1 Background and Review Material

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1 1 Background and Review Material 1.1 Vector Spaces Let V be a nonempty set and consider the following two operations on elements of V : (x,y) x + y maps V V V (addition) (1.1) (α, x) αx maps F V V (scalar multiplication) (1.2) Here, F denotes a field, typically R or C. If the following conditions are satisfied for all α,β F and x,y V 1. x + y = y + x commutativity (1.3) 2. (x + y) + z = x + (y + z) associativity (1.4) 3. there exists z V such that x + z = y existence of zero (1.5) 4. α(βx) = (αβ)x associativity of scalar multiplication (1.6) 5. (α + β)x = αx + βx existence of negative (1.7) 6. α(x + y) = αx + αy distributivity (1.8) 7. 1x = x existence of identity (1.9) we say V is a vector space (over the field F ). In particular, if F = R we call V a real vector space, while if F = C we say V is a complex vector space. The elements x V are called vectors. You might notice that (1.5) is denoted as existence of zero. Let s see why. Claim. The condition (1.5) is equivalent to the existence of the 0 element in V. Proof Sketch: i) Assume that for every pair x,y V there exists z V such that x + z = y. In particular, for each x V there is a z V such that x + z = x. It can be shown that this element z is unique, and is the element we denote by 0. ii) Now assume that there is an element z = 0 such that for each x V x + z = x. Consequent to (1.3)-(1.9) we have the following: i) If α 0 then αx = 0 x = 0. ii) If x 0 then αx = 0 α = 0. iii) 0x = x, 1x = x 1

2 Examples You should first verify that {0},R,C are each vector spaces, then verify that the following examples are vector spaces. 1) R n = {(x 1,x 2,...,x n ) x i R for i = 1,...,n} 2) C n = {(z 1,z 2,...,z n ) z j C,j = 1,...,n} 3) {(z 1,z 2,z 1 + z 2 ) z 1,z 2 C} 4) F [x] n which is the set of polynomials in the variable x up to order n (less than or equal to n) Function Spaces Let X be a nonempty set and V a vector space. Consider the set of functions mapping X to V, F = {f : X V }. F is a vector space with (f + g)(x) = f (x) + g(x) (αf )(x) = αf (x) Note that the 0-vector in F is the function identically defined to be zero: µ(x) 0 for all x X. It is also worth noting that we can regard the vector spaces R n,c n as function spaces by the following identifications: R n = {f (1,2,...,n) R} C n = {g (1,2,...,n) C} Subspaces An important concept when dealing with vector spaces is subspaces. Let V be a vector space. We call a set M V a subspace of V if for all α,β F αx + βy M for all x,y M. That is, M is a subset of V which is closed with respect to addition. Note that a vector space is a subspace of itself, however, this is usually not of much use. Examples 2

3 Let Ω be an open subset of R n. The following are subspaces of the space of all functions from Ω to C: Remarks: C(Ω) = the space of all continuous functions from Ω to C C k (Ω) = the space of all complex-valued functions which are continuous and have continuous partial derivatives up to order k C (Ω) = the space of infinitely differentiable functions defined on Ω P (Ω) = the space of all polynomials in n-variables (x 1,x 2,...,x n ) i) M,N two subspaces of V, M N, then M is a subspace of N ii) M N will always be a subspace iii) M N not necessarily a subspace Exercise Show that in R 3, W = {x = (x 1,x 2,0) T x 1,x 2 R} is a subspace of R Linear Independence and Bases A set of elements {v 1,...,v n } V is linearly independent if any linear combination of the elements equaling 0 implies that each element is 0, i.e. if n α i v i = 0 v i = 0 for each i = 1,...,n. i=1 If the above condition does not hold, we say the vectors {v 1,...,v n } are linearly dependent. If for any positive integer n there exist n linearly independent vectors in V, we say that V is infinite dimensional. If V is not infinite dimensional, then there is a positive integer n such that: a) there is a set of n vectors that are linearly independent; b) if m > n, then any m elements are linearly dependent. In this case, we call n the dimension of V. 3

