Dality Theory II: Advanced Theory

DT II c Gabriel Nagy

Notes from the Functional Analysis Course (Fall 07 - Spring 08)

Besides the weak dual topologies w# introduced in DT I, there exist many other natural topologies on dual pairs. A general recipe for constructing a large class of such topologies is outlined in sub-section A. Subsequent sections contain some important theorems due to (combinations of) Arens, Banach, Grothendieck and Mackey, which are perfect illustrations of the power of the ”Duality Machine.” The entire section is optional, as it abounds in proof-less results (left as exercises).

A. The TM-topologies

Conventions. Throughout this note K will be one of the ﬁelds R or C, and all vector spaces are over K. Whenever we say that “X and Y form a dual pair,” we will understand that one is given two linear injective maps

#: X 3 x 7−→ x# ∈ Y0, (1) #: Y 3 y 7−→ y# ∈ X 0, (2) satisfying the equality x#(y) = y#(x), ∀ x ∈ X , y ∈ Y. (3) (See DT I for the interpretation of these maps.) Remarks 1-2. Use the notations as above.

# 1. Since the w -topology on X is deﬁned by the family {py}y∈Y , given by py(x) = |y#(x)| = |x#(y)|, by Exercise 1 from from LCVS III, it follows that a non-empty set M ⊂ X is w#-bounded, if and only if,

sup |x#(y)| < ∞, ∀ y ∈ Y. (4) x∈M

Likewise, a non-empty set M ⊂ Y is w#-bounded, if and only if,

sup |y#(x)| < ∞, ∀ x ∈ X . (5) y∈M

w# 2. If M ⊂ X is w#-bounded, then so is the closure conv(bal M) . This follows again from Exercise 1 from LCVS III.

1 Exercise 1♥. Show that, if M ⊂ Y is a non-empty w#-bounded set, then its absolute

polar M is absorbing (in X ) and furthermore, its associated Minkowski functional qM is given by # qM (x) = sup |y (x)|, ∀ x ∈ X . y∈M

Notation. With M as in Exercise 1, when convenient, the Minkowski functional qM will be simply denoted by pM. (When M is a singleton {y}, pM is py, the seminorm used before.) ♥ 1 Exercise 2. Using the notations as above, show that B(pM) = M. Definition. Suppose X and Y are in a dual pairing. Given a non-empty collection M, # consisting of non-empty w -bounded subsets of Y, one constructs the TM-topology as the locally convex topology on X , defined by the family of seminorms QM = {pM : M ∈ M}. In terms of convergence, the topology TM is characterized as follows. For a net (xλ) in TM # X , the condition xλ −−→ 0 is equivalent to the condition: supy∈M |y (xλ)| → 0, ∀ M ∈ M. Likewise, for a non-empty collection N – consisting of non-empty w#-bounded subsets of X – the corresponding topology TN on Y is defined. Exercises 3-4. Given a non-empty collection M of non-empty w#-bounded subsets of S Y, consider the space hMi = span M∈M M . 3♥. Prove that the following are equivalent:

(i) TM is Hausdorﬀ; (ii) hMi is w#-dense in Y.

♥ # 4. Prove that if hMi = Y, then TM is stronger than the w -topology (on X ).

All locally convex topological vector spaces can be viewed as TM-topologies, as indicated in the following result. Proposition 1. Let (X , T) be a locally convex topological vector space, and form a dual 0 ∗ pair with either Y1 = X (the algebraic dual), or Y2 = X (the topological dual).

1 2 (i) If V is a T-neighborhood of 0 in X , then its absolute polars V in Y1 and V in Y2 1 coincide. In particular, V is contained in Y2. (ii) Every set in the collection

1 MT = {V : V neighborhood of 0 in (X , T)}

# # is w -compact (thus w -bounded) in both Y1 and in Y2.

(iii) The TMT -topology on X , deﬁned using either pairing (X with Y1, or X with Y2), coincides with T. 1 Using the notations from LCVS III, for a seminorm p on X , we deﬁne B(p) = {x ∈ X : p(x) ≤ 1}.

