FUNCTIONAL ANALYSIS NOTES (2011)


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1 FUNCTIONAL ANALYSIS NOTES (0) Mr. Andrew Pinchuck epartment of Mathematics (Pure & Applied) Rhodes University
2 Contents Introduction Linear Spaces. Introducton Subsets of a linear space Subspaces and Convex Sets Quotient Space irect Sums and Projections The Hölder and Minkowski Inequalities Normed Linear Spaces 3. Preliminaries Quotient Norm and Quotient Map Completeness of Normed Linear Spaces Series in Normed Linear Spaces Bounded, Totally Bounded, and Compact Subsets of a Normed Linear Space Finite imensional Normed Linear Spaces Separable Spaces and Schauder Bases Hilbert Spaces Introduction Completeness of Inner Product Spaces Orthogonality Best Approximation in Hilbert Spaces Orthonormal Sets and Orthonormal Bases Bounded Linear Operators and Functionals 6 4. Introduction Examples of ual Spaces The ual Space of a Hilbert Space The HahnBanach Theorem and its Consequences 8 5. Introduction Consequences of the HahnBanach Extension Theorem Bidual of a normed linear space and Reflexivity The Adjoint Operator Weak Topologies
3 6 Baire s Category Theorem and its Applications Introduction Uniform Boundedness Principle The Open Mapping Theorem Closed Graph Theorem
4 Introduction These course notes are adapted from the original course notes written by Prof. Sizwe Mabizela when he last gave this course in 006 to whom I am indebted. I thus make no claims of originality but have made several changes throughout. In particular, I have attempted to motivate these results in terms of applications in science and in other important branches of mathematics. Functional analysis is the branch of mathematics, specifically of analysis, concerned with the study of vector spaces and operators acting on them. It is essentially where linear algebra meets analysis. That is, an important part of functional analysis is the study of vector spaces endowed with topological structure. Functional analysis arose in the study of tansformations of functions, such as the Fourier transform, and in the study of differential and integral equations. The founding and early development of functional analysis is largely due to a group of Polish mathematicians around Stefan Banach in the first half of the 0th century but continues to be an area of intensive research to this day. Functional analysis has its main applications in differential equations, probability theory, quantum mechanics and measure theory amongst other areas and can best be viewed as a powerful collection of tools that have far reaching consequences. As a prerequisite for this course, the reader must be familiar with linear algebra up to the level of a standard second year university course and be familiar with real analysis. The aim of this course is to introduce the student to the key ideas of functional analysis. It should be remembered however that we only scratch the surface of this vast area in this course. We examine normed linear spaces, Hilbert spaces, bounded linear operators, dual spaces and the most famous and important results in functional analysis such as the HahnBanach theorem, Baires category theorem, the uniform boundedness principle, the open mapping theorem and the closed graph theorem. We attempt to give justifications and motivations for the ideas developed as we go along. Throughout the notes, you will notice that there are exercises and it is up to the student to work through these. In certain cases, there are statements made without justification and once again it is up to the student to rigourously verify these results. For further reading on these topics the reader is referred to the following texts: G. BACHMAN, L. NARICI, Functional Analysis, Academic Press, N.Y E. KREYSZIG, Introductory Functional Analysis, John Wiley & sons, New YorkChichesterBrisbane Toronto, 978. G. F. SIMMONS, Introduction to topology and modern analysis, McGrawHill Book Company, Singapore, 963. A. E. TAYLOR, Introduction to Functional Analysis, John Wiley & Sons, N. Y I have also found Wikipedia to be quite useful as a general reference.
5 Chapter Linear Spaces. Introducton In this first chapter we review the important notions associated with vector spaces. We also state and prove some well known inequalities that will have important consequences in the following chapter. Unless otherwise stated, we shall denote by R the field of real numbers and by C the field of complex numbers. Let F denote either R or C... efinition A linear space over a field F is a nonempty set X with two operations satisfying the following properties: [] x C y X whenever x; y X ; [] x C y y C x for all x; y X ; C W X X! X (called addition); and W F X! X (called multiplication) [3] There exists a unique element in X, denoted by 0, such that x C 0 0 C x x for all x X ; [4] Associated with each x X is a unique element in X, denoted by x, such that x C. x/ x C x 0; [5].x C y/ C z x C.y C z/ for all x; y; z X ; [6] x X for all x X and for all F; [7].x C y/ x C y for all x; y X and all F; [8]. C ˇ/ x x C ˇ x for all x X and all ; ˇ F; [9]. ˇ/ x.ˇ x/ for all x X and all ; ˇ F; [0] x x for all x X. We emphasize that a linear space is a quadruple.x; F; C; / where X is the underlying set, F a field, C addition, and multiplication. When no confusion can arise we shall identify the linear space.x; F; C; / with the underlying set X. To show that X is a linear space, it suffices to show that it is closed under addition and scalar multiplication operations. Once this has been shown, it is easy to show that all the other axioms hold.
