Math 209B Homework 2


 Vivien Jennings
 1 years ago
 Views:
Transcription
1 Math 29B Homework 2 Edward Burkard Note: All vector spaces are over the field F = R or C 4.6. Two Compactness Theorems. 4. Point Set Topology Exercise 6 The product of countably many sequentally compact spaces is sequentially compact. (Use the diagonal trick as in the proof of Theorem 4.44.) Exercise 64 Let (X, ρ) be a metric space. A function f C(X) is called Hölder continuous of exponent α (α > ) if the quantity f(x) f(y) N α (f) = sup x y ρ(x, y) α is finite. If X is compact, {f C(X) f u and N α (f) } is compact in C(X). Let F = {f C(X) f u and N α (f) }. Let f F. Then we have that N α (f) implies that f(x) f(y) ρ(x, y) α. Let ɛ > and let x X. Then to show equicontinuity, let U = B (x), hence: ɛ α f(x) f(y) (ɛ α ) α = ɛ which implies, since x was arbitrary, F is equicontinuous on X. Recall that f u = sup x X f(x). So since every function in F satisfies f u it is clear that F is uniformly bounded. So by the ArzelàAscoli Theorem I we have that, since X is compact and Hausdorff (since it is a metric space), the closure of F in C(X) is compact. So to prove the claim it remains to show that F is closed. Let {f n } be a sequence in F converging to some function f. Observe: f u = lim f n u = sup lim f n (x) x X = sup lim f n (x) x X = lim sup f n (x) x X = lim = N α (f) = sup x y lim f n (x) lim f n (y) ρ α (x, y) lim(f n (x) f n (y)) = sup x y ρ α (x, y) = sup lim (f n(x) f n (y)) x y ρ α (x, y) (f n (x) f n (y)) = lim sup x y ρ α (x, y) = lim = Hence F is closed, and therefore F is compact in C(X).
2 The StoneWeierstrass Theorem. Exercise 68 Let X and Y be compact Hausdorff spaces. The algebra generated by functions of the form f(x, y) = g(x)h(y), where g C(X) and h C(Y ), is dense in C(X Y ). Since both X and Y are compact and Hausdorff, by Tychonoff s theorem and Proposition 4. we have that X Y is compact and Hausdorff. Now consider the algebra described above. Call it A. Clearly A contains the constant functions as we can let g and h both be constants and hence we have an f(x, y) = r for all r R. Also clearly A separates points (this can be done by letting g(x) =, h(x) = 2, k(y) = 3 and given any two points (a, b) and (c, d) plugging them into f (x, y) = g(x)k(y) = 3 and f 2 (x, y) = h(x)k(y) = 6 respectively). Since A is closed by definition, we can apply the StoneWeierstrass Theorem to it to conclude that A is dense in C(X Y ). 5.. Normed Vector Spaces. Exercise 3 Complete the proof of Proposition Elements of Functional Analysis We need to verify that T := lim T n is linear and T n T. First: Hence T is linear. Now check the norm of T : T (ax + by) = lim T n (ax + by) = lim(t n (ax) + T n (by)) = lim(at n (x) + bt n (y)) = lim(at n (x)) + lim(bt n (y)) = a lim T n (x) + b lim T n (y) = at (x) + bt (y) T = sup{ T x x = } = sup{ lim n T nx x = } = sup{ lim n T nx x = } = lim n sup{ T nx x = } = lim n T n Now fix x X and let ɛ >, then since {T n } is Cauchy there is an N N such that for all m, n N: Notice: T n (x) T m (x) ɛ T n (x) T (x) = T n (x) lim m T m(x) = lim m (T n(x) T m (x)) = lim m T n(x) T m (x) < ɛ Letting ɛ we see that T n (x) T (x). So since x was arbitrary we have T n T. Exercise 6 Suppose that X is a finite dimensional vector space. Let e,..., e n be a basis for X, and define n a je j = n a j. is a norm on X. The map (a,..., a n ) n a je j is continuous from F n with the usual Euclidean topology to X with the topology defined by. c. {x X x = } is compact in the topology defined by. d. All norms on X are equivalent. (Compare any norm to.)
3 3 a j e j + b j e j = = n λ a j e j = a j + b j a j + b j a j e j + b j e j λa j = λ a j = λ a j e j Now if n a je j =, then we have n a j =, which is a sum of nonnegative numbers, hence a j = for all j, thus n a je j =. So we conclude that is a norm. Define the map η : F n X by η(a,..., a n ) = a j e j. Using the norm (a,..., a n ) = a j on F n suggested by Folland on page 53 we can see: (a,..., a n ) = a j = η(a,..., a n ) = a j e j which makes it obvious that η is continuous. c. Let K = {(a i,..., a j ) (a,..., a j ) = }. Then K is clearly compact in F n. Now since we have η(k) = {x X x = } (by the definition of η) we can see that η(k) is compact as it is the continuous image of a compact set. Thus we have our conclusion. d. Let be any norm on X. Define M = max( e,..., e n ). Then we have a j e j a j e j = a j e j M a j = M a j e j. Thus we have x M x for all x X. As this is true, we can conclude that the map sending one norm to the other, in this case use ζ : (X, ) (X, ), is continuous (this is really just a dialation of the space). Recall the set K from the previous part. Since K is compact and ζ is continuous, ζ(k) is compact. By the extreme value theorem ζ assumes a minimum value as measured by, so we may as well say assumes a minimum value on K, call it m. Since is a norm, m and since x / K, m >. Given any element x X, we can construct the element x which is in K and hence x x m which implies m x x and so: m x x M x. Since this holds for any x X we have that is equivalent to. Moreover, since was arbitrary, every norm on X is equivalent to. Exercise 7 Let X be a Banach space. If T L(X, X ) and I T < where I is the identity operator, then T is invertible; in fact, the series (I T )n converges in L(X, X ) to T. If T L(X, X ) is invertible and S T < T, then S is invertible. Thus the set of invertible operators is open in L(X, X ).
