FUNCTIONAL ANALYSIS LECTURE NOTES CHAPTER 3. BANACH SPACES


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1 FUNCTIONAL ANALYSIS LECTURE NOTES CHAPTER 3. BANACH SPACES CHRISTOPHER HEIL 1. Elementary Properties and Examples Notation 1.1. Throughout, F will denote either the real line R or the complex plane C. All vector spaces are assumed to be over the field F. Definition 1.2. Let X be a vector space over the field F. Then a seminorm on X is a function : X R such that (a) x 0 for all x X, (b) αx = α x for all x X and α F, (c) Triangle Inequality: x + y x + y for all x, y X. A norm on X is a seminorm which also satisfies: (d) x = 0 = x = 0. A vector space X together with a norm is called a normed linear space, a normed vector space, or simply a normed space. Definition 1.3. Let I be a finite or countable index set (for example, I = {1,..., N} if finite, or I = N or Z if infinite). Let w : I [0, ). Given a sequence of scalars x = (x i ) i I, set ( ) 1/p x i p w(i) p, 0 < p <, x p,w = i I x i w(i), p =, sup i I where these quantities could be infinite. Then we set } l p w {x (I) = = (x i ) i I : x p <. We call l p w(i) a weighted l p space, and often denote it just by l p w (especially if I = N). If w(i) = 1 for all i, then we simply call this space l p (I) or l p and write p instead of p,w. Date: April 11, These notes closely follow and expand on the text by John B. Conway, A Course in Functional Analysis, Second Edition, Springer,
2 2 CHRISTOPHER HEIL Exercise 1.4. Show that if 1 p then p,w defines a seminorm on l p w, and it is a norm if w(i) > 0 for all i. In particular, if I = {1,..., n} then l p w = Fn, and each choice of p and w gives a seminorm or norm on F n. Hints: The Triangle Inequality on l p is often called Minkowski s Inequality. It is easy to prove if p = 1 or p =. There are several ways to prove it for other p. One ways is to begin with x + y p p = i I x i + y i p = i I x i + y i p 1 x i + y i i I x i + y i p 1 x i + i I x i + y i p 1 y i. Then apply Hölder s Inequality to each sum using the exponent p on the first factor and p for the second (recall that p = p/(p 1)). Then divide both sides by x + y p 1 p. Definition 1.5. Let (X, Ω, µ) be a measure space. Given a measurable f : X [, ] (if F = R) or f : X C (if F = C), set ( 1/p f(x) dµ(x)) p, 0 < p <, f p = X ess sup f(x), p =, x X where these quantities could be infinite. Define { } L p (X) = f : X [, ] or C : f p <. Other notations for L p (X) are L p (µ), L p (X, µ), L p (dµ), L p (X, dµ), etc. When we write L p (R n ), it will be assumed that µ is Lebesgue measure on R n, unless specifically stated otherwise. In this case we will write dx instead of dµ(x). The space l p w (I) is a special case of Lp (X), where X = I and µ is a weighted counting measure on I. Exercise 1.6. Show that if 1 p then p is a seminorm on L p (X), and it is a norm if we identify functions that are equal almost everywhere. The Triangle Inequality on L p is often called Minkowski s Inequality, and its proof is similar to the proof of Minkowski s Inequality for l p. Exercise 1.7. Show that every subspace of a normed space is itself a normed space (using the same norm). Definition 1.8 (Distance). Let be a norm on X. Then the distance from x to y in X is d(x, y) = x y.
3 CHAPTER 3. BANACH SPACES 3 Exercise 1.9. Show that d(, ) defines a metric on X (see Appendix A). Since X has a metric and hence has an associated topology, all the standard topological notions (open/closed sets, convergence, etc.) apply to X. For convenience, we give some explicit definitions and facts relating to these topics for the setting of normed spaces. Definition 1.10 (Convergence). Let X be a normed linear space (such as an inner product space), and let {f n } n N be a sequence of elements of X. (a) We say that {f n } n N converges to f X, and write f n f, if i.e., lim f f n = 0, n ε > 0, N > 0 such that n > N = f f n < ε. (b) We say that {f n } n N is Cauchy if ε > 0, N > 0 such that m, n > N = f m f n < ε. Exercise Let X be a normed linear space. Prove the following. (a) Reverse Triangle Inequality: f g f g. (b) Continuity of the norm: f n f = f n f. (c) Continuity of vector addition: f n f and g n g = f n + g n f + g. (d) Continuity of scalar multiplication: f n f and α n α = α n f n αf. (e) All convergent sequences are bounded, and the limit of a convergent sequence is unique. (f) Every Cauchy sequence is bounded. (g) Every convergent sequence is Cauchy. Exercise Let {f n } n N be a sequence of vectors in a normed space X. Show that if f n f n+1 < 2 n for every n, then {f n } n N is Cauchy. Exercise Let l p w (I) be the weighted lp space defined in Exercise 1.3, where we assume w(i) > 0 for all i I. Let {x n } n N be a sequence of vectors in l p w(i), and let x be a vector in l p w(i). Write the components of x n and x as x n = (x n (1), x n (2),... ) and x = (x(1), x(2),... ). (a) Prove that if x n x (i.e., x x n p,w 0), then x n converges componentwise to x, i.e., for each fixed k we have lim n x n (k) = x(k). (b) Prove that if I is finite then the converse is also true, i.e., componentwise convergence implies convergence with respect to the norm p,w. (c) Prove that if I is infinite then componentwise convergence does not imply convergence in the norm of l w p (I).