6 .. efinition A real (resp. complex) linear space is a linear space over the real (resp. complex) field. A linear space is also called a vector space and its elements are called vectors...3 Examples [] For a fixed positive integer n, let X F n fx.x ; x ; : : : ; x n / W x i F; i ; ; : : : ; ng the set of all ntuples of real or complex numbers. efine the operations of addition and scalar multiplication pointwise as follows: For all x.x ; x ; : : : ; x n /; y.y ; y ; : : : ; y n / in F n and F, Then F n is a linear space over F. x C y.x C y ; x C y ; : : : ; x n C y n / x. x ; x ; : : : ; x n /: [] Let X CŒa; b f x W Œa; b! F j x is continuous g. efine the operations of addition and scalar multiplication pointwise: For all x; y X and all R, define.x C y/.t/ x.t/ C y.t/ and for all t Œa; b:. x/.t/ x.t/ Then CŒa; b is a real vector space. Sequence Spaces: Informally, a sequence in X is a list of numbers indexed by N. Equivalently, a sequence in X is a function x W N! X given by n 7! x.n/ x n. We shall denote a sequence x ; x ; : : : by x.x ; x ; : : :/.x n / : [3] The sequence space s. Let s denote the set of all sequences x.x n / of real or complex numbers. efine the operations of addition and scalar multiplication pointwise: For all x.x ; x ; : : :/, y.y ; y ; : : :/ s and all F, define Then s is a linear space over F. x C y.x C y ; x C y ; : : :/ x. x ; x ; : : :/: [4] The sequence space `. Let ` `.N/ denote the set of all bounded sequences of real or complex numbers. That is, all sequences x.x n / such that sup jx i j < : in efine the operations of addition and scalar multiplication pointwise as in example (3). Then ` is a linear space over F. [5] The sequence space `p `p.n/; p <. Let `p denote the set of all sequences x.x n / of real or complex numbers satisfying the condition jx i j p < : efine the operations of addition and scalar multiplication pointwise: For all x.x n /, y.y n / in `p and all F, define x C y.x C y ; x C y ; : : :/ x. x ; x ; : : :/: 3
7 Then `p is a linear space over F. Proof. Let x.x ; x ; : : :/, y.y ; y ; : : :/ `p. We must show that x C y `p. Since, for each i N, it follows that jx i C y i j p Œ maxfjx i j; jy i jg p p maxfjx i j p ; jy i j p g p.jx i j p C jy i j p / ; jx i C y i j p p X jx i j p C Thus, x C y `p. Also, if x.x n / `p and F, then That is, x `p.! jy i j p < : X j x i j p j j p jx i j p < : [6] The sequence space c c.n/. Let c denote the set of all convergent sequences x.x n / of real or complex numbers. That is, c is the set of all sequences x.x n / such that lim x n n! exists. efine the operations of addition and scalar multiplication pointwise as in example (3). Then c is a linear space over F. [7] The sequence space c 0 c 0.N/. Let c 0 denote the set of all sequences x.x n / of real or complex numbers which converge to zero. That is, c 0 is the space of all sequences x.x n / such that lim x n 0. efine the operations of addition and scalar multiplication n! pointwise as in example (3). Then c 0 is a linear space over F. [8] The sequence space `0 `0.N/. Let `0 denote the set of all sequences x.x n / of real or complex numbers such that x i 0 for all but finitely many indices i. efine the operations of addition and scalar multiplication pointwise as in example (3). Then `0 is a linear space over F. 4
8 . Subsets of a linear space Let X be a linear space over F; x X and A and B subsets of X and F. We shall denote by x C A W fx C a W a Ag; A C B W fa C b W a A; b Bg; A W fa W a Ag:.3 Subspaces and Convex Sets.3. efinition A subset M of a linear space X is called a linear subspace of X if (a) x C y M for all x; y M, and (b) x M for all x M and for all F. Clearly, a subset M of a linear space X is a linear subspace if and only if M C M M and M M for all F..3. Examples [] Every linear space X has at least two distinguished subspaces: M f0g and M X. These are called the improper subspaces of X. All other subspaces of X are called the proper subspaces. [] Let X R. Then the nontrivial linear subspaces of X are straight lines through the origin. [3] M fx.0; x ; x 3 ; : : : ; x n / W x i R; i ; 3; : : :; ng is a subspace of R n. [4] M fx W Œ ;! R; x continuous and x.0/ 0g is a subspace of CŒ ;. [5] M fx W Œ ;! R; x continuous and x.0/ g is not a subspace of CŒ ;. [6] Show that c 0 is a subspace of c..3.3 efinition Let K be a subset of a linear space X. The linear hull of K, denoted by lin.k/ or span.k/, is the intersection of all linear subspaces of X that contain K. The linear hull of K is also called the linear subspace of X spanned (or generated) by K. It is easy to check that the intersection of a collection of linear subspaces of X is a linear subspace of X. It therefore follows that the linear hull of a subset K of a linear space X is again a linear subspace of X. In fact, the linear hull of a subset K of a linear space X is the smallest linear subspace of X which contains K..3.4 Proposition Let K be a subset of a linear space X. Then the linear hull of K is the set of all finite linear combinations of elements of K. That is, 8 9 < = lin.k/ j x j j x ; x ; : : : ; x n K; ; ; : : : ; n F; n N : ; : 5
9 Proof. Exercise..3.5 efinition [] A subset fx ; x ; : : : ; x n g of a linear space X is said to be linearly independent if the equation x C x C C nx n 0 only has the trivial solution n 0. Otherwise, the set fx ; x ; : : : ; x n g is linearly dependent. [] A subset K of a linear space X is said to be linearly independent if every finite subset fx ; x ; : : : ; x n g of K is linearly independent..3.6 efinition If fx ; x ; : : : ; x n g is a linearly independent subset of X and X linfx ; x ; : : : ; x n g, then X is said to have dimension n. In this case we say that fx ; x ; : : : ; x n g is a basis for the linear space X. If a linear space X does not have a finite basis, we say that it is infinitedimensional..3.7 Examples [] The space R n has dimension n. Its standard basis is fe ; e ; : : : ; e n g, where, for each j ; ; : : : ; n, e j is an ntuple of real numbers with in the jth position and zeroes elsewhere; i.e., e j.0; 0; : : : ; ; 0; : : : ; 0/; where occurs in the jth position. [] The space P n of polynomials of degree at most n has dimension n C. Its standard basis is f; t; t ; : : : ; t n g. [3] The function space CŒa; b is infinitedimensional. [4] The spaces `p, with p, are infinitedimensional..3.8 efinition Let K be a subset of a linear space X. We say that (a) K is convex if x C. /y K whenever x; y K and Œ0; ; (b) K is balanced if x K whenever x K and jj ; (c) K is absolutely convex if K is convex and balanced..3.9 Remark [] It is easy to verify that K is absolutely convex if and only if x C y K whenever x; y K and jj C jj. [] Every linear subspace is absolutely convex..3.0 efinition Let S be a subset of the linear space X. The convex hull of S, denoted co.s/, is the intersection of all convex sets in X which contain S. Since the intersection of convex sets is convex, it follows that co.s/ is the smallest convex set which contains S. The following result is an alternate characterization of co.s/. 6
