The goal of this section will be to proveTheorem 4.3.9, which states that if a Legendre-⋆ form exists on a reflexive space or a space with a separable predual space then the underlying space is already isomorphic to a Hilbert space. This provides a generalization of [Harder, 2018, Theorem 1.1]. One difference is that our generalization can also be applied to spaces that are not reflexive, provided they have a separable predual space.
The other difference is thatTheorem 4.3.9 is formulated for Legendre-⋆forms and not Legendre forms.
However, we formulate the results in this section such that they also cover the situation of Legendre forms in reflexive Banach spaces. We will do this by using the assumption that a normed space X exists which is separable or reflexive. Then the Legendre-⋆ form is defined on X⋆. In the reflexive case Legendre-⋆ forms and Legendre forms coincide (seeCorollary 4.1.3) so Theorem 4.3.9also contains the result of [Harder, 2018,
One might wonder ifTheorem 4.3.9 is also true for Legendre forms in nonreflexive spaces.
AsExample 4.4.3shows, this is not the case.
We mention that the condition that X is separable or reflexive is a condition that guarantees the existence of weakly-⋆ converging subsequences of bounded sequences in X⋆, see Theorem 2.3.1. Since this property is usually required in applications, our assumption thatX is reflexive or separable is a reasonable one.
We start with some technical preparations. While investigating Legendre-⋆ forms it can be useful to restrict the Legendre-⋆ form to a weakly-⋆ closed linear subspace ofX⋆. In order to apply previous results of Legendre-⋆ forms to such a linear subspace we need to show that a Legendre-⋆ form stays a Legendre-⋆ form if restricted to such a linear subspace. Therefore, we need the following lemma.
Lemma 4.3.1. Let Xbe a normed space and letYˆ be a weakly-⋆ closed linear subspace of X⋆. Then Yˆ has a predual space, i.e. there is a spaceY such that Yˆ is isometrically isomorphic to Y⋆.
If X is separable then Y is also separable. Moreover, a sequence {yi}i∈N ⊂ Yˆ ∼= Y⋆ converges in the weak-⋆ topology ofX⋆ if and only if it converges in the weak-⋆topology of Y⋆.
Proof. Note thatYˆ⊥is a closed linear subspace ofX. Thus, the spaceX/Yˆ⊥is a normed space. We define the linear operator
ι:Yˆ →(X/Yˆ⊥)⋆, y↦→(x+Yˆ⊥↦→ ⟨y, x⟩X⋆×X).
Then it can be seen that the operatorιis well-defined. In order to show thatιis isometric, let y∈Yˆ be given. Due to ∥x+Yˆ⊥∥
X/Yˆ⊥≤ ∥x∥X we obtain
∥y∥X⋆ = sup{︂|⟨y, x⟩|⃓⃓⃓x∈X, ∥x∥X ≤1}︂
≤sup{︂|⟨y, x⟩|⃓⃓⃓x∈X, ∥x+Yˆ⊥∥
X/Yˆ⊥ ≤1}︂=∥ιy∥
(X/Yˆ⊥)⋆. For the other inequality, letε >0 be given. Then for eachx∈Xwith∥x+Yˆ⊥∥
X/Yˆ⊥ ≤1 there exists an elementx0∈Yˆ⊥ such that∥x+x0∥X ≤1 +ε. It follows that
∥ιy∥(X/Yˆ⊥)⋆ = sup{︂|⟨y, x⟩|⃓⃓⃓x∈X, ∥x+Yˆ⊥∥
X/Yˆ⊥ ≤1}︂
≤sup{︂|⟨y, x⟩|⃓⃓⃓x∈X, x0∈Yˆ⊥,∥x+x0∥X ≤1 +ε}︂
= sup{︂|⟨y, x+x0⟩|⃓⃓⃓x∈X, x0 ∈Yˆ⊥,∥x+x0∥X ≤1 +ε}︂= (1 +ε)∥y∥X⋆.
Sinceε >0 andy∈Yˆ were arbitrary, we have shown thatιis an isometry. Therefore, ιis also injective. Next, we will show that ιis surjective. Letx⋆ ∈(X/Yˆ⊥)⋆ be given.
Then we define the linear functional
y⋆ :X →R, x↦→ ⟨x⋆, x+Yˆ⊥⟩
(X/Yˆ⊥)⋆×(X/Yˆ⊥).