4 4 CHRISTOPHER HEIL It is not true in an arbitrary normed space that every Cauchy sequence must converge. Normed spaces which do have the property that all Cauchy sequences converge are given a special name. Definition 1.14 (Banach Space). A normed space X is called a Banach space if it is complete, i.e., if every Cauchy sequence is convergent. That is, {f n } n N is Cauchy in X = f X such that f n f. Exercise Show that the weighted l p space l p w (I) defined in Exercise 1.3 is a Banach space if w(i) > 0 for all i I. Hints: Consider the case I = N. Suppose that {x n } n N is a Cauchy sequence in l p w. Each x n is a sequence of scalars. Write out the components of x n as Prove that for a fixed component k we have x n = (x n (1), x n (2),... ). x m (k) x n (k) C x m x n p,w, where C is a fixed constant (determined by the weight and by k but independent of m and n). Conclude that with k fixed, {x n (k)} n N is a Cauchy sequence of scalars, and hence converges. Define x(k) = lim n x n (k) and set x = (x(1), x(2),... ). Then we have constructed a candidate limit for the sequence {x n } n N. However, so far we only have that each individual component of x n converges to the corresponding component of x, i.e., x n converges componentwise to x. This is not enough: to complete the proof you must show that x n x in the norm of l p w. Use the fact that {x n} n N is Cauchy together with the componentwise convergence to show that x x n l p w 0 as n. Compare this proof to the proof of Theorem 2.18 given below. Exercise Show that the space L p (X) defined in Example 1.5 is a Banach space if we identify functions that are equal almost everywhere. This is called the Riesz Fisher Theorem. Hint: The argument is similar in spirit but more subtle than the one used to prove that l p w (I) is a Banach space. First find a candidate limit and then show that the sequence converges in norm to this limit. The next two exercises will be useful to us later. Exercise Let X be a normed linear space. If {f n } n N is a Cauchy sequence in X and there exists a subsequence {f nk } k N that converges to f X, then f n f. Solution Choose any ε > 0. Since {f n } n N is Cauchy, there is an N such that f m f n < ε for m, 2 n > N. Also, there is a k such that n k > N and f f nk < ε. Hence for n > N we have 2 f f n f f nk + f nk f n < ε 2 + ε 2 = ε.
5 CHAPTER 3. BANACH SPACES 5 Thus f n f. Exercise Let {f n } n N be a Cauchy sequence in a normed space X. Show that there exists a subsequence {f nk } k N such that k N, f nk+1 f nk < 2 k. Solution Since {f n } n N is Cauchy, we can find an n 1 such that Then we can find an n 2 > n 1 such that m, n n 1 = f m f n < 2 1. m, n n 2 = f m f n < 2 2. Continuing in this way, we inductively construct n 1 < n 2 < such that for each k, m, n n k = f m f n < 2 k. In particular, since n k+1 > n k, we have f nk+1 f nk < 2 k. Definition 1.19 (Convergent Series). Let {f n } n N be a sequence of elements of a normed linear space X. Then the series n=1 f n converges and equals f X if the partial sums s N = N n=1 f n converge to f, i.e., if N f s N = f f n 0 as N. n=1 Definition 1.20 (Absolutely Convergent Series). Let X be a normed space and let {f n } n N be a sequence of elements of X. If f n <, then we say that the series n f n is absolutely convergent in X. n=1 Note that the definition of absolute convergence does not by itself tell us that the series n f n actually converges. That is always true if X is a Banach space, but need not be true if X is not complete. Exercise Let X be a Banach space and let {f n } n N be a sequence of elements of X. Prove that if n f n < then the series n f n does converge in X. Hint: You must show that the sequence of partial sums {s N } N N converges. Since X is a Banach space, you just have to show that this sequence is Cauchy. The converse of this exercise is also true, and is often a useful method for proving that a given normed space is a Banach space.
6 6 CHRISTOPHER HEIL Proposition Let X be a normed space. Prove that X is a Banach space if and only if every absolutely convergent series in X converges in X. Proof.. This is Exercise Suppose that every absolutely convergent series is convergent. Let {f n } n N be a Cauchy sequence in X. By Exercise 1.18, there exists a subsequence {f nk } k N such that f nk+1 f nk < 2 k for every k. The series k (f n k+1 f nk ) is absolutely convergent, because f nk+1 f nk 2 k = 1 <. k=1 By hypothesis, we conclude that k (f n k+1 f nk ) converges in X, say to f. In terms of the partial sums, this says that k f nk+1 f n1 = (f nj+1 f nj ) f as k, j=1 or f nk g = f +f n1 as k. Thus {f n } n N is a Cauchy sequence which has a subsequence that converges to g. It therefore follows from Exercise 1.17 that f n g. Therefore X is complete. k=1 Example 1.23 (The Harmonic Series). To illustrate the difference between convergence and absolute convergence, consider the onedimensional case, i.e., X = F, the field of scalars. Let x n = 1. Then n n x n = 1 n is the harmonic series, and the partial sums s n N = N n=1 this series are unbounded. Thus the harmonic series does not converge. Since s N, we usually say that 1 n diverges to infinity, and write 1 n n =. n On the other hand, consider the alternating series n ( 1)n 1. Since the terms alternate n signs and since 1 0, it follows that this series does converge to a finite scalar (in fact, it n converges to ln 2). However, it does not converge absolutely. 1 n of Definition 1.24 (Unconditionally Convergent Series). Let X be a Banach space and let {f n } n N be a sequence of elements of X. The series f = n=1 f n is said to converge unconditionally if every rearrangement of the series converges. That is, f = n=1 f n converges unconditionally if for each bijection σ : N N the series converges. n=1 f σ(n) Remark It is not obvious, but it can be shown that if n=1 f σ(n) is unconditionally convergent, then every rearrangement of the series must converge to the same sum, i.e., there is a single f such that f = n=1 f σ(n) for every permutation σ.
7 CHAPTER 3. BANACH SPACES 7 Exercise Let X be a Banach space. Prove that if a series f = n=1 f n converges absolutely, then it converges unconditionally. Remark In finite dimensions the converse to Exercise 1.26 is true, i.e., if X is finitedimensional and a series f = n=1 f n converges unconditionally, then it converges absolutely. However, this fails in infinite dimensions. Example 1.28 (The Harmonic Series Revisited). To illustrate the importance of unconditional convergence, again consider X = F and the alternating series n ( 1)n 1. We know n that this series converges, but does not converge absolutely. Now consider what happens if we change the order of summation. Let p n = 1 and 2n q n = 1, i.e., the p 2n+1 n are the positive terms from the alternative series and the q n are the absolute values of the negative terms. Each series n p n and n q n diverges. Hence there must exist an m 1 > 0 such that p p m1 > 1. Then, there must exist an m 2 > m 1 such that Continuing in this way, we see that p p m1 q 1 + p m p m2 > 2. p p m1 q 1 + p m p m2 q 2 + is a rearrangement of n ( 1)n 1 which diverges to +. n In the same way, we can construct a rearrangement which diverges to, which converges to any given real number r, or which simply oscillates without ever converging. Moreover, the same can be done for any series of real scalars which converges conditionally. The following is an equivalent formulation of unconditional convergence. Proposition Let X be a Banach space and let {f n } n N be a sequence of elements of X. Then the series f = n=1 f n converges unconditionally if and only if it converges with respect to the net of finite subsets of N, i.e., if ε > 0, finite F 0 N such that finite F F 0, f f n < ε. n F Definition 1.30 (Topology). Let X be a normed linear space. (a) The open ball in X centered at x X with radius r > 0 is B r (x) = B(x, r) = {y X : x y < r}. (b) A subset U X is open if x U, r > 0 such that B r (x) U. (c) A subset F X is closed if X \ F is open.
