Lectures Normed vector spaces. i I. finite

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1 Lectures Normed vector spaces Definition. Let K be one of the fields R or C, and let X be a K-vector space. A norm on X is a map X x x [0, ) with the following properties (i) x + y x + y, x, y X; (ii) λx = λ x, x X, λ K; (iii) x = 0 = x = 0. (Note that conditions (i) and (ii) state that. is a seminorm.) Example 2.1. Let K be either R or C. Fix some non-empty set I, and define { c K [ 0 (I) = α : I K : inf sup α(i) ] } = 0. F I F finite Remark that for a function α : I K, the fact that α belongs to c K 0 (I) is euivalent to the following condition: For every ε > 0, there exists some finite set F I, such that α(i) < ε, i I F. We euip the space c K 0 (I) with the K-vector space structure defined by point-wise addition and point-wise scalar multiplication. We also define the norm. by α = sup α(i), α c K 0 (I). When K = C, the space c C 0 (I) is simply denoted by c 0 (I). When I = N - the set of natural numbers - the space c K 0 (N) can be euivalently described as c K 0 (N) = { α = (α n ) n 1 K : lim α n = 0 }. n In this case instead of c R 0 (N) we simply write c R 0, and instead of c 0 (N) we simply write c 0. Exercise 1. Prove that. is indeed a norm on c K 0 (I). Example 2.2. Let K be either R or C, and let I be a non-empty set. We define the space fin K (I) = { α : I K : the set {i I : α(i) 0} is finite }. Then fin K (I) is a linear subspace in c K 0 (I). 63

2 64 LECTURES 9-11 Definition. Suppose X is a normed vector space, with norm.. Then there is a natural metric d on X, defined by d(x, y) = x y, x, y X. The toplogy on X, defined by this metric, is called the norm topology. Exercise 2. Let X be a normed vector space, over K(= R, C). Prove that, when euipped with the norm toplogy, X becomes a topological vetor space. That is, the maps are continuous. X X (x, y) x + y X K X (λ, x) λx X Exercise 3. Let K be one of the fields R or C, and let I be a non-empty set. Prove that fin K (I) is dense in c K 0 (I) in the norm topology. Example 2.3. Let K be one of the fields R or C, and let I be a non-empty set. Define l K (I) = { α : I K : sup α(i) < }. We euip the space l K (I) with the K-vector space structure defined by point-wise addition and point-wise scalar multiplication. We also define the norm. by α = sup α(i), α l K (I). When K = C, the space l C (I) is simply denoted by l (I). When I = N - the set of natural numbers - instead of l R (N) we simply write l R, and instead of l (N) we simply write l. Exercise 4. Prove that. is indeed a norm on l K (I). Exercise 5. Let K be one of the fields R or C, and let I be a non-empty set. Prove that c K 0 (I) is a linear subspace in l K (I), which is closed in the norm topology. In preparation for the next class of examples, we introduce the following: Definition. A map α : I K is said to be summable, if there exists some number s K such that (s) for every ε > 0 there exists some finite set F ε I such that s α(i) < ε, for all finite sets F with F ε F I. If such an s exists, then it is uniue, and it is denoted by α(i). In the case when I is finite, every map α : I K is summable, and the above notation agrees with the usual notation for the sum. Exercise 6. Assume α : I K is summable. Prove that, for every λ K, the map λα : I K is summable, and λα(i) = λ α(i). If β : I K is another summable map, prove that α + β : I K is summable, and [α(i) + β(i)] = [ α(i) ] + [ β(i) ].