4 4 Since X is a Banach space we have that it is complete, and hence by Proposition 5.4 L(X, X ) is complete. Let T L(X, X ) be such that I T <. Consider the sum (I T )n which is in L(X, X ) since it is an absolutely convergent sum (since I T < and hence I T n is a convergent geometric series) by Theorem 5.. Notice that showing I = T (I T )n proves that T exists and in fact T = (I T )n. A simple use of induction shows that T N (I T )n = I (I T ) N+. Now notice: N I (T (I T ) n ) = I (I (I T )N+ ) = (I T ) N+ I T N + which gives that I (T N (I T )n ) as N since I T <, hence I T (I T ) n = and so or equivalently T (I T ) n = I T = (I T ) n. Thus T is invertible. Let T L(X, X ) be invertible and let S L(X, X ) be such that S T < T (an equivalent statement: T S T < ). So: I ST = T (T S) T S T <. By the proof in part a, we have that ST is invertible, hence: I = (ST )(ST ) = S(T (ST ) ) and thus S = T (ST ). This proves that for any S in L(X, X ) sufficiently close to T, S is invertible, hence the set of invertible operators is open in L(X, X ). Exercise If < α, let Λ α ([, ]) be the space of Hölder continuous functions of exponent α on [, ]. That is, f Λ α ([, ]) iff f Λα <, where f(x) f(y) f Λα = f() + sup x,y [,], x y x y α. Λα is a norm that makes Λ α ([, ]) into a Banach space. Let λ α ([, ]) be the set of all f Λ α ([, ]) such that f(x) f(y) x y α as x y, for all y [, ]. If α <, λ α ([, ]) is an infinitedimensional closed subspace of Λ α ([, ]). If α =, λ α ([, ]) contains only constant functions. Exercise 2 Let X be a normed vector space and M a proper closed subspace of X. x + M = inf{ x + y y M} is a norm on X /M. For any ɛ > there exists x X such that x = and x + M ɛ. c. The projection map π(x) = x + M from X to X /M has norm. d. If X is complete, so is X /M. (Use Theorem 5.) e. The topology defined by the quotient norm is the quotient topology as defined in Exercise 28.
5 5 Let x + M, y + M X /M. Then: (x + y) + M = inf{ (x + y) + z z M} { = inf (x + 2 ) ( z + y + ) } 2 z z M {( since is a norm on X inf x + ) ( 2 z + y + ) } 2 z z M let w = z 2 = inf{ x + w w M} + inf{ y + z w M} = x + M + y + M Thus since x and y were arbitrary, satisfies the triangle inequality for all x + M, y + M X /M. Now let x + M X /M and let c R. Then: c(x + M) = inf{ c(x + y) y M} since is a norm on X = inf{ c x + y y M} = c inf{ x + y y M} = c x + M Now suppose that x + M =. Then clearly x M, and hence x + M = in X /M, because if it were not, then it would be impossible for x + M to be as we are taking the infimum over vectors in M and x / M. Thus it is a norm on X /M. As M is a proper closed subset, we can find an x X /M such that x + M = a >. Define y = ax, then y + M =. Now by the definition of infimum we have that there is some m M with y + m + ɛ 2. Thus there is a element Υ B ɛ(y + m) with Υ =. So we have y + M Υ + M Υ y + M = Υ (y + m) + M Υ (y + m) ɛ Yielding the inequality: or equivalently c. Observe: d. e. y + M Υ + M = Υ + M ɛ π(x) = x + M = ɛ Υ + M. { x if x / M if x M Hence π(x) x which means π. However part b tells us π and hence π = Linear Functionals. Exercise 7 A linear functional f on a normed vector space X is bounded iff f ({}) is closed. (Use Exercise 2) (= ) Assume that f is bounded. Then it is continuous. So since {} is closed in R and f is continuous, f ({}) is closed as desired. ( =) Suppose that f were not bounded. Then n > there is an x n X with x n = and f(x n ) > 2 n. ( ) Define y n = xn f(x. Then y x n) n = n f(x = n) f(x x n) 2 and f(y n n ) = f n f(x n) Now fix x X and consider the sequence a n = x f(x)y n. Then a n x as n since y n as n. However, f(a n ) = f(x) f(x)f(y n ) = f(x) f(x) =. Thus we have a n f ({}). This would imply that f ({}) is dense in X. So there are two choices, either f : X {} or f ({}) is not closed. Since the first choice is obviously not true, it must be that f ({}) is not closed. Hence by contrapositive we have our desired conclusion. = f(x n) f(x n) =.
