F which is a closed linear subspace of L(E, F) and D_{0} = D ∩ K. Of course it can happen that D_{0} ≠ {0} even though (H + K) ∩ Δ = {H}. We are also given a convex continuous function ψ : Δ → ℝ.

Definition 6.2. A finite family (S_{j})_{1≤j≤ℓ} will be said to be adapted to Δ if it is a basis of D and (S_{j})_{r<j≤ℓ} is a basis of D_{0}.

For λ = (λ_{1}, λ_{2}, …, λ_{ℓ}) ∈ ℝ^{ℓ} denote S_{λ} = ∑_{j} λ_{j}S_{j}. We equip ℝ^{ℓ} with its canonical euclidean structure induced by the scalar product ⟨λ, μ⟩ = ∑_{j≤ℓ} λ_{j}μ_{j}, and define Σ its unit sphere. Moreover for ξ ∈ F^{∗} define T_{ξ} ∈ L(E, ℝ^{ℓ}) by

T_{ξ}(x) = (⟨ξ, S_{1}x⟩, ⟨ξ, S_{2}x⟩, …, ⟨ξ, S_{ℓ}x⟩) = (⟨S_{1}^{∗}ξ, x⟩, ⟨S_{2}^{∗}ξ, x⟩, …, ⟨S_{ℓ}^{∗}ξ, x⟩) ∈ ℝ^{ℓ}

for x ∈ E. Let Γ be the compact convex subset {λ ∈ ℝ^{ℓ} : H +S_{λ} ∈ Δ} of ℝ^{ℓ} and χ : Γ → ℝ be then defined by χ(λ) = ψ(H + S_{λ}).

Let ϖ : ℝ^{ℓ} → ℝ^{r} be the mapping λ ↦ (λ_{1}, λ_{2}, …, λ_{r}), Γ_{0} be the subspace ker(ϖ) of ℝ^{ℓ}, and Γ_{1} = Γ_{0}^{⊥} be the subspace ϖ^{∗}(ℝ^{r}) = {λ ∈ ℝ^{ℓ} : ∀j > r λ_{j} = 0} of ℝ^{ℓ}. Then q = ϖ^{∗}ϖ : ℝ^{ℓ} → ℝ^{ℓ} is the orthogonal projection onto Γ_{1}.

Lemma 6.3. Let f be a linear functional on Γ_{0} and g be a convex continuous real function on Γ such that g(0) > 0. Then there exists a μ ∈ ℝ^{ℓ} such that f(λ) = ⟨μ, λ⟩ for all λ ∈ Γ_{0} and inf_{λ∈Γ}(g(λ) − ⟨μ, λ⟩) > 0.

Proof. Consider the two convex subsets of ℝ^{ℓ} × ℝ defined by

\{ \begin{matrix}Γ′_{0} & = \{(λ, t) : λ ∈ Γ_{0} and t = f(λ) \} \\ C_{+} & = \{(λ, t) : λ ∈ Γ and g(λ) ≤ t ≤ t^{∗} \} . \end{matrix}

where t^{∗} = sup_{λ∈Γ} g(λ). The first one is a linear subspace and the second one is compact. Moreover they are disjoint: indeed if (λ, t) ∈ Γ′_{0} ∩ C_{+} we have H + S_{λ} ∈ D_{0} ∩ Δ hence λ = 0, thus t = f(λ) = 0 < g(0) ≤ t = 0. Notice also that (λ, g(λ)) ∈ C_{+} for all λ ∈ Γ.

Then it follows from Hanh-Banach’s theorem that there exists a linear functional σ on ℝ^{ℓ} × ℝ, of the form σ(λ, t) = ρ.t − ⟨μ, λ⟩ for some ρ ∈ ℝ and μ ∈ ℝ^{ℓ}, which separates Γ′_{0} from C_{+} :

sup_{(λ,t)∈Γ′_{0}}(ρ.t − ⟨μ, λ⟩) < inf_{(λ,t)∈C_{+}}(ρ.t − ⟨μ, λ⟩) .

Since Γ′_{0} is a linear subspace we get σ(λ, t) = 0 for all (λ, t) = (λ, f(λ)) ∈ Γ′_{0} and since (λ, g(λ)) ∈ C_{+} for λ ∈ Γ we also get 0 < ρ.g(0). So ρ > 0 and replacing μ by ρ^{−1}.μ we get :

sup_{(λ,t)∈Γ′_{0}} t − ⟨μ, λ⟩ = 0 < η := inf_{λ∈Γ} g(λ) − ⟨μ, λ⟩ ,

hence g(λ) ≥ ⟨μ, λ⟩ + η for λ ∈ Γ, and f(λ) = ⟨μ, λ⟩ for λ ∈ Γ_{0}. □

Lemma 6.4. If μ_{0} ∈ ℝ^{ℓ} and δ > 0 there exists ν ∈ Γ_{1} such that :

inf_{λ∈Γ}⟨ν + μ_{0}, λ⟩ ≥ −δ

Proof. Apply Lemma 6.3 with f : λ ↦ −⟨μ_{0}, λ⟩ on Γ_{0} and g constantly equal to δ on Γ. We get μ ∈ ℝ^{ℓ} such that ⟨μ, λ⟩ = −⟨μ_{0}, λ⟩ for λ ∈ Γ_{0} and ⟨μ, λ⟩ > δ for λ ∈ Γ. Then let ν = −(μ+μ_{0}) : we get ν +μ_{0} = −μ, so ⟨ν, λ⟩ = 0 for λ ∈ Γ_{0}, hence ν ∈ Γ_{0}^{⊥} = Γ_{1}. Moreover for λ ∈ Γ : δ = g(λ) > ⟨μ, λ⟩ = −⟨ν + μ_{0}, λ⟩. Thus inf_{λ∈Γ}⟨ν + μ_{0}, λ⟩ ≥ −δ. □

Lemma 6.5. For fixed ε > 0 there exists μ̂ ∈ Γ_{1} such that ⟨λ, μ̂⟩ ≤ χ(λ) − ψ(H) + ε for all λ ∈ Γ.