8 8 CHRISTOPHER HEIL Definition 1.31 (Limit Points, Closure, Density). Let X be a normed linear space and let A X. (a) A point f A is called a limit point of A if there exist f n A with f n f such that f n f. (b) The closure of A is the smallest closed set Ā such that A Ā. Specifically, Ā = {F X : F is closed and F A}. (c) We say that A is dense in X if Ā = X. Exercise (a) The closure of A equals the union of A and all limit points of A: Ā = A {x X : x is a limit point of A} = {z X : y n A such that y n z}. (b) If X is a normed linear space, then the closure of an open ball B r (x) is the closed ball B r (x) = {y X : x y r}. (c) Prove that A is dense if and only if x X, ε > 0, y A such that x y < ε. Example The set of rationals Q is dense in the real line R. Exercise Let X be a normed linear space and let F X. Then F is closed F contains all its limit points. Solution. Suppose that F is closed but that there exists a limit point f that does not belong to F. By definition, there must exist f n F such that f n f. However, f X \ F, which is open, so there exists some r > 0 such that B(f, r) X \ F. Yet there must exist some f n with f f n < r, so this f n will belong to X \ F, which is a contradiction.. Exercise. Remark Some authors make the restriction that, when dealing with normed spaces, the terminology subspace is used only for closed subspaces. Other authors use the terminology linear manifold to denote a subspace that need not be closed. To avoid ambiguity, we will use the following terminology. Definition (a) A subset Y of a vector space X is a subspace of X if it is closed under both vector addition and scalar multiplication, i.e., if for all u, v Y and α, β F we have αu + βv Y. (b) A subset Y of a normed linear space X is a closed subspace of X if it is a subspace and it is closed with respect to the norm of X.
9 CHAPTER 3. BANACH SPACES 9 Exercise In finite dimensions, all subspaces are closed sets (this will be proved in Proposition 3.5). This exercise demonstrates that, in infinite dimensions, subspaces need not be closed sets. (a) Fix 1 p. Prove that c 00 = {x = (x 1,..., x N, 0, 0,... ) : N > 0, x 1,..., x N F} is a subspace of l p (N) that is not closed (with respect to the l p norm). Prove that c 00 is dense in l p (N) if p <, but that it is not dense in l (N). (b) Define c 0 = {x = (x k ) k=1 : lim k x k = 0}. Prove c 0 is a closed subspace of l (N) (in l norm). Prove that c 0 is the closure of c 00 (under the l norm). (c) Fix 1 p. Prove that C c (R n ), the space of continuous, compactly supported functions on R n, is a subspace of L p (R n ) that is not closed. Prove that C c (R n ) is dense in L p (R n ) if p <, but not if p =. Hints: For a continuous function we have f = sup f(x). Hence, if {f n } n N is a sequence of continuous functions in L (R n ) that converges in L norm, then it converges uniformly. From undergraduate real analysis, we know that the limit of a uniformly convergent sequence of continuous functions is continuous. (d) Let C 0 (R n ) be the set of all continuous functions f : R n F such that lim f(x) = 0. (1.1) x More precisely, (1.1) means that for every ε > 0 there exists a compact set K such that f(x) < ε for all x / K. Prove that C 0 (R n ) is a closed subspace of L (R n ) (closed in L norm). Prove that C 0 (R n ) is the closure of C c (R n ) (under the L norm). (e) Let C b (R n ) be the set of all bounded, continuous functions f : R n F. Prove that C b (R n ) is a closed subspace of L (R n ). (f) Fix 1 p, and let E be a (Lebesgue) measurable subset of R n. Let M = {f L p (R n ) : supp(f) E}. Prove that M is a closed subspace of L p (R n ). Remark (a) We took the domain in the preceding exercise to be R n just for convenience; the definitions of the spaces C c, C 0, C b can be extended to domains that are more general topological spaces. (b) Beware that some authors use the notation C 0 for the space that we are calling C c! Exercise Let X be a Banach space and let M be a subspace of X. Then M is itself a Banach space (using the norm from X) if and only if M is closed.
10 10 CHRISTOPHER HEIL Solution. Suppose that M is a Banach space. Let f be any limit point of M, i.e., suppose f n M and f n f. Then {f n } is a convergent sequence in X, and hence is Cauchy in X. Since each f n belongs to M, it is also Cauchy in M. Since M is a Banach space, {f n } must therefore converge in M, i.e., f n g for some g M. However, limits are unique, so f = g M. Therefore M contains all its limit points and hence is closed.. Exercise. Notation 1.40 (Notation for Closed Subspaces). Since we will often deal with closed subspaces of a Banach space, we declare that the notation M X means that M is a closed subspace of the Banach space X. Exercise Find examples of normed spaces that are not Banach spaces. Hint: Look for examples of normed spaces Y which are subspaces of a larger Banach space X. Remark Is every normed space Y a subspace of a larger Banach space? The answer is yes, given a normed space Y it is always possible to construct a Banach space X Y such that the norm on X, when restricted to Y, is the same as the norm on Y, and Y is dense in X with respect to that norm. This space is called the completion of Y. Definition Suppose that X is a normed linear space with respect to a norm a and also with respect to another norm b. Then we say that these norms are equivalent if there exist constants C 1, C 2 > 0 such that f X, C 1 f a f b C 2 f a. (1.2) Observe that equation (1.2) can be rearranged to read 1 C 2 f b f a 1 C 1 f b. Exercise Let X be a vector space. If a and b are two norms on X, define a b if a and b are equivalent. Prove that is an equivalence relation on the class of norms on X. The next result will show that equivalent norms define the same topology and the same convergence criterion. Proposition Let a and b be two norms on a vector space X. Then the following statements are equivalent. (a) a and b are equivalent norms. (b) a and b define the same topologies on X. That is, if U X, then U is open with respect to a if and only if it is open with respect to b.