6 6 Exercise 8 Let X be a normed vector space. If M is a closed subspace and x X /M then M + Cx is closed. (Use Theorem 5.8) Every finitedimensional subspace of X is closed. Exercise 9 Let X be an infinitedimensional normed vector space. There is a sequence {x j } in X such that x j = for all j and x j x k 2 x j inductively, using Exercised 2b and 8.) X is not locally compact. for j k. (Construct Exercise 25 If X is a Banach space and X is separable, then X is separable. (Let {f n } be a countable dense subset of X. For each n choose x n X with x n = and f n (x n ) 2 f n. Then the linear combinations of {x n } are dense in X.) Note: Separability of X does not imply separability of X The Baire Category Theorem and Its Consequences. Exercise 27 There exist meager subsets of R whose complements have Lebesgue measure zero. Exercise 29 Let Y = L (µ) where µ is counting measure on N, and let X = {f SCY {n f(n) < }, equipped with the L norm. X is a proper dense subspace of Y ; hence X is not complete. Define T : X Y by T f(n) = nf(n). Then T is closed but not bounded. c. Let S = T. Then S : Y X is bounded and surjective, but not open. c. Exercise 3 Let Y = C([, ]) and X = C ([, ]), both equipped with the uniform norm. X is not complete. The map ( d dx) : X Y is closed (see Exercise 9) but not bounded. Consider the function f(x) = x 2 defined on [, ]. By lemma 4.47, ɛ > there is a polynomial P (x) defined on R such that P () = and x P (x) u < ɛ for x [, ]. If we let Q(x) = P ( ) x 2 we can rephrase the lemma to say ɛ > there is a polynomial Q(x) such that Q ( 2) = and x 2 Q(x) u < ɛ for x [, ]. Since these Q s are polynomials they are clearly in C ([, ]), and thus we may create a sequence of them as follows: Define Q n (x) to be one of the polynomials such that x 2 Q n (x) u < n for all n N. This creates a convergent sequence {Q n } in C ([, ]) which converges to x 2 on [, ], however x 2 / C ([, ]). Thus C ([, ]) is not complete. Exercise 32 Let and 2 be norms on the vector space X such that 2. If X is complete with respect to both norms, then the norms are equivalent.
7 5.4. Topological Vector Spaces. Exercise 45 The space C (R) of all infinitely differentiable functions on R has a Fréchet space topology with respect to which f n f iff f (k) n f (k) uniformly on compact sets for all k. Exercise 47 Suppose that X and Y are Banach spaces. If {T n } L(X, Y ) and T n T weakly (or strongly), then sup N T n <. Every weakly convergent sequence in X, and every weak convergent sequence in X, is bounded (with respect to the norm). 7
THEOREMS, ETC., FOR MATH 515
THEOREMS, ETC., FOR MATH 515 Proposition 1 (=comment on page 17). If A is an algebra, then any finite union or finite intersection of sets in A is also in A. Proposition 2 (=Proposition 1.1). For every
More informationProblem Set 5. 2 n k. Then a nk (x) = 1+( 1)k
Problem Set 5 1. (Folland 2.43) For x [, 1), let 1 a n (x)2 n (a n (x) = or 1) be the base2 expansion of x. (If x is a dyadic rational, choose the expansion such that a n (x) = for large n.) Then the
More informationFolland: Real Analysis, Chapter 4 Sébastien Picard
Folland: Real Analysis, Chapter 4 Sébastien Picard Problem 4.19 If {X α } is a family of topological spaces, X α X α (with the product topology) is uniquely determined up to homeomorphism by the following
More informationLecture Notes in Advanced Calculus 1 (80315) Raz Kupferman Institute of Mathematics The Hebrew University
Lecture Notes in Advanced Calculus 1 (80315) Raz Kupferman Institute of Mathematics The Hebrew University February 7, 2007 2 Contents 1 Metric Spaces 1 1.1 Basic definitions...........................
More informationReal Analysis Notes. Thomas Goller
Real Analysis Notes Thomas Goller September 4, 2011 Contents 1 Abstract Measure Spaces 2 1.1 Basic Definitions........................... 2 1.2 Measurable Functions........................ 2 1.3 Integration..............................
More information1 Topology Definition of a topology Basis (Base) of a topology The subspace topology & the product topology on X Y 3
Index Page 1 Topology 2 1.1 Definition of a topology 2 1.2 Basis (Base) of a topology 2 1.3 The subspace topology & the product topology on X Y 3 1.4 Basic topology concepts: limit points, closed sets,
More informationconverges as well if x < 1. 1 x n x n 1 1 = 2 a nx n
Solve the following 6 problems. 1. Prove that if series n=1 a nx n converges for all x such that x < 1, then the series n=1 a n xn 1 x converges as well if x < 1. n For x < 1, x n 0 as n, so there exists
More informationREAL AND COMPLEX ANALYSIS
REAL AND COMPLE ANALYSIS Third Edition Walter Rudin Professor of Mathematics University of Wisconsin, Madison Version 1.1 No rights reserved. Any part of this work can be reproduced or transmitted in any
More informationCONTENTS. 4 Hausdorff Measure Introduction The Cantor Set Rectifiable Curves Cantor SetLike Objects...
Contents 1 Functional Analysis 1 1.1 Hilbert Spaces................................... 1 1.1.1 Spectral Theorem............................. 4 1.2 Normed Vector Spaces.............................. 7 1.2.1
More information1. Let A R be a nonempty set that is bounded from above, and let a be the least upper bound of A. Show that there exists a sequence {a n } n N
Applied Analysis prelim July 15, 216, with solutions Solve 4 of the problems 15 and 2 of the problems 68. We will only grade the first 4 problems attempted from15 and the first 2 attempted from problems
More informationReal Analysis Math 131AH Rudin, Chapter #1. Dominique Abdi
Real Analysis Math 3AH Rudin, Chapter # Dominique Abdi.. If r is rational (r 0) and x is irrational, prove that r + x and rx are irrational. Solution. Assume the contrary, that r+x and rx are rational.
