Math 172 Problem Set 5 Solutions
|
|
- Dale Spencer
- 5 years ago
- Views:
Transcription
1 Math 172 Problem Set 5 Solutions 2.4 Let E = {(t, x : < x b, x t b}. To prove integrability of g, first observe that b b b f(t b b g(x dx = dt t dx f(t t dtdx. x Next note that f(t/t χ E is a measurable function on 2 since the pull-back of the function t f(t/t to a function F (t, x = f(t/t is measurable by Corollary in the text. Hence Tonelli s theorem (Theorem implies that this last integral is equal to b t f(t t b dxdt = f(t t x b tdt = f(t dt <. This proves that g is integrable, and a nearly identical argument shows that b g(xdx = b f(tdt Consider the set E = {(x, y d+1 : y < f(x }. To show that this set is measurable, consider an increasing sequence of simple functions s i which converge to f pointwise (thus in L 1 by the dominated convergecne theorem. One easily checks that the set of points under the graph of a simple function is a measurable subset of d+1. Letting U i be the set of points under graph of s i, we have that E = i U i so that E is measurable. Moreover, since the U i are nested, we have by monotonicity that m(u i = s i m(e = f. Now the proof of the assertion is a straightforward application of Corollary 2.3.3, for we have E y = {x d : (x, y E} = {x d : f(x > y }. 1
2 Therefore, the corollary tells us that f = m(e = m(e y dy = since E y = for y <. m(e α dα 2.21a The function (x, y f(x y is measurable by Proposition in the text, and the function (x, y g(y is also measurable by Propsition in the text. Hence, as the product of measurable functions, F : (x, y f(x yg(y is also measurable on 2d. 2.21b To show that the integrability of f and g imply that F is integrable, we have ( F = f(x yg(y dxdy = f(x y g(y dx dy 2d d ( ( d ( = g(y f(x y dx dy = g(y dy f(x dx < d d d d where we have used Tonelli s theorem in the second equality and the translational invariance of the Lebesgue integral in the final equality. We thus conclude that F is integrable on 2d. 2.21c Assuming that f and g are integrable, we know from part b above that F is integrable as well. Hence a direct application of Fubini s theorem implies that F x is integrable for a.e. x, which means that the convolution (f g(x = f(x yg(ydy = d F x (ydy d is defined for a.e. x. 2.21d We have f g L 1 = f g = d d f(x yg(ydy dx f(x yg(y dydx d 2d where we have used Tonelli s theorem and the triangle inequality. By part b, we know that this last integral is equal to f L 1 g L 1. Clearly the can be changed to an = if both f and g are non-negative. 2.21e Since f(xe 2πix ξ = f(x, we have that ˆf(ξ = f(xe 2πix ξ dx f(x dx = f L 1, d d which proves that ˆf is bounded. Moreover, as in the proof of Proposition we observe that since e 2πix ξ is continuous in ξ for every x, if ξ n ξ we have that 2
3 f(xe 2πix ξn f(xe 2πix ξ. But since each function f n (x := f(xe 2πix ξn is dominated by the integrable function f, the dominated convergence theorem tells us that ˆf(ξ n = f(xe 2πix ξn dx d f(xe 2πix ξ dx = ˆf(ξ, d which proves that ˆf is continuous. Finally, we prove the identity (f g(ξ = ˆf(ξĝ(ξ. We have ( (f g(ξ = (f g(xe 2πix ξ dx = f(x yg(ydy e 2πix ξ dx d d ( d = f(x ye 2πiy ξ g(ye 2πiy ξ dy e 2πix ξ dx d ( d = f(x ye 2πi(x y ξ dx g(ye 2πiy ξ dy d d = ˆf(ξg(ye 2πiy ξ dy = ˆf(ξĝ(ξ. d where we have used part b and Fubini s theorem in the fourth inequality. Problem 2.4a As the problem suggests, consider the upper triangular matrix ( 1 a L =. 1 Clearly we have L 1 = ( 1 a 1 and so χ L(E (x, y = χ E (L 1 (x, y = χ E (x ay, y as claimed. Now since L(E is measurable (by exercise 1.8, the function χ L(E is measurable as well, and so we have ( m(l(e = χ L(E = χ L(E (x, ydx dy ( = χ E (x ay, ydx dy where we have used Tonelli s theorem to obtain the second equality. Assume for now that m(e <. Then the function χ E is integrable, and hence Fubini s theorem tells us that χ y E is integrable for almost every y. We may thus use the translational invariance of the Lebesgue integral to express the final integral above as ( χ E (x, ydx dy = χ E = m(e,, 3
