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Normed Vector Spaces Chapter II Elements of Functional Analysis Functional Analysis is generally understood a “linear algebra for infinite di- mensional vector spaces.” Most of the vector spaces that are used are spaces of (various types of) functions, therfeore the name “functional.” This chapter in- troduces the reader to some very basic results in Functional Analysis. The more motivated reader is suggested to continue with more in depth texts. 1. 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 3 x 7−→ kxk ∈ [0, ∞) with the following properties (i) kx + yk ≤ kxk + kyk, ∀ x, y ∈ X; (ii) kλxk = |λ| · kxk, ∀ x ∈ X, λ ∈ K; (iii) kxk = 0 =⇒ x = 0. (Note that conditions (i) and (ii) state that k . k is a seminorm.) Examples 1.1. Let K be either R or C, and let J be some non-empty set. A. Define `∞(J) = α : J → : sup |α(j)| < ∞ . K K j∈J We equip the space `∞(J) with the -vector space structure defined by point-wise K K addition and point-wise scalar multiplication. This vector space carries a natural norm k . k∞, defined by kαk = sup |α(j)|, α ∈ `∞(I). ∞ K j∈J When = , the space `∞(J) is simply denoted by `∞(J). When J = - the set K C C N of natural numbers - instead of `∞( ) we simply write `∞, and instead of `∞( ) R N R N we simply write `∞. B. Consider the space K c0 (J) = α : J → K : inf sup |α(j)| = 0 . F ⊂J j∈J F finite r K Remark that, if we equip J with the discrete topology, then c0 (J) is the space of those (automatically continuous) functions : J → K which have limit 0 at ∞. In particular, cK(J) is a linear subspace of `∞(J). We also equip cK(J) with the norm 0 K 0 k . k∞. As above, when K = C, we ommit it from the notation, and when J = N, we also ommit it. 1 2 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS Definition. Suppose X is a normed vector space, with norm k . k. Then there is a natural metric d on X, defined by d(x, y) = kx − yk, x, y ∈ X. The toplogy on X, defined by this metric, is called the norm topology. Exercise 1 ♦. Let X be a normed vector space, over K(= R, C). Prove that, when equipped with the norm toplogy, X becomes a topological vector space. That is, the maps X × X 3 (x, y) 7−→ x + y ∈ X K × X 3 (λ, x) 7−→ λx ∈ X are continuous. Exercise 2 ♦. Let K be one of the fields R or C, and let J be a non-empty set. A. Prove that cK(J) is a closed linear subspace in `∞(J). 0 K B. Recall (see I.2) that for a function α : J → K we denoted by [[α]] the support of α, defined as [[α]] = {j ∈ J : α(j) 6= 0}. Consider the space fin (J) = α : J → : [[α]] finite . K K Prove that fin (J) is a linear subspace in cK(J), which is dense in the K 0 norm topology. The next example is based on the notion of summability discussed in I.2. Example 1.2. Let K be either R or C, let J be a non-empty set, and let p ∈ [1, ∞) be a real number. We define `p (J) = α : J → : |α|p : J → [0, ∞) summable . K K Remark that fin (I) ⊂ `q (J). K K For α ∈ `p (J) we define K 1 p X p kαkp = |α(j)| . j∈J When = , the space `p (J) is simply denoted by `p(J). When J = - K C C N the set of natural numbers - instead of `p ( ) we simply write `p , and instead of R N R `p(N) we simply write `p. The proof of the fact that the `p spaces (1 ≤ p < ∞) are normed vector spaces is included in the following exercises. To make the formulation a little easier, we use the following terminology. Definition. Two “numbers” p, q ∈ [1, ∞] are said to be H¨olderconjugate, if 1 1 1 p + q = 1. Here we use the convention ∞ = 0. Exercise 3 ♦. Let K be one of the fields R or C, let J be a non-empty set, and let p, q ∈ [1, ∞] be two H¨older conjugate “numbers.” Consider the set q B (J) = β ∈ fin (J): kβkq ≤ 1 , K K §1. Normed vector spaces 3 and define, for functions α : J → and β ∈ fin (J), the quantity K K X hα, βi = α(j)β(j). j∈[[β]] Prove that, for α : J → K, the following are equivalent: (i) α ∈ `p (J); K (ii) sup |hα, βi| < ∞. β∈Bq (J) K Moreover, one has the equality p sup |hα, βi| = kαkp, ∀ α ∈ ` (J). K β∈Bq (J) K Hint: Use H¨older inequality (Appendix D). Exercise 4 ♦. Let K be either R or C, let J be a non-empty set, and let p ∈ [1, ∞). Prove that, when equipped with pointwise addition and scalar mul- p tiplication, then (` (J), k .kp) is a normed vector space. K Hint: Use the preceding exercise. Exercise 5 ♦. Let K be one of the fields R or C, let J be a non-empty set, and let p, q ∈ [1, ∞] be two H¨older conjugate “numbers.” A. If α ∈ `p (J) and β ∈ `q (J), then αβ ∈ `1 (J), and K K K kαβk1 ≤ kαkp · kβkq. B. Using part A, extend the pairing constructed in Exercise 3 to a bilinear map h .,. i : `p (J) × `q (J) → , defined by K K K X hα, βi = α(j)β(j), ∀ α ∈ `p (J), β ∈ `q (J). K K j∈J C. Prove that, for every α ∈ `p (J), one has the equality K q kαkp = sup {|hα, βi| : β ∈ ` (J), kβkq ≤ 1} . K Exercise 6 ♦. Let K be one of the fields R or C, and let J be a non-empty set. A. Prove that fin (J) is a dense linear subspace in `p (J), for every p ∈ [1, ∞). K K B. Why was the case p = ∞ ommitted? We now examine linear continuous maps between normed spaces. Proposition 1.1. 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 equivalent: (i) T is continuous; (ii) sup kT xk : x ∈ X, kxk ≤ 1 < ∞; (iii) sup kT xk : x ∈ X, kxk = 1 < ∞; (iv) T is continuous at 0. Proof. (i) ⇒ (ii). Assume T is continuous, but sup kT xk : x ∈ X, kxk ≤ 1 < ∞, which means there exists some sequence (xn)n≥1 ⊂ X such that (a) kxnk ≤ 1, ∀ n ≥ 1; (b) limn→∞ kT xnk = ∞. 4 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS Put −1 zn = kT xnk xn, ∀ n ≥ 1. On the one hand, we have kxnk 1 kznk = ≤ , ∀ n ≥ 1, kT xnk kT xnk which gives limn→∞ kznk = 0, i.e. limn→∞ zn = 0. Since T is assumed to be continuous, we will get (1) lim T zn = T 0 = 0. n→∞ −1 On the other hand, since T is linear, we have T zn = kT xnk T xn, so in particular we get kT znk = 1, ∀ n ≥ 1, which clearly contradicts (1). (ii) ⇒ (iii). This is obvious, since the supremum in (iii) is taken over a subset of the set used in (ii). (iii) ⇒ (iv). Let (xn)n≥1 ⊂ X be a sequence with limn→∞ xn = 0. For each n ≥ 1, define −1 kxnk xn, if xn 6= 0 un = any vector of norm 1, if xn = 0 so that we have kunk = 1 and xn = kxnkun, ∀ n ≥ 1. Since T is linear, we have (2) T xn = kxnkT un, ∀ n ≥ 1. If we define M = sup kT xk : x ∈ X, kxk = 1 , then kT unk ≤ M, ∀ n ≥ 1, so (2) will give kT xnk ≤ M · kxnk, ∀ n ≥ 1, and the condition limn→∞ xn = 0 will force limn→∞ T xn = 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 sequence (xn)n≥1 ⊂ X with limn→∞ xn = x. Put zn = xn − x, so that limn→∞ zn = 0. Then we will have limn→∞ T zn = 0, which (use the linearity of T ) means that 0 = lim kT znk = lim kT xn − T xk, n→∞ n→∞ thus proving that limn→∞ T xn = T x. Remark 1.1. Using the notations above, the quantities in (ii) and (iii) are in fact equal. Indeed, if we define M1 = sup kT xk : x ∈ X, kxk ≤ 1 , M2 = sup kT xk : x ∈ X, kxk = 1 , then as observed during the proof, we have M2 ≤ M1. Conversely, if we start with some arbitrary x ∈ X with kxk ≤ 1, then we can always write x = kxku, for some u ∈ X with kuk = 1. In particular we will get kT xk = kxk · kT uk ≤ kxk · M2 ≤ M2. §1. Normed vector spaces 5 Taking supremum in the above inequality, over all x ∈ X with kxk ≤ 1, will then give the inequality M1 ≤ M2. 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) kT k = sup kT xk : x ∈ X, kxk ≤ 1 = sup kT xk : x ∈ X, kxk = 1 When Y = K (equipped 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.
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