2 0 Proof. (i). By deﬁnition, for V a T-neighborhood of 0, its two absolute polars (in Y1 = X ∗ and in Y2 = X ) are given by

0 V 1 = {φ ∈ X : |φ(x)| ≤ 1, ∀ x ∈ V}, ∗ V 2 = {φ ∈ X : |φ(x)| ≤ 1, ∀ x ∈ V},

∗ 0 so clearly (by the inclusion X ⊂ X ) one has the inclusion V 2 ⊂ V 1 . Therefore, in order ∗ to prove that these two sets coincide, we only need to prove the inclusion V 1 ⊂ X , in other words, that every φ ∈ V 1 is T-continuous. Fix such a φ and let us show that it is continuous at 0. Equivalently this means that, for every ε > 0, the set Nε = {x ∈ X : |φ(x)| ≤ ε} is a T-neighborhood of 0. This is, however, obvious since the condition φ ∈ V 1 clearly yields the inclusion Nε ⊃ εV. # ∗ ∗ (ii). Since the w -topology on Y2 = X coincides with the w -topology, which in turn ∗ 0 2 ∗ the the induced w -topology from Y1 = X , it suffices to show that V is w -compact in ∗ Y2 = X . This, however, follows immediately from the Alaoglu-Bourbaki Theorem. (iii.) We know (see LCVS III, Theorem-Definition 1 and Remark 8), by the definition of the TMT -topology, that the collection

W = {εB(pMT ): M ∈ MT, ε > 0}

constitutes a fundamental system of neighborhoods of 0 in the TMT -topology. Therefore (by 2 Exercise 2), in order to prove that T coincides with TMT , it suﬃces to show that :

(a) for every M ∈ MT, its absolute polar M is a T-neighborhood of 0, and

(b) for every T-neighborhood U of 0, there exists M ∈ MT, such that M ⊂ U.

1 The first assertion is obvious, since every M ∈ MT is of the form M = V , for some T-neighborhood V of 0, so we obviously have the inclusion M = (V 1 ) ⊃ V. To prove (b), we use the fact (see LCVS I) that for every T-neighborhood U of 0, there exists a closed, convex, balanced T-neighborhood of 0, such that V ⊂ U. Using the duality ∗ 2 between X and Y2 = X , and the Absolute Bipolar Theorem, it follows that V = (V )), so if we take M = V 2 , we now have: M = V ⊂ U. Example 1. If one starts with a dual pairing between X and Y, and one considers M to be the collection of all finite subsets of Y, the associated TM-topology on X is precisely the w#-topology. In sub-sections B and C two other choices for M will be used to produce two new topologies. Remark 3. Suppose X and Y are in a dual pairing. Since, for two non-empty collections # M ⊂ N (of non-empty w -bounded subsets of Y), we have the inclusions QM ⊂ QN, it follows that the associated topologies satisfy the same inclusion TM ⊂ TN. Comment. In connection with the above remark, it is natural to ask the following question. Given two collections M and N, when do the topologies TM and TN coincide, or when does one have the inclusion TM ⊂ TN? Of course, we already know that, a sufficient

2 Whether MT is regarded in Y1 or in Y2, its absolute polar in X is deﬁned the same way, so we do not need subscripts. By Exercise 2, we know that B(pM) = M.

3 condition for the inclusion TM ⊂ TN is the inclusion M ⊂ N, but this is clearly not necessary. (For instance, if M is the collection of all ﬁnite subsets in Y with at least two elements, and # N is the collection of all singleton sets in Y, then TM and TN coincide with the w -topology on X , but clearly M ∩ N = ∅.) The exercise below is meant to justify the terminology that will help us answer the above questions. Exercise 5♥. Fix X and Y as above, and ﬁx some non-empty w#-bounded subset M ⊂ Y. Prove the following:

(i) pαM = |α| · pM, ∀ α ∈ K;

# (ii) the absolute bipolar (M ) is also w -bounded in Y, and p(M) = pM;

(iii) if N is a non-empty subset of M, then pN ≤ pM. Deﬁnitions. Suppose X and Y are in a dual pairing, and M is a non-empty collection of non-empty w#-bounded subsets of Y S # A. We say that M is total, if its support hMi = span M∈M M is dense in Y in the w - topology. By Exercise 3, this condition is equivalent to the fact that the TM-topology on X is Hausdorﬀ. B. We say that M is saturated, if: (i) M ∈ M ⇒ N ∈ M, ∀ N ⊂ M (non-empty); (i) M ∈ M ⇒ tM ∈ M, ∀ t > 0;

(iii) M ∈ M ⇒ (M) ∈ M;

(iv) M is directed, i.e. for any M1, M2 ∈ M, there exists M ∈ M, such that M ⊃ M1 ∪ M2. C. It is not so hard to show that the intersection of an arbitrary family of saturated collections is again saturated. Therefore, one can construct Msat to be the intersection of all saturated collections that contain M, so that the collection Msat is the smallest saturated collection that contains M. The collection Msat is called the saturation of M. (See Exercise 6 for more clariﬁcations.)