Proof. Apply Lemma 6.3 with f = 0 and g(λ) = χ(λ)−ψ(H)+ε. Notice that g(0) = ε > 0. We get μ̂ ∈ ℝ^{ℓ} such that ⟨μ̂, λ⟩ < g(λ) = χ(λ) − ψ(H) + ε for λ ∈ Γ, and ⟨μ̂, λ⟩ = 0 for λ ∈ Γ_{0}, hence μ̂ ∈ Γ_{0}^{⊥} = Γ_{1}. □

Theorem 6.6. Let E and F be infinite-dimensional separable Banach spaces, Δ a finitedimensional convex compact subset of L(E, F) containing only one compact operator H and (S_{j})_{1≤j≤ℓ} an adapted family, K ∈ L(E, V) a compact operator, F′ a separable linear subspace of F^{∗}. Then there exists a separable closed subspace F_{1} of F^{∗} containing F′ such that for comeagerly many ξ ∈ F_{1} there exists for all λ ∈ Γ_{1} a sequence (w_{n}) in E such that Kw_{n} → 0 and T_{ξ}(w_{n}) → λ.

Proof. It is clear that ξ ↦ T_{ξ} is linear and continuous from F^{∗} to L(E, ℝ^{ℓ}). For x ∈ E denote ∥ ; |x| ; ∥ = ∥Kx∥ ; this defines on E a coarser norm whose unit ball will be denoted by B_{K}. And we will say that a sequence (x_{n}) is K-null if sup_{n} ∥x_{n}∥ < +∞ and lim_{n} ∥ ; |x_{n}| ; ∥ = 0. Since F′ is separable we can replace F′ by some larger separable space F_{1} which is norming and closed in F^{∗}, and find a dense sequence (ξ_{m}) in F_{1}, so that ∥y∥ = sup_{m}{|⟨ξ_{m}, y⟩| : ∥ξ_{m}∥ ≤ 1} for all y ∈ F. Thus the set {S_{j}^{∗}ξ_{m} : 1 ≤ j ≤ ℓ, m ∈ ℕ} ⊂ E^{∗} is countable and can be enumerated in a sequence (θ_{n})_{n∈ℕ}.

Denote F_{0} the linear space generated by the ξ_{m}’s and U the compact linear operator E → ℓ^{2} defined by

Ux = (\frac{⟨θ_{n}, x⟩}{1 + 2^{n} ∥θ_{n}∥})_{n} .

Define V′ = V ×F^{ℓ−r} ×ℓ^{2}, equipped with the norm : ∥(v, (y_{j}), u)∥ = ∥v∥+∑_{j} ∥y_{j}∥+∥u∥_{2}. Replacing K by K′ : x ↦ (Kx, (S_{j}x)_{j>r}, Ux) from E to V′ which is compact too, we see that any K′-null sequence is K-null and that for all x ∈ E : |⟨θ_{n}, x⟩| ≤ 2^{n}·∥Ux∥ ≤ 2^{n}·∥K′x∥. It follows that, for all ξ ∈ F_{0} and all j ≤ ℓ, S_{j}^{∗}ξ is continuous on (E, ∥ ; |.| ; ∥). Moreover for j > r and ξ ∈ F^{∗} S_{j}^{∗}ξ is also continuous on (E, ∥ ; |.| ; ∥).

Then for ξ ∈ F_{0} there exists some constant β (depending on ξ) such that for all μ ∈ ℝ^{ℓ} and all x we have |⟨S_{μ}^{∗}ξ, x⟩| ≤ β. ∥μ∥ . ∥Kx∥. We also have ∥S_{j}x∥ ≤ ∥K′x∥ for j > r. Let E′ be the closed subspace of E^{∗} spanned by the θ_{n}’s.

Denote R = (∑_{j} ∥S_{j}∥^{2})^{1/2} : we have for all x ∈ E and λ ∈ ℝ^{ℓ} :

∥S_{λ}x∥ = ∥∑_{j} λ_{j}S_{j}x∥ ≤ ∑ |λ_{j}| . ∥S_{j}∥ . ∥x∥ ≤ (∑ λ_{j}^{2})^{1/2}(∑ ∥S_{j}∥^{2})^{1/2} ∥x∥ = R. ∥λ∥ . ∥x∥

Since Σ_{1} = Σ∩Γ_{1} is compact and none S_{λ} is compact for λ ∈ Σ_{1} it follows from Theorem 5.3 that there exists δ > 0 and an E′-null sequence (x_{n}) in the unit sphere of E such that ∥S_{λ}x_{n}∥ ≥ δ for all n and all λ ∈ Σ_{1}.

It follows from Theorem 4.4 that K′x_{n} → 0, and a fortiori Kx_{n} → 0. Passing to a subsequence we can assume moreover that ∥ ; |x_{n}| ; ∥ = ∥Kx_{n}∥ ≤ 2^{−n}.