Clearly,y⋆ ∈X⋆. It is also easy to see that y⋆ ∈Yˆ⊥⊥. Since Yˆ is weakly-⋆ closed we obtain fromTheorem 2.1.16 thatYˆ⊥⊥=Yˆ . Thus, y⋆ ∈Yˆ and we have
ιy⋆ = (x+Yˆ⊥↦→ ⟨y⋆, x⟩X⋆×X) =(︂x+Yˆ⊥ ↦→ ⟨x⋆, x+Yˆ⊥⟩
(X/Yˆ⊥)⋆×(X/Yˆ⊥)
)︂=x⋆. Sincex⋆∈(X/Yˆ⊥)⋆ was arbitrary this shows that ιis surjective.
Summarizing the results, we know thatι is an isometric isomorphism. Thus, we can defineY :=X/Yˆ⊥ and Yˆ is isometrically isomorphic to Y⋆.
The separability of Y follows from the separability of X. Forx∈X andy ∈Yˆ we have
⟨y, x⟩X⋆×X =⟨ιy, x+Yˆ⊥⟩
(X/Yˆ⊥)⋆×(X/Yˆ⊥).
This equation implies that a sequence{yi}i∈N⊂Yˆ ∼=Y⋆converges in the weak-⋆topology of X⋆ if and only if {ιyi}i∈N⊂Y⋆ ∼=Yˆ it converges in the weak-⋆ topology ofY⋆ ∼=Yˆ (with respect to the predual spaceY).
Using this lemma we can conclude the following corollary.
Corollary 4.3.2. Let X be a normed space and let Yˆ be a weakly-⋆ closed linear subspace of X⋆. IfQ:X⋆→R is a Legendre-⋆form, then its restriction to Yˆ is again a Legendre-⋆ form.
Here the weak-⋆convergent sequences in Yˆ are understood with respect to the predual spaceY of Yˆ that was described in Lemma 4.3.1.
The analogous result ofCorollary 4.3.2for Legendre forms can be found in [Harder, 2018, Lemma 4.5] without the assumption that the linear subspace is weakly-⋆ closed (or even closed).
Now that we have shown that Legendre-⋆forms are still Legendre-⋆forms if restricted to a weakly-⋆closed linear subspace, we will do the same for the property that the predual space is separable or reflexive.
Lemma 4.3.3. Let X be a normed space that is separable or reflexive and let Yˆ be a weakly-⋆closed linear subspace ofX⋆. IfX is separable or reflexive thenYˆ has a predual space Y (i.e.Yˆ is isometrically isomorphic toY⋆) that is separable or reflexive.
Proof. The existence of Y is guaranteed byLemma 4.3.1. If X is separable thenY is separable by Lemma 4.3.1. IfX is reflexive then X⋆ is reflexive. It is known that closed linear subspaces of reflexive spaces are again reflexive, see [Conway, 1985, Corollary V.4.3].
Therefore,Yˆ ∼=Y⋆ is reflexive and hence Y is reflexive.
A first Hilbertizability result was already given inProposition 2.1.34, which states that a Banach space is Hilbertizable if there exists a continuous and coercive quadratic form on that Banach space. We can use this result to show that if a Legendre-⋆ form is positive then the underlying space is Hilbertizable. This is a generalization of [Harder, 2018, Proposition 4.8]. In order to facilitate the application of the next proposition we formulate it for a sequentially weakly-⋆closed linear subspace.
Proposition 4.3.4. Let X be a normed space that is separable or reflexive. If Q is a Legendre-⋆ form on X⋆ and Q is positive on a sequentially weakly-⋆ closed linear subspace Y+⊂X⋆ thenY+ is Hilbertizable.
Proof. We follow the proof of [Harder, 2018, Proposition 4.8] but adapt it to our setting.
Assume that Q is not coercive on Y+. Then there is a sequence {yi}i∈N in Y+ with Q(yi)→0 as i→ ∞ and ∥yi∥= 1 for all i∈N.
Because{yi}i∈Nis bounded andXis separable or reflexive, we obtain fromTheorem 2.3.1 that there is a weakly-⋆ convergent subsequence of{yi}i∈N. Without loss of generality we can assume thatyi ⇀ y⋆ asi→ ∞ for somey∈X⋆. SinceY+is sequentially weakly-⋆ closed we also know that y ∈ Y+. We also have 0 ≤Q(y) ≤ lim infi→∞Q(yi) = 0. It follows thatQ(y) = 0 and therefore yi ⋆
⇀ y= 0. Applying the definition of a Legendre-⋆ form yields yi →0 which is a contradiction to ∥yi∥= 1 for i∈N.
Thus, we have shown thatQis coercive on Y+. Now we can apply Proposition 2.1.34to the closed linear subspace Y+. It follows thatY+ is Hilbertizable.