10 .3. Proposition Let S be a nonempty subset of a linear space X. Then co.s/ is the set of all convex combinations of elements of S. That is, 8 9 < = co.s/ j x j j x ; x ; : : : ; x n S; j 0 8 j ; ; : : : ; n; j ; n N : ; : Proof. Let C denote the set of all convex combinations of elements of S. That is, 8 9 < = C j x j j x ; x ; : : : ; x n S; j 0 8 j ; ; : : : ; n; j ; n N : ; : mx Let x; y C and 0. Then x i x i ; y i y i, where i ; i 0, mx i, and x i ; y i S. Thus mx x C. /y i x i C. / i y i is a linear combination of elements of S, with nonnegative coefficients, such that mx mx i C. / i i C. / i C. / : That is, x C. /y C and C is convex. Clearly S C. Hence co.s/ C. i, We now prove the inclusion C co.s/. Note that, by definition, S co.s/. Let x ; x S, 0; 0 and C. Then, by convexity of co.s/, x C x co.s/. Assume that Xn i x i co.s/ whenever x ; x ; : : : ; x n S, j 0, j ; ; : : : ; n and j. Let x ; x ; : : : ; x n S and ; ; : : : ; n be such that j 0, j ; ; : : : ; n and Xn j 0, then n. Hence for all j ; ; : : : ; n n X and j. If j x j n x n co.s/. Assume that ˇ j > 0. Then j ˇ 0 n j ˇ. By the induction assumption, X 0 Xn j x j j ˇ xj A C n x n co.s/: j ˇ xj co.s/. Hence Thus C co.s/..4 Quotient Space Let M be a linear subspace of a linear space X over F. For all x; y X, define x y.mod M / x y M: 7
11 It is easy to verify that defines an equivalence relation on X. For x X, denote by Œx W fy X W x y.mod M /g fy X W x y M g x C M; the coset of x with respect to M. The quotient space X=M consists of all the equivalence classes Œx, x X. The quotient space is also called a factor space..4. Proposition Let M be a linear subspace of a linear space X over F. For x; y X and F, define the operations Œx C Œy Œx C y and Œx Œ x: Then X=M is a linear space with respect to these operations. Proof. Exercise. Note that the linear operations on X=M are equivalently given by: For all x; y X and F,.x C M / C.y C M / x C y C M and.x C M / x C M:.4. efinition Let M be a linear subspace of a linear space X over F. The codimension of M in X is defined as the dimension of the quotient space X=M. It is denoted by codim.m / dim.x=m /. Clearly, if X M, then X=M f0g and so codim.x / 0..5 irect Sums and Projections.5. efinition Let M and N be linear subspaces of a linear space X over F. We say that X is a direct sum of M and N if X M C N and M \ N f0g: If X is a direct sum of M and N, we write X M N. In this case, we say that M (resp. N ) is an algebraic complement of N (resp. M )..5. Proposition Let M and N be linear subspaces of a linear space X over F. If X M N, then each x X has a unique representation of the form x m C n for some m M and n N. Proof. Exercise. Let M and N be linear subspaces of a linear space X over F such that X M N. codim.m / dim.n /. Also, since X M N, dim.x / dim.m /Cdim.N /. Hence dim.x / dim.m / C codim.m /: It follows that if dim.x / <, then codim.m / dim.x / dim.m /. Then The operator P W X! X is called an algebraic projection if P is linear (i.e., P. xcy/ PxCPy for all x; y X and F) and P P, i.e., P is idempotent. 8
12 .5.3 Proposition Let M and N be linear subspaces of a linear space X over F such that X M N. efine P W X! X by P.x/ m, where x m C n, with m M and n N. Then P is an algebraic projection of X onto M along N. Moreover M P.X / and N.I P/.X / ker.p/. Conversely, if P W X! X is an algebraic projection, then X M N, where M P.X / and N.I P/.X / ker.p/. Proof. Linearity of P: Let x m C n and y m C n, where m ; m M and n ; n N. For F, P. x C y/ P.. m C m / C. n C n // m C m Px C Py: Idempotency of P: Since m m C 0, with m M and 0 N, we have that Pm m and hence P x Pm m Px. That is, P P. Finally, n x m.i P/x. Hence N.I P/.X /. Also, Px 0 if and only if x N, i.e., ker.p/ N. Conversely, let x X and set m Px and n.i P/x. Then x m C n, where m M and n N. We show that this representation is unique. Indeed, if x m C n where m M and n N, then m Pu and n.i P/v for some u; v X. Since P P, it follows that Pm m and Pn 0. Hence m Px Pm C Pn Pm m. Similarly n n..6 The Hölder and Minkowski Inequalities We now turn our attention to three important inequalities. The first two are required mainly to prove the third which is required for our discussion about normed linear spaces in the subsequent chapter..6. efinition Let p and q be positive real numbers. If < p < and p C, or if p and q, or if p q and q, then we say that p and q are conjugate exponents..6. Lemma (Young s Inequality). Let p and q be conjugate exponents, with < p; q < and ; ˇ 0. Then ˇ p p C ˇq q : Proof. If p q, then the inequality follows from the fact that. ˇ/ 0. Notice also, that if 0 or ˇ 0, then the inequality follows trivially. If p 6, then consider the function f W Œ0; /! R given by p ˇq f. / p C q ˇ; for fixed ˇ > 0: Then, f 0. / p ˇ 0 when p ˇ. That is, when ˇ p q ˇ p the second derivative test to the critical point ˇ q p. f 00. /.p / p > 0; for all.0; /: > 0. We now apply 9
13 Thus, we have a global minimum at ˇ q p : It is easily verified that 0 f.ˇ q p ˇq p / f. / p C q ˇ, ˇ p p C ˇq q ; for each Œ0; /..6.3 Theorem (Hölder s Inequality for sequences). Let.x n / `p and.y n / `q, where p > and =p C =q. Then Proof. If jx k j p 0 or jx k y k j jx k j p! p X jy k j q! q : jy k j q 0, then the inequality holds. Assume that jy k j q 6 0. Then for k ; ; : : :, we have, by Lemma.6., that jx k j p 6 0 and jx k j P jx kj p p jy k j P jy kj q q p jx k j p P jx kj C jy k j q P p q jy kj : q Hence, That is, P jx ky k j P jx kj p p P jy kj q q jx k y k j jx k j p! p p C q : X jy k j q! q :.6.4 Theorem (Minkowski s Inequality for sequences). Let p > and.x n / and.y n / sequences in `p. Then jx k C y k j p! p jx k j p! p C jy k j p! p : 0
14 Proof. Let q p p. If X jx k C y k j p 0, then the inequality holds. jx k C y k j p 6 0. Then We therefore assume that jx k C y k j p jx k C y k j p jx k C y k j jx k C y k j p jx k j C jx k C y k j p jy k j jx k C y k j.p /q! q! q jx k C y k j p 4 4 jx k j p! p C jx k j p! p C! 3 p jy k j p 5! 3 p jy k j p 5 : ividing both sides by jx k C y k j p! q, we have jx k C y k j p! p! jx k C y k j p q jx k j p! p C jy k j p! p :.6.5 Exercise [] Show that the set of all n m real matrices is a real linear space. [] Show that a subset M of a linear space X is a linear subspace if and only if x C ˇy M for all x; y M and all ; ˇ F. [3] Prove Proposition.3.4 [4] Prove Proposition.4.. [5] Prove Proposition.5.. [6] Show that c 0 is a linear subspace of the linear space `. [7] Which of the following subsets are linear subspaces of the linear space CŒ ;? (a) M fx CŒ ; W x. / x./g. (b) M fx CŒ ; W Z x.t/dt g. (c) M 3 fx CŒ ; W jx.t / x.t /j jt t j for all t ; t Œ ; g.