3 CAHPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS 65 The following result characterizes summability for non-negative terms Lemma 2.1. Let K be one of the fields R or C, let I be a non-empty set, and let α : I [0, ). The following are euivalent: (i) α is{ summable; } (ii) sup α(i) : F I, finite <. Moreover, in this case we have { } sup α(i) : F I, finite = α(i). { } Proof. We denote the uantity sup α(i) : F I, finite simply by t. (i) (ii). Assume α is summable, and denote α(i) simply by s. Choose, for each ε > 0 a finite set F ε I such that s α(i) < ε, for all finite subsets F I with F F ε. Claim: For any finite set G I, and any ε > 0, one has the ineuality α(i) < s + ε. i G Indeed, if we take the finite set G F ε, then using the fact that all α s are nonnegative, we get α(i) α(i) < s + ε. i G i G F ε Using the Claim, which holds for any ε > 0, we immediately get α(i) s, for all finite subsets G I, i G so taking supremum yields t s, in particular t <. (ii) (i). Assume condition (ii) is true. We are going to show that α is summable, by proving that the number t satisfies the definition of summabilty. Consider the set { } S = α(i) : F finite subset of I, so that sup S = t <. Start with some ε > 0. Since t ε is no longer an upper bound for S, there exists some finite set F ε I, such that ε α(i) > t ε. Notice that, for any finite set F I with F F ε, we have t ε < ε α(i) α(i) t, so we immediately get t α(i) < ε.

4 66 LECTURES 9-11 Exercise 7. Let α : I [0, ) be summable. Prove that every map β : I [0, ) with β(j) α(j), j I, is summable, and β(j) α(j). Remark 2.1. It is obvious tat the above result has a version for non-positive maps as well. More explicitly, for a map α : I (, 0] the following are euivalent: (i) α is{ summable; } (ii) inf α(i) : F I, finite >. Moreover, in this case we have { } inf α(i) : F I, finite = α(i). Lemma 2.2. Let I be a non-empty set. For a function α : I C, the following are euivalent: (i) α is summable; (ii) both functions Re α, Im α : I R are summable. Moreover, in this case we have the euality α(j) = Re α(j) + i Im α(j). Proof. (i) (ii). Assume α is summable. Denote the sum α(j) simply by s. For every ε > 0 choose a finite set F ε I such that s α(j) < ε, for all finite sets F I with F F ε. Using the ineuality max { Re z, Im z } z, z C, we immediately get the ineualities Re s Re α(j) = Re [ s α(j) ] s α(j) < ε, Im s Im α(j) = Im [ s α(j) ] s α(j) < ε, for all finite sets F I with F F ε, so Re α and Im α are indeed summable and moreover, we have Im α(j) = Im s. Re α(j) = Re s and (ii) (i). Assume Re α and Im α are both summable. Denote Re α(j) by u and denote Im α(j) by v. Fix some ε > 0. Choose finite sets E ε, G ε I

5 CAHPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS 67 such that u Re α(j) < ε 2, for all finite sets E I with E E ε, j E v Im α(j) < ε 2, for all finite sets G I with G G ε. j G Put F ε = E ε G ε. Suppose F I is a finite set with F F ε. Using the inclusions F E ε and F G ε, we then get u Re α(j) < ε 2 and v Im α(j) < ε 2, so we get [u + iv] α(j) = [ u Re α(j)] + i[ v Im α(j) ] u Re α(j) + v Im α(j) < ε 2 + ε 2 = ε. This proves that α is indeed summable, and α(j) = u + iv. Exercise 8. Let K be one of the fields R or C, and let I be a non-empty set. Suppose one has two non-empty sets I 1, I 2 with I = I 1 I 2 and I 1 I 2 =. Suppose α : I K has the property that both α I1 : I 1 K and α I2 : I 2 K are summable. Prove that α is summable, and α(j) = α(j) + α(j). 1 2 Proposition 2.1. Let I be a non-empty set, let K be one of the fields R or C. For a map α : I K, the following are euivalent: (i) α is summable; (ii) α is summable. Moreover, in this case one has the ineuality (1) α(j) α(j). Proof. (i) (ii). Assume α is summable. We divide the proof in two cases: Case K = R. Define the sets I + = {j I : α(j) > 0}, I = {j I : α(j) < 0}, I 0 = {j I : α(j) = 0}. More generally, for any subset F I we define F ± = F I ± and F 0 = F I 0. (2) Claim: Both maps α I + : I + R and α I : I R are summable. Moreover, one has the euality α(j) = α(j) + α(j). +

