Banach Spaces I: Normed Vector Spaces Notes from the Functional Analysis Course (Fall 07 - Spring 08)

BS I c Gabriel Nagy

Banach spaces are the most manageable among all types of topological vector spaces, particularly due to their metric structure. Before we start treating them, we will ﬁrst treat their natural precursors to which this section is devoted. Convention. Throughout this note K will be one of the ﬁelds R or C, and all vector spaces are over K. A. Normable and Normed Vector Spaces

The starting point of our discussion is Exercise 17 from LCVS III, which we recall here as the following: Remark 1. For a locally convex topological vector space (X , T), the following are equivalent: (i) There exists a bounded neighborhood of 0;

(ii) There exists some neighborhood V of 0, such that {εV}ε>0 is a basic neighborhood system for 0. (iii) There exists one seminorm p, so that T = T(p). Moreover, any seminorm with this property is a norm, i.e. one has the implication: p(x) = 0 ⇒ x = 0. Deﬁnition. A locally convex vector space, satisfying (one of) these three equivalent conditions, is declared normable. Of course, a normable topological vector space is metrizable, by the results from LCVS III. Notations. Norms on vector spaces will be typically denoted using the symbol k . k, that is, kxk stands for (the old) p(x). On occasion the norm notation will be accompanied by subscripts or superscripts.

Remark 2. Suppose X1 and X2 are normable spaces, with their topologies deﬁned by norms k . k1 and k . k2, respectively. For a linear map T : X1 → X2, the following conditions are equivalent: (i) T is continuous; (i’) T is continuous at 0; (ii) there exists a constant C ≥ 0, such that

kT xk2 ≤ Ckxk1, ∀ x ∈ X1.

1 This statement follows directly from Theorem 4 from LCVS V.

Remarks 3-4. Suppose k . k1 and k . k2 are two norms on a vector space X , and let T1 and T2 be the locally convex topologies deﬁned by them.

3. The condition that T2 is weaker than T1 is equivalent to the existence of a constant C > 0, such that: kxk2 ≤ Ckxk1, ∀ x ∈ X .

This follows from the preceding Remark, since the condition T2 ⊂ T1 is equivalent to the continuity of the identity map I :(X , T1) → (X , T2). 4. The condition that the two topologies coincide is equivalent to the existence of two constants C, D > 0, such that

Dkxk1 ≤ kxk2 ≤ Ckxk1, ∀ x ∈ X . (1)

Two norms satisfying condition (1) are said to be equivalent. Comment. From the topological point of view, to define a normable topology on a vector space X is the same as identifying an equivalence class (in the sense defined above) in the set of all norms. When a (distinguished) norm is identified, that is, when a pair (X , k . k) is considered, we call the pair a normed vector space. Therefore, the difference between “normable” and “normed” lies in the possibility that two different norms might produce the same topology. Example 1. On a finite dimensional vector space X there exists a unique Hausdorff linear topology. Moreover, this topology is normable, so on X any two norms are equivalent. For instance, in the case when X = Kn, this unique topology is the product topology, which can be defined by the norm

n kxk∞ = max{|x1|,..., |xn|}, ∀ x = (x1, . . . , xn) ∈ K . Q If S is an infinite set, then neither the product space K (equipped with the product L S topology), nor the direct sum S K (equipped with the locally convex sum topology) is normable. (In the product space every neighborhood of 0 contains an infinite dimensional linear space, so it cannot be bounded. The direct sum is not even metrizable.) Definition. Given a normed vector space (X , k . k), the (locally convex) topology defined by the k . k is referred to as the norm topology. In terms of convergence, the condition k . k xλ −−→ x is equivalent to the condition kxλ − xk → 0. We denote (with k . k fixed), for any r > 0, the closed ball {x ∈ X : kxk ≤ r} by (X )r. Using this notation, the collection (X )r r>0 constitutes a basic neighborhood system for 0 in the norm topology. B. Methods of Constructing Norms

In this section we outline three methods for constructing norms. The ﬁrst one is simply a restatement of well known fact from the theory of metric spaces (see also Remark 1 from TVS IV).