15 [8] Show that if fm g is a family of linear subspaces of a linear space X, then M \ M is a linear subspace of X. If M and N are linear subspaces of a linear space X, under what condition(s) is M [ N a linear subspace of X?
16 Chapter Normed Linear Spaces. Preliminaries For us to have a meaningful notion of convergence it is necessary for the Linear space to have a notion distance and therefore a topology defined on it. This leads us to the definition of a norm which induces a metric topology in a natural way... efinition A norm on a linear space X is a realvalued function kk W X! R which satisfies the following properties: For all x; y X and F, N. kxk 0; N. kxk 0 x 0; N3. kxk jjkxk; N4. kx C yk kxk C kyk (Triangle Inequality). A normed linear space is a pair.x; k k/, where X is a linear space and k k a norm on X. The number kxk is called the norm or length of x. Unless there is some danger of confusion, we shall identify the normed linear space.x; k k/ with the underlying linear space X... Examples (Examples of normed linear spaces.) [] Let X F. For each x X, define kxk jxj. Then.X; k k/ is a normed linear space. We give the proof for X C. Properties N N3 are easy to verify. We only verify N4. Let x; y C. Then kx C yk jx C yj.x C y/.x C y/.x C y/.x C y/ xx C yx C xy C yy jxj C xy C xy C jyj jxj C Re.xy/ C jyj jxj C jxyj C jyj jxj C jxjjyj C jyj jxj C jxjjyj C jyj.jxj C jyj/.kxk C kyk/ : Taking the positive square root both sides yields N4. 3
17 [] Let n be a natural number and X F n. For each x.x ; x ; : : : ; x n / X, define kxk p kxk max jx i j: in jx i j p! p ; for p < ; and Then.X; k k p / and.x; k k / are normed linear spaces. We give a detailed proof that.x; k k p / is a normed linear space for p <. N. For each i n, jx i j 0 ) N. For any x X, jx i j p 0 ) jx i j p! p 0 ) kxk p 0: kxk p 0 N3. For any x X and any F, jx i j p! p 0 jx i j p 0 for all i ; ; 3; : : :; n x i 0 for all i ; ; 3; : : :; n x 0: kxk p jj jx i j p! p jx i j p! p jj p jjkxk p : jx i j p! p N4. For any x; y X, kx C yk p jx i C y i j p! p jx i j p! p C kxk p C kyk p : jy i j p! p.by Minkowski 0 s Inequality/ [3] Let X BŒa; b be the set of all bounded realvalued functions on Œa; b. For each x X, define kxk sup jx.t/j: atb Then.X; k k / is a normed linear space. We prove the triangle inequality: For any t Œa; b and any x; y X, jx.t/ C y.t/j jx.t/j C jy.t/j sup jx.t/j C sup jy.t/j kxk C kyk : atb atb 4
18 Since this is true for all t Œa; b, we have that kx C yk sup jx.t/ C y.t/j kxk C kyk : atb [4] Let X CŒa; b. For each x X, define kxk sup jx.t/j atb 0 Z b a jx.t/j dta Then.X; k k / and.x; k k / are normed linear spaces. : [5] Let X `p; p <. For each x.x i / X, define kxk p X in jx i j p! p : Then.X; k k p / is a normed linear space. [6] Let X `; c or c 0. For each x.x i / X, define Then X is a normed linear space. kxk kxk sup jx i j: in [7] Let X L.C n / be the linear space of all n n complex matrices. For A L.C n /, let.a/.a/ ii be the trace of A. For A L.C n /, define v v kak p ux.a A/ t n ux.a/ ki.a/ ki t n j.a/ ki j ; where A is the conjugate transpose of the matrix A. Notation Let a be an element of a normed linear space.x; k k/ and r > 0. B.a; r/ fx X j kx ak < rg.open ball with centre a and radius r/i BŒa; r fx X j kx ak rg.closed ball with centre a and radius r/i S.a; r/ fx X j kx ak rg.sphere with centre a and radius r/: 5
19 y y y x x x k.x; y/k k.x; y/k < k.x; y/k Equivalent Norms..3 efinition Let kk and kk 0 be two different norms defined on the same linear space X. We say that kk is equivalent to k k 0 if there are positive numbers and ˇ such that kxk kxk 0 ˇkxk; for all x X:..4 Example Let X F n. For each x.x ; x ; : : : ; x n / X, let kxk jx i j; kxk jx i j! ; and kxk max in jx ij: We have seen that k k ; k k and k k are norms on X. We show that these norms are equivalent. Equivalence of k k and k k : Let x.x ; x ; : : : ; x n / X. For each k ; ; : : : ; n, jx k j jx i j ) Also, for k ; ; : : : ; n, max jx k j kn jx i j kxk kxk : jx k j max jx k j kxk ) kn jx i j kxk nkxk kxk nkxk : Hence, kxk kxk nkxk. We now show that k k is equivalent to k k. Let x.x ; x ; : : : ; x n / X. For each k ; ; : : : ; n, jx k j kxk ) jx k j.kxk / ) jx i j.kxk / n.kxk / kxk p nkxk : 6
20 Also, for each k ; ; : : : ; n, jx k j! = jx i j kxk ) max jx k j kxk kxk kxk : kn Consequently, kxk kxk p nkxk, which proves equivalence of the norms k k and k k. It is, of course, obvious now that all the three norms are equivalent to each other. We shall see later that all norms on a finitedimensional normed linear space are equivalent...5 Exercise Let N.X / denote the set of norms on a linear space X. For k k and k k 0 in N.X /, define a relation ' by k k ' k k 0 if and only if k k is equivalent to k k 0 : Show that ' is an equivalence relation on N.X /, i.e., ' is reflexive, symmetric, and transitive. Open and Closed Sets..6 efinition A subset S of a normed linear space.x; k k/ is open if for each s S there is an > 0 such that B.s; / S. A subset F of a normed linear space.x; k k/ is closed if its complement X n F is open...7 efinition Let S be a subset of a normed linear space.x; k k/. We define the closure of S, denoted by S, to be the intersection of all closed sets containing S. It is easy to show that S is closed if and only if S S. Recall that a metric on a set X is a realvalued function d W X X! R which satisfies the following properties: For all x; y; z X, M. d.x; y/ 0; M. d.x; y/ 0 x y; M3. d.x; y/ d.y; x/; M4. d.x; z/ d.x; y/ C d.y; z/... Theorem (a) If.X; k k/ is a normed linear space, then d.x; y/ kx yk defines a metric on X. Such a metric d is said to be induced or generated by the norm k k. Thus, every normed linear space is a metric space, and unless otherwise specified, we shall henceforth regard any normed linear space as a metric space with respect to the metric induced by its norm. (b) If d is a metric on a linear space X satisfying the properties: For all x; y; z X and for all F,.i/ d.x; y/ d.x C z; y C z/ (Translation Invariance).ii/ d.x; y/ jjd.x; y/ (Absolute Homogeneity); then defines a norm on X. kxk d.x; 0/ 7