6 68 LECTURES 9-11 Denote the sum α(j) simply by s. Start by choosing some finite set F I such that s α(j) < 1, for all finite sets G I with G F. j G Let E I + be a finite subset. Then the set Ẽ = E F will be a finite subset of I with Ẽ F,, so we will have s α(j) < 1, j Ẽ so we get α(j) j E [ ] [ α(j) = α(j) + α(j) α(j) j E F + j E F + 0 F 0 F = [ α(j) ] [ ] [ ] α(j) < s + 1 α(j). j Ẽ In particular this gives { } sup α(j) : E I +, finite s + 1 [ ] α(j), j E so by Lemma??, the map α I + : I + [0, ) is indeed summable. The fact that the map α I : I (, 0] is summable is proven the exact same way. The euality (2) follows from Exercise?? Having proven the Claim, we notice now that the map α I : I [0, ) is also summable. Using Exercise??, it is clear then that the map α : I [0, ) is summable, simply because all the three maps α I + = α I +, α I = α I, and α I 0 = 0 are all summable. Case K = C. By Lemma?? we know that the maps Re α, Im α : I R are summable. In particular, using the real case, we get the fact that the maps Re α, Im α : I [0, ) are summable. Using the obvious ineuality z Re z + Im z, z C, we get α(j) Re α(j) + Im α(j) Re α(j) + Im α(j), for every finite subset F I. Then we get { } sup α(j) : F I, finite Re α(j) + Im α(j) <, so α : I [0, ) is indeed summable. Having proven the implication (i) (ii), let us prove the ineuality (1). If s denotes the sum α(j), then for every ε > 0 there exists F ε I finite such that s α(j) < ε, for all finite sets F I with F F ε. ] =

7 CAHPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS 69 In particular, we get s ε + α(j) ε + α(j) ε + α(j). ε ε Since this ineuality holds for all ε > 0, we then get s α(j). ε (ii) (i). Assume now α : I [0, ) is summable. Case K = R. It is obvious that α J : J [0, ) is summable, for any subset J I. In particular, using the notations from the proof of (i) (ii), it follows that α I + = α I +, α I = α I, and α I 0 = 0 are all summable. Then the summability of α follows from Exercise??. Case K = C. Using the ineuality max { Re z, Im z } z, z C, combined with Exercise??, it follows that both maps Re z, Im z : I [0, ) are summable. Using the real case it then follows that both maps Re α, Im α : I R are summable. Then the summability of α follows from Lemma??. The following result shows that summability is essentially the same as the summability of series. Proposition 2.2. Suppose α : I K is summable. Then the support set is at most countable. [[α]] = {j I : α(j) 0} Proof. For every integer n 1, we define the set J n = {j I : α(j) 1 n}. Since α is summable, the sets J n, n 1 are all finite. The desired result then follows from the obvious euality [[α]] = n=1 J n. We are now ready to discuss our next class of examples. Example 2.4. Let K be either R or C, let I be a non-empty set, and let p [1, ) be a real number. We define l p K (I) = { α : I K : α p : I [0, ) summable }. For α l p K (I) we define [ ] 1 α p = α(j) p p. When K = C, the space l C (I) is simply denoted by l (I). When I = N - the set of natural numbers - instead of l R (N) we simply write l R, and instead of l (N) we simply write l. In order to show that the l p spaces (1 p < ) are normed vector spaces, we will need several preliminary results. The first result we are going to need is the (classical) Hölder ineuality.