Lemma 6.7. Let ξ ∈ F_{1}. If q ∘ T_{ξ}(B_{K}) is not the whole Γ_{1}, there exists μ ∈ Σ_{1} and β > 0

such that |⟨S_{μ}^{∗}ξ, x⟩| ≤ β. ∥ ; |x| ; ∥ holds for all x ∈ E.

Proof. Consider the symmetric convex set C = ϖ ∘ T_{ξ}(B_{K}) ⊊ ℝ^{r}. If we had ϖ^{−1}(\overline{C}) = ℝ^{ℓ}, we would necessarily have \overline{C} = ℝ^{r} hence C = ℝ^{r}. Thus there exists λ_{0} ∈ ℝ^{ℓ} \ ϖ^{−1}(\overline{C}), and following Hahn-Banach’s theorem, some μ ∈ ℝ^{ℓ} such that sup_{ϖ(λ)∈\overline{C}}⟨μ, λ⟩ < β := ⟨μ, λ_{0}⟩.

We necessarily have μ ≠ 0 and ϖ(μ) = 0. Replace μ by \frac{μ}{∥μ∥} and β by \frac{β}{∥μ∥}: then μ ∈ Σ_{1}.

It follows that, for all x ∈ B_{K} we have : ⟨S_{μ}^{∗}ξ, x⟩ = ∑_{j} μ_{j}⟨ξ, S_{j}x⟩ = ⟨μ, T_{ξ}(x)⟩ ≤ β, and

by homogeneity |⟨S_{μ}^{∗}ξ, x⟩| ≤ β. ∥Kx∥ = β. ∥ ; |x| ; ∥. □

By compactness of Σ_{1}, one can find a finite family (μ_{k})_{k≤N} in Σ_{1} such that Σ_{1} is covered

by the N balls B(μ_{k}, \frac{δ}{4R}). Then consider the set

M_{k,m} = \{ξ ∈ F_{1} : ∃μ ∈ Σ_{1} ∥μ − μ_{k}∥ ≤ \frac{δ}{4R} and ∀x ∈ E |⟨S_{μ}^{∗}ξ, x⟩| ≤ m ∥Kx∥ \} .

The set

N_{k,m} = \{(μ, ξ) ∈ Σ_{1} × F_{1} : ∥μ − μ_{k}∥ ≤ \frac{δ}{4R} and ∀x ∈ B_{K} |⟨ξ, S_{μ}x⟩| ≤ m. ∥Kx∥ \}

is clearly closed in Σ_{1} × F_{1} and the second projection π_{2} : (μ, ξ) → ξ is a perfect mapping since Σ_{1} is compact. Thus M_{k,m} = π_{2}(N_{k,m}) is closed in F_{1}. Let M be the set of those ξ ∈ F_{1} for which ϖ ∘ T_{ξ}(B_{K}) ≠ ℝ^{r}. It follows from what precedes that M ⊂ ⋃_{k,m} M_{k,m}.

We now show that each M_{k,m} has empty interior in F_{1} = \overline{F_{0}} : indeed if not it would exist some ξ_{0} ∈ F_{0} and ρ > 0 such that F_{1} ∩ B(ξ_{0}, ρ) ⊂ M_{k,m}. For any χ in the unit ball of F_{1} let ξ = ξ_{0} + ρ.χ : then ξ ∈ B(ξ_{0}, ρ). Since ξ_{0} ∈ F_{0}, there exists some constant β such

that |⟨S_{μ}^{∗}ξ_{0}, x⟩| ≤ β ∥ ; |x| ; ∥ for all μ ∈ Σ and x ∈ E, hence

⟨S_{μ}^{∗}ξ, x_{n}⟩ ≥ ρ.⟨χ, S_{μ_{k}}x_{n}⟩ − ρ. |⟨χ, S_{μ−μ_{k}}x_{n}⟩| − |⟨S_{μ}^{∗}ξ_{0}, x_{n}⟩|

≥ ρ.⟨χ, S_{μ_{k}}x_{n}⟩ − ρ ∥S_{μ−μ_{k}}∥ . ∥x_{n}∥ − β ∥ ; |x_{n}| ; ∥

≥ ρ.⟨χ, S_{μ_{k}}x_{n}⟩ − ρR ∥μ − μ_{k}∥ − β ∥ ; |x_{n}| ; ∥

≥ ρ.⟨χ, S_{μ_{k}}x_{n}⟩ − ρ\frac{δ}{4} − β.2^{−n}

Choose n ∈ ℕ such that (m + β).2^{−n} < \frac{δρ}{4}. We can then find χ ∈ F_{1} such that ∥χ∥ ≤ 1

and ⟨χ, S_{μ_{k}}x_{n}⟩ > \frac{3}{4} ∥S_{μ_{k}}x_{n}∥, hence ρ.⟨χ, S_{μ_{k}}x_{n}⟩ ≥ \frac{3ρ}{4}. ∥S_{μ_{k}}x_{n}∥ ≥ \frac{3ρδ}{4} .

We then get

⟨S_{μ}^{∗}ξ, x_{n}⟩ − m. ∥Kx_{n}∥ ≥ ⟨S_{μ}^{∗}ξ, x_{n}⟩ − m.2^{−n} ≥ \frac{3ρδ}{4} − \frac{ρδ}{4} − (m + β).2^{−n} > \frac{ρδ}{4} > 0

what shows that ξ ∉ M_{k,n}, a contradiction. It follows that each M_{k,m} is nowhere dense and that M is meager. In particular Z = F_{1} \ M is dense (and even comeager) in F_{1}. And for every ξ ∈ Z, n ∈ ℕ and λ ∈ Γ_{1}, 2^{n}.λ ∈ Γ_{1} ⊂ T_{ξ}(B_{k}).