4 as claimed in the problem. This argument extends easily to an upper (lower unitriangular d d matrix. In particular, suppose L us upper unitriangular. One easily shows that its inverse is also upper unitriangular, so we may write 1 a 12 a 1d L 1 1 a 2d = We thus have χ L(E (x 1,..., x d = χ E (L 1 (x 1,..., x d = χ E (y 1,..., y d, y i = x i + d j=i+1 Hence, by Tonelli s theorem, we have ( m(l(e = χ (y2,...,y d E (x 1 + a 12 x 2 + a 1d x d dx 1 dx 2 dx d. a ij x j. Once again by translational invariance, we have that this integral is equal to ( χ E (x 1, y 2,..., y d dx 1 dx 2 dx d. Another d 1 applications of Tonelli s theorem and translational invariance gives that this integral is equal to d χ E = m(e. An essentially identical argument shows that m(l(e = m(e for lower unitriangular matrices, so we are done with the first part of this problem. Problem 2.4b Suppose we had an LDU decomposition (which is a slight variant of the standard LU decomposition from linear algebra L = L 1 L 2 where L 1 is an upper (lower unitriangular matrix, is a diagonal matrix, and L 2 is a lower (upper unitriangular matrix as in the problem. We write a 11 a 22 =..... a dd It s easy to see that (E = E δ where δ = (a 11, a 22,..., a dd. Hence we know from (an easy generalization of exercise 1.7 that m( (E = a 11 a 22 a dd m(e = det m(e 4
5 and hence m(l(e = m(l 1 L 2 (E = m( L 2 (E = det m(l 2 (E = det m(e. Since det = det L, this proves the claim in this case. In general, this decomposition might fail to exist, but we may always find a permutation matrix P such that P L = L 1 L 2. Since a permutation just amounts to a relabeling of the axes, it is easy to see that m(p L(E = m(l(e, so we are finished as long as m(e <. If not, replace E in the argument above with E n = E B n where B n is the closed ball of radius n. This implies either m(l(e n or m(l(e n = for all n, depending on det L. Either way, the general statement holds by monotonicity. Problem 5 (i. It s easy to see that f per, g per are measurable functions on d. For example, we know that for every a, the set E = f 1 ([, a is a measurable subset of := [ π, π] d. Hence fper([, 1 a is the union of all translates of E by 2πn, where n = (n 1,..., n d has integer coordinates. Thus since f per, g per are measurable, we argue exactly as in exercise 2.21a above to conclude that the function (x, y f per (x yg per (y is measurable on I 2d. Problem 5 (ii. We first note that a straightforward generalization of exercise 2.3 (see problem set 3 implies that f per (x y dx = f(x dx. (.1 Therefore, as exercise 2.21b above, we have ( f per (x yg per (y dxdy = f per (x y dx g(y dy I 2d I ( d ( = f(x dx g(y dy = f L1 ( g L1 ( <, where in the first equality we have used Fubini s theorem and in the second we have used (.1. Problem 5 (iii. Define a function F : 2d by F (x, y = f(x yg(yχ I 2d. We know from (ii that F is integrable, so Fubini s theorem tells us that F x is integrable for a.e. x. This means that F x (ydy = f per (x yg per (ydy = f g(x d is well-defined for a.e. x as claimed. 5
6 Problem 5 (iv. We have f g L1 ( = f g = f per (x yg per (ydy dx f per (x yg per (y dydx where we have used the standard triangle inequaltiy for the Lebesgue integral. Now by applying Tonelli s theorem and (ii above, we see that this last expression is equal to f L1 ( g L1 (, proving the first claim. Now observe that the sign which appears in the triangle equality above can be replaced by an = sign if both f per and g per (and hence both f and g are non-negative a.e. Problem 5 (v. We showed in (iii that f g is defined a.e., and we claim that at such points we have (f g(x = f per (x yg per (ydy = f per (ug per (x udu. To see this, observe that since f g L 1 by assumption, Fubini s theorem implies that we may express this integral as π = π π π xd π x1 π x d +π xd +π = x d π f per (x 1 y 1,..., x d y d g per (y 1,..., y d dy 1 dy d x 1+π x1+π x 1 π f per (u 1,..., u d g per (x 1 u 1,..., x d u d ( 1 d du 1 du d f per (u 1,..., u d g per (x 1 u 1,..., x d u d du 1 du d = f per (ug per (x udu as claimed, where the last equality follows from (an easy generalization of the translational invariance established in exercise 2.3. This last integral, where it is defined, is precisely the definition of (g f(x, and we know that it exists a.e. Hence f g = g f away from a set of measure zero, and therefore we must have f g g f L 1 ( =. Problem 5 (vi. We first show that if f per C 1, then h (f g = ( h f g, where h is the directional derivative in the direction of h d. For any h d, consider a sequence = t n h where {t n } is a decreasing sequence t n. We have (f g(x + (f g(x = [ fper (x + y f per (x y ] g per (ydy. (.2 6