11 CHAPTER 3. BANACH SPACES 11 (c) a and b define the same convergence criterion. That is, if {f n } n N is a sequence in X and f X, then lim f f n a = 0 lim f f n b = 0. n n Proof. (b) (a). Assume that statement (b) holds. Let Br a (f) and Br(f) b denote the open balls of radius r centered at f X with respect to a and b. Since B1 a (0) is open with respect to a, the hypothesis that statement (b) holds implies that B1 a (0) is open with respect to b. Therefore, since 0 B1 a (0), there must exist some r > 0 such that Br(0) b B1(0). a Now choose any f X and any ε > 0. Then so (r ε) f f b B b r (0) Ba 1 (0), (r ε) f a < 1. f b Rearranging, this implies (r ε) f a < f b. Since this is true for every ε, we conclude that r f a f b. A symmetric argument, interchanging the roles of the two norms, shows that there exists an s > such that f b s f a for every f X. Hence the two norms are equivalent The remaining implications are exercises. Hints on (c) (a): Suppose that statement (c) holds. Show that this implies that the function ν : X R given by ν(f) = f b is continuous with respect to the norm a. Since ( 1, 1) is an open subset of R, it follows that ν 1 ( 1, 1) is an open subset of X (with respect to a ). Since 0 ν 1 ( 1, 1), there must exist an r > 0 such that Br(0) b ν 1 ( 1, 1). But ν 1 ( 1, 1) = B1(0), a so the remainder of the proof proceeds exactly like the proof of (b) (a). 2. Linear Operators on Normed Spaces Definition 2.1 (Notation for Operators). Let X, Y be vector spaces. Let T : X Y be a function (= operator = transformation) mapping X into Y. We write either T (f) or T f to denote the image of an element f X. (a) T is linear if T (αf + βg) = αt (f) + βt (g) for every f, g X and α, β F. (b) T is injective if T (f) = T (g) implies f = g. (c) The kernel or nullspace of T is ker(t ) = {f X : T (f) = 0}. (c) The range of T is range(t ) = {T (f) : f X}. (d) The rank of T is the dimension of its range, i.e., rank(t ) = dim(range(t )). particular, T is finiterank if range(t ) is finitedimensional. (d) T is surjective if range(t ) = Y. In
12 12 CHRISTOPHER HEIL (e) T is a bijection if it is both injective and surjective. (f) We use the notation I or I X to denote the identity map of a space X onto itself. Definition 2.2 (Continuous and Bounded Operators). Let X, Y be normed linear spaces, and let L: X Y be a linear operator. (a) L is continuous at a point f X if f n f in X implies Lf n Lf in Y. (b) L is continuous if it is continuous at every point, i.e., if f n f in X implies Lf n Lf in Y for every f. (c) L is bounded if there exists a finite K 0 such that f X, Lf K f. Note that Lf is the norm of Lf in Y, while f is the norm of f in X. (d) The operator norm of L is L = sup Lf. f =1 (e) We let B(X, Y ) denote the set of all bounded linear operators mapping X into Y, i.e., B(X, Y ) = {L: X Y : L is bounded and linear}. If X = Y then we write B(X) = B(X, X). (f) If Y = F then we say that L is a functional. The set of all bounded linear functionals on X is the dual space of X, and is denoted X = B(X, F) = {L: X F : L is bounded and linear}. Another common notation for the dual space is X. Note that since the norm on F is just absolutely value, the operator norm of a linear functional L X = B(X, F) is L = sup Lf. f =1 Exercise 2.3. Show that if T : X Y is linear and continuous, then ker(t ) is a closed subspace of X and that range(t ) is a subspace of Y. Must range(t ) be a closed subspace? Exercise 2.4. Let X, Y be normed linear spaces. Let L: X Y be a linear operator. (a) L is injective if and only if ker L = {0}. (b) If L is a bijection then L 1 : Y X is also a linear bijection. (c) L is bounded if and only if L <. (d) If L is bounded then Lf L f for every f X (note that three different meanings of the symbol appear in this statement!).
13 CHAPTER 3. BANACH SPACES 13 (e) If L is bounded then L is the smallest value of K such that Lf K f holds for all f X. Lf (f) L = sup Lf = sup f 1 f 0 f. Exercise 2.5. Show that B(X, Y ) is a subspace of the vector space V consisting of ALL functions A: X Y. Moreover, show that the operator norm is a norm on the space B(X, Y ), i.e., (a) 0 L < for all L B(X, Y ), (b) L = 0 if and only if L = 0 (the zero operator that sends every element of X to the zero vector in Y ), (c) αl = α L for every L B(X, Y ) and every α F, (d) L + K L + K for every L, K B(X, Y ). The preceding exercise shows that B(X, Y ) is a normed linear space, and we will show in Theorem 2.18 that it is a Banach space if Y is a Banach space. However, in addition to vector addition and scalar multiplication operations, there is a third operation that we can perform with functions: composition. Exercise 2.6. Prove that the operator norm is submultiplicative, i.e., prove if A B(X, Y ) and B B(Y, Z), then BA B(X, Z) and BA B A. (2.1) In particular, when X = Y = Z, we see that B(X) is closed under compositions. The space B(X) is an example of an algebra. Exercise 2.7. Let F n be ndimensional Euclidean space over F, under the Euclidean norm, and let Y be any normed linear space. Prove that if L: F n Y is linear, then L is bounded. Hint: If x = (x 1,..., x n ) F n then x = x 1 e 1 + +x n e n where {e 1,..., e n } is the standard basis for F n. Use the Triangle and the CauchySchwarz Inequalities. Solution Given x F n, we have n Lx = L(x k e k ) k=1 n x k Le k k=1 ( n ) 1/2 ( n ) 1/2 x k 2 Le k 2 = C x, k=1 where C = ( n k=1 Le k 2) 1/2. Hence L is bounded. Remark 2.8. We will prove later that if X is any finitedimensional vector space and is any norm on X, then any linear function L: X Y is bounded. To do this we will use the fact (that we will prove later) that all norms on a finitedimensional space are equivalent. k=1
14 14 CHRISTOPHER HEIL The next lemma is a standard fact about continuous functions. L 1 (U) denotes the inverse image of U Y, i.e., L 1 (U) = {f X : Lf U}. Exercise 2.9. Let X, Y be normed linear spaces. Let L: X Y be linear. Then ( ) L is continuous U open in Y = L 1 (U) open in X. Solution. Suppose that L is continuous and that U is an open subset of Y. We will show that X \ L 1 (U) is closed by showing that it contains all its limit points. Suppose that f is a limit point of X \L 1 (U). Then there exist f n X \L 1 (U) such that f n f. Since L is continuous, this implies Lf n Lf. However, f n / L 1 (U), so Lf n / U, i.e., Lf n Y \ U, which is a closed set. Therefore Lf Y \ U, and hence f X \ L 1 (U). Thus X \ L 1 (U) is closed, so L 1 (U) is open.. Exercise. Theorem 2.10 (Equivalence of Bounded and Continuous Linear Operators). Let X, Y be normed linear spaces, and let L: X Y be a linear mapping. Then the following statements are equivalent. (a) L is continuous at some f X. (b) L is continuous at f = 0. (c) L is continuous. (d) L is bounded. Proof. (c) (d). Suppose that L is continuous but unbounded. Then L =, so there must exist f n X with f n = 1 such that Lf n n. Set g n = f n /n. Then g n 0 = g n = f n /n 0, so g n 0. Since L is continuous and linear, this implies Lg n L0 = 0, and therefore Lg n 0 = 0. But Lg n = 1 n Lf n 1 n n = 1 for all n, which is a contradiction. Hence L must be bounded. (d) (c). Suppose that L is bounded, so L <. Suppose that f X and that f n f. Then f n f 0, so i.e., Lf n Lf. Thus L is continuous. Lf n Lf = L(f n f) L f n f 0, The remaining implications are exercises. Definition 2.11 (Isometries and Isometric Isomorphisms). Let X, Y be normed linear spaces and let L: X Y be linear.