2 Remark 5. If (X , k . k) is a normed vector space and T denotes its norm topology, then for every linear subspace Y ⊂ X , the restriction of k . k to Y is a norm, and furthermore, the corresponding norm topology on Y coincides with the induced topology T Y . The second method is an easy application of Exercise 2 from LCVS III (or Lemma 2 from TVS IV). Proposition-Deﬁnition 1. Suppose (X , k . k) is a normed space and Y ⊂ X is a linear subspace, which is closed in the norm topology. If we consider the quotient space X /Y and we denote by π : X → X /Y the quotient map, then the map k . kˆ: X /Y → [0, ∞) given by

kvkˆ= inf{kxk : x ∈ X , π(x) = v}, v ∈ X /Y defines a norm on X /Y. Moreover, the norm topology defined by k . kˆ coincides with the quotient topology. The norm k . kîs referred to as the quotient norm. Proof. From Exercise 2 in LCVS III we know that k . kˆ is a seminorm, which defines the quotient topology. Since the quotient topology is Hausdorff, it follows that k . kîs in fact a norm. The third method is a bit more elaborate, but quite effective, as shown by the examples below. It uses dual pairs, as introduced in DT I. Proposition-Definition 2. Suppose X and Y are in a dual pairing, defined by means of two injective linear maps1

#: X 3 x 7−→ x# ∈ Y0 #: Y 3 y 7−→ y# ∈ X 0 satisfying: x#(y) = y#(x), ∀ x ∈ X , y ∈ Y. Suppose M ⊂ Y is some non-empty absorbing set, and deﬁne the map k . kM : X → [0, ∞] by

# kxkM = sup |x (y)|. y∈M

(i) The set XM = {x ∈ X : kxkM < ∞} is a linear subspace in X .

(ii) k . kM is a norm, when restricted to XM.

2 # (iii) If we consider the absolute polar M = {x ∈ X : |x (y)| ≤ 1, ∀ y ∈ M}, then we

S have the equalities: XM = span(M ) = t>0 tM .

(iv) The unit ball (XM)1 coincides with M.

# (v) The norm topology on XM is stronger than the induced w -topology.

Proof. Before we prove each of the statements (i)-(v) we ﬁrst show that k . kM has the following properties3

1 Here X 0 and Y0 denote the algebraic duals of X and Y. 2 # From DT I we know that M is convex, balanced, and closed in the w -topology. 3 We use the usual conventions: number + ∞ = ∞ + ∞ = ∞ and (positive number) · ∞ = ∞.

3 (a) kx1 + x2kM ≤ kx1kM + kx2kM, ∀ x1, x2 ∈ M;

(b) if α ∈ K, α 6= 0, then kαxkM = |α| · kxkM, ∀ x ∈ X ;

Property (a) is quite trivial, since for every y ∈ M we have the inequalities

# # # # # |(x1 + x2) (y)| = |x1 (y) + x2 (y)| ≤ |x1 (y)| + |x2 (y)| ≤ kx1kM + kx2kM.

For property (b), we ﬁrst notice that, for every y ∈ M we have the inequality |(αx)#(y)| = # # |α · x (y)| = |α| · |x (y)| ≤ |α| · kxkM, so taking supremum yields

kαxkM ≤ |α| · kxkM, ∀ x ∈ X , α 6= 0. (2)

If we use the above inequality with α−1 in place of α, and αx in place of x, we also get −1 −1 kxkM = kα (αx)kM ≤ |α | · kαxkM, which yields

|α| · kxkM ≤ kαxkM, ∀ x ∈ X , α 6= 0, (3) and then (b) follows from (2) and (3). Finally, to prove (c) we use the fact that M is absorbing in Y. Start with x such that kxkM = 0. On the one hand, by the definition of k . kM, we know that x#(y) = 0, ∀ y ∈ M. (4) On the other hand, since M is absorbing, for every y ∈ Y there exists ρ > 0 such that ρy ∈ M, so by (4) we have x#(y) = ρ−1x#(ρy) = 0, thus proving that x# = 0. Finally, the injectivity of x 7−→ x# forces x = 0. Statements (i)-(ii) are now obvious from (a)-(c). Statement (iv) is obvious from the definition of k . kM. Clearly (iii) follows from (iv). Finally to prove (v), we start with some w# net (xλ)λ∈Λ in XM, which converges in norm to 0, and we show that xλ −−→ 0, which means that # xλ (y) → 0, ∀ y ∈ Y. (5) # First of all, since we have the inequality |xλ (y)| ≤ kxλkM, ∀ y ∈ M, the condition (5) is clearly satisfied if y ∈ M. By linearity we now have

# xλ (ρy) → 0, ∀ y ∈ M, ρ > 0, and then the desired conclusion (5) follows from the fact that M is absorbing in Y. Remarks 6-7. Suppose X and Y are in a dual pairing, and M ⊂ Y is non-empty and absorbing.

# 6. The set M is w -bounded, if and only if XM = X . In this case k . kM deﬁnes a norm on the whole space X , and the norm topology TM on X , deﬁned by k . kM, is weaker than the strong topology (see DT II).