So there is some x_{n} ∈ B_{K} such that ϖ ∘ T_{ξ}(x_{n}) = ϖ(2^{n}.λ). Then w_{n} = 2^{−n}x_{n} satisfies ∥ ; |w_{n}| ; ∥ ≤ 2^{−n} and ϖ(T_{ξ}w_{n} − λ) = 0, hence S_{j}^{∗}ξ = λ_{j} for j ≤ r. Moreover, for j > r, ∥S_{j}^{∗}ξw_{n}∥ ≤ ∥ ; |w_{n}| ; ∥ → 0. Thus T_{ξ}w_{n} → λ. And this completes the proof of Theorem 6.6. □

7. The minimax theorem

Lemma 7.1. Let E be a separable infinite-dimensional Banach space, F be a Banach space, Δ be a finite-dimensional convex compact subset of L(E, F) containing only one compact operator H, (S_{j})_{1≤j≤ℓ} be an adapted family, Y be a non-empty convex open subset of the normed space V and π : E → V be a compact operator with dense range, X_{0} = π^{−1}(Y), F′ be a separable subspace of F^{∗} and ψ : Δ → ℝ be a convex continuous function. Then there exists a separable closed subspace F_{1} of F^{∗} containing F′ and a dense subset Z of F_{1} such that for all ξ ∈ Z

sup_{x∈X_{0}} inf_{λ∈Γ}(⟨ξ, (H + S_{λ})x⟩ + χ(λ)) ≥ sup_{x∈X_{0}}⟨ξ, Hx⟩ + ψ(H) .

Proof. Let ξ ∈ Z, λ ∈ Γ, x_{0} ∈ X_{0}, y_{0} = π(x_{0}) ∈ Y and ε > 0, and fix μ̂ ∈ Γ_{1} such that ⟨λ, μ̂⟩ ≤ χ(λ) − ψ(H) + \frac{ε}{2} for all λ ∈ Γ, following Lemma 6.5. Choose also by Lemma 6.4

some ν ∈ Γ_{1} such that inf_{λ∈Γ}⟨T_{ξ}(x_{0}) + ν, λ⟩ ≥ −\frac{ε}{2}.

Define K the operator x ↦ (Hx, πx) from E to the product space W = F × V normed by ∥(y, u)∥ = ∥y∥ + ∥u∥ and recall that χ(λ) = ψ(H + S_{λ}); Since H and π are compact, K is compact too and by Theorem 6.6 one can find a separable closed subspace F_{1} of F^{∗} containing F′ and a dense subset Z of F_{1} such that for fixed ξ ∈ Z and μ ∈ Γ_{1} there is a sequence (w_{n}) in E such that ∥Kw_{n}∥ → 0 and that μ_{n} = T_{ξ}(w_{n}) → ν − μ̂.

Then with θ = S_{λ}^{∗}ξ , x_{n} = x_{0} + w_{n} satisfies

⟨θ, x_{n}⟩ = ⟨S_{λ}^{∗}ξ, x_{n}⟩ = ⟨ξ, S_{λ}(x_{0} + w_{n})⟩ = ⟨λ, T_{ξ}(x_{0})⟩ + ⟨λ, T_{ξ}(w_{n})⟩

= ⟨λ, T_{ξ}(x_{0})⟩ + ⟨λ, μ_{n}⟩ → ⟨λ, T_{ξ}(x_{0}) + ν⟩ − ⟨λ, μ̂⟩ ≥ −⟨λ, μ̂⟩ − \frac{ε}{2}

and πx_{n} = y_{0} + πw_{n} → y_{0}. Because Y is open in V we have πx_{n} ∈ Y for n large enough, hence x_{n} ∈ X_{0}. Moreover

⟨ξ, Hx_{n}⟩ = ⟨ξ, Hx_{0}⟩ + ⟨ξ, Hw_{n}⟩ → ⟨ξ, Hx_{0}⟩ ,

from what it follows that for λ ∈ Γ,

⟨ξ, (H + S_{λ})x_{n}⟩ + χ(λ) = ⟨ξ, Hx_{n}⟩ + ⟨ξ, S_{λ}x_{n}⟩ + χ(λ)

≥ ⟨ξ, Hx_{n}⟩ + ⟨θ, x_{n}⟩ + ⟨λ, μ̂⟩ + ψ(H) − \frac{ε}{2}

≥ ⟨ξ, Hx_{n}⟩ + ψ(H) − ε → ⟨ξ, Hx_{0}⟩ + ψ(H) − ε

thus

sup_{x∈X_{0}} inf_{λ∈Γ}(⟨ξ, (H + S_{λ})x⟩ + χ(λ)) ≥ ⟨ξ, Hx_{0}⟩ + ψ(H) − ε

and since this holds for all x_{0} ∈ X_{0} and ε > 0 we get

sup_{x∈X_{0}} inf_{λ∈Γ}(⟨ξ, (H + S_{λ})x⟩ + χ(λ)) ≥ sup_{x∈X_{0}}⟨ξ, Hx⟩ + ψ(H)

what is the wanted inequality. □

Lemma 7.2. Let E and F be separable infinite-dimensional Banach spaces, Δ be a finitedimensional convex compact subset of L(E, F) containing only one compact operator H, (S_{j})_{1≤j≤ℓ} be an adapted family, Y be a non-empty convex open subset of the normed space V and π : E → V be a compact operator with dense range, X_{0} = π^{−1}(Y), φ : F → ℝ be a convex continuous and coercive function and ψ : Δ → ℝ be a convex continuous function. Then

sup_{π(x)∈Y} inf_{λ∈Γ} φ(Hx + S_{λ}x) + ψ(H + S_{λ}) ≥ ψ(H) + sup_{π(x)∈Y} φ(Hx) .