7 The mean value theorem tells us that the absolute value of the integrand is bounded above by the function (sup h f g per which is integrable on. Since the pointwise limit of the integrand as n is ( h f per (x yg per (y, and the limit on the left hand side of (.2 is clearly h (f g, the dominated convergence theorem tells us that h (f g = ( h f g. Since f per is bounded and g is integrable, it is easy to see that ( h f g is bounded. We can show that this function is (uniformly continuous as well, for since h f per is uniformly continuous, for every ɛ > there exists a δ > such that x x < δ h f per (x y h f per (x y < ɛ. Therefore if x x < δ, by the triangle inequality ( h f g(x ( h f g(x f per (x y f per (x y g(y dy < ɛ g L 1. Since h (f g = ( h f g is bounded and continous for any h, we conclude that f g is C 1 on the interior of. Hence (f g per is C 1 on the set d \( +2πZ d For x + 2π ˆm where ˆm Z d, we have (f g per (x + (f g(x h (f g per (x = lim (f g(x + 2π ˆm (f g(x = lim h ( n fper (x + 2π ˆm f per (x = lim g(ydy = lim ( fper (x + f per (x g(ydy by the periodicity of f per. Hence we may apply the same argument at x as for points interior to, and therefore (f g per is C 1 at x, and therefore on all of d. Of course, if f per C k then we apply the argument above and a the standard inductive argument to find that (f g per C k. Moreover, if we instead have g C k, we use part (v and the argument above to ensure that all x-derivatives fall on g, and so the dominated convergence theorem again implies that f g C k. 7
Solutions: Problem Set 4 Math 201B, Winter 2007
Solutions: Problem Set 4 Math 2B, Winter 27 Problem. (a Define f : by { x /2 if < x
More informationII. FOURIER TRANSFORM ON L 1 (R)
II. FOURIER TRANSFORM ON L 1 (R) In this chapter we will discuss the Fourier transform of Lebesgue integrable functions defined on R. To fix the notation, we denote L 1 (R) = {f : R C f(t) dt < }. The
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 informationREAL ANALYSIS I HOMEWORK 4
REAL ANALYSIS I HOMEWORK 4 CİHAN BAHRAN The questions are from Stein and Shakarchi s text, Chapter 2.. Given a collection of sets E, E 2,..., E n, construct another collection E, E 2,..., E N, with N =
More informationTools from Lebesgue integration
Tools from Lebesgue integration E.P. van den Ban Fall 2005 Introduction In these notes we describe some of the basic tools from the theory of Lebesgue integration. Definitions and results will be given
More informationMath 104: Homework 7 solutions
Math 04: Homework 7 solutions. (a) The derivative of f () = is f () = 2 which is unbounded as 0. Since f () is continuous on [0, ], it is uniformly continous on this interval by Theorem 9.2. Hence for
More informationReal Analysis: Homework # 12 Fall Professor: Sinan Gunturk Fall Term 2008
Eduardo Corona eal Analysis: Homework # 2 Fall 2008 Professor: Sinan Gunturk Fall Term 2008 #3 (p.298) Let X be the set of rational numbers and A the algebra of nite unions of intervals of the form (a;
More informationThus f is continuous at x 0. Matthew Straughn Math 402 Homework 6
Matthew Straughn Math 402 Homework 6 Homework 6 (p. 452) 14.3.3, 14.3.4, 14.3.5, 14.3.8 (p. 455) 14.4.3* (p. 458) 14.5.3 (p. 460) 14.6.1 (p. 472) 14.7.2* Lemma 1. If (f (n) ) converges uniformly to some
More informationMATH 411 NOTES (UNDER CONSTRUCTION)
MATH 411 NOTES (NDE CONSTCTION 1. Notes on compact sets. This is similar to ideas you learned in Math 410, except open sets had not yet been defined. Definition 1.1. K n is compact if for every covering
More informationHomework 11. Solutions
Homework 11. Solutions Problem 2.3.2. Let f n : R R be 1/n times the characteristic function of the interval (0, n). Show that f n 0 uniformly and f n µ L = 1. Why isn t it a counterexample to the Lebesgue
More informationbe the set of complex valued 2π-periodic functions f on R such that
. Fourier series. Definition.. Given a real number P, we say a complex valued function f on R is P -periodic if f(x + P ) f(x) for all x R. We let be the set of complex valued -periodic functions f on
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 information1 Fourier Integrals on L 2 (R) and L 1 (R).