We mention that it can be shown that sequentially weakly-⋆ closed linear subspaces are weakly-⋆ closed if the predual space is separable, see [Conway, 1985, Corollary V.12.7].
However, we do not rely on this result.
Having discussed positive Legendre-⋆forms, we move to nonpositive Legendre-⋆ forms.
The next lemma is a generalization of [Harder, 2018, Proposition 4.9]. However, our proof has to use a different strategy because we are interested in applying the result for linear subspaces that are not necessarily weakly-⋆ closed.
Lemma 4.3.5. Let X be a normed space that is separable or reflexive. If Q is a Legendre-⋆ form on X⋆ andY ⊂X⋆ is a linear subspace such that Qis nonpositive on Y then Y is finite-dimensional.
Proof. Due to the lower semi-continuity ofQit can be seen thatQ is also nonpositive on the closure ofY. Thus, we can without loss of generality assume thatY is closed.
We will show that the unit ball inY is sequentially compact (with respect to the norm topology). Let{yi}i∈N be a sequence inY such that∥yi∥ ≤1 for alli∈N. BecauseX is separable or reflexive we obtain fromTheorem 2.3.1that there is a bounded subsequence that converges weakly-⋆to an elementy ∈X⋆ with ∥y∥ ≤1. Without loss of generality we haveyi ⋆
⇀ y as i→ ∞. Using the sequential weak-⋆ lower semi-continuity of Q we first obtain
Q(yi−y)≤lim inf
j→∞ Q(yi−yj)≤0 ∀i∈N and then
0 =Q(0)≤lim inf
i→∞ Q(yi−y)≤lim sup
i→∞ Q(yi−y)≤0.
The convergence Q(yi−y) →Q(0) follows. Applying the definition of the Legendre-⋆ form yields thatyi−y→0 asi→ ∞. SinceY is closed it follows that y∈Y. Therefore, the sequence {yi}i∈N has a convergent subsequence in Y. Thus, the unit ball in Y is sequentially compact and therefore compact. According to [Rudin, 1991, Theorem 1.22]
this implies that the linear subspaceY is finite-dimensional.
We have seen that in Hilbert spaces we have the decomposition of the underlying space into Q-orthogonal linear subspacesY+, Y−, Y0 such thatQis positive onY+, negative onY−, and vanishes onY0, seeTheorem 4.2.2. Thus, it might be a good idea to construct similar linear subspaces for our situation. The next lemma is a straightforward generalization of [Harder, 2018, Lemma 4.11] to Legendre-⋆forms. It constructs a maximal linear subspace Y−⊂X⋆.
Lemma 4.3.6. LetX be a normed space that is separable or reflexive. Suppose thatQ is a Legendre-⋆ form on X⋆. Then there is a maximal linear subspaceY−⊂X⋆ (with respect to set inclusion) with the property thatQis negative on Y−.
Proof. We follow the proof of [Harder, 2018, Lemma 4.11]. Suppose there is no maximal linear subspace with the desired property. Then we can construct a sequence of linear subspaces{Yi}i∈N such thatYi ⊊Yi+1 andQis negative onYi for all i∈N. If we define the unionY :=⋃︁i∈NYi, then we obtain thatY is an infinite-dimensional linear subspace such thatQis negative on Y. By Lemma 4.3.5this is a contradiction.
Now that we have constructed a maximal linear subspaceY− such thatQ is negative on Y−, we have to construct a linear subspace Y+ such that Q is positive onY+ and Y ⊥QY . Recall that the operatorT ∈L(X⋆, X⋆⋆) is the unique symmetric operator
Lemma 4.3.7. LetX be a normed space that is separable or reflexive. Suppose that Q is a Legendre-⋆ form onX⋆ and that T is injective. Then there exist closed linear subspaces Y−, Y+ ⊂X⋆ with X⋆ = Y−+̇Y+, Y− ⊥Q Y+ such that Q is positive on Y+ and negative onY−.
Proof. We can use Lemma 4.3.6 to obtain a maximal linear subspace Y− ⊂X⋆ (with respect to set inclusion) with the property thatQ is negative onY−. We define Y+ :=
{x∈X⋆|x⊥Qy ∀y∈Y−}. We remark that Y+ is a linear subspace and because Qis continuous, this linear subspace is also closed. It is also clear that Y− ⊥Q Y+ holds. Let x∈Y+∩Y− be given. Then we can use Y− ⊥Q Y+ to obtain
4Q(x) =Q(x+x) =Q(x) +Q(x) = 2Q(x)
and thereforeQ(x) = 0. Since x∈Y− andQ is negative onY− this impliesx= 0. Thus, we have shown that Y+∩Y− ={0} holds.
