FUNCTIONAL ANALYSIS LECTURE NOTES CHAPTER 3. BANACH SPACES

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "FUNCTIONAL ANALYSIS LECTURE NOTES CHAPTER 3. BANACH SPACES"

Transcription

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 semi-norm 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 semi-norm 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 semi-norm 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 semi-norm 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 semi-norm 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 one-dimensional 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 finite-rank if range(t ) is finite-dimensional. (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 n-dimensional 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 Cauchy-Schwarz 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 finite-dimensional 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 finite-dimensional 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 left-shift 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 right-shift 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 well-defined, 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 well-defined. 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. Finite-Dimensional Normed Spaces In this section we will prove some basic facts about finite-dimensional spaces. First, recall that a finite-dimensional 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 finite-dimensional 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 finite-dimensional 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 finite-dimensional space are equivalent. Theorem 3.3. If X is a finite-dimensional 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 real-valued 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 finite-dimensional 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 finite-dimensional 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 finite-dimensional 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 finite-dimensional 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 finite-dimensional 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 finite-dimensional subspace of X, then M is closed. Exercise 3.7. Let X be a finite-dimensional 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 finite-dimensional. Hints: Suppose that X is infinite-dimensional. 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 infinite-dimensional. 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 finite-dimensional. (b) Prove that if X is infinite-dimensional, 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 1-1 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 well-defined. 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 well-defined. 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 semi-normed 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 well-defined. (b) is a semi-norm 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.

FUNCTIONAL ANALYSIS LECTURE NOTES: QUOTIENT SPACES

FUNCTIONAL ANALYSIS LECTURE NOTES: QUOTIENT SPACES FUNCTIONAL ANALYSIS LECTURE NOTES: QUOTIENT SPACES CHRISTOPHER HEIL 1. Cosets and the Quotient Space Any vector space is an abelian group under the operation of vector addition. So, if you are have studied

More information

BANACH AND HILBERT SPACE REVIEW

BANACH AND HILBERT SPACE REVIEW BANACH AND HILBET SPACE EVIEW CHISTOPHE HEIL These notes will briefly review some basic concepts related to the theory of Banach and Hilbert spaces. We are not trying to give a complete development, but

More information

Mathematics Course 111: Algebra I Part IV: Vector Spaces

Mathematics Course 111: Algebra I Part IV: Vector Spaces Mathematics Course 111: Algebra I Part IV: Vector Spaces D. R. Wilkins Academic Year 1996-7 9 Vector Spaces A vector space over some field K is an algebraic structure consisting of a set V on which are

More information

Metric Spaces. Chapter 7. 7.1. Metrics

Metric Spaces. Chapter 7. 7.1. Metrics Chapter 7 Metric Spaces A metric space is a set X that has a notion of the distance d(x, y) between every pair of points x, y X. The purpose of this chapter is to introduce metric spaces and give some

More information

Mathematical Methods of Engineering Analysis

Mathematical Methods of Engineering Analysis Mathematical Methods of Engineering Analysis Erhan Çinlar Robert J. Vanderbei February 2, 2000 Contents Sets and Functions 1 1 Sets................................... 1 Subsets.............................

More information

Notes on metric spaces

Notes on metric spaces Notes on metric spaces 1 Introduction The purpose of these notes is to quickly review some of the basic concepts from Real Analysis, Metric Spaces and some related results that will be used in this course.

More information

FUNCTIONAL ANALYSIS LECTURE NOTES CHAPTER 2. OPERATORS ON HILBERT SPACES

FUNCTIONAL ANALYSIS LECTURE NOTES CHAPTER 2. OPERATORS ON HILBERT SPACES FUNCTIONAL ANALYSIS LECTURE NOTES CHAPTER 2. OPERATORS ON HILBERT SPACES CHRISTOPHER HEIL 1. Elementary Properties and Examples First recall the basic definitions regarding operators. Definition 1.1 (Continuous

More information

x if x 0, x if x < 0.

x if x 0, x if x < 0. Chapter 3 Sequences In this chapter, we discuss sequences. We say what it means for a sequence to converge, and define the limit of a convergent sequence. We begin with some preliminary results about the

More information

1 if 1 x 0 1 if 0 x 1

1 if 1 x 0 1 if 0 x 1 Chapter 3 Continuity In this chapter we begin by defining the fundamental notion of continuity for real valued functions of a single real variable. When trying to decide whether a given function is or

More information

I. GROUPS: BASIC DEFINITIONS AND EXAMPLES

I. GROUPS: BASIC DEFINITIONS AND EXAMPLES I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called

More information

MA651 Topology. Lecture 6. Separation Axioms.