4 7. If M is w#-compact, convex, balanced (thus w#-bounded), then the norm topology TM is weaker than the Mackey topology (see DT II). In particular, the correspondence #: Y 3 y 7−→ y# ∈ X 0 establishes a topological linear isomorphism:

# ∼ ∗ ∗ (Y, w ) −→ ((X , TM) , w ).

Comment. As we shall see a little later, all normed spaces can be constructed according to the recipe indicated in Remark 7.

C. Examples: the `p-spaces

In this section we specialize the construction outlined in Proposition-Definition 2 to the following situation (all notations will be fixed throughout the entire sub-section): • S will denote a non-empty set (the interesting case will be when S is infinite); Q •X will denote the product space S K; L •Y will denote the direct sum S K. The spaces X and Y will be put in a dual pairing by

# # X x (y) = y (x) = xsys, ∀ x = (xs)s∈S ∈ X , y = (ys)s∈S ∈ Y. s∈S

(By construction, for every y = (ys)s∈S ∈ Y, the set supp y = {s ∈ S : ys 6= 0} is ﬁnite, so the above sum has only ﬁnitely many non-zero terms.)

Notations. For every “number” p ∈ [1, ∞], we define the set Mp ⊂ Y as follows:  {y ∈ Y : maxs∈S |ys| ≤ 1}, if p = 1    p P p−1 Mp = {y ∈ Y : s∈S |ys| ≤ 1}, if 1 < p < ∞    P {y ∈ Y : s∈S |ys| ≤ 1}, if p = ∞ L (By the definition of Y as S K, the sums used in the definitions of Mp, p ∈ (1, ∞], involve only finitely many non-zero terms.) Using the notations from Proposition-Definition 2, we have, for each p ∈ [1, ∞] a vector space X , which from now on we denote by `p (S), equipped with a norm k . k , which Mp K Mp from now on we will simply denote by k . kp. When K = C, the field will be omitted from the notation. When S = N the set will also be omitted from the notation. The explicit presentation of the normed spaces (`p (S), k . k ), p ∈ [1, ∞] is as follows. K p Q Proposition 3. Use the notations as above, and fix some element x = (xs)s∈S ∈ S K. (i) The condition that x belongs to `∞(S) is equivalent to the condition: sup |x | < ∞. K s∈S s Moreover, in this case one has the equality

kxk∞ = sup |xs|. (6) s∈S

5 (ii) If 1 ≤ p < ∞, the condition that x belongs to `p (S) is equivalent to the condition: K P p s∈S |xs| < ∞. Moreover, in this case one has the equality

1/p X p kxkp = |xs| . (7) s∈S

Q Proof. (i). For every x ∈ S K, denote the quantity sups∈S |xs| ∈ [0, ∞] by Q∞(x). To prove the “only if” implication, assume ﬁrst x ∈ `∞(S), so K # kxk∞ = sup |x (y)| < ∞. (8) y∈M∞ L In particular, if we consider, for every t ∈ S the element δt = (δts)s∈S ∈ S K (here δts denotes the Kronecker symbol, deﬁned to be 1 if s = t, and 0, otherwise), then obviously # x (δt) = xt, and δt ∈ M∞, so by (8) we get |xt| ≤ kxk∞, ∀ t ∈ S, so

Q∞(x) ≤ kxk∞, (9) thus proving the desired implication. Conversely, assume Q (x) < ∞, and let us prove that x ∈ `∞(S). If we start with an ∞ K arbitrary y = (ys)s∈S ∈ M∞, then using the inequalities |xs| ≤ Q∞(x), we have # X X X |x (y)| = xsys ≤ |xs| · |ys| ≤ Q∞(x) · |ys| ≤ Q∞(x), s∈supp y s∈supp y s∈supp y so taking supremum (over all y ∈ M ) not only shows that x ∈ `∞(S), but also proves the ∞ K inequality kxk∞ ≤ Q∞(x), which combined with (9) yields the desired equality (6). P p (ii). Denote, for p ∈ [1, ∞), the quantity s∈S |xs| ∈ [0, ∞] by Qp(x). (By convention, P p p we deﬁne s∈S |xs| = ∞, if the S-tuple (|xs| )s∈S is not summable.) With this notation, the desired equality (7) reads

Q (x) = kxk p, ∀ x ∈ `p (S). (10) p p K We divide now the proof into two cases: (a) 1 < p < ∞; (b) p = 1. p 1 1 (a) Denote the number by q, so that + = 1, and the set M is deﬁned as p − 1 p q p

M X q Mp = {y = (ys)s∈S ∈ K : |ys| ≤ 1}. (11) S s∈S

To prove the “only if” implication, ﬁx some element x ∈ `p (S), so K # kxkp = sup |x (y)| < ∞, (12) y∈Mp

6 and let us show that Qp(x) < ∞. Let us deﬁne, for every ﬁnite set F ⊂ S, the quantity P p L Qp,F (x) = s∈F |xs| , and the element yF = (yF s)s∈S ∈ S K as follows:  |x |p  s , if s ∈ F and x 6= 0  F 1/q s xsQp (x) yF s =   0, if either s 6∈ F , or xs = 0. Remark that, by construction, using the identity qp − q = p, we have

 p |xs|  , if s ∈ F and xs 6= 0 q  Qp,F (x) |yF s| =   0, if either s 6∈ F , or xs = 0.