Proof. Put Φ(λ, x) = φ(Hx + S_{λ}x) and χ(λ) = ψ(H + S_{λ}). It follows from lemma 6.1 that, for some dense subset Z of some closed separable linear subspace F_{1} of F^{∗}, we have

Φ(λ, x) = sup_{ξ∈Z}⟨ξ, Hx + S_{λ}x⟩ − φ^{∗}(ξ) ,

hence

inf_{λ∈Γ}(Φ(λ, x) + χ(λ)) = inf_{λ∈Γ} sup_{ξ∈Z}(⟨ξ, Hx + S_{λ}x⟩ − φ^{∗}(ξ) + χ(λ))

≥ sup_{ξ∈Z} inf_{λ∈Γ}(⟨ξ, Hx + S_{λ}x⟩ − φ^{∗}(ξ) + χ(λ))

≥ sup_{ξ∈Z}(−φ^{∗}(ξ) + inf_{λ∈Γ}(⟨ξ, (H + S_{λ})x⟩ + χ(λ))) ,

and after Lemma 7.1,

sup_{π(x)∈Y} inf_{λ∈Γ}(Φ(λ, x) + χ(λ)) ≥ sup_{π(x)∈Y} sup_{ξ∈Z}(−φ^{∗}(ξ) + inf_{λ∈Γ}(⟨ξ, (H + S_{λ})x⟩ + χ(λ)))

≥ sup_{ξ∈Z}(−φ^{∗}(ξ) + sup_{π(x)∈Y} inf_{λ∈Γ}(⟨ξ, (H + S_{λ})x⟩ + χ(λ)))

≥ sup_{ξ∈Z}(−φ^{∗}(ξ) + ψ(H) + sup_{π(x)∈Y} ⟨ξ, Hx⟩)

≥ ψ(H) + sup_{π(x)∈Y} (sup_{ξ∈Z}⟨ξ, Hx⟩ − φ^{∗}(ξ))

= ψ(H) + sup_{π(x)∈Y} φ(Hx)

what is the wanted inequality. □

Theorem 7.3. Let E and F be separable infinite-dimensional Banach spaces, Δ be a finitedimensional compact convex subset of L(E, F), φ : F → ℝ be a convex continuous and coercive function, X be a convex subset of E whose interior is non-empty for the topology β(E, E^{∗}) and ψ : Δ → ℝ be a convex continuous function. Assume that Δ contains at most one compact operator. Then the following holds :

sup_{x∈X} inf_{T ∈Δ}(φ(T x) + ψ(T)) = inf_{T ∈Δ} sup_{x∈X}(φ(T x) + ψ(T)) .

Proof. By lemma 4.6, we know that there exists a normed space V, a compact linear mapping π : E → V with dense range and a convex open subset Y of V such that π^{−1}(Y) ⊂ X ⊂ π^{−1}(\overline{Y}). It is then easy to see that the supremum over X is equal to the supremum over π^{−1}(Y), and we will only prove the statement when X = π^{−1}(Y). The inequality

sup_{x∈X} inf_{T ∈Δ}(φ(T x) + ψ(T)) ≤ inf_{T ∈Δ} sup_{x∈X}(φ(T x) + ψ(T))

is standard. So we have only to prove the converse inequality.

Following Corollary 5.6, it is enough to consider the case where the convex compact set Δ contains exactly one compact operator H.

Then by Corollary 5.8, we have

inf_{T ∈Δ} sup_{x∈X}(φ(T x) + ψ(T)) = sup_{x∈X}(φ(Hx) + ψ(H)) ,

and it follows from Lemma 7.2 that

sup_{π(x)∈Y} φ(Hx) + ψ(H) ≤ sup_{π(x)∈Y} inf_{λ∈Γ} φ(Hx + S_{λ}x) + ψ(H + S_{λ})

= sup_{π(x)∈Y} inf_{T ∈Δ} φ(T x) + ψ(T),

and this completes the proof. □

The following statement extends Theorem 7.3 by removing the hypotheses of separability on E and F.

Theorem 7.4. Let E and F be infinite-dimensional Banach spaces, Δ be a compact convex finite-dimensional subset of L(E, F), φ : F → ℝ be a convex continuous and coercive function, X be a convex subset of E whose interior is non-empty for the topology β(E, E^{∗}) and ψ : Δ → ℝ be a convex continuous function. Assume that Δ contains at most one compact operator. Then the following holds :

sup_{x∈X} inf_{T ∈Δ}(φ(T x) + ψ(T)) = inf_{T ∈Δ} sup_{x∈X}(φ(T x) + ψ(T)) .

Proof. If E is separable, we can assume F is also separable : indeed G = \overline{∑_{j} S_{j}(E)} is a separable Banach space, and φ_{|G} is convex continuous and coercive.

From now on assume E is not separable. The function (x, T) ↦ φ(T x) + ψ(T) is continuous on X × Δ and Δ is compact.

