Appendix 1. Introduction to Banach Spaces A1.1 Linear Vector Spaces and Norms Definition A1.1 A linear vector space (LVS) X over a field F is a set with two binary operations: addition (a map of X x X ~ X) and numerical (or scalar) multiplication (a map of F x X ~ X), which satisfy the commutative, associative, and distributive properties. We are primarily interested in spaces of functions and will take F to be either lR or cc. Examples A1.2. (1) X = lRn with the operations (+, .), the usual componentwise operations, and with F =R (2) X = C([O, I D, all cqntinuous, complex-valued functions on the interval [0, I]. Addition is pointwise,thatis, if f, g E X then (j+g)(x) = f(x)+g(x), and scalar multiplication for F = C is also pointwise. (3) X = II', where II' consists of all infinite sequences of complex numbers: x == (Xl, ... , xn, ... ), Xi E C such that L~l IXi I" < 00, for 1 :::: p < 00. Again, addition and scalar multiplication are defined componentwise. Problem A1.1. Show that II' is an LVS. (Hint: Use the Minkowski inequality: 286 Appendix 1. Introduction to Banach Spaces Definition A1.3. Let X be an LVS over F, F = lR or <C. A norm II . lion X is a map II . II : X ~ lR+ U {O} such that (i) IIx II 2: 0 and if IIx II = 0 then x = 0 (positive definiteness); (ii) II Ax II = IAlllxll, A E F (homogeneity); (iii) IIx + yll :::: IIxll + Ilyll, x, y E X (triangle inequality). Examples AI.4. (1) X = lRn and define IIxll = (2::;'=1 Ixd2)1/2, x E lRn. This is a norm, called the Euclidean norm, and has the interpretation of the length of the vector x. (2) X = C([O, 1]) can be equipped with many norms, for example, the LP- 1 )I/P norms: Ilfllp == ( fa If(xWdx ,f E x, 1:::: p < oo,andthesup-norm (p = 00): Ilflloo == sup If(x)l, f E X. XE[O,IJ (3) X = IP has a norm given by IIxll == (2::~1 IxilP)I/P, x EX. Problem A1.2. Show that the maps X ~ lR+ claimed to be norms in Examples A1.4 are norms. Definition A1.S. An LVS X with a norm II . II is a normed linear vector space (NLVS), that is, an NLVS X is a pair (X, II . II) where X is an LVS and II . II is a norm on X. Al.2 Elementary Topology in Normed Vector Spaces Elementary topology studies relations between certain families of subsets of a set X. A topology for a set X is built out of a distinguished family of subsets, called the open sets. In an NLVS X, there is a standard construction of these open sets. Definition A1.6. Let X be an NLVS, and let a E X. An open ball about a of radius r, denoted Br(a), is defined by Br(a) == {x E Xlllx - all < r}. In analogy with the Euclidean metric on lRn (see (1), Examples A1.4), we in­ terpret Ilx - all as "the distance from x to a," and so Br(a) consists of all points x E X lying within distance r from a. The open balls in X are the fundamental building blocks of open sets. Definition AI.7. A subset E C X, X an NLVS, is open iffor each x E E there exists an r > 0 (depending on x) such that Br(x) C E. A neighborhood ofx EX is an open set containing x. Al.2. Elementary Topology in Normed Vector Spaces 287 Example A1.S. Let X = JRI1 with the Euclidean metric. Then simply a ball of radius E about the origin. Note that BE(O) does not include its boundary, that is, {x I (2:::'=1 x?) 1/2 = E}. It is easy to check that BE(O) is an open set. If we were to include the boundary, the resulting set would not be open. DefinitionA1.9.Asubset Be X, X anNLVS, isclosedifX\B == {x E Xix rj. B} (the complement of B in X) is open. Sequences provide a convenient tool for studying many properties of subsets of an NLVS X, including open and closed sets. Definition A1.l0. Let X be an NLVS. (I) A sequence {XI1 l in X converges to x E X if Ilxn - x II ~ 0 as n ~ 00. (2) A sequence (x" l in X is Cauchy if Ilxn - Xm II ~ 0 as n, m ~ 00. Problem Al.3. Prove that (I) every convergent sequence is a Cauchy sequence, (2) every