More informationSummer JumpStart Program for Analysis, 2012 SongYing Li. 1 Lecture 7: Equicontinuity and Series of functions
Summer JumpStart Program for Analysis, 0 SongYing Li Lecture 7: Equicontinuity and Series of functions. Equicontinuity Definition. Let (X, d) be a metric space, K X and K is a compact subset of X. C(K)
More informationMcGill University Math 354: Honors Analysis 3
Practice problems McGill University Math 354: Honors Analysis 3 not for credit Problem 1. Determine whether the family of F = {f n } functions f n (x) = x n is uniformly equicontinuous. 1st Solution: The
More informationREVIEW OF ESSENTIAL MATH 346 TOPICS
REVIEW OF ESSENTIAL MATH 346 TOPICS 1. AXIOMATIC STRUCTURE OF R Doğan Çömez The real number system is a complete ordered field, i.e., it is a set R which is endowed with addition and multiplication operations
More informationMH 7500 THEOREMS. (iii) A = A; (iv) A B = A B. Theorem 5. If {A α : α Λ} is any collection of subsets of a space X, then
MH 7500 THEOREMS Definition. A topological space is an ordered pair (X, T ), where X is a set and T is a collection of subsets of X such that (i) T and X T ; (ii) U V T whenever U, V T ; (iii) U T whenever
More information2 (Bonus). Let A X consist of points (x, y) such that either x or y is a rational number. Is A measurable? What is its Lebesgue measure?
MA 6454A (Real Analysis), Dr. Chernov Homework assignment 1 (Due 9/5). Prove that every countable set A is measurable and µ(a) = 0. 2 (Bonus). Let A consist of points (x, y) such that either x or y is
More informationIntroduction to Topology
Introduction to Topology Randall R. Holmes Auburn University Typeset by AMSTEX Chapter 1. Metric Spaces 1. Definition and Examples. As the course progresses we will need to review some basic notions about
More informationThe ArzelàAscoli Theorem
John Nachbar Washington University March 27, 2016 The ArzelàAscoli Theorem The ArzelàAscoli Theorem gives sufficient conditions for compactness in certain function spaces. Among other things, it helps
More informationMost Continuous Functions are Nowhere Differentiable
Most Continuous Functions are Nowhere Differentiable Spring 2004 The Space of Continuous Functions Let K = [0, 1] and let C(K) be the set of all continuous functions f : K R. Definition 1 For f C(K) we
More informationReal Analysis Problems
Real Analysis Problems Cristian E. Gutiérrez September 14, 29 1 1 CONTINUITY 1 Continuity Problem 1.1 Let r n be the sequence of rational numbers and Prove that f(x) = 1. f is continuous on the irrationals.
More informationPROBLEMS. (b) (Polarization Identity) Show that in any inner product space
1 Professor Carl Cowen Math 54600 Fall 09 PROBLEMS 1. (Geometry in Inner Product Spaces) (a) (Parallelogram Law) Show that in any inner product space x + y 2 + x y 2 = 2( x 2 + y 2 ). (b) (Polarization
More informationLocally convex spaces, the hyperplane separation theorem, and the KreinMilman theorem
56 Chapter 7 Locally convex spaces, the hyperplane separation theorem, and the KreinMilman theorem Recall that C(X) is not a normed linear space when X is not compact. On the other hand we could use semi
More information16 1 Basic Facts from Functional Analysis and Banach Lattices
16 1 Basic Facts from Functional Analysis and Banach Lattices 1.2.3 Banach Steinhaus Theorem Another fundamental theorem of functional analysis is the Banach Steinhaus theorem, or the Uniform Boundedness
More informationExercise Solutions to Functional Analysis
Exercise Solutions to Functional Analysis Note: References refer to M. Schechter, Principles of Functional Analysis Exersize that. Let φ,..., φ n be an orthonormal set in a Hilbert space H. Show n f n
More informationLecture 5  Hausdorff and GromovHausdorff Distance
Lecture 5  Hausdorff and GromovHausdorff Distance August 1, 2011 1 Definition and Basic Properties Given a metric space X, the set of closed sets of X supports a metric, the Hausdorff metric. If A is
More informationContinuity of convex functions in normed spaces
Continuity of convex functions in normed spaces In this chapter, we consider continuity properties of realvalued convex functions defined on open convex sets in normed spaces. Recall that every infinitedimensional
More information1/12/05: sec 3.1 and my article: How good is the Lebesgue measure?, Math. Intelligencer 11(2) (1989),
Real Analysis 2, Math 651, Spring 2005 April 26, 2005 1 Real Analysis 2, Math 651, Spring 2005 Krzysztof Chris Ciesielski 1/12/05: sec 3.1 and my article: How good is the Lebesgue measure?, Math. Intelligencer
More informationMath 421, Homework #9 Solutions
Math 41, Homework #9 Solutions (1) (a) A set E R n is said to be path connected if for any pair of points x E and y E there exists a continuous function γ : [0, 1] R n satisfying γ(0) = x, γ(1) = y, and
More informationCHAPTER I THE RIESZ REPRESENTATION THEOREM
CHAPTER I THE RIESZ REPRESENTATION THEOREM We begin our study by identifying certain special kinds of linear functionals on certain special vector spaces of functions. We describe these linear functionals
More informationAn introduction to some aspects of functional analysis
An introduction to some aspects of functional analysis Stephen Semmes Rice University Abstract These informal notes deal with some very basic objects in functional analysis, including norms and seminorms
More informationHonours Analysis III
Honours Analysis III Math 354 Prof. Dmitry Jacobson Notes Taken By: R. Gibson Fall 2010 1 Contents 1 Overview 3 1.1 padic Distance............................................ 4 2 Introduction 5 2.1 Normed
More informationSOME QUESTIONS FOR MATH 766, SPRING Question 1. Let C([0, 1]) be the set of all continuous functions on [0, 1] endowed with the norm
SOME QUESTIONS FOR MATH 766, SPRING 2016 SHUANGLIN SHAO Question 1. Let C([0, 1]) be the set of all continuous functions on [0, 1] endowed with the norm f C = sup f(x). 0 x 1 Prove that C([0, 1]) is a
More informationCHAPTER V DUAL SPACES
CHAPTER V DUAL SPACES DEFINITION Let (X, T ) be a (real) locally convex topological vector space. By the dual space X, or (X, T ), of X we mean the set of all continuous linear functionals on X. By the
More information2) Let X be a compact space. Prove that the space C(X) of continuous realvalued functions is a complete metric space.