MA651 Topology. Lecture 6. Separation Axioms. MA651 Topology. Lecture 6. Separation Axioms. This text is based on the following books: Fundamental concepts of topology by Peter O Neil Elements of Mathematics: General Topology by Nicolas Bourbaki Counterexamples

More information

1 Norms and Vector Spaces

1 Norms and Vector Spaces 008.10.07.01 1 Norms and Vector Spaces Suppose we have a complex vector space V. A norm is a function f : V R which satisfies (i) f(x) 0 for all x V (ii) f(x + y) f(x) + f(y) for all x,y V (iii) f(λx)

More information

Basic Concepts of Point Set Topology Notes for OU course Math 4853 Spring 2011

Basic Concepts of Point Set Topology Notes for OU course Math 4853 Spring 2011 Basic Concepts of Point Set Topology Notes for OU course Math 4853 Spring 2011 A. Miller 1. Introduction. The definitions of metric space and topological space were developed in the early 1900 s, largely

More information

Undergraduate Notes in Mathematics. Arkansas Tech University Department of Mathematics

Undergraduate Notes in Mathematics. Arkansas Tech University Department of Mathematics Undergraduate Notes in Mathematics Arkansas Tech University Department of Mathematics An Introductory Single Variable Real Analysis: A Learning Approach through Problem Solving Marcel B. Finan c All Rights

More information

Metric Spaces. Chapter 1

Metric Spaces. Chapter 1 Chapter 1 Metric Spaces Many of the arguments you have seen in several variable calculus are almost identical to the corresponding arguments in one variable calculus, especially arguments concerning convergence

More information

Chapter 1. Metric Spaces. Metric Spaces. Examples. Normed linear spaces

Chapter 1. Metric Spaces. Metric Spaces. Examples. Normed linear spaces Chapter 1. Metric Spaces Metric Spaces MA222 David Preiss d.preiss@warwick.ac.uk Warwick University, Spring 2008/2009 Definitions. A metric on a set M is a function d : M M R such that for all x, y, z

More information

Math 4310 Handout - Quotient Vector Spaces

Math 4310 Handout - Quotient Vector Spaces Math 4310 Handout - Quotient Vector Spaces Dan Collins The textbook defines a subspace of a vector space in Chapter 4, but it avoids ever discussing the notion of a quotient space. This is understandable

More information

Metric Spaces Joseph Muscat 2003 (Last revised May 2009)

Metric Spaces Joseph Muscat 2003 (Last revised May 2009) 1 Distance J Muscat 1 Metric Spaces Joseph Muscat 2003 (Last revised May 2009) (A revised and expanded version of these notes are now published by Springer.) 1 Distance A metric space can be thought of

More information

NOTES ON LINEAR TRANSFORMATIONS

NOTES ON LINEAR TRANSFORMATIONS NOTES ON LINEAR TRANSFORMATIONS Definition 1. Let V and W be vector spaces. A function T : V W is a linear transformation from V to W if the following two properties hold. i T v + v = T v + T v for all

More information

REAL ANALYSIS LECTURE NOTES: 1.4 OUTER MEASURE

REAL ANALYSIS LECTURE NOTES: 1.4 OUTER MEASURE REAL ANALYSIS LECTURE NOTES: 1.4 OUTER MEASURE CHRISTOPHER HEIL 1.4.1 Introduction We will expand on Section 1.4 of Folland s text, which covers abstract outer measures also called exterior measures).

More information

1. Prove that the empty set is a subset of every set.

1. Prove that the empty set is a subset of every set. 1. Prove that the empty set is a subset of every set. Basic Topology Written by Men-Gen Tsai email: b89902089@ntu.edu.tw Proof: For any element x of the empty set, x is also an element of every set since

More information

CHAPTER II THE LIMIT OF A SEQUENCE OF NUMBERS DEFINITION OF THE NUMBER e.