P q so s∈S |yF s| is either equal to 0, if xs = 0, ∀ s ∈ F , or equal to 1 otherwise. In any case, it follows that yF belongs to Mp, and we have  0, if xs = 0, ∀ s ∈ F  # X  x (yF ) = xsyF s P p s∈F |xs| s∈F  1−1/q 1/p  1/q = Qp,F (x) = Qp,F (x) , otherwise  Qp,F (x) By (12) the above estimate yields

p X p kxkp ≥ Qp,F (x) = |xs| , s∈F so taking supremum (over all ﬁnite subsets F ⊂ S) we obtain the inequality

p X p kxkp ≥ |xs| = Qp(x), (13) s∈S thus proving the desired implication. Conversely, assume Q (x) < ∞, and let us prove that x ∈ `p (S). If we start with an p K 4 arbitrary y = (ys)s∈S ∈ Mq, then using the Elementary H¨olderInequality and (11), we have

1/p 1/q # X X p X q 1/p |x (y)| = xsys ≤ |xs| · |ys| ≤ Qp(x) , s∈supp y s∈supp y s∈supp y

so taking supremum (over all y ∈ M ) not only shows that x ∈ `p (S), but also proves the p K inequality 1/p kxkp ≤ Qp(x) ,

4 The Elementary H¨olderInequality states that, for any x1, . . . , xn, y1, . . . , yn ∈ C and any two real 1 1 Pn Pn p1/p Pn q1/q numbers p, q > 1 satisfying + = 1, one has the inequality xjyj ≤ |xj| · |yj| . p q j=1 j=1 j=1 p p−1 q q−1 Moreover, equality holds, if and only if the n-tuples (x1|x1| , . . . , xn|xn| ) and (y1|y1| , . . . , yn|yn| ) are proportional.

7 which combined with (13) yields the desired equality (10). (b) In this case (p = 1), the set M1 is deﬁned as

M q Mp = {y = (ys)s∈S ∈ K : sup |ys| ≤ 1}. (14) s∈S S

To prove the “only if” implication, ﬁx some element x ∈ `1 (S), so K # kxkp = sup |x (y)| < ∞, (15) y∈M1

and let us show that Q1(x) < ∞. As in (a), we deﬁne, for every ﬁnite set F ⊂ S, the the L element yF = (yF s)s∈S ∈ S K as:   |xs|  , if s ∈ F and xs 6= 0 xs yF s =   0, if either s 6∈ F , or xs = 0.

By construction, we have |yF s| ≤ 1, ∀ s ∈ S, so yF belongs to M1, and we have

# X X x (yF ) = xsyF s = |xs|. s∈F s∈F P By (15) the above estimate yields kxk1 ≥ s∈F |xs|, so taking supremum (over all ﬁnite subsets F ⊂ S) we obtain the inequality X kxk1 ≥ |xs| = Q1(x), (16) s∈S thus proving the desired implication. Conversely, assume Q (x) < ∞, and let us prove that x ∈ `1 (S). If we start with an 1 K arbitrary y = (ys)s∈S ∈ M1, then |ys| ≤ 1, ∀ s ∈ S, so we have

# X X X X |x (y)| = xsys ≤ |xs| · |ys| ≤ |xs| ≤ |xs| = Q1(x),

s∈supp y s∈supp y s∈supp y s∈S

so taking supremum (over all y ∈ M ) not only shows that x ∈ `1 (S), but also proves the 1 K inequality kxk1 ≤ Q1(x),

which combined with (16) yields the desired equality kxk1 = Q1(x). Comment. The space `∞(S) has been introduced earlier in LCVS IV. One important K subspace in `∞(S) is c (S), which was deﬁned in LCVS IV in two diﬀerent ways: K 0,K (a) the closure of L in `∞(S) in the norm topology, or equivalently S K K Q (b) the space of all S-tuples x = (xs)s∈S ∈ S K, such that

8 • for every ε > 0, there exists a ﬁnite subset F ⊂ S, such that |xs| < ε, ∀ s ∈ S rF . Exercise 1♥. Prove that, if 1 ≤ p < ∞, then L is dense in `p (S) in the norm S K K topology. Exercise 2. Prove that if 1 ≤ p < q < ∞, and S is inﬁnite, then