8 70 LECTURES 9-11 Exercise 9. Let > 1 and let u, v 0. Define the function f : [0, 1] R by f(t) = ut + v(1 t ) 1, t [0, 1]. Prove that max f(t) = t [0,1] (up + v p ) 1 p, where p =. Prove that, unless u = v = 0, there exists a uniue s [0, 1] such 1 that f(s) = max f(t). t [0,1] Hint: ( t Analyze the derivative: f ) 1 p (t) = u v, t (0, 1). 1 t Lemma 2.3 (Hölder s ineuality). Let a 1, a 2,..., a n, b 1, b 2,..., b n be non-negative numbers. Let p, > 1 be real number with the property 1 p + 1 = 1. Then: n ( n ) 1 ( p n ) 1 (3) a j b j. a p j Moreover, one has euality only when the seuences (a p 1,..., ap n) and (b 1,..., b n) are proportional. Proof. The proof will be carried on by induction on n. The case n = 1 is trivial. Case n = 2. Assume (b 1, b 2 ) (0, 0). (Otherwise everything is trivial). Define the number Notice that r [0, 1], and we have r = b 1 (b 1 + b 2 )1/. b j b 2 (b 1 + b 2 )1/ = (1 r ) 1/. Notice also that, upon dividing by (b 1 + b 2 )1/, the desired ineuality (4) a 1 b 1 + a 2 b 2 (a p 1 + ap 2 ) 1 p (b 1 + b 2 ) 1 reads a 1 r + a 2 (1 r ) 1/ (a p 1 + ap 2 )1/p, and it follows immediately from the exercise, applied to the function f(t) = a 1 t + a 2 (1 t ) 1/, t [0, 1]. Let us examine when euality holds. If a 1 = a 2 = 0, the euality obviosuly holds, and in this case (a 1, a 2 ) is clearly proportional to (b 1, b 2 ). Assume (a 1, a 2 ) (0, 0). Put a p/ 1 s = (a p 1 +, ap 2 )1/ and notice that (1 s ) 1/ = ( ) 1/ 1 ap 1 a p/ a p = ap 2 (a p 1 +, ap 2 )1/

9 CAHPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS 71 so we have p f(s) = a a 1+ p 2 (a p 1 + ap 2 ) 1 = ap 1 + ap 2 (a p 1 + ap 2 ) 1 = (a p 1 + ap 2 )1 1 = (a p 1 + ap 2 ) 1 p = max t [0,1] f(t). By the exercise, it follows that we have euality in (4) precisely when r = s, i.e. or euivalently Obviously this forces b 1 (b 1 + b 2 ) 1 b 1 b 1 + b 2 b 2 p a1 = (a p 1 + ap 2 ), 1 = ap 1 a p 1 +. ap 2 b 1 + = ap 2 b 2 a p 1 +, ap 2 so indeed (a p 1, ap 2 ) and (b 1, b 2 ) are proportional. Having proven the case n = 2, we now proceed with the proof of: The implication: Case n = k Case n = k + 1. Start with two seuences (a 1, a 2,..., a k, a k+1 ) and (b 1, b 2,..., a k, b k+1 ). Define the numbers ( k a = a p j ) 1 p and b = ( k b j ) 1. Using the assumption that the case n = k holds, we have k+1 ( k ) 1 ( p k ) 1 (5) a j b j + ak+1 b k+1 = ab + a k+1 b k+1. a p j Using the case n = 2 we also have ( k+1 (6) ab + a k+1 b k+1 (a p + a p k+1 ) 1 p (b + b k+1 ) 1 = b j a p j ) 1 ( p k+1 b j ) 1, so combining with (5) we see that the desired ineuality (3) holds for n = k + 1. Assume now we have euality. Then we must have euality in both (5) and in (6). On the one hand, the euality in (5) forces (a p 1, ap 2,..., ap k ) and (b 1, b 2,..., b k ) to be proportional (since we assume the case n = k). On the other hand, the euality in (6) forces (a p, a p k+1 ) and (b, b k+1 ) to be proportional (by the case n = 2). Since k k a p = a p j and b = b j, it is clear that (a p 1, ap 2,..., ap k, ap k+1 ) and (b 1, b 2,..., b k, b k+1 ) are proportional. Definition. Two numbers p, [1, ) are said to be Hölder conjugate, if 1 p + 1 = 1. Here we use the convention 1 = 0. Proposition 2.3. Let K be one of the fields R or C, let I be a non-empty set, and let p, [1, ] be two Hölder conjugate numbers. If α l p K (I) and β l K (I), then αβ l 1 K (I), and αβ 1 α p β.