4 If A = {v 1,...,v n }, we denote the set of all finite linear combinations of vectors in A by spana = {α 1 v 1 + α 2 v α k v k α i F,k N}. If A is a subset of a vector space V, you should check that spana is a subspace of V. A set of vectors A V is called a basis of V if A is linearly independent and spana = V. If the elements {v k } n k=1 form a basis for V, then it can readily be checked that any x V can be written as n x = α k v k. 2 Normed Spaces k=1 For vector spaces, and various other spaces, it is quite useful to have some notion of distance. This section is devoted to reviewing the basics about normed spaces. A normed space (or normed linear space) is a vector space V on which there is defined a function : V F called the norm such that the following hold: a) x 0 and x = 0 if and only if x = 0 b) αx = α x c) x + y x + y. In this case, we often write (V, ) is a normed space, unless it should be apparent which norm corresponds to V. Note that a normed space is the same as a metric space with the metric given by d(x,y) = x y. Example ) We define a norm on the real vector space R n. Let x = (x 1,x 2,...,x n ) t R n, then 2 n x 2 = x i 2. (2.1) 2) More generally, for 1 p we define the p-norms for x R n by i=1 i=1 1 n p x p = x i p, for 1 p < (2.2) x = max 1 i n x i (2.3) 4

5 As an exercise, you should show that p is indeed a norm for p = 1,2,. Note that when n = 1 all of the p-norms coincide with the absolute value. It can be shown that x = lim p x p. Example 2.2. For p [1, ], we define the sequence spaces l p (the little l-p spaces) as l p = {x = (x n ) n 1 : x l p < } where the corresponding p-norms are given by 1/p x l p = x n p n=1 x l = sup n 1 for 1 p <, (2.4) x n for p =. (2.5) The l p spaces consist of all infinite sequences of complex numbers which are bounded with respect to the above norms. We can define similar norms on function spaces. We must be careful, however, as some of these notions require technical details that we do not have time to cover in this course. If any of these details become necessary, we will cover them as they arise. Example 2.3. For the space of continuous functions over the closed interval [a,b], denoted C[a,b], we define the maximum norm by f = max f (x), f a x b C[a,b]. In addition to being able to discuss the relative size of a vector or the distance between two vectors, norms allow us to discuss the convergence of sequences. Definition 2.4. Let (V, ) be a normed space. We say that a sequence (x n ) in V converges to an element x V, if for every ε > 0 there exists a number δ(ε) such that for every n δ(ε) we have x n x < ε. In this case, we may write lim n x n = x or x n x. Definition 2.5. Let 1 and 2 be norms on a vector space V. We say that the norms 1 and 2 are equivalent if there exist positive constants α,β such that for all x V. α x 1 x 2 β x 1 An important fact, which we may or may not prove, depending on time is the following. 5

6 Claim. Let V be a finite dimensional vector space. Then any two norms 1, 2 on V are equivalent. This statement is not true for an infinite dimensional space. Norms also allow us to define the concepts of continuity, open/closed sets, extreme values and a host of other notions we have from calculus/real analysis. Exercise 2.1. Define a normed space as the vector space V = C[0,1] with the norm defined by Consider the sequence Show that however ( 1 1/p v p = v(x) dx) p, 1 p < v = sup v(x). 0 x 1 0 u n (x) = { 1 nx, 0 x 1 n 0, 1 n < x 1 u n p = (n(p + 1)) 1/p, 1 p <, u n = 1. Thus, the sequence {u n } converges in the normed spaces (V, p ) for 1 p <, but divereges in the normed space (V, ). Exercise 2.2. Show that the following two norms are equivalent on C[0,1]: 2.1 Banach Spaces 1 f a = f (a) + f (x) dx f b = f (x) dx f (x) dx. Definition 2.6. A sequence {x n } in a normed space, V, is called a Cauchy sequence if lim x n x m = 0. m,n Further, if every Cauchy sequence in V is convergent to a x V, we say the normed space is complete. A complete, normed space is called a Banach space. 6

7 Example 2.7. Let P([0,1]) be the space of all polynomials on [0,1] with the norm of uniform convergence p = max p(x). Define [0,1] p n (x) = 1 + x + x2 xn + + 2! n! for n = 1,2,... Then {p n } P([0,1]) is a Cauchy sequence. Does {p n } converge to an element of P([0,1])? Example 2.8 (Examples of Banach spaces). 1) The spaces of square-summable sequences, l 2, with norm is a Banach space. 1/2 x l 2 = x n 2 2) The space of continuous functions on [a,b], C([a,b]), with norm is a Banach space. n=1 f = max f (x) [a,b] 3) Bypassing some formalities (at our own risk), we may define the big L-p spaces, L p (R). For real p > 1, we denote the space of complex-valued (locally integrable) functions f such that f p is integrable ( f p L 1 (R)) by L p (R). Analogous to the spaces l p, we define a norm on L p (R) by f p = ( f p ) 1/p. Formally, the integral should be understood in the sense of Lebesgue, which is a detail we will not have time to cover. However, if we are lucky, the Lebesgue integral will coincide with the Riemann integral. 2.2 Operators on normed spaces Given two spaces V,W, an operator L from V W is a mapping which assigns to each element of V a unique element in W. The domain of L, D(L), is the following subset of V D(L) = {v V : L(v) exists }, 7