♥ # Exercise 6. Prove that the collection BY , of all non-empty w -bounded subsets in Y, is saturated. Therefore, the collection used in the saturation construction in Deﬁnition C above is non-empty. Exercise 7*. Suppose that M is directed (as in Deﬁnition B), and all sets in M are w#-closed, convex and balanced. Prove that

sat M = {N= 6 ∅ : there exist M ∈ M and t > 0, such that N ⊂ tM}.

With this terminology, the above mentioned questions are answered as indicated in the exercises below. Exercises 8-11. Suppose X and Y are in a dual pairing, and a non-empty collection M of non-empty w#-bounded subsets of Y is given.

4 8. Prove that TM = TMsat .

sat 9. Prove that the collection W = {M : M ∈ M } constitutes a basic system of TM-neighborhoods of 0 system. In fact W is the collection of all TM-closed, convex, balanced TM-neighborhoods of 0. 10 Prove that for a seminorm p on X , the following are equivalent:

(i) p is TM-continuous; sat (ii) there exists M ∈ M , such that p = pM; (ii’) there exists a unique M ∈ Msat, which is also w#-closed, balanced and convex, such that p = pM.

11. Prove that, if N is another non-empty collection of non-empty w#-bounded subsets of Y, then the following conditions are equivalent:

(i) TN ⊂ TM; (ii) Nsat ⊂ Msat.

sat sat In particular, TN = TM ⇔ N = M .

B. The Strong Topology and the Banach-Mackey Theorem

In this sub-section we consider the TM-topology, for the largest admissible collection M.

# Notation-Deﬁnition. Suppose X and Y are in a dual pairing. Denote by BY the # # collection of all w -bounded subsets in Y. The TM-topology on X , associated with M = BY , is called the strong (dual) topology, and is denoted by s#. # # Remark 4. The collection BY is total and saturated. Since all ﬁnite sets belong to BY , by Remark 3, it follows that the s#-topology is stronger than the w#-topology. Comment. In preparation for Theorem 1 below, we review the notion of self-completeness introduced in LCVS III C. Given a vector space Y, a non-empty, convex, balanced set B ⊂ X was declared self-complete, if: T (i) t>0 tB = {0},

(ii) (B, dB) is a complete metric space, where the metric dB is deﬁned as follows. Start with the space Z = span B, in which B is absorbing, then using (i) it follows that the Minkwoski functional associated to B is in fact a 0 0 norm, hereafter denoted by k . kB. The metric dB on Z is then deﬁned by dB(z, z ) = kz−z kB. We know that self-completeness of B implies the completeness of the metric space (Z, dB). Using the above terminology, we have the following interesting and very useful technical result. (The reader is urged to review Exercises ?? from LCVS III.)

5 Theorem 1 (Banach-Mackey Strong Boundedness Principle). Suppose X and Y are in a dual pairing. If B ⊂ X is convex, balanced, w#-closed, w#-bounded, and self-complete, then B is bounded in the s#-topology, that is:

# # sup |y (x)| < ∞, ∀ M ∈ BY . (6) y∈M x∈B

Proof. Use the notations as above. Since condition (6) states that supx∈B pM(x) < ∞, # # ∀ M ∈ BY , by Exercise 1 from LCVS III this condition is indeed equivalent to the s - boundedness of B.. norm Let TB be the locally convex topology on Z deﬁned by the (semi)norm k . kB. From LCVS III (Exercise 15 and Remarks 14-15), we know that:

norm # (a) the topology TB on Z is stronger than the induced w -topology (from X );

norm (b) (Z, TB ) is Frechet.

# Fix some non-empty w -bounded set M ⊂ Y. Consider now the linear maps θy = # y Z : Z → K, x ∈ M. Since all these maps are continuous with respect to the induced # norm w -topology (from X ), by (a) it follows that all θy, y ∈ M are also TB -continuous. Since M ⊂ M is w#-bounded, by Remark 1 we have (5), so in particular (specializing norm to x ∈ Z), we now have: supy∈M |θy(x)| < ∞, ∀ z ∈ Z. Since (Z, TB ) is Frechet, by the Equi-Continuity Principle (see TVS IV), it follows that the collection Θ = {θy}y∈M is norm equi-continuous. In particular, there exists a TB -neighborhood U of 0 in Z, such that

|θy(z)| ≤ 1, ∀ z ∈ U, y ∈ M. (7)

Choose r > 0, such that U ⊃ {z ∈ Z : kzkB ≤ r}. With this choice of r, let us now observe that if we start with arbitrary y ∈ M and x ∈ B, then kxkB ≤ 1, so rx ∈ U, # which by (7) yields 1 ≥ |θy(rx)| = r|y (x)|, thus proving the desired conclusion, explicitly: # supy∈M |y (x)| ≤ 1/r. x∈B Corollary 1. Given X and Y in a dual pairing as above, B ⊂ X is convex, balanced, and w#-compact, then B is bounded in the s#-topology. Proof. By Exercise 16 from LCVS III, we know that every non-empty, convex, balanced, w#-compact set B ⊂ X is self-complete, so everything follows from Theorem 1.