21 Proof. (a) We show that d.x; y/ kx yk defines a metric on X. To that end, let x; y; z X. M. d.x; y/ kx yk 0 by N. M. d.x; y/ 0 kx yk 0 x y 0 by N x y: M3. d.x; y/ kx yk k. /.y x/k j jky xk by N3 ky xk d.y; x/: M4. d.x; z/ kx zk k.x y/ C.y z/k kx yk C ky zk by N4 d.x; y/ C d.y; z/: (b) Exercise. It is clear from Theorem.., that a metric d on a linear space X is induced by a norm on X if and only if d is translationinvariant and positive homogeneous.. Quotient Norm and Quotient Map We now want to introduce a norm on a quotient space. Let M be a closed linear subspace of a normed linear space X over F. For x X, define kœxk W inf kyk: yœx If y Œx, then y x M and hence y x C m for some m M. Hence kœxk inf kyk inf kx C mk inf kx mk d.x; M /: yœx mm mm.. Proposition Let M be a closed linear subspace of a normed linear space X over F. The quotient space X=M is a normed linear space with respect to the norm Proof. kœxk W inf kyk; where Œx X=M: yœx N. It is clear that for any x X, kœxk d.x; M / 0. N. For any x X, kœxk 0 d.x; M / 0 x M M x C M M Œ0: N3. For any x; y X and F n f0g, kœxk kœxk d.x; M / inf ym kx jj inf zm kx zk jjd.x; M / jjkœxk: yk inf x ym y 8
22 N4. Let x; y X. Then kœx C Œyk kœx C yk d.x C y; M / inf zm kx C y inf kx C y.z C z /k z ;z M inf k.x z / C.y z /k z ;z M inf kx z k C ky z k z ;z M inf kx z M z k C inf z M ky z k zk d.x; M / C d.y; M / kœxk C kœyk: The norm on X=M as defined in Proposition.. is called the quotient norm on X=M. Let M be a closed subspace of the normed linear space X. The mapping Q M from X! X=M defined by Q M.x/ x C M; x X; is called the quotient map (or natural embedding) of X onto X=M..3 Completeness of Normed Linear Spaces Now that we have established that every normed linear space is a metric space, we can deploy on a normed linear space all the machinery that exists for metric spaces..3. efinition Let.x n / n be a sequence in a normed linear space.x; k k/. (a).x n / n is said to converge to x if given > 0 there exists a natural number N N./ such that Equivalently,.x n / n converges to x if If this is the case, we shall write kx n xk < for all n N: lim kx n xk 0: n! x n! x or lim x n x: n! Convergence in the norm is called norm convergence or strong convergence. (b).x n / n is called a Cauchy sequence if given > 0 there exists a natural number N N./ such that kx n x m k < for all n; m N: Equivalently,.x n / is Cauchy if lim kx n x m k 0: n;m! 9
23 In the following lemma we collect some elementary but fundamental facts about normed linear spaces. In particular, it implies that the operations of addition and scalar multiplication, as well as the norm and distance functions, are continuous..3. Lemma Let C be a closed set in a normed linear space.x; k k/ over F, and let.x n / be a sequence contained in C such that lim n! x n x X. Then x C. Proof. Exercise..3.3 Lemma Let X be a normed linear space and A a nonempty subset of X. [] jd.x; A/ d.y; A/j kx yk for all x; y X ; [] j kxk kyk j kx yk for all x; y X ; [3] If x n! x, then kx n k! kxk; [4] If x n! x and y n! y, then x n C y n! x C y; [5] If x n! x and n!, then nx n! x; [6] The closure of a linear subspace in X is again a linear subspace; [7] Every Cauchy sequence is bounded; [8] Every convergent sequence is a Cauchy sequence. Proof. (). For any a A, d.x; A/ kx ak kx yk C ky ak; so d.x; A/ kx yk C d.y; A/ or d.x; A/ d.y; A/ kx yk: Interchanging the roles of x and y gives the desired result. () follows from () by taking A f0g. (3) is an obvious consequence of (). (4), (5) and (8) follow from the triangle inequality and, in the case of (5), the absolute homogeneity. (6) follows from (4) and (5). (7). Let.x n / be a Cauchy sequence in X. Choose n so that kx n x n k for all n n. By (), kx n k C kx n k for all n n. Thus kx n k maxf kx k; kx k; kx 3 k; : : : ; kx n k; C kx n kg for all n. (8) Let.x n / be a sequence in X which converges to x X and let > 0. Then there is a natural number N such that kx n xk < for all n N. For all n; m N, kx n x m k kx n xk C kx x m k < C : Thus,.x n / is a Cauchy sequence in X..3.4 Proposition Let.X; kk/ be a normed linear space over F. A Cauchy sequence in X which has a convergent subsequence is convergent. 0
24 Proof. Let.x n / be a Cauchy sequence in X and.x nk / its subsequence which converges to x X. Then, for any > 0, there are positive integers N and N such that kx n x m k < for all n; m N and kx nk xk < for all k N : Let N maxfn ; N g. If k N, then since n k k, kx k xk kx k x nk k C kx nk xk < C : Hence x n! x as n!..3.5 efinition A metric space.x; d/ is said to be complete if every Cauchy sequence in X converges in X..3.6 efinition A normed linear space that is complete with respect to the metric induced by the norm is called a Banach space..3. Theorem Let.X; k k/ be a Banach space and let M be a linear subspace of X. Then M is complete if and only if the M is closed in X. Proof. Assume that M is complete. We show that M is closed. To that end, let x M. Then there is a sequence.x n / in M such that kx n xk! 0 as n!. Since.x n / converges, it is Cauchy. Completeness of M guarantees the existence of an element y M such that kx n yk! 0 as n!. By uniqueness of limits, x y. Hence x M and, consequently, M is closed. Assume that M is closed. We show that M is complete. Let.x n / be a Cauchy sequence in M. Then.x n / is a Cauchy sequence in X. Since X is complete, there is an element x X such that kx n xk! 0 as n!. But then x M since M is closed. Hence M is complete..3.7 Examples [] Let p <. Then for each positive integer n,.f n ; k k p / is a Banach space. [] For each positive integer n,.f n ; k k / is a Banach space. [3] Let p <. The sequence space `p is a Banach space. Because of the importance of this space, we give a detailed proof of its completeness. The classical sequence space `p is complete. Proof. Let.x n / be a Cauchy sequence in `p. We shall denote each member of this sequence by x n.x n./; x n./; : : :/: Then, given > 0, there exists an N./ N N such that kx n x m k p For each fixed index i, we have jx n.i/ x m.i/j p! p jx n.i/ x m.i/j < for all n; m N: < for all n; m N:
25 That is, for each fixed index i,.x n.i// there exists x.i/ F such that is a Cauchy sequence in F. Since F is complete, x n.i/! x.i/ as n! : efine x.x./; x./; : : :/. We show that x `p, and x n! x. To that end, for each k N, That is,! p kx jx n.i/ x m.i/j p kx n x m k p jx n.i/ x m.i/j p! p kx jx n.i/ x m.i/j p < p ; for all k ; ; 3; : : :: Keep k and n N fixed and let m!. Since we are dealing with a finite sum, kx jx n.i/ x.i/j p p : Now letting k!, then for all n N, < : jx n.i/ x.i/j p p ;.:3:7:/ which means that x n x `p. Since x n `p, we have that x.x x n / C x n `p. It also follows from (.3.7.) that x n! x as n!. [4] The space `0 of all sequences.x i / with only a finite number of nonzero terms is an incomplete normed linear space. It suffices to show that `0 is not closed in ` (and hence not complete). To that end, consider the sequence.x i / with terms x.; 0; 0; 0; : : :/ x.; ; 0; 0; 0; : : :/ x 3.; ; ; 0; 0; 0; : : :/ : x n.; ; ; : : : ; ; 0; 0; 0; : : :/ n : This sequence.x i / converges to x.; ; ; : : : ; ; n ; ; : : :/: n nc Indeed, since x x n.0; 0; 0; : : :; 0; ; ; : : :/, it follows that n nc kx n xk That is, x n! x as n!, but x 6 `0: kn! 0 as n! : k
26 [5] The space C Œ ; of continuous realvalued functions on Œ ; with the norm 0 Z x.t/ dta is an incomplete normed linear space. To see this, it suffices to show that there is a Cauchy sequence in C Œ ; which converges to an element which does not belong to C Œ ;. Consider the sequence.x n / C Œ ; defined by 8 0 if t 0 ˆ< x n.t/ nt if 0 t n ˆ: if n t : = y x n.t/ n 0 t We show that.x n / is a Cauchy sequence in C Œ ;. To that end, for positive integers m and n such that m > n, kx n x m k Z Œx n.t/ x m.t/ dt Z =m Œnt mt dt C Z =n Œ nt dt 0 =m Z =m Œm t mnt C n t dt C Z =n Œ nt C n t dt 0.m mn C n / t3 3 ˇ m mn C n 3m n =m 0.m n/ 3m n =m C t nt C n t3 3 ˇˇˇˇ =n =m! 0 as n; m! : efine x.t/ 0 if t 0 if 0 < t : 3
27 Then x 6 C Œ ;, and Z kx n xk Œx n.t/ x.t/ dt Zn 0 Œnt dt! 0 as n! : 3n That is, x n! x as n!..4 Series in Normed Linear Spaces Let.x n / be a sequence in a normed linear space.x; k k/. To this sequence we associate another sequence.s n / of partial sums, where s n x k..4. efinition Let.x n / be a sequence in a normed linear space.x; k k/. If the sequence.s n / of partial sums converges to s, then we say that the series x k converges and that its sum is s. In this case we write x k s. The series x k is said to be absolutely convergent if kx k k <. We now give a series characterization of completeness in normed linear spaces..4. Theorem A normed linear space.x; k k/ is a Banach space if and only if every absolutely convergent series in X is convergent. Proof. Let X be a Banach space and suppose that kx j k <. We show that the series x j converges. To that end, let > 0 and for each n N, let s n jk C kx j k <. Then, for all m > n > K, we have ks m mx s n k x j mx x j x j nc x j. nc Let K be a positive integer such that mx kx j k kx j k nc kx j k < : Hence the sequence.s n / of partial sums forms a Cauchy sequence in X. Since X is complete, the sequence.s n / converges to some element s X. That is, the series x j converges. Conversely, assume that.x; k k/ is a normed linear space in which every absolutely convergent series converges. We show that X is complete. Let.x n / be a Cauchy sequence in X. Then there is an n N such that kx n x m k < whenever m > n. Similarly, there is an n N with n > n such that kx n x m k < whenever m > n. Continuing in this way, we get natural numbers n < n < such 4 K C
28 that kx nk x m k < whenever m > n k k. In particular, we have that for each k N, kx nkc x nk k < k. For each k N, let y k x nkc x nk. Then Hence, the series follows that ky k k kx nkc x nk k < ky k k <. That is, the series y k is absolutely convergent, and hence, by our assumption, k : y k is convergent in X. That is, there is an s X such that s j s j jx y k jx y k! s as j!. It jx j! Œx nkc x nk x njc x n! s: j! Hence x njc! s C xn. Thus, the subsequence x nk of.xn / converges in X. But if a Cauchy sequence has a convergent subsequence, then the sequence itself also converges (to the same limit as the subsequence). It thus follows that the sequence.x n / also converges in X. Hence X is complete. We now apply Theorem.4. to show that if M is a closed linear subspace of a Banach space X, then the quotient space X=M, with the quotient norm, is also a Banach space..4. Theorem Let M be a closed linear subspace of a Banach space X. Then the quotient space X=M is a Banach space when equipped with the quotient norm. Proof. Let.Œx n / be a sequence in X=M such that y j M such that It now follows that kx j kœx j k <. For each j N, choose an element kx j y j k kœx j k C j : y j k <, i.e., the series.x j y j / is absolutely convergent in X. Since X is complete, the series.x j y j / converges to some element z X. We show that the series Œx j converges to Œz. Indeed, for each n N, Œx j Œz x j 5 Œz 4 x j z5 inf mm x j z m x j z y j.x j y j / z! 0 as n! : 5