It follows that the function Φ : x ↦ inf_{T ∈Δ}(φ(T x) + ψ(T)) is continuous on X. So there is some countable set D ⊂ X such that sup_{x∈X} Φ(x) = sup_{x∈D} Φ(x). And for any separable subspace E_{0} ⊂ E containing D we also have

sup_{x∈X} Φ(x) = sup_{x∈D} Φ(x) ≤ sup_{x∈X∩E_{0}} Φ(x) ≤ sup_{x∈X} Φ(x)

hence sup_{x∈X∩E_{0}} Φ(x) = sup_{x∈X} Φ(x).

Assume that for all T ≠ H in Δ, T is not compact : then for all T ∈ Δ there is δ > 0 and a sequence (x_{n}) in the unit ball of E such that ∥T x_{n} − T x_{p}∥ ≥ 3δ if n ≠ p. If S ∈ Δ and ∥S − T ∥ < δ we have for n ≠ p :

∥Sx_{n} − Sx_{p}∥ ≥ ∥T x_{n} − T x_{p}∥ − ∥S − T ∥ . ∥x_{n}∥ − ∥S − T ∥ . ∥x_{p}∥ > 3δ − δ − δ = δ .

It follows that Δ \ {H} is covered by balls B(T, δ_{T}) such that exists a sequence x_{n}^{(T)} in the unit ball of E with ∥Sx_{n}^{(T)} − Sx_{p}^{(T)}∥ > δ_{T} for n ≠ p and S ∈ B(T, δ_{T}). By Lindelöf’s theorem Δ \ {H} is covered by some countable subfamily : so there is some countable set Γ_{1} ⊂ E such that for each T ∈ Δ \ {H} one can find an infinite subset J of Γ_{1} satisfying inf_{x≠y, x,y∈J} ∥T x − T y∥ > 0. It follows that for every separable subspace E_{1} of E containing Γ_{1} and every T ∈ Δ \ {H} the restriction T_{|E_{1}} is not compact.

There is a countable D′ ⊂ X such that

sup_{x∈D′} φ(Hx) + ψ(H) = sup_{x∈X} φ(Hx) + ψ(H)

Choosing a separable subspace E_{1} containing D ∪ Γ_{1} ∪ D′, and applying Theorem 7.3 to E_{1}, we get

sup_{x∈X} inf_{T ∈Δ}(φ(T x) + ψ(T)) = sup_{x∈X∩E_{1}} inf_{T ∈Δ}(φ(T x) + ψ(T))

= inf_{T ∈Δ} sup_{x∈X∩E_{1}}(φ(T x) + ψ(T)) = inf_{T ∈Δ} sup_{x∈X}(φ(T x) + ψ(T))

what completes the proof of the general case. □

8. Counter-examples

The hypothesis made on the set Δ to contain at most one compact operator could seem quite artificial. In fact without this hypothesis on Δ the statement of Theorem 7.3 becomes false, as shown by the following example, extracted from [SR], where S and T are compact.

Example 8.1. There exist two continuous linear operators S and T from the Hilbert space E = ℓ^{2} to a Banach space F isomorphic to ℓ^{2}, X a convex subset of E whose interior is non-empty for β(E, E^{∗}), a compact interval J = [0, 2] and a convex continuous and coercive function φ : F → ℝ such that

sup_{x∈X} inf_{λ∈J}(φ(T x − λSx)) < inf_{λ∈J} sup_{x∈X}(φ(T x − λSx)) .

If one replaces the segment {T − λS : λ ∈ J} ⊂ L(E, F) by some 2-dimensional simplex Δ = {λ_{0}T_{0} + λ_{1}T_{1} + λ_{2}T_{2} : λ_{j} ≥ 0, ∑ λ_{j} = 1}, even adding the hypotheses that the mapping x ↦ (T_{j}x)_{j≤2} is an isomorphism from E onto its range in F^{3} and that X is a convex subset of E with non-empty interior for β(E, E^{∗}), the minimax equality can no longer hold. The construction is similar as in Example 8.1:

Example 8.2. There exist two continuous linear operators S and T from the Hilbert space E = ℓ^{2} to a Banach space F, X a convex subset of E whose interior is non-empty for β(E, E^{∗}), a compact interval J ⊂ ℝ and a convex continuous and coercive function φ : F → ℝ such that

sup_{x∈X} inf_{λ∈J}(φ(T x − λSx)) < inf_{λ∈J} sup_{x∈X}(φ(T x − λSx))

We begin by exhibiting an explicit example (the same as in [SR]) of a pair A, B of operators between normed spaces which do not satisfy the minimax equality of Asplund and Pták.

Define the two-dimensional normed spaces E and F as the linear space ℝ^{2} equipped respectively with the norms ∥.∥_{1} : (x, y) ↦ |x| + |y| and ∥.∥_{∞} : (x, y) ↦ max(|x| , |y|).

Denote by A ∈ L(E, F) the operator whose matrix in the canonical bases is (\begin{matrix}3 & 1 \\ −1 & 0 \end{matrix}) and

by I the identity mapping.

Lemma 8.3. We have : inf_{λ∈ℝ} sup_{∥z∥_{1}≤1} ∥Az − λz∥_{∞} = inf_{λ∈ℝ} ∥A − λI∥ = \frac{3}{2}.

Lemma 8.4. We have : sup_{∥z∥_{1}≤1} inf_{λ∈ℝ} ∥Az − λz∥_{∞} ≤ \frac{5}{4} < \frac{3}{2}. More precisely, for each

z in the unit ball of E there is a λ^{∗} ∈ [0, 2] such that ∥Az − λ^{∗}z∥_{∞} ≤ \frac{5}{4}.

