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Functional Analysis Functional Analysis Lyudmila Turowska Department of Mathematical Sciences Chalmers University of Technology and the University of Gothenburg These lectures concern the theory of normed spaces and bounded linear oper- ators between normed spaces. We shall define these terms and study their prop- erties. You can consult the following books. 1. G. Folland: Real Analysis. Modern Techniques and their Applications, John Wiley & Sons, 1999, Chapters 5-7 and parts of Chapter 4. 2 1 Vector Spaces Definition 1.1. A complex vector space (or a vector space over the field C) (or a complex linear space) is a set V with addition V × V ! V; (x; y) 7! x + y the sum of x and y; and scalar multiplication C × V ! V; (λ, x) 7! λx the product of λ and x; satisfying the following rules: x + (y + z) = (x + y) + z 8x; y; z 2 V; x + y = y + x 8x; y 2 V; 90 2 V : 0 + x = x 8x 2 V; 8x 2 V 9 − x 2 V : x + (−x) = 0; λ(x + y) = λx + λy 8x; y 2 V; 8λ 2 C; (λ + µ)x = λx + µx 8x 2 V; 8λ, µ 2 C; λ(µx) = (λµ)x 8x 2 V; 8λ, µ 2 C; 1x = x 8x 2 V: It is easy to see that the usual arithmetic holds in V . Remark 1.2. Similarly, we can define a real vector space by replacing C with R in the definition. n Examples 1.3. 1. C = f(x1; : : : ; xn): xi 2 C 8ig, the complex vector space of n-tuples of complex numbers with coordinatewise addition n (x1; : : : ; xn) + (y1; : : : ; yn) = (x1 + y1; : : : ; xn + yn); (x1; : : : ; xn); (y1; : : : ; yn) 2 C and coordinatewise scalar multiplication n λ(x1; : : : ; xn) = (λx1; : : : ; λxn); λ 2 C; (x1; : : : ; xn) 2 C : 2. An infinite-dimensional complex vector space: N C = f(x1; : : : ; xn;:::): xi 2 C 8ig 3 is the complex vector space of sequences of complex numbers with coordinatewise addition and scalar multiplication: x + y = (x1 + y1; : : : ; xn + yn;:::); λx = (λx1; : : : ; λxn;:::); N x = (x1; : : : ; xn;:::); y = (y1; : : : ; yn;:::) 2 C ; λ 2 C: 3. The complex vector space of all continuous complex-valued functions on [0; 1] C[0; 1] = ff : [0; 1] ! C : f is continuous on [0; 1]g with pointwise addition (f + g)(t) = f(t) + g(t); f; g 2 C[0; 1]; t 2 [0; 1]; and scalar multiplication (λf)(t) = λf(t); f 2 C[0; 1]; λ 2 C; t 2 [0; 1]: These operations are well-defined due to the following fact (from the basic analysis course): If f; g are continuous functions on [0; 1] and λ 2 C then (f + g) and λf are continuous on [0; 1]. C f n 2 C g × 4. Mn( ) = (aij)i;j=1 : aij ; i; j = 1; : : : ; n , the complex vector space of n n matrices with complex entries with addition n n n (aij)i;j=1 + (bij)i;j=1 = (aij + bij)i;j=1; and scalar multiplication n n λ(aij)i;j=1 = (λaij)i;j=1: (So we have addition and scalar multiplication by entries.) 1.1 Vector Subspaces Definition 1.4. A non-empty subset of W of a vector space V is called a vector subspace (or a linear subspace) if 1. for every x; y 2 W; x + y 2 W ; 2. for every x 2 W and every λ 2 C (or R), λx 2 W . Remark 1.5. A vector subspace W is itself a vector space with respect to the addition and scalar multiplication defined on V . 4 Question 1.6. Let D be the closed unit disc, that is D = fz 2 C : jzj ≤ 1g: It is obvious that D is a subset of C. Is D a vector subspace of C? No. Conditions (1) and (2) of the definition of a vector subspace are not satisfied: 1. For z1 = z2 = 1, we have z1 + z2 = 2, so jz1 + z2j = 2. Thus z1 + z2 2= D. 2. For z = i, we have jλzj = jλj for every λ 2 C. Thus λz2 = D when λ is such that jλj > 1. R f 2 1 g Example 1.7. Show that W = f C[0; 1] : 0 f(x) dx = 0 is a linear subspace of C[0; 1]. Proof. 