29 Hence, every absolutely convergent series in X=M is convergent, and so X=M is complete..5 Bounded, Totally Bounded, and Compact Subsets of a Normed Linear Space.5. efinition A subset A of a normed linear space.x; k k/ is bounded if A BŒx; r for some x X and r > 0. It is clear that A is bounded if and only if there is a C > 0 such that kak C for all a A..5. efinition Let A be a subset of a normed linear space.x; k k/ and > 0. A subset A X is called an net for A if for each x A there is an element y A such that kx yk <. Simply put, A X is an net for A if each element of A is within an distance to some element of A. A subset A of a normed linear space.x; k k/ is totally bounded (or precompact) if for any > 0 there is a finite net F X for A. That is, there is a finite set F X such that A [ B.x; /: xf The following proposition shows that total boundedness is a stronger property than boundedness..5.3 Proposition Every totally bounded subset of a normed linear space.x; k k/ is bounded. Proof. This follows from the fact that a finite union of bounded sets is also bounded. The following example shows that boundedness does not, in general, imply total boundedness..5.4 Example Let X ` and consider B B.X / fx X j kxk g, the closed unit ball in X. Clearly, B is bounded. We show that B is not totally bounded. Consider the elements of B of the form: for j N, e j.0; 0; : : : ; 0; ; 0; : : :/, where occurs in the jth position. Note that ke i e j k p for all i j. Assume that an net B X existed for 0 < <. Then for each j N, there is an element y j B such that ke j y j k <. This says that for each j N, there is an element y j B such that y j B.e j ; /. But the balls B.e j ; / are disjoint. Indeed, if i 6 j, and z B.e i ; / \ B.e j ; /, then by the triangle inequality p p kei e j k ke i zk C kz e j k < < p ; which is absurd. Since the balls B.e j ; / are (at least) countably infinite, there can be no finite net for B. In our definition of total boundedness of a subset A X, we required that the finite net be a subset of X. The following proposition suggests that the finite net may actually be assumed to be a subset of A itself. 6
30 .5.5 Proposition A subset A of a normed linear space.x; k k/ is totally bounded if and only if for any > 0 there is a finite set F A such that A [ xf B.x; /: Proof. Exercise. We now give a characterization of total boundedness..5. Theorem A subset K of a normed linear space.x; k k/ is totally bounded if and only if every sequence in K has a Cauchy subsequence. Proof. Assume that K is totally bounded and let.x n / be an infinite sequence in K. There is a finite set of points fy ; y ; : : : ; y r g in K such that K r[ B.y j ; /: At least one of the balls B.y j ; /; j ; ; : : : ; r, contains an infinite subsequence.x n/ of.x n /. Again, there is a finite set fy ; y ; : : : ; y s g in K such that K s[ B.y j ; /: At least one of the balls B.y j ; /; j ; ; : : :; s, contains an infinite subsequence.x n / of.x n /. Continuing in this way, at the mth step, we obtain a subsequence.x nm / of.x n.m / / which is contained in a ball of the form B y mj ;. m Claim: The diagonal subsequence.x nn / of.x n / is Cauchy. Indeed, if m > n, then both x nn and x mm are in the ball of radius n. Hence, by the triangle inequality, kx nn x mm k < n! 0 as n! : Conversely, assume that every sequence in K has a Cauchy subsequence and that K is not totally bounded. Then, for some > 0, no finite net exists for K. Hence, if x K, then there is an x K such that kx x k. (Otherwise, kx yk < for all y K and consequently fx g is a finite net for K, a contradiction.) Similarly, there is an x 3 K such that kx x 3 k and kx x 3 k : Continuing in this way, we obtain a sequence.x n / in K such that kx n x m k for all m n. Therefore.x n / cannot have a Cauchy subsequence, a contradiction..5.6 efinition A normed linear space.x; k k/ is sequentially compact if every sequence in X has a convergent subsequence..5.7 Remark It can be shown that on a metric space, compactness and sequential compactness are equivalent. Thus, it follows, that on a normed linear space, we can use these terms interchangeably. 7
31 .5. Theorem A subset of a normed linear space is sequentially compact if and only if it is totally bounded and complete. Proof. Let K be a sequentially compact subset of a normed linear space.x; k k/. We show that K is totally bounded. To that end, let.x n / be a sequence in K. By sequential compactness of K,.x n / has a subsequence.x nk / which converges in K. Since every convergent sequence is Cauchy, the subsequence.x nk / of.x n / is Cauchy. Therefore, by Theorem.5., K is totally bounded. Next, we show that K is complete. Let.x n / be a Cauchy sequence in K. By sequential compactness of K,.x n / has a subsequence.x nk / which converges in K. But if a subsequence of a Cauchy sequence converges, so does the full sequence. Hence.x n / converges in K and so K is complete. Conversely, assume that K is a totally bounded and complete subset of a normed linear space.x; k k/. We show that K is sequentially compact. Let.x n / be a sequence in K. By Theorem.5.,.x n / has a Cauchy subsequence.x nk /. Since K is complete,.x nk / converges in K. Hence K is sequentially compact..5.8 Corollary A subset of a Banach space is sequentially compact if and only if it is totally bounded and closed. Proof. Exercise..5.9 Corollary A sequentially compact subset of a normed linear space is closed and bounded. Proof. Exercise. We shall see that in finitedimensional spaces the converse of Corollary.5.9 also holds..5.0 Corollary A closed subset F of a sequentially compact normed linear space.x; k k/ is sequentially compact. Proof. Exercise..6 Finite imensional Normed Linear Spaces The theory for finitedimensional normed linear spaces turns out to be much simpler than that of their infinitedimensional counterparts. In this section we highlight some of the special aspects of finitedimensional normed linear spaces. The following Lemma is crucial in the analysis of finitedimensional normed linear spaces..6. Lemma Let.X; k k/ be a finitedimensional normed linear space with basis fx ; x ; : : : ; x n g. Then there is a constant m > 0 such that for every choice of scalars ; ; : : : ; n, we have m j jj j x j : 8