University of Bergen General Functional Analysis Problems with solutions 6 ) Prove that is unique in any normed space. Solution of ) Let us suppose that there are 2 zeros and 2. Then = + 2 = 2 + = 2. 2)
More informationFilters in Analysis and Topology
Filters in Analysis and Topology David MacIver July 1, 2004 Abstract The study of filters is a very natural way to talk about convergence in an arbitrary topological space, and carries over nicely into
More informationReal Analysis. Jesse Peterson
Real Analysis Jesse Peterson February 1, 2017 2 Contents 1 Preliminaries 7 1.1 Sets.................................. 7 1.1.1 Countability......................... 8 1.1.2 Transfinite induction.....................
More informationCHAPTER II THE HAHNBANACH EXTENSION THEOREMS AND EXISTENCE OF LINEAR FUNCTIONALS
CHAPTER II THE HAHNBANACH EXTENSION THEOREMS AND EXISTENCE OF LINEAR FUNCTIONALS In this chapter we deal with the problem of extending a linear functional on a subspace Y to a linear functional on the
More informationFUNCTIONAL ANALYSIS LECTURE NOTES: COMPACT SETS AND FINITEDIMENSIONAL SPACES. 1. Compact Sets
FUNCTIONAL ANALYSIS LECTURE NOTES: COMPACT SETS AND FINITEDIMENSIONAL SPACES CHRISTOPHER HEIL 1. Compact Sets Definition 1.1 (Compact and Totally Bounded Sets). Let X be a metric space, and let E X be
More information7 About Egorov s and Lusin s theorems
Tel Aviv University, 2013 Measure and category 62 7 About Egorov s and Lusin s theorems 7a About SeveriniEgorov theorem.......... 62 7b About Lusin s theorem............... 64 7c About measurable functions............
More informationThe weak topology of locally convex spaces and the weak* topology of their duals
The weak topology of locally convex spaces and the weak* topology of their duals Jordan Bell jordan.bell@gmail.com Department of Mathematics, University of Toronto April 3, 2014 1 Introduction These notes
More informationMATH 4200 HW: PROBLEM SET FOUR: METRIC SPACES
MATH 4200 HW: PROBLEM SET FOUR: METRIC SPACES PETE L. CLARK 4. Metric Spaces (no more lulz) Directions: This week, please solve any seven problems. Next week, please solve seven more. Starred parts of
More informationWhat to remember about metric spaces
Division of the Humanities and Social Sciences What to remember about metric spaces KC Border These notes are (I hope) a gentle introduction to the topological concepts used in economic theory. If the
More informationOPERATOR THEORY ON HILBERT SPACE. Class notes. John Petrovic
OPERATOR THEORY ON HILBERT SPACE Class notes John Petrovic Contents Chapter 1. Hilbert space 1 1.1. Definition and Properties 1 1.2. Orthogonality 3 1.3. Subspaces 7 1.4. Weak topology 9 Chapter 2. Operators
More informationBanach Spaces II: Elementary Banach Space Theory
BS II c Gabriel Nagy Banach Spaces II: Elementary Banach Space Theory Notes from the Functional Analysis Course (Fall 07  Spring 08) In this section we introduce Banach spaces and examine some of their
More informationMATH 140B  HW 5 SOLUTIONS
MATH 140B  HW 5 SOLUTIONS Problem 1 (WR Ch 7 #8). If I (x) = { 0 (x 0), 1 (x > 0), if {x n } is a sequence of distinct points of (a,b), and if c n converges, prove that the series f (x) = c n I (x x n
More informationSOLUTIONS TO SOME PROBLEMS
23 FUNCTIONAL ANALYSIS Spring 23 SOLUTIONS TO SOME PROBLEMS Warning:These solutions may contain errors!! PREPARED BY SULEYMAN ULUSOY PROBLEM 1. Prove that a necessary and sufficient condition that the
More informationSequences. Chapter 3. n + 1 3n + 2 sin n n. 3. lim (ln(n + 1) ln n) 1. lim. 2. lim. 4. lim (1 + n)1/n. Answers: 1. 1/3; 2. 0; 3. 0; 4. 1.