(i) `p (S) `q (S) c (S). K ( K ( 0,K (ii) kxk ≤ kxk ≤ kxk , ∀ x ∈ `p (S). In particular, the inclusion maps `p (S) ,→ `q (S) ∞ q p K K K and `p (S) ,→ c (S) are continuous. K 0,K Q Remark 8. For an S-tuples x = (xs)s∈S ∈ S K, the condition that x belongs to `1 (S) is equivalent (see P I) to the condition that x is summable, i.e. the net of finite K P sums ΣF (x) , defined by ΣF (x) = xs, is convergent. If we denote, for every F ∈Pfin(S) s∈F x ∈ `1 (S), the sum P x by Σ (x), this gives rise to a linear functional Σ : `1 (S) → , K s∈S s S S K K which is continuous in the norm topology, since it obviously satisfies the inequality

|Σ (x)| ≤ kxk , ∀ x ∈ `1 (S). (17) S 1 K

The map ΣS will be referred to as the summation functional. Definition. Two “numbers” p, q ∈ [1, ∞] are said to be Hölderconjugate to each other, 1 1 if either (a) p = 1 and q = ∞, or (b) p = ∞ and q = 1, or (c) p, q > 1 and + = 1. p q Proposition 4. Suppose p, q ∈ [1, ∞] are Hölderconjugate to each other.

(i) Given two S-tuples x = (x ) ∈ `p (S) and y = (y ) ∈ `q (S), their point-wise s s∈S K s s∈S K product xy = (x y ) belongs to `1 (S), and furthermore one has the inequality s s s∈S K

kxyk1 ≤ kxkp · kykq. (18)

(ii) The map Φ: `p (S) × `q (S) 3 (x, y) 7−→ Σ (xy) ∈ establishes a dual pairing between K K S K `p (S) and `q (S). K K (iii) If we deﬁne, for every x ∈ `p (S) the map x# : `q (S) 3 y 7−→ Φ(x, y) ∈ , then x# is K K K continuous in the norm topology, and satisﬁes

kxk = sup |x#(y)| : y ∈ `q (S) . (19) p K 1

(iv) If p > 1, then the map

#:(`p (S), w#) 3 x 7−→ x# ∈ (`q (S), k . k )∗, w∗ K K q is a topological linear isomorphism.

Proof. The three instances of the H´olderconjugacy condition will be labeled as in the preceding deﬁnition.

9 (i). Fix x = (x ) ∈ `p (S) and y = (s ) ∈ `q (S), as well as some ﬁnite set F ⊂ S. s s∈S K s s∈S K In case (a) we have p = 1 and q = ∞, so |ys| ≤ kyk∞, ∀ s ∈ S, which immediately yields

X X X |xsys| ≤ |xs| · kyk∞ ≤ |xs| · kyk∞ = kxk1 · kyk∞. (20) s∈F s∈F s∈S

Likewise, in case (b), when p = ∞ and q = 1, we have |xs| ≤ kxk∞, ∀ s ∈ S, which immediately yields

X X X |xsys| ≤ kxk∞ · |ys| ≤ kxk∞ · |ys| = kxk∞ · kyk1. (21) s∈F s∈F s∈S

Finally, in case (c) we use the Elementary H¨olderInequality, which yields

1/p 1/q X X X p X q |xsys| = |xs| · |ys| ≤ |xs| · |ys| ≤ s∈F s∈F s∈F s∈F (22) 1/p 1/q X p X q ≤ |xs| · |ys| = kxkp · kykq. s∈S s∈S P So in any of the three cases, the inequalities (20), (21) and (22) show that |xsys| ≤ kxkp· s∈F P kykq, so by taking the supremum over all finite subsets F ⊂ S, it follows that s∈S |xsys| ≤ kxkp · kykq, thus proving (i). (ii). By symmetry (for the other condition simply flip p and q), it suffices to show that, if x ∈ `p (S) is such that K Φ(x, y) = 0, ∀ y ∈ `q (S), (23) K then x = 0. But this is quite obvious, by using elements of the form y = δ ∈ L ⊂ `q (S), t S K K introduced in the preceding proof, since Φ(x, δt) = xt, ∀ t ∈ S, so condition (23) forces xt = 0, ∀ t ∈ S. (iii). The fact that x# : `q (S) → is norm-continuous is immediate from (i). To prove K K (19), we first observe that the unit ball `q (S) clearly contains the set M previously K 1 p considered in the definition of the norm k . kp, simply because M Mp = {y ∈ K : kykq ≤ 1}, (24) S