C. The Mackey Topology

In this sub-section we consider the TM-topology, associated with another natural choice of the collection M, motivated by Proposition 1. # Notation-Deﬁnition. Suppose X and Y are in a dual pairing. We denote by KY the # collection of all non-empty convex, balanced, w -compact sets in Y). The TM-topology, # associated with the collection M = KY , is called the Mackey topology (on X ) associated with

6 the dual pairing, and is denoted by m#. Similarly, the Mackey topology m# on Y can be constructed. Remark 5. Since all w#-compact sets are w#-bounded (see TVS I, Exercise 16), we # # # have the inclusion KY ⊂ BY , so the Mackey topology is weaker than the s -topology. On # the other hand, for every y ∈ Y, the set Ky = {αy : α ∈ K, |α| ≤ 1} is clearly w -compact, # convex, and balanced, thus a member of KY , therefore the singleton {y} is a member of the # sat # saturation (KY ) . This shows that the seminorms py, y ∈ Y are all m -continuous, so the Mackey topology is stronger than the w#-topology. Exercise 12. Use the notations as above.

# (i) Prove that KY is directed.

# sat (ii) Prove that a non-empty set M belongs to the saturation (KY ) , if and only if its # absolute bipolar (A) is w -compact.

# # (iii) Prove that a seminorm q on X is m -continuous, if and only if there exists M ∈ KY , such that q ≤ pM. The Mackey topology has a remarkable property, that will be discussed in Theorem 2 below. In preparation for that result, we introduce the following terminology. Deﬁnition. Suppose X and Y are in a dual pairing. A locally convex topology T on X is said to be compatible with the dual pairing, if one has the equality (X , T)∗ = (X , w#)∗. Exercises 13-15. Suppose a dual pairing between X and Y is given, together with a locally convex topology T on X . 13. Prove that the following conditions are equivalent:

(i) (X , w#)∗ ⊂ (X , T)∗; (ii) for every y ∈ Y, the linear functional y# : X → K is T-continuous; (iii) T is stronger than w#, i.e. one has the inclusion T ⊃ w#. 14. Assume T satisﬁes one of the conditions (i)-(iii) from the preceding Exercise. Consider, by condition (ii), the linear map

T : Y 3 y 7−→ y# ∈ (X , T)∗.

Prove that the following are equivalent:

(i) T is compatible with the dual pairing; (ii) the map T is surjective; (ii’) the map T is a linear isomorphism; (iii) when we equip Y with the w#-topology, and the dual space (X , T)∗ with the weak* topology, the map T is a topological linear isomorphism.

15. Prove that the w#-topology on X is the weakest among all topologies compatible with the dual pairing.

7 Exercise 16. Start with a locally convex topological vector space (X , T), and we use ∗ the dual pairing between X and Y = (X , T) . Prove that both T and wT (the weak topology implemented by T), are compatible with this dual pairing. Comment. Since for a dual pairing between X and Y, there always exists a smallest locally convex topology on X , which is compatible with the pairing (namely w#), it is natural to ask whether there exists a strongest one. Remarkably, this question has an aﬃrmative answer: Theorem 2 (Mackey-Arens). Suppose X and Y are in a dual pairing. The Mackey topology m# (on X ) is the strongest among all locally convex topologies that are compatible with the (given) dual pairing between X and Y.

Proof. The ﬁrst step is to prove that m# is compatible with the dual pairing. By Remark 5, we already know that m# ⊃ w#, so we have the inclusion (X , w#)∗ ⊂ (X , m#)∗. Therefore, by Exercise 14, in order to prove that m# is compatible with the dual pairing, it suﬃces to show that every linear m#-continuous functional φ : X → K can be represented as φ = y#, for some y ∈ Y. Fix such a φ, and note that, using Exercise 12, applied to the seminorm # |φ|, there exists M ∈ KY , such that |φ(x)| ≤ pM(x), ∀ x ∈ X . Equivalently (by Exercise 1), φ is dominated by the Minkowski functional of the absolute polar of M, that is:

|φ(x)| ≤ qM (x), ∀ x ∈ X . (8)

0 # Consider now the dual pairing between X and its algebraic dual X , and denote by w0 the corresponding weak dual topologies. Since the linear map