Chapter 3 Sequences Both the main elements of calculus (differentiation and integration) require the notion of a limit. Sequences will play a central role when we work with limits. Definition 3.. A Sequence
More informationMA651 Topology. Lecture 9. Compactness 2.
MA651 Topology. Lecture 9. Compactness 2. This text is based on the following books: Topology by James Dugundgji Fundamental concepts of topology by Peter O Neil Elements of Mathematics: General Topology
More informationMATHS 730 FC Lecture Notes March 5, Introduction
1 INTRODUCTION MATHS 730 FC Lecture Notes March 5, 2014 1 Introduction Definition. If A, B are sets and there exists a bijection A B, they have the same cardinality, which we write as A, #A. If there exists
More informationFinitedimensional spaces. C n is the space of ntuples x = (x 1,..., x n ) of complex numbers. It is a Hilbert space with the inner product
Chapter 4 Hilbert Spaces 4.1 Inner Product Spaces Inner Product Space. A complex vector space E is called an inner product space (or a prehilbert space, or a unitary space) if there is a mapping (, )
More informationMath 117: Continuity of Functions
Math 117: Continuity of Functions John Douglas Moore November 21, 2008 We finally get to the topic of ɛ δ proofs, which in some sense is the goal of the course. It may appear somewhat laborious to use
More informationNotes on Integrable Functions and the Riesz Representation Theorem Math 8445, Winter 06, Professor J. Segert. f(x) = f + (x) + f (x).
References: Notes on Integrable Functions and the Riesz Representation Theorem Math 8445, Winter 06, Professor J. Segert Evans, Partial Differential Equations, Appendix 3 Reed and Simon, Functional Analysis,
More informationSome Background Material
Chapter 1 Some Background Material In the first chapter, we present a quick review of elementary  but important  material as a way of dipping our toes in the water. This chapter also introduces important
More informationBounded uniformly continuous functions
Bounded uniformly continuous functions Objectives. To study the basic properties of the C algebra of the bounded uniformly continuous functions on some metric space. Requirements. Basic concepts of analysis:
More informationBanach Spaces V: A Closer Look at the w and the w Topologies
BS V c Gabriel Nagy Banach Spaces V: A Closer Look at the w and the w Topologies Notes from the Functional Analysis Course (Fall 07  Spring 08) In this section we discuss two important, but highly nontrivial,
More informationEigenvalues and Eigenfunctions of the Laplacian
The Waterloo Mathematics Review 23 Eigenvalues and Eigenfunctions of the Laplacian Mihai Nica University of Waterloo mcnica@uwaterloo.ca Abstract: The problem of determining the eigenvalues and eigenvectors
More informationA Brief Introduction to Functional Analysis
A Brief Introduction to Functional Analysis Sungwook Lee Department of Mathematics University of Southern Mississippi sunglee@usm.edu July 5, 2007 Definition 1. An algebra A is a vector space over C with
More informationNOTES ON EXISTENCE AND UNIQUENESS THEOREMS FOR ODES
NOTES ON EXISTENCE AND UNIQUENESS THEOREMS FOR ODES JONATHAN LUK These notes discuss theorems on the existence, uniqueness and extension of solutions for ODEs. None of these results are original. The proofs
More informationContinuity. Chapter 4
Chapter 4 Continuity Throughout this chapter D is a nonempty subset of the real numbers. We recall the definition of a function. Definition 4.1. A function from D into R, denoted f : D R, is a subset of
More informationL p Functions. Given a measure space (X, µ) and a real number p [1, ), recall that the L p norm of a measurable function f : X R is defined by
L p Functions Given a measure space (, µ) and a real number p [, ), recall that the L p norm of a measurable function f : R is defined by f p = ( ) /p f p dµ Note that the L p norm of a function f may
More informationMath 320: Real Analysis MWF 1pm, Campion Hall 302 Homework 8 Solutions Please write neatly, and in complete sentences when possible.
Math 320: Real Analysis MWF pm, Campion Hall 302 Homework 8 Solutions Please write neatly, and in complete sentences when possible. Do the following problems from the book: 4.3.5, 4.3.7, 4.3.8, 4.3.9,
More informationx 0 + f(x), exist as extended real numbers. Show that f is upper semicontinuous This shows ( ɛ, ɛ) B α. Thus
Homework 3 Solutions, Real Analysis I, Fall, 2010. (9) Let f : (, ) [, ] be a function whose restriction to (, 0) (0, ) is continuous. Assume the onesided limits p = lim x 0 f(x), q = lim x 0 + f(x) exist
More informationC ALGEBRAS MATH SPRING 2015 PROBLEM SET #6
C ALGEBRAS MATH 113  SPRING 2015 PROBLEM SET #6 Problem 1 (Positivity in C algebras). The purpose of this problem is to establish the following result: Theorem. Let A be a unital C algebra. For a A,
More informationIndeed, if we want m to be compatible with taking limits, it should be countably additive, meaning that ( )
Lebesgue Measure The idea of the Lebesgue integral is to first define a measure on subsets of R. That is, we wish to assign a number m(s to each subset S of R, representing the total length that S takes
More informationExercise 1. Let f be a nonnegative measurable function. Show that. where ϕ is taken over all simple functions with ϕ f. k 1.