Next, we argue that X⋆ = Y++Y−. Let x ∈ X⋆ be given. We recall that due to Lemma 4.3.5 the linear subspace Y− has to be finite-dimensional, i.e. n = dimY− for some n∈N. Thus, we can find a pairwise Q-orthogonal basis {yi}ni=1 of Y− with the property thatQ(yi) =−1 for alli∈ {1, . . . , n}. We definexˆ :=x+∑︁nj=1⟨T x, yj⟩yj. Then fori∈ {1, . . . , n}we have⟨T xˆ, yi⟩=⟨T x, yi⟩+⟨T x, yi⟩Q(yi) = 0 which is equivalent to xˆ⊥Qyi. Since {yi}ni=1 is a basis of Y−, it follows thatxˆ⊥QY−. Thus,xˆ∈Y+ and from the definition ofxˆ it can be seen thatx ∈xˆ +Y−⊂Y−+Y+, completing the proof of X⋆ =Y−+̇Y+.
It remains to show that Q is positive on Y+. Let y ∈ Y+ be given. We assume that Q(y) < 0. Then it follows from y ⊥Q Y− that Q is negative on the linear subspace Y−+̇ lin(y). This would be a contradiction to the maximality ofY−. Thus, we know that Q is nonnegative onY+.
Now let y∈Y+ be given such that Q(y) = 0. Then for α∈R and x∈X⋆ we can use (2.11) to obtain
2α⟨T y, x⟩X⋆⋆×X⋆ =Q(αy+x)−Q(αy)−Q(x) =Q(αy+x)−Q(x).
Since we have shown X⋆=Y−+Y+ earlier, there are elementsx−∈Y−, x+∈Y+ such that x=x−+x+. Then we can conclude
2α⟨T y, x⟩X⋆⋆×X⋆=Q(αy+x−+x+)−Q(x−+x+) =Q(αy+x+)−Q(x+) from theQ-orthogonalityx−⊥Qx+. Since αy+x+∈Y+, we know thatQ(αy+x+)≥0 holds. Then the inequality
2α⟨T y, x⟩X⋆⋆×X⋆≥ −Q(x+)
follows. However, sinceα∈Rwas arbitrary, this is only possible if ⟨T y, x⟩X⋆⋆×X⋆= 0.
Because x ∈ X⋆ was arbitrary, this means that y ∈ ker(T) and therefore y = 0 by assumption. Thus, Qis positive on Y+.
We move to a proof of Hilbertizability under the assumption that T is injective. An analogous result for Legendre forms can be found in [Harder, 2018, Proposition 4.14].
However, our proof is very different.
Lemma 4.3.8. LetX be a normed space that is separable or reflexive. Suppose thatQ is a Legendre-⋆ form onX⋆ and that T is injective. Then X⋆ is Hilbertizable.
Proof. Our first step is to show that we can without loss of generality assume thatX is a Banach space. ByLemma 2.1.5there is a Banach spaceXˆ such that there is an isometric isomorphism ι ∈ L(Xˆ⋆, X⋆) and {ιxi}i∈N ⊂ X⋆ converges weakly-⋆ for all weakly-⋆ convergent sequences {xi}i∈N ⊂Xˆ⋆. Then Qˆ : Xˆ⋆ → Rdefined via Qˆ (x) := Q(ιx) for x∈Xˆ is a quadratic form. Since for all weakly-⋆ convergent sequences {xi}i∈N ⊂Xˆ⋆ the image sequence{ιxi}i∈N converges weakly-⋆, it follows that Qˆ is a Legendre-⋆ form.
It can also be seen that the unique symmetric operator Tˆ ∈L(Xˆ⋆, Xˆ⋆⋆) that satisfies Qˆ (x) =⟨Tˆx, x⟩for all x∈Xˆ is injective. Because the Hilbertizability ofXˆ⋆ implies the Hilbertizability ofX⋆ we can without loss of generality assume thatX is a Banach space.
UsingLemma 4.3.7yields a decomposition X⋆ =Y−+̇Y+ with closed linear subspaces Y−, Y+⊂X⋆ such that Qis negative onY−, positive onY+, andY−⊥Q Y+. Our goal is to applyProposition 4.3.4 to the linear subspace Y+. Thus, we have to show thatY+ is sequentially weakly-⋆ closed.