# # 0 # T :(Y, w ) 3 y 7−→ y ∈ (X , w0 )

# 0 is clearly continuous, it follows that T (M) is w0 -compact in X . Since T (M) is balanced, by the Absolute Bipolar Theorem, we have the equality

T (M) = [T (M)]. (9)

(The inner absolute polar is taken in X , the outer absolute polar is taken in X 0.) By construction, the absolute polar T (M), in X , is given by:

# T (M) = {x ∈ X : |y (x)| ≤ 1, ∀ y ∈ M} = M,

where the second absolute polar is computed relative to the dual pairing between X and Y. In particular, by condition (8) we have |φ(x)| ≤ 1, ∀ x ∈ M = T (M), so φ belongs to [T (M)]. But now we are done, since (9) forces φ to belong to T (M), which means that there exists some y ∈ M, such that ψ = y#. To ﬁnish the proof, we start with an arbitrary locally convex topology T on X , which is compatible with the dual pairing between X and Y, and we show that T is weaker than the Mackey topology m#. Denote the topological dual space (X , w#)∗ = (X , T)∗ simply by X ∗. The topological linear isomorphism T :(Y, w#) → (X ∗, w∗) establishes a bijective correspondence # Θ: KY 3 M 7−→ T (M) ∈ KX ∗ ,

8 ∗ where KX ∗ denotes the collection of all non-empty, convex, balanced w -compact subsets in X ∗. Therefore, the Mackey topology on X , coming from its dual pairing with Y, coincides ∗ with the Mackey topology coming from the dual pairing with X , which is the TM-topology ∗ associated with M = KX . By Proposition 1, we know that T coincides with the TMT - topology, where MT = {V : V neighborhood of 0 in (X , T)}. ∗ We know (also by Proposition 1) that all sets in MT are w -compact, as well as (see DT I) convex and balanced, so we have the inclusion MT ⊂ KX ∗ , which then by Remark 3 gives the desired inclusion T ⊂ m#. The following result is a beautiful illustration of the power of the “Duality Machine.” Although it might be possible to prove it in diﬀerent ways, the proof presented here is the most elegant one. Theorem 3 (Mackey). Suppose T and T0 are two locally convex Hausdorﬀ topologies on a vector space X , such that (X , T)∗ = (X , T0)∗. Then for every subset B ⊂ X , the following conditions are equivalent: (i) B is T-bounded;

(ii) B is T0-bounded. Proof. Denote the topological dual space simply by Y, and put it in the standard dual pairing with X , so that both T and T0 are compatible with this dual pairing. In particular, by the Mackey-Arens Theorem, we have the inclusions

w# ⊂ T, T0 ⊂ m#, so it suﬃces to prove that, for B ⊂ X , the following are equivalent:

(i’) B is w#-bounded;

(ii’) B is m#-bounded. Since w# ⊂ m#, the implication (ii0) ⇒ (i0) is trivial. Assume now B is w#-bounded, and let us prove that it is also m#-bounded. Using the deﬁnition of the Mackey topology – as # the the locally convex topology deﬁned by the seminorms pM, M ∈ KY – and Exercise 1 # from LCVS III, this amounts to showing that supx∈B pM(x) < ∞, ∀ M ∈ KY , that is:

# # sup sup |y (x)| < ∞, ∀ M ∈ KY . x∈B y∈M The above condition, however, is immediate from Corollary 1.

D. Completing the TM-topology

Convention. Throughout the entire sub-section, we ﬁx two vector spaces X and Y that # are in a dual pairing. The corresponding weak dual topologies will be denoted by wX (the # weak dual topology on X ) and wY (the weak dual topology on Y).

9 The ﬁrst instance of the calculation of the completion is contained in the following result. Proposition 2. Let w0 denote the weak dual topology on the algebraic dual Y0 deﬁned3 by its dual pairing with Y. (i) The locally convex topological vector space (Y0, w0) is complete.