Real Variables, Fall 2014 Problem set 3 Solution suggestions xercise 1. Let f be a nonnegative measurable function. Show that f = sup ϕ, where ϕ is taken over all simple functions with ϕ f. For each n
More informationIntroduction to Empirical Processes and Semiparametric Inference Lecture 08: Stochastic Convergence
Introduction to Empirical Processes and Semiparametric Inference Lecture 08: Stochastic Convergence Michael R. Kosorok, Ph.D. Professor and Chair of Biostatistics Professor of Statistics and Operations
More informationReal Analysis April f (t) dt. Show that g(x) 0 as x.
eal Analysis April Part A. Let f L ( and let g(x x f (t dt. Show that g(x as x. Let {x n } be a sequence in such that x n as n, and define h n (t : χ [xn, f (t. Now, given ε >, there is some N such that
More information3 COUNTABILITY AND CONNECTEDNESS AXIOMS
3 COUNTABILITY AND CONNECTEDNESS AXIOMS Definition 3.1 Let X be a topological space. A subset D of X is dense in X iff D = X. X is separable iff it contains a countable dense subset. X satisfies the first
More informationA Précis of Functional Analysis for Engineers DRAFT NOT FOR DISTRIBUTION. JeanFrançois Hiller and KlausJürgen Bathe
A Précis of Functional Analysis for Engineers DRAFT NOT FOR DISTRIBUTION JeanFrançois Hiller and KlausJürgen Bathe August 29, 22 1 Introduction The purpose of this précis is to review some classical
More information1 Takehome exam and final exam study guide
Math 215  Introduction to Advanced Mathematics Fall 2013 1 Takehome exam and final exam study guide 1.1 Problems The following are some problems, some of which will appear on the final exam. 1.1.1 Number
More informationWe are going to discuss what it means for a sequence to converge in three stages: First, we define what it means for a sequence to converge to zero
Chapter Limits of Sequences Calculus Student: lim s n = 0 means the s n are getting closer and closer to zero but never gets there. Instructor: ARGHHHHH! Exercise. Think of a better response for the instructor.
More informationReal Analysis  Notes and After Notes Fall 2008
Real Analysis  Notes and After Notes Fall 2008 October 29, 2008 1 Introduction into proof August 20, 2008 First we will go through some simple proofs to learn how one writes a rigorous proof. Let start
More informationare Banach algebras. f(x)g(x) max Example 7.4. Similarly, A = L and A = l with the pointwise multiplication
7. Banach algebras Definition 7.1. A is called a Banach algebra (with unit) if: (1) A is a Banach space; (2) There is a multiplication A A A that has the following properties: (xy)z = x(yz), (x + y)z =
More informationRecall that if X is a compact metric space, C(X), the space of continuous (realvalued) functions on X, is a Banach space with the norm
Chapter 13 Radon Measures Recall that if X is a compact metric space, C(X), the space of continuous (realvalued) functions on X, is a Banach space with the norm (13.1) f = sup x X f(x). We want to identify
More information2.2 Annihilators, complemented subspaces
34CHAPTER 2. WEAK TOPOLOGIES, REFLEXIVITY, ADJOINT OPERATORS 2.2 Annihilators, complemented subspaces Definition 2.2.1. [Annihilators, PreAnnihilators] Assume X is a Banach space. Let M X and N X. We
More informationContents. 2 Sequences and Series Approximation by Rational Numbers Sequences Basics on Sequences...
Contents 1 Real Numbers: The Basics... 1 1.1 Notation... 1 1.2 Natural Numbers... 4 1.3 Integers... 5 1.4 Fractions and Rational Numbers... 10 1.4.1 Introduction... 10 1.4.2 Powers and Radicals of Rational
More informationExamples of Dual Spaces from Measure Theory
Chapter 9 Examples of Dual Spaces from Measure Theory We have seen that L (, A, µ) is a Banach space for any measure space (, A, µ). We will extend that concept in the following section to identify an
More information1.2 Fundamental Theorems of Functional Analysis
1.2 Fundamental Theorems of Functional Analysis 15 Indeed, h = ω ψ ωdx is continuous compactly supported with R hdx = 0 R and thus it has a unique compactly supported primitive. Hence fφ dx = f(ω ψ ωdy)dx
More informationHOMEWORK ASSIGNMENT 6
HOMEWORK ASSIGNMENT 6 DUE 15 MARCH, 2016 1) Suppose f, g : A R are uniformly continuous on A. Show that f + g is uniformly continuous on A. Solution First we note: In order to show that f + g is uniformly
More informationThe Way of Analysis. Robert S. Strichartz. Jones and Bartlett Publishers. Mathematics Department Cornell University Ithaca, New York
The Way of Analysis Robert S. Strichartz Mathematics Department Cornell University Ithaca, New York Jones and Bartlett Publishers Boston London Contents Preface xiii 1 Preliminaries 1 1.1 The Logic of
More informationCopyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
.1 Limits of Sequences. CHAPTER.1.0. a) True. If converges, then there is an M > 0 such that M. Choose by Archimedes an N N such that N > M/ε. Then n N implies /n M/n M/N < ε. b) False. = n does not converge,
More informationCHAPTER VIII HILBERT SPACES
CHAPTER VIII HILBERT SPACES DEFINITION Let X and Y be two complex vector spaces. A map T : X Y is called a conjugatelinear transformation if it is a reallinear transformation from X into Y, and if T (λx)
More informationMaths 212: Homework Solutions
Maths 212: Homework Solutions 1. The definition of A ensures that x π for all x A, so π is an upper bound of A. To show it is the least upper bound, suppose x < π and consider two cases. If x < 1, then
More informationLecture 4 Lebesgue spaces and inequalities
Lecture 4: Lebesgue spaces and inequalities 1 of 10 Course: Theory of Probability I Term: Fall 2013 Instructor: Gordan Zitkovic Lecture 4 Lebesgue spaces and inequalities Lebesgue spaces We have seen how
More informationCHAPTER 1. Metric Spaces. 1. Definition and examples
CHAPTER Metric Spaces. Definition and examples Metric spaces generalize and clarify the notion of distance in the real line. The definitions will provide us with a useful tool for more general applications
More informationAfter taking the square and expanding, we get x + y 2 = (x + y) (x + y) = x 2 + 2x y + y 2, inequality in analysis, we obtain.