Lemma 8.5. There exist two Banach spaces E_{2} and F_{2} which are isomorphic to the Hilbert space ℓ^{2} and operators A_{2} and B_{2} in L(E_{2}, F_{2}) such that

sup_{∥z∥_{E_{2}} ≤1} inf_{λ∈[0,2]} ∥A_{2}z − λB_{2}z∥_{F_{2}} ≤ \frac{5}{4} < \frac{3}{2} ≤ inf_{λ∈ℝ} sup_{∥z∥_{E_{2}} ≤1} ∥A_{2}z − λB_{2}z∥_{F_{2}} .

They are Lemmas 16 to 18 in [SR].

Proof. Since E_{2} is isomorphic to ℓ^{2}, we can find a one-to-one compact operator π : E = ℓ^{2} → E_{2} with dense range, define F = F_{2}, T = A_{2} ∘ π and S = B_{2} ∘ π and φ = ∥.∥_{F_{2}}, choose J = [0, 2] and Y as the unit ball of E_{2}, then define X = π^{−1}(Y). So X has non-empty interior for β(E, E^{∗}) and for any λ ∈ J we have

sup_{x∈X} ∥T x − λSx∥ = sup_{y∈Y} ∥(A_{2} − λB_{2})y∥ = sup_{y∈Y ∩π(ℓ^{2})} ∥(A_{2} − λB_{2})y∥

Take T_{0} = T, T_{1} = T − 2S, and choose for T_{2} any isomorphism from E = ℓ^{2} onto F. Since T_{2} is an isomorphism, it is clear that x ↦ (T_{j}x)_{j≤2} is an isomorphism from E onto its range in F^{3}.

So any U ∈ Δ has the form U = λ_{2}T_{2} + (1 − λ_{2})(T − λS) for some λ ∈ J = [0, 2] and λ_{2} ∈ [0, 1]. It follows that

inf_{U∈Δ} φ(Ux) ≤ inf_{λ∈J} φ(T x − λSx)

hence

sup_{x∈X} inf_{U∈Δ} φ(Ux) ≤ sup_{x∈X} inf_{λ∈J} φ(T x − λSx) ≤ \frac{5}{4}

And by Lemma 5.5, since T − λS is compact for any λ ∈ J,

sup_{x∈X} φ(λ_{2}T_{2}x + (1 − λ_{2})(T x − λSx)) = +∞

for every λ_{2} ∈]0, 1]. Thus

inf_{U∈Δ} sup_{x∈X} φ(Ux) = inf_{λ∈J} sup_{x∈X} φ(T x − λSx) ≥ \frac{3}{2}

and this shows that sup_{x∈X} inf_{U∈Δ} φ(Ux) < inf_{U∈Δ} sup_{x∈X} φ(Ux). □

The previous example shows that Δ should contain at most one compact operator for granting the validity of the minimax equality.

We now prove that the interior of X for β(E, E^{∗}) has to be non-empty for guaranteeing the validity of Theorem 7.3 for all (F, Δ, φ), at least for X being symmetric.

Assume E is an infinite-dimensional Banach space and X is a convex symmetric open subset of E. Then, as in Lemma 4.6, the Minkowski functional p of X is a semi-norm on E ; we denote V the separated completion of (E, p) and π : E → V the canonical mapping. It follows from Theorem 4.7 that π is compact if and only if X is a neighborhood of 0 for β(E, E^{∗}).

Assume now that X is not a neighborhood of 0 for β(E, E^{∗}) : so π is not compact and in particular V is infinite-dimensional. Choose two linearly independent continuous linear functionals g_{1} and g_{2} on V : the continuous linear mapping g : V → ℝ^{2} defined by g(u) = (⟨g_{1}, u⟩, ⟨g_{2}, u⟩) is onto. Let V_{0} = ker(g) and F be V_{0} × ℝ^{2} equipped with the norm (u, λ) ↦ ∥(u, λ)∥_{F} = α ∥u∥ + ∥λ∥_{1} where ∥λ∥_{1} = |λ_{1}| + |λ_{2}| and α a constant to be defined later on. Denote P : V → V_{0} any linear projection.

Then the mapping j : V → F defined by x ↦ (P x, gx) is an isomorphism. If B_{V} denotes the unit ball of V then C = \overline{g(B_{V})} is a compact convex symmetric subset of ℝ^{2}, with nonempty interior since its spans ℝ^{2}, hence the unit ball for some norm ∥ ; |.| ; ∥ on ℝ^{2}. Since ∥ ; |.| ; ∥ and ∥.∥_{1} are not both euclidean on ℝ^{2}, it follows from [AP] that the minimax inequality is strict for L((ℝ^{2}, ∥ ; |.| ; ∥), (ℝ^{2}, ∥.∥_{1}) it is :

β = sup_{∥ ; |u| ; ∥≤1} inf_{t∈ℝ} ∥S_{0}u + tS_{1}u∥_{1} < inf_{t∈ℝ} sup_{∥ ; |u| ; ∥≤1} ∥S_{0}u + tS_{1}u∥_{1} = γ

for some two linear operators S_{0} and S_{1} from ℝ^{2} to ℝ^{2}. Define now two linear operators from V to F by :

Ŝ_{0}x = (P x, S_{0}gx)  ;  Ŝ_{1}x = (0, S_{1}gx)

and two operators from E to F by S̃_{0} = Ŝ_{0} ∘ π and S̃_{1} = Ŝ_{1} ∘ π. The continuous convex coercive function φ on F will be the norm ∥.∥_{F}

Lemma 8.6. For every t ∈ ℝ the mapping S̃_{0} + tS̃_{1} is not compact.