# 0 0 (ii) The linear map #:(X , wX ) → (Y , w ) is continuous, and admits a linear continuous ˜ # 0 0 extension to a homeomorphism T :(X , wgX ) → (Y , w ). (See the beginning of the proof on what “extension” means.) ˜ # # Proof. A word of explanation is due here. We denote by (X , wgX ) the completion of (X , wX ), ˜ and by JX : X ,→ X the standard inclusion. The phrase “T extends #” means: (T ◦JX )(x) = x#, ∀ x ∈ X . 0 0 (i). The fact that every Cauchy net (φλ)λ∈Λ in (Y , w ) is convergent is pretty clear. Indeed, by the Cauchy condition (see LCVS III) it follows that, for every y ∈ Y, the net φλ(y) λ∈Λ is Cauchy in K, thus convergent to some φ(y) ∈ K. By the uniqueness of the limit, it follows that the map y 7−→ φ(y) is linear. (ii). Let us consider the linear subspace Z = Range # ⊂ Y0, and let us denote the 0 # induced topology w Z simply by T. By the definition of the w -topology, it is obvious that # #:(X , wX ) → (Z, T) is a linear homeomorphism. Therefore the extension of # to the ˜ ˜ # ˜ ˜ completions #:(X , wgX ) → (Z, T) is a linear homeomorphism. According to Theorem 1 ˜ ˜ ∼ w0 from TVS IV, there is a linear homeomorphism S :(Z, T) −→ (Z , T), such that (S ◦JZ )z = 0 0 w 0 z, ∀ z ∈ Z. (Here T denotes the induced topology w w0 on the closure Z of Z in Y , Z ˜ ˜ and JZ : Z ,→ Z denotes the standard inclusion.) Since # ◦ JX = JZ ◦ #, we see that ˜ ˜ # w0 when we consider the linear homeomorphism T = S ◦ #:(X , wgX ) → (Z , T), we clearly have the identity T ◦ JX = (S ◦ JZ ) ◦ # = #, so T indeed extends #. To finish the 0 proof, all we need to do is to show that the space Zw = Range T coincides with Y0, i.e. to show that Z = Range # is w0-dense in Y0. Fix some φ ∈ Y0, and let us construct a net in Range # which is w0-convergent to φ. Let us denote by fd(Y) the set of all finite dimensional linear subspaces of Y, which becomes a directed set, when equipped with in inclusion relation: L1 L2 ⇔ L1 ⊃ L2. (For L1, L2 ∈ fd(Y), the vector space L = L1 + L2 is finite dimensional, and contains both L1 and L2.) Let us choose, for every L ∈ fd(Y), a # ∗ linear continuous functional ψL ∈ (Y, wY ) , such that ψL = φ . (This is possible, using L L the Hahn-Banach Theorem, combined with the fact that φ L is continuous with respect to # the induced wY -topology. Of course, the functional ψL is not unique.) By duality (see DT # I), there exists some xL ∈ X , such that ψL = xL . To summarize, we can construct a net (xL)L∈fd(Y) ⊂ X , such that # xL (y) = φ(y), ∀ L ∈ fd(Y), y ∈ L. (10) # Of course, the above condition implies that limL∈fd(Y) xL (y) = φ(y), ∀ y ∈ Y, which means # w0 precisely that xL −→ φ.

0 3 0 # 0 w Of course, w is the w -topology on Y , that comes from its pairing with Y, which means that φλ −→ φ ⇔ φλ(y) → φ(y), ∀ y ∈ Y. We use this notation in order to avoid future confusions.

10 Convention. For the remainder of this sub-section, we assume that all non-empty # collections M, mentioned hereafter, consist of non-empty wY -bounded subsets of Y, and moreover have the following properties:

• all ﬁnite subsets of Y belong to M, so in particular the TM-topology on X is stronger # than the wX -topology;

• M is saturated, so all TM-continuous seminorms on X are of the form pM, with M ∈ M. ˜ ˜ In what follows we are interested in characterizing the completion (X , TM) of the locally convex space (X , TM). To avoid any future confusions, the underlying space of this completion will be denoted occasionally by X ∼TM .

Comment. We also know (see Exercise 9) that the collection W = {M : M ∈ M} # constitutes a basic system of neighborhoods of 0 in (X , TM). Since all sets in W are wX - closed, using Bourbaki’s Embedding Lemma (see TVS IV, Lemma 1), we see that, if we # consider the identity map I :(X , TM) → (X , w ), it follows that its extension to the 4 ∼T ∼ # completions I˜ : X M → X wX is injective. If we compose I˜with the linear homeomorphism ∼ # ∼ 0 T : X wX −→Y , we now obtain an injective linear continuous map

˜ ∼TM ˜ 0 0 Γ = T ◦ I :(X , TM) ,→ (Y , w ).