10 72 LECTURES 9-11 Proof. Using Lemma??, it suffices to prove the ineuality (7) α(j)β(j) α p β, for every finite set F I. Fix for the moment a finite subset F I. Assume p, (1, ), using Hölder s ineuality we have [ ] 1 [ p (8) α(j)β(j) = α(j) β(j) α(j) p β(j) ] 1. Notice however that so we get α(j) p α(j) p = ( ) p, α p β(j) β(j) ( ), = β [ α(j) ] 1 [ p p ] 1 α p and β(j) β, so when we go back to (8) we immediately get the desired ineuality (7) In the case when p = 1, we immediately have α(j)β(j) [ α(j) ] [ max β(j) ] [ ] [ ] α(j) β(j) = α 1 β. sup The case p = is proven in the exact same way. Remark 2.2. Suppose p, [1, ] are Hölder conjugate numbers. For any α l p K (I) and β l K (I), the map αβ is summable (by Proposition??). In particular, one can define the number α, β = α(j)β(j) K. As a conseuence we get the ineuality α, β α p β, α l p K (I), β l K (I). Notations. Let K be either R or C, let I be a non-empty set, and let [1, ] be a real number. We define B K (I) = { α fink(i) : α 1 }. (remark that fin K (I) l K (I), for all [1, ].) Theorem 2.1 (Dual definition of l p spaces). Let p, (1, ) be Hölder conjugate numbers, let K be one of the fields R or C, and let I be a non-empty set. For a function α : I K, the following are euivalent: (i) α l p K (I); (ii) sup α, β <. β B K (I)

11 CAHPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS 73 Moreover, one has the euality (9) sup β B K (I) α, β = α p, α l p K (I). Proof. It will be convenient to introduce several notations. Given a function α : I K, and a finite set F I, we define the function βα F : I K, as follows: α(i) 1+ p βα F (i) = α(i) ( if i F and α(i) 0 α(j) p) 1/ 0 if i F or α(i) = 0 Notice that [[βα F ]] F, and unless βα F is identically zero, we have βα F (i) = 1. i [[βα F ]] So in any case we have βα F B K (I). Notice also that, unless βf α is identically zero, we have α, βα F = p α(i)βα F (i) = α(i) 1+ ( = α(i) p α(j) p) 1/ ( = α(j) p) 1/ (10) = ( α(i) p) 1 1 = ( α(i) p) 1/p. It is clear that the euality (10) actually holds even when βα F is identicallyzero. To make the exposition a bit clearer, we denote the uantity sup α, β β B K (I) simply by α. We now proceed with the proof of the Theorem. (i) (ii). Assume α l p K (I). In order to prove (ii) it suffices to prove the ineuality (11) α α p. Start with some arbitrary β B K (I). Using Hölder ineuality we have α, β = α(j)β(j) α(j) β(j) j [[β]] j [[β]] j [[β]] j [[β]] ( ) 1/p ( ) 1/ ( α(j) p β(j) sup F I finite [ α(i) p]) 1/p = α p. Since this ineuality holds for all β B K (I), the ineuality (11) follows. (ii) (i). Assume now α <. In order to prove condition (i) it suffices to prove that (12) α(i) p α p, for every finite subset F I. By (10) we know that for every finite subset F I we have (13) α(i) p α p = α, βα F p.

12 74 LECTURES 9-11 In particular we get the fact that α, βα F = α, βα F, and the fact that βα F belongs to B K (I), combined with (13) will give ( ) p α(i) p = α, βα F p sup α, β = α p. β B K (I) Having proven the euivalence (i) (ii), let us now observe that (9) is an immediate conseuence of (11) and (12). Exercise 10. Prove that Theorem 9.1 holds also in the cases (p, ) = (1, ) and (p, ) = (, 1). Corollary 2.1. Let K be either R or C, let I be a non-empty set, and let p 1. (i) When euipped with point-wise addition and scalar multiplication, the set l p K (I) is a K-vector space. (ii) The map l p K (I) α α p [0, ) is a norm. Proof. Let be the Hölder conjugate of p. If α l p K (I), and λ K, then λα, β = λ α, β, β fin K (I), so we get sup λα, β = λ sup α, β, β B K (I) β B K (I) which gives the fact that λα l p K (I), as well as the euality λα p = λ α p. If α 1, α 2 l p K (I), then α 1 + α 2, β = α 1, β + α 2, β, β fin K (I), so we get sup α 1 + α 2, β = sup α1, β + α 2, β β B K (I) β B K (I) ( sup α1, β + α 2, β ) β B K (I) sup α 1, β + sup α 2, β, β B K (I) β B K (I) which gives the fact that α 1 + α 2 l p K (I), as well as the ineuality α 1 + α 2 p α 1 p + α 2 p. The implication α p = 0 α = 0 is obvious. Exercise 11. Let p 1 be a real number, let K be one of the fields R or C, and let I be a non-empty set. Prove that fin K (I) is a dense linear subspace in l p K (I). Remark 2.3. Let p, [1, ] be Hölder conjugate. Then the map l p K (I) l K (I) (α, β) α, β K is bilinear, in the sense that for any γ l p K (I) and any η l K (I), the maps l p K (I) α α, η K, l K (I) β γ, β K are linear. These facts follow immediately from Exercise?? We now examine linear continuous maps between normed spaces.