Proof. The projection q : (u, λ) ↦ u is continuous from F to V_{0}. If S̃_{0} + tS̃_{1} was compact for some t ∈ ℝ, so would be q ∘ (S̃_{0} + tS̃_{1}) = P ∘ π and so would be also π since (I −P) ∘ π has finite rank 2, but this contradicts the assumption that X has empty interior for β(E, E^{∗}). □

Lemma 8.7. There exists an integer n_{0} such that

β′ = sup_{∥ ; |u| ; ∥≤1} inf_{|t|≤n_{0}} ∥S_{0}u + tS_{1}u∥_{1} < γ .

Proof. For all t ∈ ℝ the function u ↦ ∥S_{0}u + tS_{1}u∥_{1} is (|t| . ∥S_{1}∥)-lipschitz on C: it follows that the function f_{n} : u ↦ inf_{|t|≤n} ∥S_{0}u + tS_{1}u∥_{1} is also (n. ∥S_{1}∥)-lipschitz on C hence continuous on the compact set C. Moreover the sequence (f_{n}) is obviously nonincreasing and for each u ∈ C we have lim_{n} f_{n}(u) = inf_{n} f_{n}(u) = β < γ. It follows then from Dini’s theorem that the convergence is uniform on C and that for some n_{0} we have

f_{n_{0}}(u) ≤ β′ = \frac{β + γ}{2} < γ for all u ∈ C. □

Let Δ = {S̃_{0} + tS̃_{1} : |t| ≤ n_{0}}. This set is a segment, hence compact and convex, and contains none compact operator.

Lemma 8.8. We have inf_{T ∈Δ} sup_{x∈X} φ(T x) ≥ γ.

Proof. Indeed if T ∈ Δ there is t ∈ [−n_{0}, n_{0}] such that T = S̃_{0} + tS̃_{1}, and if x ∈ X then ∥πx∥ ≤ 1 ; let v = πx. Thus

φ(T x) = ∥Ŝ_{0}v + tŜ_{1}v∥_{F} = α ∥P (Ŝ_{0}v + tŜ_{1}v)∥ + ∥S_{0}v + tS_{1}v∥_{1} ≥ ∥S_{0}v + tS_{1}v∥_{1} .

So

sup_{x∈X} φ(T x) ≥ sup_{∥v∥≤1} ∥S_{0}v + tS_{1}v∥_{1} ≥ inf_{|t|≤n_{0}} sup_{v∈B_{V}} ∥S_{0}v + tS_{1}v∥_{1}

≥ inf_{t∈ℝ} sup_{v∈B_{V}} ∥S_{0}v + tS_{1}v∥_{1} = γ ,

and we get inf_{T ∈Δ} sup_{x∈X} φ(T x) ≥ γ. □

Lemma 8.9. We have sup_{x∈X} inf_{T ∈Δ} φ(T x) < γ.

Proof. Again for x ∈ X and T ∈ Δ we have ∥v∥ ≤ 1 for v = πx and T = S̃_{0} + tS̃_{1} for some t ∈ [−n_{0}, n_{0}]. Thus

φ(T x) = ∥Ŝ_{0}v + tŜ_{1}v∥_{F} = α ∥P (Ŝ_{0}v + tŜ_{1}v)∥ + ∥S_{0}v + tS_{1}v∥_{1}
≤ α ∥P ∥ (∥Ŝ_{0}∥ + n_{0} ∥Ŝ_{1}∥) + ∥S_{0}v + tS_{1}v∥_{1} .

Let M = ∥P ∥ (∥Ŝ_{0}∥ + n_{0} ∥Ŝ_{1}∥) and fix α < \frac{γ − β′}{M}. Thus

inf_{T ∈Δ} φ(T x) ≤ Mα + inf_{|t|≤n_{0}} ∥S_{0}v + tS_{1}v∥_{1} ,

hence

sup_{x∈X} inf_{T ∈Δ} φ(T x) = sup_{v∈B_{V}} inf_{T ∈Δ} φ(T x) ≤ Mα + sup_{v∈B_{V}} inf_{|t|≤n_{0}} ∥S_{0}v + tS_{1}v∥_{1} ≤ Mα + β′ < γ ,

what is the wanted inequality. □

The two previous lemmas show that the minimax equality does not hold as soon as X is not assumed to be a β(E, E^{∗})-neighborhood of 0 in E (at least for X symmetric).

References

1. E. Asplund, V. Pták, A Minimax Inequality for Operators and a Related Numerical Range, Acta Math. 126 (1971), 53–62.

2. J. Bourgain, D.H. Fremlin, M Talagrand, Pointwise compact sets of Baire-measurable functions, Amer. J. Math. 100 (1978), no. 4, 845–886.

3. A. Dvoretzky, C.A. Rogers, Absolute and unconditional convergence in normed linear spaces. Proc. Nat. Acad. Sci. U. S. A. 36, (1950). 192–197

4. B. Ricceri, A minimax Theorem in infinite-dimensional Topological Vector Spaces, Linear and Nonlinear Analysis 2, no. 1, (2016), 47–52

5. J. Saint Raymond A minimax theorem for linear operators, Minimax Theory and its Applications, 1, no. 2 (2016), 291–305

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