This allows us to identify the underlying space X ∼TM with the linear subspace Range Γ in the algebraic dual space Y0. With this identiﬁcation, the problem of calculating the completion of (X , TM) is now focused on the following two issues:

(a) identify the space Range Γ, ˜ (b) identify which topology on Range Γ corresponds to TM. It is desirable for both solutions to be presented intrinsically inside Y0. Before we attack these two problems, we need to point out that the map Γ admits a fairly easy description, as follows. Suppose one starts with some vector v ∈ X ∼TM . Then T˜ M ∼TM there exists a Cauchy net (xλ)λ∈Λ in (X , TM), so that J(xλ) −−→ v, in X . The Cauchy condition is # # lim sup sup |xλ (y) − xλ (y)| = 0, ∀ M ∈ M. λ∈Λ 1 2 λ1,λ2 λ y∈M # In particular, using singletons, it follows that xλ (y) λ∈Λ is Cauchy in K, for every y ∈ Y, 0 # so we can deﬁne φ ∈ Y by φ(y) = limλ xλ (y), and this functional is precisely Γ(v). In preparation for the answers to questions (a) and (b) above we ﬁrst prove the following technical result. Proposition 3. Assume X , Y, and M are as above, and M is some set in M. Every # net (yλ)λ∈Λ ⊂ M has a subnet (yη(σ))σ∈Σ, which is Cauchy in (Y, wY ).

# # 4 ∼wX Here X denotes the underlying space for the completion of (X , wX ).

11 Proof. Replacing M with a larger set (use the fact that M is saturated), we can in fact assume that M is in fact w#-closed, balanced and convex. If we take its absolute bipolar V = M in X , we also have the equality

M = V . (11)

Consider now the linear homeomorphism

# # ∗ ∗ #:(Y, wY ) → ((X , wX ) , w ), (12)

# ∗ ∗ and the obvious inclusion of topological dual spaces (X , wX ) ⊂ (X , TM) , which is a con- # sequence of the fact that the topology TM is stronger than wX -topology. If we equip the ∗ ∗ topological dual space (X , TM) with its weak dual topology, which we denote by wM, then ∗ # ∗ ∗ it is trivial that the weak dual topology w on (X , wX ) coincides with the induced wM- ∗ topology. So now, if we view the map (12) as a map which takes values in (X , TM) we obtain a linear homeomorphism

# ∼ ∗ #:(Y, wY ) −→ (Range #, wM Range #). (13)

Using the equality (11) we clearly have the inclusion

∗ #(M) ⊂ {φ ∈ (X , TM) : |φ(x)| ≤ 1, ∀ x ∈ V}. (14)

Denote the set on the right-hand side of (14) by C. The point here is that, since V is a neighborhood of 0 in (X , TM), by the Alaoglu-Bourbaki Theorem it follows that C is compact ∗ ∗ in ((X , TM) , wM), so if we start with a net (yλ)λ∈Λ ⊂ M, there exists a subnet (yη(σ))σ∈Σ, # ∗ such that (yη(σ))σ∈Σ is convergent in the wM-topology to some φ ∈ C. In particular, the # ∗ ∗ double net ((yη(σ1) − yη(σ2)) )(σ1,σ2)∈Σ×Σ is convergent to 0 in ((X , TM) , wM), so using the # homeomorphism (13), the double net (yη(σ1) −yη(σ2))(σ1,σ2)∈Σ×Σ is convergent to 0 in (Y, wY ), # i.e. (yη(σ))σ∈Σ is Cauchy in (Y, wY ). One key consequence of Proposition 3 is stated below as Corollary 2 below. To facilitate the exposition, both for Corollary 2 and for Theorem 4, let us introduce the following no- # tation. Given M ∈ M, we are going to denote from now on the induced topology wY M # simply by wM. Corollary 2. Assume X , Y and M are as above. Assume φ : Y → K is a linear map, that satisﬁes the following condition:

# • φ M : M → K is wM-continuous, for every M ∈ M. Then φ(M) is bounded, for every M ∈ M. Proof. Fix some M ∈ M, and choose some M0 ∈ M, which is convex, balanced, and satisﬁes the inclusion M1 ⊃ 2M. In particular, one also has the inclusion

M + (−1)M ⊂ M1. (15)

12 In order to prove that φ(M) ⊂ K is bounded, we are going to show that every net in φ(M) has a convergent subnet. (This will imply that φ(M) is relatively compact, thus bounded in K.) Start with some net (αλ)λ∈Λ ⊂ φ(M), written as αλ = φ(yλ), where (yλ)λ∈Λ is a net in M. By Proposition 2, there exists a subnet (yη(σ))σ∈Σ of (yλ)λ∈Λ (deﬁned by a # directed map η :Σ → Λ), which is Cauchy in (Y, wY ). In particular, the double net # (yη(σ1) − yη(σ2))(σ1,σ2)∈Σ×Σ is convergent to 0 in (Y, wY ). By our choice of M1 this net is also # # convergent to 0 in (M1, w ) so by linearity and the w -continuity assumption on φ M1 M1 M1 it follows that lim φ(yη(σ1)) − φ(yη(σ2)) = lim φ(yη(σ1) − yη(σ2)) = φ(0) = 0, (σ1,σ2)∈Σ×Σ (σ1,σ2)∈Σ×Σ which means that the net (αη(σ))σ∈Σ = φ(yη(σ)) σ∈Σ is Cauchy, thus convergent in K. We are now in position to introduce candidates for the solutions of questions (a) and (b) above. Notations. Let X , Y, and M be as above. Using these notations, we deﬁne the space