convergent sequence is uniformly bounded, that is, if {xn l is conver- gent, then there exists an M, 0 < M < 00, such that Ilxn II :::: M for all n = 1,2, .... (Hints: (1) Use an E/2-argument and the identity: X il - Xm = Xn - X + X - X m . (2) Use the triangle inequality to show that I Ilxn II - Ilxm II I:::: Ilxn - Xm II·) A nice feature of NLVS is that sequences can be used to characterize closed sets. Proposition Al.ll. A subset E c X, X an NLVS, is closed iffor any convergent sequence (xnl in E, limn->ooxn E E. Proof. (1) Suppose E is closed and lim l1 -+oc Xn = x rj. E but Xn E E. Since X \ E is open, there is an E > 0 such that BE(x) n E = </>. But X I1 ~ x, so for this E, there is an N such that n > N =? Ilx n - xII < E =? Xn E BE(x). So we have a contradiction, and x E E. 288 Appendix 1. Introduction to Banach Spaces (2) Conversely, suppose E contains the limit of all its convergent sequences. Suppose X \ E is not open; then there is a point Y E X \ E such that for each E > 0, BE(y) n E =I ¢. Let En == lin and choose Yn E BE,Jy) n E. Then {Yn} is a sequence in E and limn-+oo Yn = y. But y tI E, hence we get a contradiction (i.e., X \ E is open so E is closed). 0 Definition A1.12. If M c X, X an NLVS, then M, the closure of M in X, is M U {limits of all nonconstant convergent sequences in M}. The set ofpoints that are limits in X of nonconstant sequences in M are the accumulation points or cluster points of M. Remark Al.13. By definition, M :J M and M is closed. M is, in fact, the smallest closed set containing M. Remark A1.14. Exercise (1) in Problem A1.3 is significant because the converse is not true in every NLVS X. That is, if {xn} is a Cauchy sequence in X, it is not necessarily true that {xn} is convergent. As an example, consider X = C([O, 1]) in the II . III" 1:s p < 00, norm (see Example AlA (2)). We construct a sequence fn E X by 0, O:SX:S~, .l<x<.l+.l f,,(X) = I n(x-~), 2- -2 n' 1, .l+.l<x<1.2 n -- Clearly, fn E C([O, 1]) and, for n ::::: m and p = 1, Ilfn-fmll! = fo!If,,(X)-fm(X)ldX 1+1 1+1- l1l h2 11 (n - m)(x - 1/2)dx + h:l [1 - m(x - 1/2)]dx 2 2 11 ~ (~-~), 2 m n and so {fn} is a Cauchy sequence. It is easy to check that for any p, {fn} is Cauchy. Now for any x E [0, 1], we can compute limn-+ oo fn(x). The limit function is 0, 0 < x :s ~, f(x) = I 1, ~ < x :s 1. This limit function f tI C([O, 1]) (i.e., f is discontinuous), and so C([O, 1]) does not contain the limit of all its Cauchy sequences! Later we will discuss the manner in which we can extend C([O, 1]) to a larger NLVS (called the completion) to include the limit of all the Cauchy sequences in C([O, 1]). AI.3 Banach Spaces Definition A1.15.An NLVS X is complete ifevery Cauchy sequence in X converges to an element of X. A complete NLVS is called a Banach space. Al.3. Banach Spaces 289 Remark A1.16. Every finite-dimensional NLVS is complete and hence a Banach space. This follows from the fact that both ffi. and C are complete and from the fact that every finite-dimensional LVS is isomorphic to ffi.N or C N for some N. For this reason, the term "Banach space" is usually used to denote an infinite-dimensional NLVS. Examples A1.17. (1) X = ffi.N with the Euclidean norm II . lie is complete. (2) X = C([O, I)) is complete with the sup-norm II . 1100, but it is not complete with the p-norm for 1 :s p < 00; see Remark A1.l4. (3) [P is complete; we prove this in Proposition AU8. Problem AI.4. Prove statements (l) and (2) in Examples A 1.17. (Hint: For (2), use standard results on uniform convergence.) Proposition A1.18. The NLVS IP as defined in (3) of Examples Al.2, with the II . lip-norm, 1 :s p < 00, is a Banach space. Proof. Let (x(n)), x(ll) == (xiII), xiII), ... ) E [P, be a Cauchy sequence, that is, (ALl) as m, n -+ 00. We must show (1) that (x(Il)} is convergent to some x, and (2) that x E [P.