CHAPTER II THE LIMIT OF A SEQUENCE OF NUMBERS DEFINITION OF THE NUMBER e. CHAPTER II THE LIMIT OF A SEQUENCE OF NUMBERS DEFINITION OF THE NUMBER e. This chapter contains the beginnings of the most important, and probably the most subtle, notion in mathematical analysis, i.e.,

More information

Linear Maps. Isaiah Lankham, Bruno Nachtergaele, Anne Schilling (February 5, 2007)

Linear Maps. Isaiah Lankham, Bruno Nachtergaele, Anne Schilling (February 5, 2007) MAT067 University of California, Davis Winter 2007 Linear Maps Isaiah Lankham, Bruno Nachtergaele, Anne Schilling (February 5, 2007) As we have discussed in the lecture on What is Linear Algebra? one of

More information

Vector and Matrix Norms

Vector and Matrix Norms Chapter 1 Vector and Matrix Norms 11 Vector Spaces Let F be a field (such as the real numbers, R, or complex numbers, C) with elements called scalars A Vector Space, V, over the field F is a non-empty

More information

Vector Spaces II: Finite Dimensional Linear Algebra 1

Vector Spaces II: Finite Dimensional Linear Algebra 1 John Nachbar September 2, 2014 Vector Spaces II: Finite Dimensional Linear Algebra 1 1 Definitions and Basic Theorems. For basic properties and notation for R N, see the notes Vector Spaces I. Definition

More information

Finite dimensional topological vector spaces

Finite dimensional topological vector spaces Chapter 3 Finite dimensional topological vector spaces 3.1 Finite dimensional Hausdorff t.v.s. Let X be a vector space over the field K of real or complex numbers. We know from linear algebra that the

More information

Chapter 5. Banach Spaces

Chapter 5. Banach Spaces 9 Chapter 5 Banach Spaces Many linear equations may be formulated in terms of a suitable linear operator acting on a Banach space. In this chapter, we study Banach spaces and linear operators acting on

More information

PART I. THE REAL NUMBERS

PART I. THE REAL NUMBERS PART I. THE REAL NUMBERS This material assumes that you are already familiar with the real number system and the representation of the real numbers as points on the real line. I.1. THE NATURAL NUMBERS

More information

LINEAR ALGEBRA W W L CHEN

LINEAR ALGEBRA W W L CHEN LINEAR ALGEBRA W W L CHEN c W W L Chen, 1997, 2008 This chapter is available free to all individuals, on understanding that it is not to be used for financial gain, and may be downloaded and/or photocopied,

More information

Ri and. i=1. S i N. and. R R i

Ri and. i=1. S i N. and. R R i The subset R of R n is a closed rectangle if there are n non-empty closed intervals {[a 1, b 1 ], [a 2, b 2 ],..., [a n, b n ]} so that R = [a 1, b 1 ] [a 2, b 2 ] [a n, b n ]. The subset R of R n is an

More information

3. Equivalence Relations. Discussion

3. Equivalence Relations. Discussion 3. EQUIVALENCE RELATIONS 33 3. Equivalence Relations 3.1. Definition of an Equivalence Relations. Definition 3.1.1. A relation R on a set A is an equivalence relation if and only if R is reflexive, symmetric,

More information

1 Normed vector spaces and convexity

1 Normed vector spaces and convexity B4a Banach spaces lectured by Dmitry Belyaev based on notes by B. Kirchheim and CJK Batty 1 Normed vector spaces and convexity Mods analysis on the real line derives many of its basic statements (algebra

More information

Introduction to Topology

Introduction to Topology Introduction to Topology Tomoo Matsumura November 30, 2010 Contents 1 Topological spaces 3 1.1 Basis of a Topology......................................... 3 1.2 Comparing Topologies.......................................

More information

Continued Fractions and the Euclidean Algorithm

Continued Fractions and the Euclidean Algorithm Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction

More information

The Dirichlet Unit Theorem

The Dirichlet Unit Theorem Chapter 6 The Dirichlet Unit Theorem As usual, we will be working in the ring B of algebraic integers of a number field L. Two factorizations of an element of B are regarded as essentially the same if

More information

16.3 Fredholm Operators

16.3 Fredholm Operators Lectures 16 and 17 16.3 Fredholm Operators A nice way to think about compact operators is to show that set of compact operators is the closure of the set of finite rank operator in operator norm. In this

More information

Appendix A. Appendix. A.1 Algebra. Fields and Rings

Appendix A. Appendix. A.1 Algebra. Fields and Rings Appendix A Appendix A.1 Algebra Algebra is the foundation of algebraic geometry; here we collect some of the basic algebra on which we rely. We develop some algebraic background that is needed in the text.