32 Proof. If j jj 0, then j 0 for all j ; ; : : : ; n and the inequality holds for any m > 0. Assume that condition j jj 0. We shall prove the result for a set of scalars f ; ; : : : ; ng that satisfy the j jj. Let A f. ; ; : : : ; n/ F n j j jj g: Since A is a closed and bounded subset of F n, it is compact. efine f W A! R by f. ; ; : : : ; n/ j x j : Since for any. ; ; : : : ; n/ and.ˇ; ˇ; : : : ; ˇn/ in A jf. ; ; : : : ; n/ f.ˇ; ˇ; : : : ; ˇn/j ˇ. j max jn j x j j x j kx j k ˇj x j ˇj/x j j j ˇj x j ˇ j j ˇjj; ˇjjkx j k f is continuous on A. Since f is a continuous function on a compact set A, it attains its minimum on A, i.e., there is an element. ; ; : : : ; n / A such that f. ; ; : : : ; n / infff. ; ; : : : ; n/ j. ; ; : : : ; n/ Ag: Let m f. ; ; : : : ; n /. Since f 0, it follows that m 0. If m 0, then j x j 0 ) j x j 0: Since the set fx ; x ; : : : ; x n g is linearly independent, j 0 for all j ; ; : : : ; n. This is a contradiction since. ; ; : : : ; n / A. Hence m > 0 and consequently for all. ; ; : : : ; n/ A, 0 < m f. ; ; : : : ; n/ m j jj j x j : Now, let f ; ; : : : ; ng be any collection of scalars and set ˇ j jj. If ˇ 0, then the 9
33 inequality holds vacuously. If ˇ > 0, then ˇ ; ˇ ; : : : ; n ˇ That is, m j x j j jj j ˇ xj j x j. ˇ f ˇ ; ˇ ; : : : ; n ˇ A and consequently ˇ mˇ m j jj:.6. Theorem Let X be a finitedimensional normed linear space over F. Then all norms on X are equivalent. Proof. Let fx ; x ; : : : ; x n g be a basis for X and k k 0 and k k be any two norms on X. For any x X there is a set of scalars f ; ; : : : ; ng such that x jx j. By Lemma.6., there is an m > 0 such that By the triangle inequality where M max kx j k 0. Hence jn kxk 0 M m kxk m kxk 0 j jj jx j kxk: j jjkx j k 0 M j jj; ) m M kxk 0 kxk kxk 0 kxk where m M : Interchanging the roles of the norms k k 0 and k k, we similarly get a constant ˇ such that kxk ˇkxk 0. Hence, kxk 0 kxk ˇkxk 0 for some constants and ˇ..6. Theorem Every finitedimensional normed linear space.x; k k/ is complete. Proof. Let fx ; x ; : : : ; x n g be a basis for X and let.z k / be a Cauchy sequence in X. Then, given any > 0, there is a natural number N such that Also, for each k N, z k kz k z`k < for all k; ` > N: kj x j. By Lemma.6., there is an m > 0 such that m j kj `j j kz k z`k: Hence, for all k; ` > N and all j ; ; : : : ; n, j kj `j j m kz k z`k < m : 30
34 That is, for each j ; ; : : : ; n,. kj / k is a Cauchy sequence of numbers. Since F is complete, kj! j as k! for each j ; ; : : : ; n. efine z jx j. Then z X and kz k zk kj x j jx j. kj j/x j j kj jjkx j k! 0 as k!. That is, the sequence.z k / converges to z X. hence X is complete..6. Corollary Every finitedimensional normed linear space X is closed. Proof. Exercise..6.3 Theorem In a finitedimensional normed linear space.x; k k/, a subset K X is sequentially compact if and only if it is closed and bounded. Proof. We have seen (Corollary.5.9), that a compact subset of a normed linear space is closed and bounded. Conversely, assume that a subset K X is closed and bounded. We show that K is compact. Let fx ; x ; : : : ; x n g be a basis for X and let.z k / be any sequence in K. Then for each k N, z k kj x j. Since K is bounded, there is a positive constant M such that kz k k M for all k N. By Lemma.6., there is an m > 0 such that m j kj j kj x j kz kk M: It now follows that j kj j M for each j ; ; : : : ; n, and for all k N. That is, for each fixed j m ; ; : : : ; n, the sequence. kj / k of numbers is bounded. Hence the sequence. kj / k has a subsequence. kr j/ which converges to j for j ; ; : : : ; n. Setting z jx j, we have that kz kr zk kr jx j jx j j kr j jjkx j k! 0 as r! : That is, z kr! z as r!. Since K is closed, z K. Hence K is compact..6.3 Lemma (Riesz s Lemma). Let M be a closed proper linear subspace of a normed linear space.x; k k/. Then for each 0 < <, there is an element z X such that kzk and Proof. Choose x X n M and define ky zk > for all y M: d d.x; M / inf mm kx mk: 3
35 Since M is closed, d > 0. By definition of infimum, there is a m M such that Take z d kx mk < d C d d. C /: m x. Then kzk and for any y M, km xk ky zk m x y C ky.km xk/ C m xk km xk km xk d km xk > d d. C / C C > : We now give a topological characterization of the algebraic concept of finite dimensionality..6.4 Theorem A normed linear space.x; k k/ is finitedimensional if and only its closed unit ball B.X / fx X j kxk g is compact. Proof. Assume that.x; k k/ is finitedimensional normed linear space. Since the ball B.X / is closed and bounded, it is compact. Assume that the closed unit ball B.X / fx X j kxk g is compact. Then B.X / is totally bounded. Hence there is a finite net fx ; x ; : : : ; x n g in B.X /. Let M linfx ; x ; : : : ; x n g. Then M is a finitedimensional linear subspace of X and hence closed. Claim: M X. If M is a proper subspace of X, then, by Riesz s Lemma there is an element x 0 B.X / such that d.x 0 ; M / >. In particular, kx 0 x k k > for all k ; ; : : : ; n: However this contradicts the fact that fx ; x ; : : : ; x n g is a net in B.X /. Hence M X and, consequently, X is finitedimensional. We now give another argument to show that boundedness does not imply total boundedness. Let X ` and B.X / fx X j kxk g. It is obvious that B.X / is bounded. We show that B.X / is not totally bounded. Since X is complete and B.X / is a closed subset of X, B.X / is complete. If B.X / were totally bounded, then B.X / would, according to Theorem.6, be compact. By Theorem.6.4, X would be finitedimensional. But this is false since X is infinitedimensional..7 Separable Spaces and Schauder Bases.7. efinition (a) A subset S of a normed linear space.x; k k/ is said to be dense in X if S X ; i.e., for each x X and > 0, there is a y S such that kx yk <. (b) A normed linear space.x; k k/ is said to be separable if it contains a countable dense subset..7. Examples [] The real line R is separable since the set Q of rational numbers is a countable dense subset of R. [] The complex plane C is separable since the set of all complex numbers with rational real and imaginary parts is a countable dense subset of C. [3] The sequence space `p, where p <, is separable. Take M to be the set of all sequences with rational entries such that all but a finite number of the entries are zero. (If 3
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