13 CAHPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS 75 Proposition 2.4. Let K be either R or C, let X and Y be normed K-vector spaces, and let T : X Y be a K-linear map. The following are euivalent: (i) T is continuous; (ii) sup { T x : x X, x 1 } < ; (iii) sup { T x : x X, x = 1 } < ; (iv) T is continuous at 0. Proof. (i) (ii). Assume T is continuous, but sup { T x : x X, x 1 } <, which means there exists some seuence (x n ) n 1 X such that (a) x n 1, n 1; (b) lim n T x n =. Put z n = T x n 1 x n, n 1. On the one hand, we have z n = x n T x n 1 T x n, n 1, which gives lim n z n = 0, i.e. lim n z n = 0. Since T is assumed to be continuous, we will get (14) lim n T z n = T 0 = 0. On the other hand, since T is linear, we have T z n = T x n 1 T x n, so in particular we get T z n = 1, n 1, which clearly contradicts (14). (ii) (iii). This is obvious, since the supremum in (iii) is taken over a subset of the set used in (ii). (iii) (iv). Let (x n ) n 1 X be a seuence with lim n x n = 0. For each n 1, define x n 1 x n, if x n 0 u n = any vector of norm 1, if x n = 0 so that we have u n = 1 and x n = x n u n, n 1. Since T is linear, we have (15) T x n = x n T u n, n 1. If we define M = sup { T x : x X, x = 1 }, then T u n M, n 1, so (15) will give T x n M x n, n 1, and the condition lim n x n = 0 will force lim n T x n = 0. (iv) (i). Assume T is continuous at 0, and let us prove that T is continuous at any point. Start with some arbitrary x X and an arbitrary seuence (x n ) n 1 X with lim n x n = x. Put z n = x n x, so that lim n z n = 0. Then we will have lim n T z n = 0, which (use the linearity of T ) means that 0 = lim n T z n = lim n T x n T x,

14 76 LECTURES 9-11 thus proving that lim n T x n = T x. Remark 2.4. Using the notations above, the uantities in (ii) and (iii) are in fact eual. Indeed, if we define M 1 = sup { T x : x X, x 1 }, M 2 = sup { T x : x X, x = 1 }, then as observed during the proof, we have M 2 M 1. Conversely, if we start with some arbitrary x X with x 1, then we can always write x = x u, for some u X with u = 1. In particular we will get T x = x T u x M 2 M 2. Taking supremum in the above ineuality, over all x X with x 1, will then give the ineuality M 1 M 2. Notations. Let K be either R or C, and let X and Y be normed K-vector spaces. We define L(X, Y) = { T : X Y : T K-linear and continuous }. For T L(X, Y) we define (see the above remark) T = sup { T x : x X, x 1 } = sup { T x : x X, x = 1 } When Y = K (euipped with the absolute value as the norm), the space L(X, K) will be denoted simply by X, and will be called the topological dual of X. Proposition 2.5. Let K be either R or C, and let X and Y be normed K-vector spaces. (i) The space L(X, Y) is a K-vector space. (ii) For T L(X, Y) we have (16) T = min { C 0 : T x C x, x X }. In particular one has (17) T x T x, x X. (iii) The map L(X, Y) T T [0, ) is a norm. Proof. The fact that L(X, Y) is a vector space is clear. (ii). Assume T L(X, Y). We begin by proving (17). Start with some arbitrary x X, and write it as x = x u, for some u X with u = 1. Then by definition we have T u T, and by linearity we have T x = x T u x T. To prove the euality (16) let us define the set C T = { C 0 : T x C x, x X }. On the one hand, by (17) we know that T C T. On the other hand, if we take an arbitrary C C T, then for every u X with u = 1, we will have T u C u = C, so taking supremum, over all u with u = 1, will immediately give T C. Since we now have T C, C C T,