0 # Z = {φ ∈ Y : φ M wM-continuous, ∀ M ∈ M}. Clearly Z is a linear subspace in Y0. By Corollary 2, for every φ ∈ Z and every M ∈ M, the quantity qM(φ) = supy∈M |φ(y)| is finite. It is clear that, for every M ∈ M, the map qM : Z 3 φ 7−→ qM(φ) ∈ [0, ∞) defines a seminorm on M, which satisfies the identity

# qM(x ) = pM, ∀ x ∈ X .

0 Using the collection QM = {qM : M ∈ M} we endow Z with a locally convex Hausdorﬀ topology, which we denote by S. It is pretty obvious that, by the properties of M, the 0 collection of seminorms QM is directed, with respect to the ordinary order relation. In particular: (∗) for every linear continuous functional Φ ∈ (Z, S)∗, there exists M ∈ M, such that

|Φ(φ)| ≤ qM(φ), ∀ φ ∈ Z. (16)

With this terminology, Proposition 2 has the following far reaching generalization. Theorem 4 (Grothendieck). Use the notations as above. (i) The locally convex topological vector space (Z, S) is complete.

(ii) The map #: X → Y has Range # ⊂ Z. Furthermore, when we equip Range # with the induced S-topology, the map

#:(X , TM) → (Range #, S Range #) (17) is a linear homeomorphism.

(iii) If we consider the standard inclusion J : X ,→ X ∼TM , the map Γ: X ∼TM → Y0 has the following properties:

13 • Range Γ = Z; • (Γ ◦ J)(x) = x#, ∀ x ∈ X ;

∼TM ˜ • Γ:(X , TM) → (Z, S) is a linear homeomorphism. Proof. Statements (i) and (ii) are pretty obvious. To prove statement (iii) we will argue a bit indirectly, ﬁrst proving the following

Claim 1; For every y ∈ Y, the map θy : Z 3 φ 7−→ φ(y) ∈ K is linear and S-continuous. ∗ 5 Furthermore the map Θ: Y 3 y 7−→ θy ∈ (Z, S) is linear .

This is pretty obvious, since for every y ∈ Y the map θy satisfies (16) with M = {y}. Claim 2: Range # is S-dense in Z. Using annihilator spaces, it suffices to show that: whenever Φ ∈ (Z, S)∗ is a linear continuous functional, such that Φ Range # = 0, it follows that Φ is identically 0. Fix such a functional, and use property (∗) in the Comment that preceded the Theorem to find some M ∈ M, satisfying (16). Enlarging if necessary (use the fact that M is saturated), we can assume that M is convex and balanced. Using Claim 1, we can consider then the convex balanced set Θ(M) ⊂ Z∗, and then using the dual pairing between Z and Z∗ we see that condition (16) forces Φ to belong to the absolute bi-polar (Θ(M)). By the Absolute w∗ Bipolar Theorem, it follows that Φ lies in the w∗-closure Θ(M) , which means that there ∗ w ∗ exists a net (yλ)λ∈Λ ⊂ M, such that Θ(yλ) −→ Φ, in Z , that is:

φ(yλ) → Φ(φ), ∀ φ ∈ Z. (18)

# # By our assumption, we know that Φ(x ) = 0, so we get x (yλ) → 0, ∀ x ∈ X , which means, # wM of course, that yλ −−→ 0, and then by the deﬁnition of Z we also get

φ(yλ) → φ(0) = 0, ∀ φ ∈ Z, so going back to (18) it follows that Φ is indeed identically 0. Having proved Claim 2, we now employ Theorem 1 from TVS IV (argue as in in the proof ∼TM ˜ of Proposition 4), to obtain the existence of some linear homeomorphism T :(X , TM) → (Z, S), such that T ◦ J = #. Of course, we can also regard T as a linear continuous map ∼TM ˜ 0 0 T :(X , TM) → (Y , w ), and this forces the equality T = Γ, which proves all the desired properties.

5 # ∗ ∗ The map Θ is not continuous. Its restrictions Θ M :(M, wM) → (Z , w ) are continuous, for all M ∈ M.