1) It is obvious that W is a subset of C[0; 1], i.e., W ⊂ C[0; 1]. − R2) W is a non-emptyR subset. For example, f0(t) = 2t 1 belongs to W , since 1 1 − t2 − j1 0 f0(t) dt = 0 2t 1 dt = (2 2 t) 0 = 0. 3) For every f; g 2 W , we have Z Z 1 1 (f + g)(t) dt = f(t) + g(t) dt 0 Z0 Z 1 1 = f(t) dt + g(t) dt (by the linearity of integrals) 0 0 = 0 + 0 = 0: Thus f + g 2 W . 4) For every f 2 W and λ 2 C, we have Z Z 1 1 (λf)(t) dt = λf(t) dt 0 0Z 1 = λ f(t) dt (by the linearity of integrals) 0 = 0: Thus λf 2 W . Hence W is a vector subspace of C[0; 1]. 5 2 1 Example 1.8. Let ` be the set of all complex sequences x = (xn)n=1 which are square summable, i.e. satisfy X1 2 jxnj < 1; n=1 we write X1 2 2 ` = f(x1; : : : ; xn;:::): xi 2 C 8i and jxnj < 1g: n=1 2 N N Show that ` is a vector subspace of C . Recall that C = f(x1; : : : ; xn;:::): xi 2 C 8ig with x + y = (x1 + y1; : : : ; xn + yn;:::) and λx = (λx1; : : : ; λxn;:::). Proof. 1). It is obvious that `2 is a subset of CN. P 1 1 1 2 2 1 1 2). The sequence x = (1; 2 ; 3 ;:::; n ;:::) ` since the series n=1 n2 converges. Thus `2 is non-empty. 1 1 2 2 3). We have to prove that, for any x = (xn)n=1; y = (yn)n=1 ` , x + y = 2 P(x1 + y1; : : : ; xn + yn;:::) belongs to ` . Thus we have to show that the series 1 j j2 n=1 xn + yn converges. We will prove this in Theorem 1.11. 1 2 2 2 C 4). For any x = (xn)n=1 ` and λ we have λx = (λx1; : : : ; λxn;:::). Note that X1 X1 2 2 2 jλxnj = jλj jxnj n=1 n=1 X1 2 2 = jλj jxnj < 1; n=1 so λx 2 `2. Hence `2 is a vector subspace of CN, and `2 is a complex vector space. The Cauchy-Schwarz Inequality 1.9. Xn Xn Xn 2 1 2 1 ≤ 2 2 akbk ( ak) ( bk) k=1 k=1 k=1 for all ak; bk 2 R; ak; bk ≥ 0; k = 1; 2; : : : ; n; n 2 N. 6 It follows from ! 2 Xn Xn Xn 1 Xn Xn a b = a2 b2 − (a b − b a )2: k k k k 2 i j i j k=1 k=1 k=1 i=1 j=1 P 2 2 1 Lemma 1.10. For any x; y ` , the series n=1 xnyn converges absolutely and ( ) 1 ( ) 1 X1 X1 2 X1 2 2 2 jxnynj ≤ jxnj jynj : n=1 n=1 n=1 Proof. Using the Cauchy-Schwarz inequality, for every k 2 N, we have Xk Xk jxnynj = jxnjjynj n=1 n=1 ( ) 1 ( ) 1 Xk 2 Xk 2 2 2 ≤ jxnj jynj by the Cauchy-Schwarz inequality n=1 n=1 ( ) 1 ( ) 1 X1 2 X1 2 2 2 ≤ jxnj jynj : n=1 n=1 The latter expression is a finite number independent of k, and so the series X1 jxnynj n=1 converges and ( ) 1 ( ) 1 X1 X1 2 X1 2 2 2 jxnynj ≤ jxnj jynj : n=1 n=1 n=1 Theorem 1.11. For any x; y 2 `2, 0 1 ( ) 1 ( ) 1 2 X1 X1 2 X1 2 2 @ 2 2 A jxn + ynj ≤ jxnj + jynj n=1 n=1 n=1 and x + y 2 `2. 7 Proof. For every n 2 N, 2 jxn + ynj = (xn + yn)(xn + yn) = xnxn + ynxn + xnyn + ynyn) 2 2 = jxnj + xnyn + xnyn + jynj 2 2 ≤ jxnj + 2jxnynj + jynj : Hence, for every k 2 N, Xk Xk 2 2 2 jxn + ynj ≤ (jxnj + 2jxnynj + jynj ) n=1 n=1 Xk Xk Xk 2 2 = jxnj + 2 jxnynj + jynj n=1 n=1 n=1 X1 X1 X1 2 2 ≤ jxnj + 2 jxnynj + jynj n=1 n=1 n=1 ! 1 ! 1 X1 X1 2 X1 2 X1 2 2 2 2 ≤ jxnj + 2 jxnj jynj + jynj 0n=1 n=1 n=11 n=1 ( ) 1 ( ) 1 2 X1 2 X1 2 @ 2 2 A = jxnj + jynj : n=1 n=1 The latter expression is a finite number independent of k, so the series X1 2 jxn + ynj n=1 converges and 0 1 ( ) 1 ( ) 1 2 X1 X1 2 X1 2 2 @ 2 2 A jxn + ynj ≤ jxnj + jynj : n=1 n=1 n=1 The theorem is proved. 8 2 Normed Spaces Suppose we have a vector space V , the generalisation of Rn, and we want to know about size of vectors x 2 V , and about the distance between two vectors x; y 2 V . Suppose we would also like to estimate the distance from a vector x 2 V to a vector subspace W ⊂ V . To deal with this kind of problem we define a norm on V .
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