Let{yi}i∈N be a sequence inY+ that converges weakly-⋆ to an element x∈X⋆. Due to the decomposition of X⋆ =Y−+̇Y+ there arex−∈Y−, x+ ∈Y+ such thatx=x++x−. Let us assume thatx−̸= 0. Then we have Q(x−)<0. We definea:= lim infi→∞Q(yi) andα:=Q(x−)−1(a−Q(x+))/2−1. SinceX is a Banach space, the sequence{yi}i∈N is bounded. BecauseQ is continuous and{Q(yi)}i∈N is bounded, we have 0≤a < ∞. Using the sequential weak-⋆ lower semi-continuity of Q for the weak-⋆ convergence yi+αx− ⋆
⇀ x++ (1 +α)x− results in
Q(x++ (1 +α)x−)≤lim inf
i→∞ Q(yi+αx−). If we also use theQ-orthogonalities yi⊥Qx−,x+⊥Qx− we obtain
Q(x+) + (1 +α)2Q(x−)≤lim inf
i→∞ Q(yi) +Q(αx−) =a+α2Q(x−). Then the inequality
Q(x+)−a≤ −(1 + 2α)Q(x−) =Q(x+)−a+Q(x−)
follows, which is a contradiction to the condition that Q(x−)<0. Hence, x− = 0 and we haveyi ⇀ x⋆ =x+ ∈Y+. Since{yi}i∈N⊂Y+ was an arbitrary weakly-⋆ convergent sequence, we have shown that Y+ is sequentially weakly-⋆ closed.
Thus, we can applyProposition 4.3.4and obtain thatY+ is Hilbertizable. It follows from Lemma 2.1.2 thatX⋆ is Hilbertizable.
Now we are able to prove the main theorem of this chapter. It is a generalization of [Harder, 2018, Theorem 1.1] to nonreflexive Banach spaces with a separable predual space. The difference of this theorem to Lemma 4.3.8is that we can drop the assumption that T is injective.
Theorem 4.3.9. LetX be a normed space that is separable or reflexive. Suppose that Q is a Legendre-⋆ form onX⋆. Then X⋆ is Hilbertizable.
Proof. We follow the proof of [Harder, 2018, Theorem 1.1] but adapt it to our setting.
Since Q(x) = 0 for all x ∈ kerT, it follows from Lemma 4.3.5 that kerT is a finite-dimensional linear subspace. ByLemma 2.1.3 (b) there exists a weakly-⋆ closed linear subspace Y of X⋆ withX⋆ =Y +̇ kerT.
Let us consider the restriction of Q to Y and the corresponding symmetric operator TY :Y →Y⋆. We will show that kerTY ={0}. Lety∈kerTY be given. Then for each yˆ∈Y,x0 ∈kerT we have
0 =⟨TYy, yˆ⟩Y⋆×Y = 12(︁Q(y+yˆ)−Q(y)−Q(yˆ))︁
=⟨T y, yˆ⟩X⋆⋆×X⋆+⟨T x0, y⟩X⋆⋆×X⋆=⟨T y, yˆ +x0⟩X⋆⋆×X⋆,
where we used(2.11). Thus, ⟨T y, x⟩X⋆⋆×X⋆ = 0 for allx∈X⋆=Y +̇ kerT and therefore y∈kerT ∩Y ={0}.
Since Y is weakly-⋆ closed, Corollary 4.3.2 and Lemma 4.3.3 allow us to transfer the assumptions to the linear subspaceY. Thus, we can applyLemma 4.3.8for the linear subspace Y and the predual space ofY. It follows thatY is Hilbertizable. Finally, we can applyLemma 2.1.2 toX⋆=Y +̇ kerT, which completes the proof.
An important implication of this theorem is that we should not try to apply results where a Legendre-⋆form appears in spaces that have a separable predual space but are not Hilbertizable. Likewise, we should not try to apply results where a Legendre form appears in reflexive spaces that are not Hilbertizable. Such spaces include ℓ1,ℓ∞, L∞(Ω) for the nonreflexive case as well as Lp(Ω) and ℓp for p ∈(1,∞)\ {2} for the reflexive case.
There are still some cases where we do not know whether Legendre forms or Legendre-⋆
forms can exist. For example it is open whether Legendre forms can exist in L1(Ω).
Another question is what happens to Legendre-⋆ forms in nonreflexive spaces if we
remove the condition that the predual space is separable. In such a case it might be more reasonable to redefine Legendre-⋆ forms using weakly-⋆ converging nets instead of weakly-⋆ converging sequences. It is an open question whether Theorem 4.3.9 can be generalized to this setting.