More information

Let H and J be as in the above lemma. The result of the lemma shows that the integral

Let H and J be as in the above lemma. The result of the lemma shows that the integral Let and be as in the above lemma. The result of the lemma shows that the integral ( f(x, y)dy) dx is well defined; we denote it by f(x, y)dydx. By symmetry, also the integral ( f(x, y)dx) dy is well defined;

More information

Elementary Number Theory We begin with a bit of elementary number theory, which is concerned

Elementary Number Theory We begin with a bit of elementary number theory, which is concerned CONSTRUCTION OF THE FINITE FIELDS Z p S. R. DOTY Elementary Number Theory We begin with a bit of elementary number theory, which is concerned solely with questions about the set of integers Z = {0, ±1,

More information

THE BANACH CONTRACTION PRINCIPLE. Contents

THE BANACH CONTRACTION PRINCIPLE. Contents THE BANACH CONTRACTION PRINCIPLE ALEX PONIECKI Abstract. This paper will study contractions of metric spaces. To do this, we will mainly use tools from topology. We will give some examples of contractions,

More information

Math 5311 Gateaux differentials and Frechet derivatives

Math 5311 Gateaux differentials and Frechet derivatives Math 5311 Gateaux differentials and Frechet derivatives Kevin Long January 26, 2009 1 Differentiation in vector spaces Thus far, we ve developed the theory of minimization without reference to derivatives.

More information

GROUPS ACTING ON A SET

GROUPS ACTING ON A SET GROUPS ACTING ON A SET MATH 435 SPRING 2012 NOTES FROM FEBRUARY 27TH, 2012 1. Left group actions Definition 1.1. Suppose that G is a group and S is a set. A left (group) action of G on S is a rule for

More information

1 Sets and Set Notation.

1 Sets and Set Notation. LINEAR ALGEBRA MATH 27.6 SPRING 23 (COHEN) LECTURE NOTES Sets and Set Notation. Definition (Naive Definition of a Set). A set is any collection of objects, called the elements of that set. We will most

More information

1 VECTOR SPACES AND SUBSPACES

1 VECTOR SPACES AND SUBSPACES 1 VECTOR SPACES AND SUBSPACES What is a vector? Many are familiar with the concept of a vector as: Something which has magnitude and direction. an ordered pair or triple. a description for quantities such

More information

MEASURE AND INTEGRATION. Dietmar A. Salamon ETH Zürich

MEASURE AND INTEGRATION. Dietmar A. Salamon ETH Zürich MEASURE AND INTEGRATION Dietmar A. Salamon ETH Zürich 12 May 2016 ii Preface This book is based on notes for the lecture course Measure and Integration held at ETH Zürich in the spring semester 2014. Prerequisites

More information

MATH10212 Linear Algebra. Systems of Linear Equations. Definition. An n-dimensional vector is a row or a column of n numbers (or letters): a 1.

MATH10212 Linear Algebra. Systems of Linear Equations. Definition. An n-dimensional vector is a row or a column of n numbers (or letters): a 1. MATH10212 Linear Algebra Textbook: D. Poole, Linear Algebra: A Modern Introduction. Thompson, 2006. ISBN 0-534-40596-7. Systems of Linear Equations Definition. An n-dimensional vector is a row or a column

More information

LEARNING OBJECTIVES FOR THIS CHAPTER

LEARNING OBJECTIVES FOR THIS CHAPTER CHAPTER 2 American mathematician Paul Halmos (1916 2006), who in 1942 published the first modern linear algebra book. The title of Halmos s book was the same as the title of this chapter. Finite-Dimensional

More information

Group Theory. Contents

Group Theory. Contents Group Theory Contents Chapter 1: Review... 2 Chapter 2: Permutation Groups and Group Actions... 3 Orbits and Transitivity... 6 Specific Actions The Right regular and coset actions... 8 The Conjugation

More information

Duality of linear conic problems

Duality of linear conic problems Duality of linear conic problems Alexander Shapiro and Arkadi Nemirovski Abstract It is well known that the optimal values of a linear programming problem and its dual are equal to each other if at least

More information

Section 1.1. Introduction to R n

Section 1.1. Introduction to R n The Calculus of Functions of Several Variables Section. Introduction to R n Calculus is the study of functional relationships and how related quantities change with each other. In your first exposure to