15 CHAPTER 3. BANACH SPACES 15 (a) If Lf = f for all f X then L is called an isometry or is said to be normpreserving. (b) An isometry L: X Y that is a bijection is called an isometric isomorphism. In this case we say that X and Y are isometrically isomorphic. Exercise (a) Suppose that L: X Y is an isometry. Prove that L is injective and find L. (b) Find an example of an isometry that is not surjective. Contrast this with the fact that if A: C n C n is linear, then A is injective if and only if it is surjective. (c) Prove that if L: X Y is an isometric isomorphism, then L 1 : Y X is also an isometric isomorphism. Exercise 2.13 (Unilateral Shift Operators). Fix 1 p. (a) Define L: l p (N) l p (N) by L(x) = (x 2, x 3,... ) for x = (x 1, x 2,... ) l p (N). Prove that this leftshift operator is bounded, linear, surjective, not injective, and is not an isometry. Find L. (b) Define R: l p (N) l p (N) by R(x) = (0, x 1, x 2, x 3,... ) for x = (x 1, x 2,... ) l p (N). Prove that this rightshift operator is bounded, linear, injective, not surjective, and is an isometry. Find R. (c) Compute LR and RL. Contrast this computation with the fact that in finite dimensions, if A, B : C n C n are linear maps (hence correspond to multiplication by n n matrices), then AB = I implies BA = I and conversely. Definition 2.14 (Topological Isomorphisms). Let X, Y be normed linear spaces. If L: X Y is a linear bijection such that both L and L 1 are bounded, then L is called a topological isomorphism. In this case we say that X and Y are topologically isomorphic. Remark We will see later that if X and Y are Banach spaces and L: X Y is a bounded bijection, then L 1 is automatically bounded and hence L is a topological isomorphism. Thus, when X and Y are Banach spaces, every continuous invertible map is a topological isomorphism. Sometimes the abbreviation isomorphism or invertible map is used to mean a topological isomorphism, but it should be noted that these terms are ambiguous. Exercise Prove that if L: X Y is a topological isomorphism, then 1 f X, f Lf L f. L 1 Exercise Let X be a Banach space and Y a normed linear space. Suppose that L: X Y is bounded and linear. Prove that if there exists c > 0 such that Lf c f for all f X, then L is injective and range(l) is closed.
16 16 CHRISTOPHER HEIL The next theorem shows that B(X, Y ) is a Banach space whenever Y is a Banach space. Theorem If X is a normed space and Y is a Banach space, then B(X, Y ) is a Banach space. Proof. Assume that {A n } n N is a sequence of operators A n B(X, Y ) that is Cauchy in operator norm. For any given f X, we have A m f A n f A m A n f, so we conclude that {A n f} n N is a Cauchy sequence of vectors in Y. Since Y is complete, this sequence must converge, say Af n g Y. Define Af = g. This gives us a candidate limit operator A. Exercise: Show that A defined in this way is a linear operator. To show that A is bounded, first recall that all Cauchy sequences are bounded. Hence we must have C = sup A n <. If f X, then since A n f Af we have Af = lim A n f sup n n N A n f sup A n f = C f. n N Hence A is bounded, and A C. Finally, we must show that A n A in operator norm. Fix any ε > 0. Then there exists an N such that m, n > N = A m A n < ε 2. Choose any f X with f = 1. Then since A m f Af, there exists an m > N such that Hence for any n > N we have Af A m f < ε 2. Af A n f Af A m f + A m f A n f Af A m f + A m A n f < ε 2 + ε 2 = ε. Taking the supremum over all unit vectors, we conclude that A A n ε for all n > N. Thus A n A. Corollary If X is a normed space, then its dual space X = B(X, F) is a Banach space. The next exercise deals with the problem of extending an operator defined only a dense subspace to the entire space. Exercise 2.20 (Extension of Bounded Operators). Let Y be a dense subspace of a normed space X, and let Z be a Banach space. (a) Suppose that L: Y Z is a bounded linear operator. Show that there exists a unique bounded linear operator L: X Z whose restriction to Y is L. Prove that L = L. (b) Prove that if L: Y range(l) is a topological isomorphism, then L: X range(l) is also.
17 CHAPTER 3. BANACH SPACES 17 (c) Prove that if L: Y range(l) is an isometry, L: X range(l) is also. Solution (a) Fix any f X. Since Y is dense in X, there exist g n Y such that g n f. Since L is bounded, we have Lg m Lg n L g m g n. But {g n } n N is Cauchy in X, so this implies that {Lg n } n N is Cauchy in Z. Since Z is a Banach space, we conclude that there exists an h Z such that Lg n h. Define Lf = h. To see that L is welldefined, suppose that we also had g n f for some g n Y. Then Lg n Lg n L g n g n 0. Since Lg n h, it follows that Lg n = Lg n +(Lg n Lg n ) h + 0 = h. Thus L is welldefined. To see that L is linear, suppose that f, g X are given and c F. Then there exist f n, g n Y such that f n f and g n g. Since cf n + g n cf + g and L(cf n + g n ) = clf n + Lg n c Lf + Lg, by definition we have that L(cf + g) = c Lf + Lg. To see that L is an extension of L, suppose that g Y is fixed. If we set g n = g, then g n g and Lg n Lg, so by definition we have Lg = Lg. Hence the restriction of L to Y is L. Consequently, L = sup Lf f X, f =1 sup Lf = f Y, f =1 sup Lf = L. f Y, f =1 Finally, suppose that f X. Then there exist g n Y such that g n f and Lg n Lf, so Lf = lim Lg n lim L g n = L f. n n Hence L L. Combining this with the opposite inequality derived above, we conclude that L = L. (b) Suppose that L: Y range(y ) is a topological isomorphism. We already know that L: X range( L) is bounded. We need to show that L is injective, that L 1 : range( L) X is bounded, and that range( L) = range(l). Fix any f X. Then there exist g n Y such that g n f and Lg n Lf. Since L is a topological isomorphism, we have by Exercise 2.16 that g n L 1 Lg n. Hence Lf = lim Lg n n lim n g n L 1 = Consequently, L is injective and for any h range( L) we have f L 1. L 1 h L 1 L( L 1 h) = L 1 h. Therefore L 1 : range( L) X is bounded. It remains only to show that the range of L is the closure of the range of L. If f X, then by definition there exist g n Y such that g n f and Lg n Lf. Hence Lf range(l), so range( L) range(l).