so if we denote the right-hand side of (19) by Up(x), we have the inequality kxkp ≤ Up(x). The inequality Up(x) ≤ kxkp is also obvious by (18) (iv). The injectivity of the map # is obvious, by part (ii). To prove its surjectivity, start with an arbitrary linear norm-continuous functional φ : `q (S) → , and let us show the K K existence of some x ∈ `p (S), such that φ = x#. First of all, by the norm-continuity of φ, K there exists a constant C > 0, such that

|φ(y)| ≤ Ckyk , ∀ y ∈ `q (S). (25) q K

10 Deﬁne now, for every t ∈ S, the number x = φ(δ ) ∈ , where δ ∈ L ⊂ `q (S) is the t t K t S K K element considered in the proof of Proposition 3. Note that, if we start with an arbitrary L P element y ∈ S K, then y = s∈supp y ysδt, so we have X X xtyt = ytφ(δt) = φ(y). t∈S t∈supp y

In particular, using (24) and (25) we get

X xtyt = |φ(y)| ≤ C, ∀ y ∈ Mp,

t∈S

so taking supremum over all y ∈ M forces x to belong to `p (S), as well as the inequality p K kxk ≤ C. Now that x belongs to `p (S) we clearly have the equality p K

# M φ(y) = x (y), ∀ y ∈ K, S

so the norm-continuity of both φ and x#, combined with Exercise 2, forces φ = x#. The statement that the map # : `p (S) → (`q (S), k . k )∗ is continuous, when we equip `p (S) with K K q K the w#-topology, and the target space (`q (S), k . k )∗ with the w∗-topology, is tautologic. K q Comment. In connection with statement (iv) from Proposition 4, it should be noted that the case p = 1 is omitted. Although the map

#: `1 (S) 3 x 7−→ x# ∈ (`∞(S), k . k )∗ (26) K K q is legitimately defined, injective, and continuous, when we equip `1 (S) with the w#-topology, K and the target space (`∞(S), k . k )∗ with the w∗-topology, it is not surjective. In order to K q clarify this phenomenon, another dual pairing involving `1 (S) needs to be introduced, for K which Proposition 4 has the following analogue. Proposition 5. Consider the dual pairing map Φ: `1 (S)×`∞(S) 3 (x, y) 7−→ Σ (xy) ∈ K K S , defined above, and let Φ denote the restriction of Φ to `1 (S) × c (S). K 0 K 0,K (i) The map Φ establishes a dual pairing between `1 (S) and c (S). 0 K 0,K (ii) If we define, for every x ∈ `1 (S) the map x# : c (S) 3 y 7−→ Φ (x, y) ∈ , then x# K 0,K 0 K is continuous in the norm topology, and satisfies

# kxk1 = sup |x (y)| : y ∈ c0,K(S) 1 . (27)

(iii) The the map

#:(`1 (S), #) 3 x 7−→ x# ∈ (c (S), k . k )∗, ∗ (28) K w 0,K q w is a topological linear isomorphism.

11 Proof. (i). On the one hand, we already know that, if y ∈ c0,K(S) is such that Φ0(x, y) = 0, ∀ x ∈ `1 (S), then y = 0. (This follows from statement (ii) in Proposition 4.) On the other K hand, if x ∈ `1 (S) is such that Φ (x, y) = 0, ∀ y ∈ c (S), then arguing exactly as in the K 0 0,K proof of Proposition 4 (ii) we also have xt = Φ0(x, δt) = 0, ∀ t ∈ S, so again x = 0. (ii). The proof of the equality (27) is virtually identical to the one given for Proposition 4 (iii), simply because the set M1, which is characterized as M M1 = {y ∈ K : kyk∞ ≤ 1}, (29) S is contained in the unit ball c0,K(S) 1. (iii). Except for the surjectivity of (28), all other features are proven in the exact same way as in Proposition 4 (iv). To prove surjectivity, we start with some linear continuous norm-continuous functional φ : c (S) → , and we prove the existence of some x ∈ `1 (S), 0K K K such that φ = x#. First of all, by the norm-continuity of φ, there exists a constant C > 0, such that |φ(y)| ≤ Ckyk∞ = C · sup |ys| , ∀ y ∈ c0,K(S). (30) s∈S

Deﬁne now, as in the proof of Proposition 4 (iv), for every t ∈ S, the number xt = φ(δt) ∈ K. L P Note that, if we start with an arbitrary element y ∈ S K, then y = s∈supp y ysδt, so we have X X xtyt = ytφ(δt) = φ(y). t∈S t∈supp y In particular, using (29) and (30) we get