More information

IRREDUCIBLE OPERATOR SEMIGROUPS SUCH THAT AB AND BA ARE PROPORTIONAL. 1. Introduction

IRREDUCIBLE OPERATOR SEMIGROUPS SUCH THAT AB AND BA ARE PROPORTIONAL. 1. Introduction IRREDUCIBLE OPERATOR SEMIGROUPS SUCH THAT AB AND BA ARE PROPORTIONAL R. DRNOVŠEK, T. KOŠIR Dedicated to Prof. Heydar Radjavi on the occasion of his seventieth birthday. Abstract. Let S be an irreducible

More information

SOLUTIONS TO EXERCISES FOR. MATHEMATICS 205A Part 3. Spaces with special properties

SOLUTIONS TO EXERCISES FOR. MATHEMATICS 205A Part 3. Spaces with special properties SOLUTIONS TO EXERCISES FOR MATHEMATICS 205A Part 3 Fall 2008 III. Spaces with special properties III.1 : Compact spaces I Problems from Munkres, 26, pp. 170 172 3. Show that a finite union of compact subspaces

More information

Math212a1010 Lebesgue measure.

Math212a1010 Lebesgue measure. Math212a1010 Lebesgue measure. October 19, 2010 Today s lecture will be devoted to Lebesgue measure, a creation of Henri Lebesgue, in his thesis, one of the most famous theses in the history of mathematics.

More information

Geometric Transformations

Geometric Transformations Geometric Transformations Definitions Def: f is a mapping (function) of a set A into a set B if for every element a of A there exists a unique element b of B that is paired with a; this pairing is denoted

More information

Finite Fields and Error-Correcting Codes

Finite Fields and Error-Correcting Codes Lecture Notes in Mathematics Finite Fields and Error-Correcting Codes Karl-Gustav Andersson (Lund University) (version 1.013-16 September 2015) Translated from Swedish by Sigmundur Gudmundsson Contents

More information

Functional analysis and its applications

Functional analysis and its applications Department of Mathematics, London School of Economics Functional analysis and its applications Amol Sasane ii Introduction Functional analysis plays an important role in the applied sciences as well as

More information

T ( a i x i ) = a i T (x i ).

T ( a i x i ) = a i T (x i ). Chapter 2 Defn 1. (p. 65) Let V and W be vector spaces (over F ). We call a function T : V W a linear transformation form V to W if, for all x, y V and c F, we have (a) T (x + y) = T (x) + T (y) and (b)

More information

Numerical Analysis Lecture Notes

Numerical Analysis Lecture Notes Numerical Analysis Lecture Notes Peter J. Olver 5. Inner Products and Norms The norm of a vector is a measure of its size. Besides the familiar Euclidean norm based on the dot product, there are a number

More information

THE DIMENSION OF A VECTOR SPACE

THE DIMENSION OF A VECTOR SPACE THE DIMENSION OF A VECTOR SPACE KEITH CONRAD This handout is a supplementary discussion leading up to the definition of dimension and some of its basic properties. Let V be a vector space over a field

More information

Linear Algebra I. Ronald van Luijk, 2012

Linear Algebra I. Ronald van Luijk, 2012 Linear Algebra I Ronald van Luijk, 2012 With many parts from Linear Algebra I by Michael Stoll, 2007 Contents 1. Vector spaces 3 1.1. Examples 3 1.2. Fields 4 1.3. The field of complex numbers. 6 1.4.

More information

Matrix Representations of Linear Transformations and Changes of Coordinates

Matrix Representations of Linear Transformations and Changes of Coordinates Matrix Representations of Linear Transformations and Changes of Coordinates 01 Subspaces and Bases 011 Definitions A subspace V of R n is a subset of R n that contains the zero element and is closed under

More information

The Ideal Class Group

The Ideal Class Group Chapter 5 The Ideal Class Group We will use Minkowski theory, which belongs to the general area of geometry of numbers, to gain insight into the ideal class group of a number field. We have already mentioned

More information

THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS

THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS KEITH CONRAD 1. Introduction The Fundamental Theorem of Algebra says every nonconstant polynomial with complex coefficients can be factored into linear

More information

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS. + + x 2. x n. a 11 a 12 a 1n b 1 a 21 a 22 a 2n b 2 a 31 a 32 a 3n b 3. a m1 a m2 a mn b m