18 18 CHRISTOPHER HEIL On the other hand, suppose that h range(l) Then there exist g n Y such that Lg n h. Since L 1 is bounded and L extends L, we conclude that g n = L 1 (Lg n ) L 1 (h). Hence f = L 1 (h), so f range( L). 3. FiniteDimensional Normed Spaces In this section we will prove some basic facts about finitedimensional spaces. First, recall that a finitedimensional vector space has a finite basis, which gives us a natural notion of coordinates of a vector, which in turn yields a linear bijection of X onto F n for some n. Example 3.1 (Coordinates). Let X be a finitedimensional vector space over F. Then X has a finite basis, say B = {e 1,..., e n }. Every element of X can be written uniquely in this basis, say, x = c 1 (x) e c n (x) e n, x X. Define the coordinates of x with respect to the basis B to be c 1 (x) [x] B =.. c n (x) Then the mapping T : X F n given by x [x] B is, by definition of basis, a linear bijection of X onto F n. Since we already know how to construct many norms on F n, by transferring these to X we obtain a multitude of norms for X. Exercise 3.2 (l p w Norms on X). Let X be a finitedimensional vector space over F and let B = {e 1,..., e n } be any basis. Fix any 1 p and any weight w : {1,..., n} (0, ). Using the notation of Example 3.1, given x X define ( n ) 1/p x p,w = c k (x) p w(k) p, 1 p <, [x]b p,w = k=1 max c k (x) w(k), p =. k Note that while we use the same symbol p,w to denote a function on X and on F n, by context it has different meanings depending on whether it is being applied to an element of X or to an element of F n. Prove the following. (a) p,w is a norm on X. (b) x [x] B is a isometric isomorphism of X onto F n (using the norm p,w on X and the norm p,w on F n ).
19 CHAPTER 3. BANACH SPACES 19 (c) Let {x n } n N be a sequence of vectors in X and let x X. Prove that x n x with respect to the norm p,w on X if and only if the coordinate vectors [x n ] B converge componentwise to the coordinate vector [x] B. (d) X is complete in the norm p,w. Now we can show that all norms on a finitedimensional space are equivalent. Theorem 3.3. If X is a finitedimensional vector space over F, then any two norms on X are equivalent. Proof. Let B = {e 1,..., e n } be any basis for X, and let be the norm on X defined in Exercise 3.2. Since equivalence of norms is an equivalence relation, it suffices to show that an arbitrary norm on X is equivalent to. Using the notation of Exercise 3.1, given x X we can write x uniquely as x = c 1 (x) e c n (x) e n. Therefore, n ( n (max x c k (x) e k e k ) c k (x) ) = C 2 x, k=1 k=1 where C 2 = n k=1 e k is a nonzero constant independent of x. It remains to show that there is a constant C 1 > 0 such that C 1 x x for every x. First, let D = {x X : x = 1} be the l unit circle in X. Exercise: Show that D is compact (with respect to the norm ). Hints: Suppose that {x n } n N is a sequence of vectors in D. Then for each n, we have c k (x n ) = 1 for some k {1,..., n}. Hence there must be some k such that c k (x nj ) = 1 for infinitely many j. Since {c 1 (x nj )} j N is an infinite sequence of scalars in the compact set {c F : c 1}, we can select a subsequence whose first coordinates converge. Repeat for each coordinate, and remember that the kth coordinate is always 1. Hence we can construct a subsequence that converges to x D with respect to the l norm. Our next goal is to show that D is also compact with respect to the norm. Let {x n } n N be any sequence of vectors in D. Since D is compact with respect to, there exists a subsequence {x nk } k N and an x D such that x x nk 0. But then x x nk C 2 x x nk 0, so {x nk } k N is a subsequence that converges to x X with respect to. Hence D is compact with respect to. Now, is a continuous function with respect to the convergence criteria defined by (this is part (b) of Exercise 1.11). The set D is compact with respect to the topology defined by. A realvalued continuous function on a compact set must achieve a maximum and minimum on that set. Hence, there must exist constants m and M such that m x M for all x D. Since x D if and only if x = 1, this implies that x X, m x x M x. If we had m = 0, then this would imply that there is an x D such that x = 0. But then x = 0, which implies x = 0, contradicting the fact that x D. Hence m > 0, so we can take C 1 = m. k
20 20 CHRISTOPHER HEIL Consequently, from now on we need not specify the norm on a finitedimensional vector space X we can take any norm that we like whenever we need it. Exercise 3.4. Let F m n be the space of all m n matrices with entries in F. F m n is naturally isomorphic to F mn. Prove that if a is any norm on F m n, b is any norm on F n k, and c is any norm on F m k, then there exists a constant C > 0 such that A F m n, B F n k, AB c C A a B b. Proposition 3.5. If M is a subspace of a finitedimensional vector space X, then M is closed. Proof. Let be any norm on X. Suppose that x n M and that x n y X. Set M 1 = span{m, y} = {m + cy : m M, c F}. Then M 1 is finitedimensional subspace of X. Moreover, every element z M 1 can be written uniquely as z = m(z) + c(z)y where m(z) M and c(z) is a scalar. For z M 1 define z M1 = m(z) + c(z). Exercise: Show that M1 is a norm on M 1. Since is also a norm on M 1 and all norms on a finitedimensional space are equivalent, we conclude that there is a constant C > 0 such that z M1 C z for all z M 1. Since c(x n ) = 0 for every n, we therefore have c(y) = c(y) c(x n ) = c(y x n ) m(y x n ) + c(y x n ) Therefore c(y) = 0, so y M. = y x n M1 C y x n 0. If X is any normed vector space and M is a finitedimensional subspace of X, then a proof identical to the one used in the preceding proposition, except using the given norm on X, shows that M is closed. Exercise 3.6. Let X be a normed linear space. If M is a finitedimensional subspace of X, then M is closed. Exercise 3.7. Let X be a finitedimensional normed space, and let Y be a normed linear space. Prove that if L: X Y is linear, then L is bounded. The following lemma will be needed for Exercise 3.9.