15 CAHPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS 77 we clearly get T = min C T. (iii). Let T, S L(X, Y). Using (17), we have (T + S)x = T x + Sx T x + Sx ( T + S ) x, x X. Then using (16) we get T + S T + S. If T L(X, Y) and λ K, then the euality (λt )x = λ T x, x X will immediately give λt = λ T. Finally if T L(X, Y) has T = 0, then using (17) one immediately gets T = 0. Notation. Let I be a non-empty set, let K be one of the fields R or C, and let p [1, ]. Let be the Hölder conjugate of p. For every element α l p K (I) we define the map θ α : l K (I) K by θ α (β) = α, β = α(i)β(i), β l K (I). We know that θ α is linear, and by Remark 9.2, we have θα (β) α p β, β l K (I), so θ α is continuous, and we have the ineuality (18) θ α α p. Proposition 2.6. Using the above notations, but assuming p (1, ], the map Θ : l p K (I) α θ α ( l K (I)) is a linear isomorphism of K-vector spaces. Moreover, Θ is isometric, in the sense that (19) Θα = α p, α l p K (I). Proof. We begin by proving (19). Since we have the inclusion it follows that {β l K (I) : β 1} B K (I), (20) θ α = sup { θα (β) : β l K (I), β 1 } sup { θα (β) : β B K (I)}. We know however (see Theorem 9 and Exercise 7) that so using (20) we get sup { θα (β) : β B K (I)} = α p, θ α α p. Combining this with (18) yields the desired euality. The fact that Θ is linear is pretty obvious. Notice now that since Θ is isometric, it is clear that Θ is injective, so the only thing we need to prove is the fact that Θ is surjective. Start with an arbitrary linear continuous map φ : l K (I) K. For every i I we define the function δ i : I K by δ i (j) = { 1 if j = i 0 if j i

16 78 LECTURES 9-11 It is clear that δ i l K (I), for all i I. (In fact δi fin K (I).) We define α : I K by α(i) = φ(δ i ), i I. Notice that, for every β fin K, we have (21) α(i)β(i) = β(i)φ(δ i ) = φ ( β(i)δ i) = φ(β), i β where β = {i I : β(i) 0}. (Since β fin K (I), the set β is finite.) Using Hölder s ineuality, the above computation shows that α, β φ β, β fin K (I). By Theorem 9.1 and Exercise 7, this proves that α l p K (I). Going back to (21) we now have θ α (β) = φ(β), β fin K (I). Since both θ α and φ are continuous, and fin K (I) is dense in l K (I) (by Exercise 10), it follows that φ = θ α. Remark 2.5. In the case p = 1, the map Θ : l 1 K(I) α θ α ( l K (I) ) is still isometric, but it is no longer surjective, unless I is finite. The explanation is the fact that when I is infinite, the subspace fin K (I) is not dense in l K (I). For example, if we take 1 l K (I) to be the constant function 1, then it is pretty obvious that 1 β 1, β fin K (I). The above euality can be immediately extended to (22) λ1 + β λ, λ K, β fin K (I). If we then consider the subspace fin K (I) = {λ1 + β : β fin K (I), λ K}, we see that the map φ 0 : fin K (I) λ1 + β λ K is linear, continuous, and has the property that (23) (24) φ 0 fink (I) = 0, φ 0(1) = 1, φ 0 (γ) γ, γ fin K (I). Using the Hahn-Banach Theorem, we can then extend φ 0 to a linear map φ : l K (I) K which will still satisfy (23) and (24), in particular we have φ ( l K (I)). Notice however that if we had φ = θ α, for some α l 1 K (I), then we must have α(i) = φ(δ i ) = 0, for all i I, so this would force φ = 0, which is impossible, since φ(1) = 1. Exercise 12. Use the notations above. For every α l 1 K (I), define σ α = θ α c K 0 (I) : ck 0 (I) K. Prove that σ α is linear and continuous. Prove that the map Σ : l 1 K(I) α σ α ( c K 0 (I) ) is an isometric linear isomorphism of K-vector spaces.