X xtyt = |φ(y)| ≤ C, ∀ y ∈ Mp,

t∈S so taking supremum over all y ∈ M forces x to belong to `1 (S), as well as the inequality 1 K kxk ≤ C. Now that x belongs to `1 (S) we clearly have the equality 1 K # M φ(y) = x (y), ∀ y ∈ K, S # L so the norm-continuity of both φ and x , combined with the fact that S K is dense (by # deﬁnition) in (c0,K(S), k ., k∞), forces φ = x . Exercise 3. Prove that the map (26) is not surjective. Speciﬁcally, prove that if φ :

`∞(S) → K is a linear norm-continuous functional, such that φ = 0, then the condition K c0,K(S) that φ = x# for some x ∈ `1 (S) is equivalent to the condition that φ = 0. In particular, K if φ = 0, but φ is not identically 0, then φ cannot be represented as x#, for any c0,K(S) x ∈ `1 (S). K D. Linear Continuous Maps

In this sub-section we take a closer look at the space L(X , Y) of all linear continuous op- erators from X into Y, in the special case when both spaces are normed spaces. Throughout

12 the entire sub-section, it is understood that both X and Y come equipped with distinguished norms, both denoted by k . k. (When one wants to be speciﬁc about the normed vector space structure, a notation like L(X , k . k), (Y, k . k0) should be employed.) Proposition-Deﬁnition 6. Suppose T : X → Y is linear and continuous.

(i) Among all non-negative constants C, satisfying5

kT xk ≤ Ckxk, ∀ x ∈ X , (31)

then there exists a smallest one CT .

(ii) The constant CT , introduced above, is also given by CT = sup kT xk : x ∈ (X )1 . (32)

The constant CT will be denoted from now on by kT k, and will be referred to as the operator norm of T .

Proof. We are going to define CT by (32), and then show that CT satisfies property (i). First of all, we know that there exists at least one non-negative constant C satisfying (31), so any such C satisfies the inequalities

kT xk ≤ C, ∀ x ∈ (X )1, so the constant CT defined in (32) is indeed finite, and satisfies the inequality

CT ≤ inf{C ≥ 0 : C satisﬁes (31)}.

This means that, in order to finish the proof, all we need to do is show that CT itself satisfies (31). Start with some x ∈ X , and notice that the inequality is trivial, if x = 0, so we can −1 assume that x 6= 0. If we define x1 = kxk x, the kx1k = 1, so we by linearity have the inequality −1 −1 CT ≥ kT x1k = k(kxk )T xk = kxk · kT x}, which clearly gives (31) for C = CT .

Comment. With the notation CT = kT k, using condition (i) we have the inequality

kT xk ≤ kT k · kxk, ∀ T ∈ L(X , Y), x ∈ X . (33)

This is referred to as the Operator Norm Inequality. In practice, Proposition-Deﬁnition 6 is used for producing estimates for the operator norm, because what the above result really states is the equivalence (for a given non-negative constant C) between the condition (31) and the inequality kT k ≤ C. Exercises 4-7. Use the notations as above

4♥. Prove that the map L(X , Y) 3 T 7−→ kT k ∈ [0, ∞) deﬁnes a norm on L(X , Y).

5 At least one such constant exists, by Remark 2.

13 5♥. Prove that, if Z is another normed vector space, then

kST k ≤ kSk · kT k, ∀ T ∈ L(X , Y),S ∈ L(Y, Z).

(The norms kST k, kT k, and kSk are computed in L(X , Z), L(X , Y), and L(Y, Z), respectively.)

6♥. Prove that the operator norm topology on L(X , Y) is stronger than the strong operator Tso k . k topology Tso. (Recall that Tλ −−→ T means: Tλ −−→ T x (in Y), ∀ x ∈ X .)

♥ 7. Let X and Y be normed vector spaces and assume (Tλ)λ∈Λ ⊂ L(X , Y) is bounded in norm. Suppose A ⊂ X is such that span A is dense in X . (Such a set is called total in X . Prove that the following are equivalent:

Tso (i) Tλ −−→ 0;

(ii) Tλa → 0 (in Y in norm), for all a ∈ A.

Comment. As a special case of the operator norm, we can consider Y = K, equipped with the modulus | . | as norm. The space L(X , K) is the topological dual of (X , k . k), which is denoted by X ∗. (If the norm specification is needed, this space will be denoted by (X , k . k)∗.) As a special case of Proposition-Definition 6, we see that X ∗ is naturally equipped with the norm, defined by kφk = sup |φ(x)| : x ∈ (X )1 . This norm is referred to as the dual norm. Definitions. For an operator T ∈ L(X , Y), the following features will be interesting, when present.

A. T is said to be contractive, if: kT xk ≤ kxk, ∀ x ∈ X . By Proposition-Deﬁnition 6, this is equivalent to the condition kT k ≤ 1.

B. T is said to be isometric, if: kT xk = kxk, ∀ x ∈ X . By Proposition-Deﬁnition 6, this is implies the equality kT k = 1, provided X is not zero-dimensional.