21 CHAPTER 3. BANACH SPACES 21 Lemma 3.8 (F. Riesz s Lemma). Let M be a proper, closed subspace of a normed space X. Then for each ε > 0, there exists g X with g = 1 such that dist(g, M) = inf g f > 1 ε. f M Proof. Choose any u X \ M. Since M is closed, we have a = dist(u, M) = inf u f > 0. f M Fix δ > 0 small enough that that a u v < a + δ. Set a a+δ > 1 ε. By definition of infimum, there exists v M such g = u v u v, and note that g = 1. Given f M we have h = v + u v f M, so g f = u v u v f u h u v = u v > a a + δ > 1 ε. Exercise 3.9. Let X be a normed linear space. Let B = {x X : x 1} be the closed unit ball in X. Prove that if B is compact, then X is finitedimensional. Hints: Suppose that X is infinitedimensional. Given any nonzero e 1 with e 1 1, by Lemma 3.8 there exists e 2 X \span{e 1 } with e 2 1 such that e 2 e 1 > 1. Continue in 2 this way to construct vectors e k such that {e 1,..., e n } are independent for any n. Conclude that X is infinitedimensional. Definition We say that a normed linear space X is locally compact if for each f X there exists a compact K X with nonempty interior K such that f K. In other words, X is locally compact if for every f X there is a neighborhood of f that is contained in a compact subset of X. For example, F n is locally compact. With this terminology, we can reword Exercise 3.9 as follows. Exercise Let X be a normed linear space. (a) Prove that if X is locally compact, then X is finitedimensional. (b) Prove that if X is infinitedimensional, then no nonempty open subset of X has compact closure.
22 22 CHRISTOPHER HEIL 4. Quotients and Products of Normed Spaces Any vector space is an abelian group under the operation of vector addition. So, if you are familiar with the basic notions of abstract algebra, the concept of a coset will be familiar to you. However, even if you have not studied abstract algebra, the idea of a coset in a vector space is very natural. Example 4.1 (Cosets in R 2 ). Consider the vector space X = R 2. Let M be any onedimensional subspace of R 2, i.e., M is a line in R 2 through the origin. A coset of M is simply a rigid translate of M by a vector in R 2. For concreteness, let us specifically consider the case where M is the x 1 axis in R 2, i.e., M = {(x 1, 0) : x 1 R}. Then given a vector y = (y 1, y 2 ) R 2, the coset y + M is the set y + M = {y + m : m M} = {(y 1 + x 1, y 2 + 0) : x 1 R} = {(x 1, y 2 ) : x 1 R}, which is the horizontal line at height y 2. This is not a subspace of R 2, but it is a rigid translate of the x 1 axis. Note that there are infinitely many different choices of y that give the same coset. Furthermore, we have the following facts for this particular setting. (a) Two cosets are either identical or entirely disjoint. (b) The union of all the cosets is all of R 2. (c) The set of distinct cosets is a partition of R 2. The preceding example is entirely typical. Definition 4.2 (Cosets). Let M be a subspace of a vector space X. Then the cosets of M are the sets f + M = {f + m : m M}, f M. Exercise 4.3. Let X be a vector space, and let M be a subspace of X. Given f, g M, define f g if f g M. Prove the following. (a) is an equivalence relation on X. (b) The equivalence class of f under the relation is [f] = f + M. (c) If f, g M then either f + M = g + M or (f + M) (g + M) =. (d) f + M = g + M if and only if f g M. (e) f + M = M if and only if f M. (f) If f X and m M then f + M = f + m + M. (g) The set of distinct cosets of M is a partition of X. Definition 4.4 (Quotient Space). If M is a subspace of a vector space X, then the quotient space X/M is X/M = {f + M : f X}.
23 CHAPTER 3. BANACH SPACES 23 Since two cosets of M are either identical or disjoint, the quotient space X/M is simply the set of all the distinct cosets of M. Example 4.5. Again let M = {(x 1, 0) : x 1 R} be the x 1 axis in R 2. Then, by Example 4.1, we have that R 2 /M = {y + M : y R 2 } = {(x 1, 0) + M : x 1 R}, i.e., R 2 /M is the set of all horizontal lines in R 2. Note that R 2 /M is in 11 correspondence with the set of distinct heights, i.e., there is a natural bijection of R 2 /M onto R. This is a special case of a more general fact that we will explore. Next we define two natural operations on the set of cosets: addition of cosets and multiplication of a coset by a scalar. These are defined formally as follows. Definition 4.6. Let M be a subspace of a vector space X. Given f, g X, define addition of cosets by (f + M) + (g + M) = (f + g) + M. Given f X and c F, define scalar multiplication by c(f + M) = cf + M. Remark 4.7. Before proceeding, we must show that these operations are actually welldefined. After all, there need not be just one f that determines the coset f + M how do we know that if we choose different vectors that determine the same cosets, we will get the same result when we compute (f + g) + M? We must show that f 1 + M = f 2 + M and g 1 + M = g 2 + M then (f 1 + g 1 ) + M = (f 2 + g 2 ) + M in order to know that Definition 4.6 makes sense. Proposition 4.8. If M is a subspace of a vector space X, then the addition of cosets of M given in Definition 4.6 is welldefined. Proof. Suppose that f 1 + M = f 2 + M and g 1 + M = g 2 + M. Then by Exercise 4.3(d) we know that f 1 f 2 = k M and g 1 g 2 = l M. If h (f 1 + g 1 ) + M then we have h = f 1 + g 1 + m for some m M. Hence h = (f 2 + k) + (g 2 + l) + m = (f 2 + g 2 ) + (k + l + m) (f 2 + g 2 ) + M. Thus (f 1 + g 1 ) + M (f 2 + g 2 ) + M, and the converse inclusion is symmetric. Exercise 4.9. Show that scalar multiplication is likewise welldefined. Now we can show that the quotient space is actually a vector space under the operations just defined. Proposition If M is a subspace of a vector space X, then X/M is a vector space with respect to the operations given in Definition 4.6.