8 while the range of L is the following subset of W R(L) = {w W : w = L(v) for some v V }. Another important set related to an operator, is the nullspace N (L) = {v V : L(v) = 0}. Given two operators L and M from V W, we define the addition of operators by (L + M)(v) = L(v) + M(v). This defines a new operator, L + M, with D(L + M) = D(L) D(M). Likewise, for α K we define scalar multiplication of an operator by (αl)(v) = αl(v). Again, this defines a new operator αl : V W with domain D(L). Definition 2.9. An operator L : V W is one-to-one (or 1-1) if L(v 1 ) = L(v 2 ) v 1 = v 2, v 1,v 2 D(L). An operator which is 1-1 is also said to be injective. If R(L) = W, we say L is surjective. Furthermore, if L is both injective and surjective, we say L is bijective. Example Let V = C[0,1] = W be the vector space of continuous functions on [0,1]. Define the differentiation operator by d dx f = f, f C[0,1]. In this case, we have D( d dx ) = C1 [0,1] C[0,1]. Is d dx an injection and/or surjection? For operators on normed spaces, properties such as continuity and boundedness are defined in terms of vectors and sequences of vectors in the domain of the operator. Definition Let V, W be two normed spaces. An operator L : V W is continuous at v D(L) V if for a sequence {v n } D(L) we have v n v in V L(v n ) L(v) in W. We say that L is continuous if it is continuous for all v D(L). The operator L is said to be bounded if for any γ > 0, there is a R > 0 such that for v D(L) we have v γ L(v) R. 8

9 Exercise 2.3. Let V = C[0,1] be the space of complex-valued continuous functions defined on the closed interval [0,1]. Define an operator L by L(v)(t) = v(0) + t 0 v(s) ds. What is the domain of L? How about the range? Show that the function v(t) = ae t is a fixed point of L for any a C. Analogous to functions, a fixed point of an operator L is an element of D(L) such that L(v) = v. Definition If L : V W satisfies the following properties: L(αv 1 + βv 2 ) = αl(v 1 ) + βl(v 2 ) for all v 1,v 2 V, α,β K we say that L is linear. We denote the set of all linear operators mapping V to W by L(V,W ) and by L(V ) whenever V = W. In this case, we often use the notation Lv in place of L(v). Proposition Suppose L is a linear operator from a vector space V to a vector space W. Then a) the continuity of L over the whole space V is equivalent to continuity at any one point (usually taken to be v = 0.) b) L is bounded if an only if there exists a constant α 0 such that Lv W α v V for all v V. c) L is continuous on V if and only if L is bounded on V. As a consequence of Proposition 2.13, we define a norm on L(V,W ) by Lv L V,W = sup W. (2.6) v 0 v V This norm is usually referred to as the operator norm. The following theorem is obtained as a consequence. Theorem The set L(V,W ) with the norm (2.6) is a normed space. Proof. Exercise. In fact, as we will discover soon, under certain conditions the space L(V,W ) is a Banach space. The following property of the operator norm can sometimes be useful Lv W L V,W v V for all v V. 9

10 Example For a normed space V, the identity operator I : V V, defined by is an element of L(V ) and L = 1. Iv = v, Example Let V = W = C[a,b] with the norm. Let k C([a,b] [a,b]) and define an operator K : C[a,b] C[a,b] by (Kv)(x) = b a k(x, y)v(y) dy. The operator K is an example of a linear integral operator, which we will be studying this semester. The function k(, ) is called the kernel (or kernel function) of the integral operator. Definition A mapping f from a subset B of a normed space V into V is called a contraction (or contraction mapping) if there exists a positive number α < 1 such that f (u) f (v) α u v for all u,v B. The following theorem, sometimes referred to as the Banach Lemma, is quite powerful for a number of reasons. The result is useful for error analysis, approximating a problem by a nearby problem, and can be useful for computing the inverse of (I L). Theorem 2.18 (Banach Lemma/Geomtric Series Theorem). Let V be a Banach space and suppose L L(V ). If L < 1 (2.7) then the operator I L is a bijection on V, its inverse is a bounded linear operator defined by and its norm satisfies (I L) 1 = (I L) 1 L n, (2.8) n=0 1 1 L. (2.9) 10