18.103 Fall 2013 1 Fourier Integrals on L 2 () and L 1 (). The first part of these notes cover 3.5 of AG, without proofs. When we get to things not covered in the book, we will start giving proofs. The
More informationMeasure and Integration: Solutions of CW2
Measure and Integration: s of CW2 Fall 206 [G. Holzegel] December 9, 206 Problem of Sheet 5 a) Left (f n ) and (g n ) be sequences of integrable functions with f n (x) f (x) and g n (x) g (x) for almost
More informationLebesgue Integration on R n
Lebesgue Integration on R n The treatment here is based loosely on that of Jones, Lebesgue Integration on Euclidean Space We give an overview from the perspective of a user of the theory Riemann integration
More informationBernstein s inequality and Nikolsky s inequality for R d
Bernstein s inequality and Nikolsky s inequality for d Jordan Bell jordan.bell@gmail.com Department of athematics University of Toronto February 6 25 Complex Borel measures and the Fourier transform Let
More informationMATH MEASURE THEORY AND FOURIER ANALYSIS. Contents
MATH 3969 - MEASURE THEORY AND FOURIER ANALYSIS ANDREW TULLOCH Contents 1. Measure Theory 2 1.1. Properties of Measures 3 1.2. Constructing σ-algebras and measures 3 1.3. Properties of the Lebesgue measure
More informationMATH 6337: Homework 8 Solutions
6.1. MATH 6337: Homework 8 Solutions (a) Let be a measurable subset of 2 such that for almost every x, {y : (x, y) } has -measure zero. Show that has measure zero and that for almost every y, {x : (x,
More informationSOLUTIONS OF SELECTED PROBLEMS
SOLUTIONS OF SELECTED PROBLEMS Problem 36, p. 63 If µ(e n < and χ En f in L, then f is a.e. equal to a characteristic function of a measurable set. Solution: By Corollary.3, there esists a subsequence
More informationTHE INVERSE FUNCTION THEOREM
THE INVERSE FUNCTION THEOREM W. PATRICK HOOPER The implicit function theorem is the following result: Theorem 1. Let f be a C 1 function from a neighborhood of a point a R n into R n. Suppose A = Df(a)
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 informationFolland: Real Analysis, Chapter 8 Sébastien Picard
Folland: Real Analysis, Chapter 8 Sébastien Picard Problem 8.3 Let η(t) = e /t for t >, η(t) = for t. a. For k N and t >, η (k) (t) = P k (/t)e /t where P k is a polynomial of degree 2k. b. η (k) () exists
More informationIndeed, the family is still orthogonal if we consider a complex valued inner product ( or an inner product on complex vector space)
Fourier series of complex valued functions Suppose now f is a piecewise continuous complex valued function on [, π], that is f(x) = u(x)+iv(x) such that both u and v are real valued piecewise continuous
More informationLecture 4: Fourier Transforms.