C. T is said to be bounded from below, if there exists some constant D > 0, such that:

kT xk ≥ Dkxk, ∀ x ∈ X . (34)

This condition forces T to be injective.

Example 1. Let S be some (inﬁnite) non-empty set. If p, q ∈ [1, ∞] are H¨older conjugate, and we equip the topological dual space `q (S)∗ with the dual norm, then the linear map K #: `p (S) 3 x 7−→ x# ∈ `q (S)∗, (35) K K considered in Proposition 4, is isometric, by property (iii). By property (iv) from Proposition 4, we also know that (35) is a linear isomorphism, if 1 < p ≤ ∞.

14 Example 2. In the case p = 1, we use Proposition 5 instead, by which the map

#: `1 (S) 3 x 7−→ x# ∈ c (S)∗, (36) K 0,K

∗ is an isometric linear isomorphism, when the topological dual space c0,K(S) is equipped with the dual norm. Proposition 7. If X is a normed vector space, then for every x ∈ X , there exists φ ∈ X ∗, with kφk ≤ 1, such that φ(x) = kxk. Proof. The case x = 0 is trivial, so we can assume that x 6= 0. Consider the linear subspace Kx, and the linear map ψ : Kx 3 αx 7−→ αkxk ∈ K. It is pretty obvious that

|ψ(y)| ≤ kyk, ∀ y ∈ Kx. Use then the Hahn-Banach Theorem (for seminorms) to deduce the existence of a linear map φ : X → K, such that |φ(y)| ≤ kyk, ∀ y ∈ X , (37) and such that φ(y) = ψ(y), ∀ y ∈ Kx, so in particular we get φ(x) = kxk. Of course, condition (37) shows precisely that kφk ≤ 1, and we are done. Corollary. If we deﬁne, for every x ∈ X , the linear functional x# : X ∗ 3 φ 7−→ φ(x) ∈ K, then:

(i) for every x ∈ X , the linear functional x# : X ∗ → K is norm-continuous; (ii) when we equip the topological dual X ∗∗ = (X ∗, k . k)∗ with the dual norm, the map

#: X 3 x 7−→ x# ∈ X ∗∗

is linear and isometric.

Proof. (i). The norm continuity of x# is clear, since

|x#(φ)| = |φ(x)| ≤ kxk · kφk, ∀ x ∈ X , φ ∈ X ∗.

The above inequality also shows that kx#k ≤ kxk, ∀ x ∈ X . (ii). The linearity of # is trivial. To prove that this map is isometric, all we need to do is to show that ∗ sup |φ(x)| : φ ∈ (X )1 = kxk. One inequality (≤) has already been discussed above. The second inequality (≥ 1) follows immediately from Proposition 7. The Exercise below is a “poor man’s version” of the Inverse Mapping Principle. Exercise 8♥. Suppose X and Y are normed vector spaces, and T : X → Y is linear, continuous, and injective. Let Z = Range T and let S : Z → X be the inverse of the bijective linear map T : X −→Z∼ . Show that, when we equip Z with the induced norm from Y, the following conditions are equivalent.

15 (i) S : Z → X is norm continuous.

(ii) T is bounded from below, i.e. there exists D > 0, satisfying condition (34).

Show that in this case, there exists a largest positive constant DT satisfying (34), namely −1 DT = kSk . Exercises 9-10. Let S be an inﬁnite set.

9. For p ∈ [1, ∞), let J denote the inclusion map `p (S) ,→ c (S). By Exercise 2, we p K 0,K know that Jp is a contraction. Prove that Jp is not bounded from below. 10. For 1 ≤ p < q < ∞, let J denote the inclusion map `p (S) ,→ `q (S). Again by pq K K Exercise 2, we know that Jpq is a contraction. Prove that Jpq is not bounded from below.

We conclude with a note on the topological transpose. Assume X and Y are normed vector spaces. Recall (see LCVS V) that for an operator T ∈ L(X , Y) and any topological vector space Z, the transpose of T is deﬁned as the linear operator

T ∗ : L(Y, Z) 3 S 7−→ ST ∈ L(X , Z).

We also know that, when the spaces L(X , Z) and L(Y, Z) are equipped with the strong ∗ operator topology Tso, then T is continuous. In particular, when Z = K, one has a linear continuous operator T ∗ :(Y∗, w∗) → (X ∗, w∗). Exercise 11♥. Use the notations as above. Prove that, if Z is a normed vector space, and T ∈ L(X , Y), then T ∗ : L(Y, Z) → L(X , Z) is also continuous, when both the source and the target spaces are equipped with the operator norm topologies. Furthermore, if Z is non-zero, one has kT ∗k = kT k.