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS. + + x 2. x n. a 11 a 12 a 1n b 1 a 21 a 22 a 2n b 2 a 31 a 32 a 3n b 3. a m1 a m2 a mn b m MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 1. SYSTEMS OF EQUATIONS AND MATRICES 1.1. Representation of a linear system. The general system of m equations in n unknowns can be written a 11 x 1 + a 12 x 2 +

More information

5. Linear algebra I: dimension

5. Linear algebra I: dimension 5. Linear algebra I: dimension 5.1 Some simple results 5.2 Bases and dimension 5.3 Homomorphisms and dimension 1. Some simple results Several observations should be made. Once stated explicitly, the proofs

More information

MATH 304 Linear Algebra Lecture 20: Inner product spaces. Orthogonal sets.

MATH 304 Linear Algebra Lecture 20: Inner product spaces. Orthogonal sets. MATH 304 Linear Algebra Lecture 20: Inner product spaces. Orthogonal sets. Norm The notion of norm generalizes the notion of length of a vector in R n. Definition. Let V be a vector space. A function α

More information

n k=1 k=0 1/k! = e. Example 6.4. The series 1/k 2 converges in R. Indeed, if s n = n then k=1 1/k, then s 2n s n = 1 n + 1 +...

n k=1 k=0 1/k! = e. Example 6.4. The series 1/k 2 converges in R. Indeed, if s n = n then k=1 1/k, then s 2n s n = 1 n + 1 +... 6 Series We call a normed space (X, ) a Banach space provided that every Cauchy sequence (x n ) in X converges. For example, R with the norm = is an example of Banach space. Now let (x n ) be a sequence

More information

So let us begin our quest to find the holy grail of real analysis.

So let us begin our quest to find the holy grail of real analysis. 1 Section 5.2 The Complete Ordered Field: Purpose of Section We present an axiomatic description of the real numbers as a complete ordered field. The axioms which describe the arithmetic of the real numbers

More information

MATH 304 Linear Algebra Lecture 9: Subspaces of vector spaces (continued). Span. Spanning set.

MATH 304 Linear Algebra Lecture 9: Subspaces of vector spaces (continued). Span. Spanning set. MATH 304 Linear Algebra Lecture 9: Subspaces of vector spaces (continued). Span. Spanning set. Vector space A vector space is a set V equipped with two operations, addition V V (x,y) x + y V and scalar

More information

Lebesgue Measure on R n

Lebesgue Measure on R n 8 CHAPTER 2 Lebesgue Measure on R n Our goal is to construct a notion of the volume, or Lebesgue measure, of rather general subsets of R n that reduces to the usual volume of elementary geometrical sets

More information

Quotient Rings and Field Extensions

Quotient Rings and Field Extensions Chapter 5 Quotient Rings and Field Extensions In this chapter we describe a method for producing field extension of a given field. If F is a field, then a field extension is a field K that contains F.

More information

Matrix Norms. Tom Lyche. September 28, Centre of Mathematics for Applications, Department of Informatics, University of Oslo

Matrix Norms. Tom Lyche. September 28, Centre of Mathematics for Applications, Department of Informatics, University of Oslo Matrix Norms Tom Lyche Centre of Mathematics for Applications, Department of Informatics, University of Oslo September 28, 2009 Matrix Norms We consider matrix norms on (C m,n, C). All results holds for

More information

and s n (x) f(x) for all x and s.t. s n is measurable if f is. REAL ANALYSIS Measures. A (positive) measure on a measurable space

and s n (x) f(x) for all x and s.t. s n is measurable if f is. REAL ANALYSIS Measures. A (positive) measure on a measurable space RAL ANALYSIS A survey of MA 641-643, UAB 1999-2000 M. Griesemer Throughout these notes m denotes Lebesgue measure. 1. Abstract Integration σ-algebras. A σ-algebra in X is a non-empty collection of subsets

More information

Linear Algebra. A vector space (over R) is an ordered quadruple. such that V is a set; 0 V ; and the following eight axioms hold:

Linear Algebra. A vector space (over R) is an ordered quadruple. such that V is a set; 0 V ; and the following eight axioms hold: Linear Algebra A vector space (over R) is an ordered quadruple (V, 0, α, µ) such that V is a set; 0 V ; and the following eight axioms hold: α : V V V and µ : R V V ; (i) α(α(u, v), w) = α(u, α(v, w)),