Lecture 1: August 25 Introduction. Topology grew out of certain questions in geometry and analysis about 100 years ago. As Wikipedia puts it, the motivating insight behind topology is that some geometric
More informationMath 361: Homework 1 Solutions
January 3, 4 Math 36: Homework Solutions. We say that two norms and on a vector space V are equivalent or comparable if the topology they define on V are the same, i.e., for any sequence of vectors {x
More informationMATH NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS
MATH. 4433. NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS TOMASZ PRZEBINDA. Final project, due 0:00 am, /0/208 via email.. State the Fundamental Theorem of Algebra. Recall that a subset K
More informationExercises to Applied Functional Analysis
Exercises to Applied Functional Analysis Exercises to Lecture 1 Here are some exercises about metric spaces. Some of the solutions can be found in my own additional lecture notes on Blackboard, as the
More informationAn Introduction to Complex Analysis and Geometry John P. D Angelo, Pure and Applied Undergraduate Texts Volume 12, American Mathematical Society, 2010
An Introduction to Complex Analysis and Geometry John P. D Angelo, Pure and Applied Undergraduate Texts Volume 12, American Mathematical Society, 2010 John P. D Angelo, Univ. of Illinois, Urbana IL 61801.
More informationIn English, this means that if we travel on a straight line between any two points in C, then we never leave C.
Convex sets In this section, we will be introduced to some of the mathematical fundamentals of convex sets. In order to motivate some of the definitions, we will look at the closest point problem from
More informationAliprantis, Border: Infinitedimensional Analysis A Hitchhiker s Guide
aliprantis.tex May 10, 2011 Aliprantis, Border: Infinitedimensional Analysis A Hitchhiker s Guide Notes from [AB2]. 1 Odds and Ends 2 Topology 2.1 Topological spaces Example. (2.2) A semimetric = triangle
More informationMetric Spaces. Exercises Fall 2017 Lecturer: Viveka Erlandsson. Written by M.van den Berg
Metric Spaces Exercises Fall 2017 Lecturer: Viveka Erlandsson Written by M.van den Berg School of Mathematics University of Bristol BS8 1TW Bristol, UK 1 Exercises. 1. Let X be a nonempty set, and suppose
More informationIowa State University. Instructor: Alex Roitershtein Summer Homework #5. Solutions
Math 50 Iowa State University Introduction to Real Analysis Department of Mathematics Instructor: Alex Roitershtein Summer 205 Homework #5 Solutions. Let α and c be real numbers, c > 0, and f is defined
More informationCHAPTER 7. Connectedness
CHAPTER 7 Connectedness 7.1. Connected topological spaces Definition 7.1. A topological space (X, T X ) is said to be connected if there is no continuous surjection f : X {0, 1} where the two point set
More informationSummary of Real Analysis by Royden
Summary of Real Analysis by Royden Dan Hathaway May 2010 This document is a summary of the theorems and definitions and theorems from Part 1 of the book Real Analysis by Royden. In some areas, such as
More informationThree hours THE UNIVERSITY OF MANCHESTER. 24th January
Three hours MATH41011 THE UNIVERSITY OF MANCHESTER FOURIER ANALYSIS AND LEBESGUE INTEGRATION 24th January 2013 9.45 12.45 Answer ALL SIX questions in Section A (25 marks in total). Answer THREE of the
More informationA LITTLE REAL ANALYSIS AND TOPOLOGY
A LITTLE REAL ANALYSIS AND TOPOLOGY 1. NOTATION Before we begin some notational definitions are useful. (1) Z = {, 3, 2, 1, 0, 1, 2, 3, }is the set of integers. (2) Q = { a b : aεz, bεz {0}} is the set
More informationTHE INVERSE FUNCTION THEOREM FOR LIPSCHITZ MAPS
THE INVERSE FUNCTION THEOREM FOR LIPSCHITZ MAPS RALPH HOWARD DEPARTMENT OF MATHEMATICS UNIVERSITY OF SOUTH CAROLINA COLUMBIA, S.C. 29208, USA HOWARD@MATH.SC.EDU Abstract. This is an edited version of a
More informationContribution of Problems
Exam topics 1. Basic structures: sets, lists, functions (a) Sets { }: write all elements, or define by condition (b) Set operations: A B, A B, A\B, A c (c) Lists ( ): Cartesian product A B (d) Functions
More information