24 24 CHRISTOPHER HEIL Proof. Addition of cosets is commutative because (f + M) + (g + M) = (f + g) + M = (g + f) + M = (g + M) + (f + M). The zero vector in X/M is the coset 0+M = M, because (f +M)+(0+M) = (f +0)+M = f + M. Exercise: Show that the remaining axioms of a vector space are satisfied. Definition 4.11 (Codimension). If M is a subspace of a vector space X, then the codimension of M is the dimension of X/M, i.e., codim(m) = dim(x/m). Example Let C(R) be space of continuous functions on R, and let P be the subspace containing the polynomials. Given f C(R), the coset f + P is f + P = {f + p : p is a polynomial}. Further, f + P = g + P if and only if f g is a polynomial. Thus, f + P can be thought of as f modulo the polynomials, i.e., it is the equivalence class obtained by identifying functions which differ by a polynomial. In the same way, a coset f + M can be thought of as the equivalence class obtained by identifying vectors which differ by an element of M. We can imagine the mapping that takes f to f + M as collapsing information modulo M. 1 Definition If M is a subspace of a vector space X, then the canonical projection or the canonical mapping of X onto X/M is π : X X/M defined by π(f) = f + M, f X. Exercise Let M be a subspace of a vector space X. (a) Prove that the canonical projection π is linear. (b) Prove that π is surjective and ker(π) = M. (c) Prove that if E X, then the inverse image of π(e) is π 1( π(e) ) = E + M = {u + m : u E, m M}. Solution (c) Suppose that u E and m M are given. Then π(u + m) = u + m + M = u + M = π(u) π(e). Hence u + m π 1( π(e) ). Now suppose that v π 1( π(e) ). Then, by definition, π(v) π(e) = {u + M : u E}. Hence v + M = π(v) = u + M for some u E. But then m = v u M, so v = u + m with u E and m M. 1 Conway calls this map Q, but I prefer to call it π for projection.
25 CHAPTER 3. BANACH SPACES 25 We will mostly be interested in the case where X is a normed space. The following result shows that X/M is a seminormed space in general, and is a normed space if M is closed. Proposition Let M be a subspace of a normed linear space X. Given f X, define Then the following statements hold. (a) is welldefined. (b) is a seminorm on X/M. f + M = dist(f, M) = (c) If M is closed, then is a norm on X/M. inf f m. m M Proof. (a) Exercise. Hint: Show that if f 1 + M = f 2 + M then {f 1 m : m M} = {f 2 m : m M}. (b) Exercise. (c) Suppose that M is closed, and that f + M = 0. Then inf m M f m = 0. Hence there exist vectors g n M such that f g n 0 as n. But M is closed, so this implies f M. By Exercise 4.3(e), we therefore have f + M = M = 0 + M, which is the zero vector in X/M. Now we derive some basic properties of the canonical projection π of X onto X/M. Proposition Let M be a closed subspace of a normed linear space X. following statements hold. (a) π(f) = f + M f for each f X. Then the (b) Let Br X (f) denote the open ball of radius r in X centered at f, and let BX/M r (f + M) denote the open ball of radius r in X/M centered at f + M. Then for any f X and r > 0 we have π ( Br X (f) ) = Br X/M (f + M). (c) W X/M is open in X/M if and only if π 1 (W ) = {f X : f + M W } is open in X. (d) π is an open mapping, i.e., if U is open in X then π(u) is open in X/M. Proof. (a) Choose any f X. Since 0 is one of the elements of M, we have π(f) = f + M = inf f m f 0 = f. m M (b) First consider the case f = 0 and r > 0. Suppose that g + M π ( Br X (0) ). Then g + M = h + M for some h Br X (0), i.e., h < r. Hence g + M = h + M h < r, so g + M Br X/M (0 + M). Now suppose that g + M Br X/M (0 + M). Then inf m M g m = g + M < r. Hence there exists m M such that g m < r. Thus g m Br X (0), so g + M = g m + M = π(g m) π ( B X r (0)).
26 26 CHRISTOPHER HEIL Exercise: Show that statement (b) holds for an arbitrary f X. (c). Part (a) implies that π is continuous. Hence π 1 (W ) must be open in X if W is open in X/M.. Suppose that W is a subset of X/M such that π 1 (W ) is open in X. We must show that W is open in X/M. Choose any point f + M W. Then f π 1 (W ), which is open in X. Hence, there exists an r > 0 such that Br X(f) π 1 (W ). By part (b) we therefore have Br X/M (f + M) = π ( Br X (f) ) π ( π 1 (W ) ) = W. Therefore W is open. (d) Suppose that U is an open subset of X. Then by Exercise 4.14(c), we have π 1( π(u) ) = U + M = {u + m : u U, m M} = (U + m). But each set U + m, being the translate of the open set U, is itself open. Hence π 1( π(u) ) is open, since it is a union of open sets. Part (c) therefore implies that π(u) is open in X/M. Exercise Let M be a closed subspace of a normed space X, and let π be the canonical projection π of X onto X/M. Prove that π = 1. Hint: Lemma 3.8. Now we can prove that if X is a Banach space, then X/M inherits a Banach space structure from X. Theorem If M is a closed subspace of a Banach space X, then X/M is a Banach space. Proof. We have already shown that X/M is a normed space, so we must show that it is complete in that norm. Suppose that {f n + M} n N is a Cauchy sequence in X/M. It would be convenient if this implies that {f n } n N is a Cauchy sequence in X, but this need not be the case. For, the vectors f n are not unique in general: if we replace f n by any vector f n + m with m M, then we obtain the same coset. We will show that by choosing an appropriate subsequence {f nk } k N and replacing the f nk by appropriate vectors that determine the same cosets f nk +M, we can create a sequence in X that is Cauchy and hence converges, and then use this to show that the original sequence of cosets {f n + M} n N converges in X/M. We begin by applying Exercise 1.18: there exists a subsequence {f nk + M} k N such that m M k N, (f nk+1 f nk ) + M = (f nk+1 + M) (f nk + M) < 2 k. Now we seek to create vectors g k M so that {f nk g k } k N will converge in X. Note that the cosets determined by f nk and by f nk g k are identical. Set g 1 = 0. Then inf (f n 1 g 1 ) (f n2 g) = g M inf (f n 1 f n2 ) + g = (f n1 f n2 ) + M < 1 g M 2.
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