More information

MATH 110 Spring 2015 Homework 6 Solutions

MATH 110 Spring 2015 Homework 6 Solutions MATH 110 Spring 2015 Homework 6 Solutions Section 2.6 2.6.4 Let α denote the standard basis for V = R 3. Let α = {e 1, e 2, e 3 } denote the dual basis of α for V. We would first like to show that β =

More information

CHAPTER 5. Product Measures

CHAPTER 5. Product Measures 54 CHAPTER 5 Product Measures Given two measure spaces, we may construct a natural measure on their Cartesian product; the prototype is the construction of Lebesgue measure on R 2 as the product of Lebesgue

More information

2 Complex Functions and the Cauchy-Riemann Equations

2 Complex Functions and the Cauchy-Riemann Equations 2 Complex Functions and the Cauchy-Riemann Equations 2.1 Complex functions In one-variable calculus, we study functions f(x) of a real variable x. Likewise, in complex analysis, we study functions f(z)

More information

God created the integers and the rest is the work of man. (Leopold Kronecker, in an after-dinner speech at a conference, Berlin, 1886)

God created the integers and the rest is the work of man. (Leopold Kronecker, in an after-dinner speech at a conference, Berlin, 1886) Chapter 2 Numbers God created the integers and the rest is the work of man. (Leopold Kronecker, in an after-dinner speech at a conference, Berlin, 1886) God created the integers and the rest is the work

More information

Sec 4.1 Vector Spaces and Subspaces

Sec 4.1 Vector Spaces and Subspaces Sec 4. Vector Spaces and Subspaces Motivation Let S be the set of all solutions to the differential equation y + y =. Let T be the set of all 2 3 matrices with real entries. These two sets share many common

More information

Mikołaj Krupski. Regularity properties of measures on compact spaces. Instytut Matematyczny PAN. Praca semestralna nr 1 (semestr zimowy 2010/11)

Mikołaj Krupski. Regularity properties of measures on compact spaces. Instytut Matematyczny PAN. Praca semestralna nr 1 (semestr zimowy 2010/11) Mikołaj Krupski Instytut Matematyczny PAN Regularity properties of measures on compact spaces Praca semestralna nr 1 (semestr zimowy 2010/11) Opiekun pracy: Grzegorz Plebanek Regularity properties of measures

More information

MA106 Linear Algebra lecture notes

MA106 Linear Algebra lecture notes MA106 Linear Algebra lecture notes Lecturers: Martin Bright and Daan Krammer Warwick, January 2011 Contents 1 Number systems and fields 3 1.1 Axioms for number systems......................... 3 2 Vector

More information

Separation Properties for Locally Convex Cones

Separation Properties for Locally Convex Cones Journal of Convex Analysis Volume 9 (2002), No. 1, 301 307 Separation Properties for Locally Convex Cones Walter Roth Department of Mathematics, Universiti Brunei Darussalam, Gadong BE1410, Brunei Darussalam

More information

ZORN S LEMMA AND SOME APPLICATIONS

ZORN S LEMMA AND SOME APPLICATIONS ZORN S LEMMA AND SOME APPLICATIONS KEITH CONRAD 1. Introduction Zorn s lemma is a result in set theory that appears in proofs of some non-constructive existence theorems throughout mathematics. We will

More information

INDISTINGUISHABILITY OF ABSOLUTELY CONTINUOUS AND SINGULAR DISTRIBUTIONS

INDISTINGUISHABILITY OF ABSOLUTELY CONTINUOUS AND SINGULAR DISTRIBUTIONS INDISTINGUISHABILITY OF ABSOLUTELY CONTINUOUS AND SINGULAR DISTRIBUTIONS STEVEN P. LALLEY AND ANDREW NOBEL Abstract. It is shown that there are no consistent decision rules for the hypothesis testing problem

More information

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS Systems of Equations and Matrices Representation of a linear system The general system of m equations in n unknowns can be written a x + a 2 x 2 + + a n x n b a

More information

The fundamental group of the Hawaiian earring is not free (International Journal of Algebra and Computation Vol. 2, No. 1 (1992), 33 37) Bart de Smit

The fundamental group of the Hawaiian earring is not free (International Journal of Algebra and Computation Vol. 2, No. 1 (1992), 33 37) Bart de Smit The fundamental group of the Hawaiian earring is not free Bart de Smit The fundamental group of the Hawaiian earring is not free (International Journal of Algebra and Computation Vol. 2, No. 1 (1992),

More information

Functional Analysis. Alexander C. R. Belton

Functional Analysis. Alexander C. R. Belton Functional Analysis Alexander C. R. Belton Copyright c Alexander C. R. Belton 2004, 2006 Hyperlinked and revised edition All rights reserved The right of Alexander Belton to be identified as the author

More information

The Rational Numbers

The Rational Numbers Math 3040: Spring 2011 The Rational Numbers Contents 1. The Set Q 1 2. Addition and multiplication of rational numbers 3 2.1. Definitions and properties. 3 2.2. Comments 4 2.3. Connections with Z. 6 2.4.

More information

POWER SETS AND RELATIONS

POWER SETS AND RELATIONS POWER SETS AND RELATIONS L. MARIZZA A. BAILEY 1. The Power Set Now that we have defined sets as best we can, we can consider a sets of sets. If we were to assume nothing, except the existence of the empty

More information

Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm.

Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. We begin by defining the ring of polynomials with coefficients in a ring R. After some preliminary results, we specialize

More information

Linear Algebra Notes for Marsden and Tromba Vector Calculus

Linear Algebra Notes for Marsden and Tromba Vector Calculus Linear Algebra Notes for Marsden and Tromba Vector Calculus n-dimensional Euclidean Space and Matrices Definition of n space As was learned in Math b, a point in Euclidean three space can be thought of

More information

The determinant of a skew-symmetric matrix is a square. This can be seen in small cases by direct calculation: 0 a. 12 a. a 13 a 24 a 14 a 23 a 14

The determinant of a skew-symmetric matrix is a square. This can be seen in small cases by direct calculation: 0 a. 12 a. a 13 a 24 a 14 a 23 a 14 4 Symplectic groups In this and the next two sections, we begin the study of the groups preserving reflexive sesquilinear forms or quadratic forms. We begin with the symplectic groups, associated with

More information

Mathematics Review for MS Finance Students

Mathematics Review for MS Finance Students Mathematics Review for MS Finance Students Anthony M. Marino Department of Finance and Business Economics Marshall School of Business Lecture 1: Introductory Material Sets The Real Number System Functions,

More information

Section 6.1 - Inner Products and Norms

Section 6.1 - Inner Products and Norms Section 6.1 - Inner Products and Norms Definition. Let V be a vector space over F {R, C}. An inner product on V is a function that assigns, to every ordered pair of vectors x and y in V, a scalar in F,

More information

Theta Functions. Lukas Lewark. Seminar on Modular Forms, 31. Januar 2007

Theta Functions. Lukas Lewark. Seminar on Modular Forms, 31. Januar 2007 Theta Functions Lukas Lewark Seminar on Modular Forms, 31. Januar 007 Abstract Theta functions are introduced, associated to lattices or quadratic forms. Their transformation property is proven and the

More information

Ideal Class Group and Units

Ideal Class Group and Units Chapter 4 Ideal Class Group and Units We are now interested in understanding two aspects of ring of integers of number fields: how principal they are (that is, what is the proportion of principal ideals

More information

3. Prime and maximal ideals. 3.1. Definitions and Examples.

3. Prime and maximal ideals. 3.1. Definitions and Examples. COMMUTATIVE ALGEBRA 5 3.1. Definitions and Examples. 3. Prime and maximal ideals Definition. An ideal P in a ring A is called prime if P A and if for every pair x, y of elements in A\P we have xy P. Equivalently,

More information

3. INNER PRODUCT SPACES

3. INNER PRODUCT SPACES . INNER PRODUCT SPACES.. Definition So far we have studied abstract vector spaces. These are a generalisation of the geometric spaces R and R. But these have more structure than just that of a vector space.

More information

Chapter 7. Permutation Groups

Chapter 7. Permutation Groups Chapter 7 Permutation Groups () We started the study of groups by considering planar isometries In the previous chapter, we learnt that finite groups of planar isometries can only be cyclic or dihedral

More information

MODULAR ARITHMETIC. a smallest member. It is equivalent to the Principle of Mathematical Induction.

MODULAR ARITHMETIC. a smallest member. It is equivalent to the Principle of Mathematical Induction. MODULAR ARITHMETIC 1 Working With Integers The usual arithmetic operations of addition, subtraction and multiplication can be performed on integers, and the result is always another integer Division, on

More information