1 Definition. Lecture 4: Fourier Transforms. We now come to Fourier transforms, which we give in the form of a definition. First we define the spaces L 1 () and L 2 (). Definition 1.1 The space L 1 ()
More informationPart II Probability and Measure
Part II Probability and Measure Theorems Based on lectures by J. Miller Notes taken by Dexter Chua Michaelmas 2016 These notes are not endorsed by the lecturers, and I have modified them (often significantly)
More informationAnalysis Comprehensive Exam Questions Fall F(x) = 1 x. f(t)dt. t 1 2. tf 2 (t)dt. and g(t, x) = 2 t. 2 t
Analysis Comprehensive Exam Questions Fall 2. Let f L 2 (, ) be given. (a) Prove that ( x 2 f(t) dt) 2 x x t f(t) 2 dt. (b) Given part (a), prove that F L 2 (, ) 2 f L 2 (, ), where F(x) = x (a) Using
More informationANALYSIS QUALIFYING EXAM FALL 2017: SOLUTIONS. 1 cos(nx) lim. n 2 x 2. g n (x) = 1 cos(nx) n 2 x 2. x 2.
ANALYSIS QUALIFYING EXAM FALL 27: SOLUTIONS Problem. Determine, with justification, the it cos(nx) n 2 x 2 dx. Solution. For an integer n >, define g n : (, ) R by Also define g : (, ) R by g(x) = g n
More information3 (Due ). 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 645-4A (Real Analysis), Dr. Chernov Homework assignment 1 (Due ). Show that the open disk x 2 + y 2 < 1 is a countable union of planar elementary sets. Show that the closed disk x 2 + y 2 1 is a countable
More informationCourse 212: Academic Year Section 1: Metric Spaces
Course 212: Academic Year 1991-2 Section 1: Metric Spaces D. R. Wilkins Contents 1 Metric Spaces 3 1.1 Distance Functions and Metric Spaces............. 3 1.2 Convergence and Continuity in Metric Spaces.........
More information1.5 Approximate Identities
38 1 The Fourier Transform on L 1 (R) which are dense subspaces of L p (R). On these domains, P : D P L p (R) and M : D M L p (R). Show, however, that P and M are unbounded even when restricted to these
More information6.2 Fubini s Theorem. (µ ν)(c) = f C (x) dµ(x). (6.2) Proof. Note that (X Y, A B, µ ν) must be σ-finite as well, so that.
6.2 Fubini s Theorem Theorem 6.2.1. (Fubini s theorem - first form) Let (, A, µ) and (, B, ν) be complete σ-finite measure spaces. Let C = A B. Then for each µ ν- measurable set C C the section x C is
More informationSobolev spaces. May 18
Sobolev spaces May 18 2015 1 Weak derivatives The purpose of these notes is to give a very basic introduction to Sobolev spaces. More extensive treatments can e.g. be found in the classical references
More informationMath 172 Problem Set 8 Solutions
Math 72 Problem Set 8 Solutions Problem. (i We have (Fχ [ a,a] (ξ = χ [ a,a] e ixξ dx = a a e ixξ dx = iξ (e iax e iax = 2 sin aξ. ξ (ii We have (Fχ [, e ax (ξ = e ax e ixξ dx = e x(a+iξ dx = a + iξ where
More informationNAME: MATH 172: Lebesgue Integration and Fourier Analysis (winter 2012) Final exam. Wednesday, March 21, time: 2.5h
NAME: SOLUTION problem # 1 2 3 4 5 6 7 8 9 points max 15 20 10 15 10 10 10 10 10 110 MATH 172: Lebesgue Integration and Fourier Analysis (winter 2012 Final exam Wednesday, March 21, 2012 time: 2.5h Please
More informationTOPICS IN FOURIER ANALYSIS-III. Contents
TOPICS IN FOUIE ANALYSIS-III M.T. NAI Contents 1. Fourier transform: Basic properties 2 2. On surjectivity 7 3. Inversion theorem 9 4. Proof of inversion theorem 10 5. The Banach algebra L 1 ( 12 6. Fourier-Plancheral
More information3 hours UNIVERSITY OF MANCHESTER. 22nd May and. Electronic calculators may be used, provided that they cannot store text.
3 hours MATH40512 UNIVERSITY OF MANCHESTER DYNAMICAL SYSTEMS AND ERGODIC THEORY 22nd May 2007 9.45 12.45 Answer ALL four questions in SECTION A (40 marks in total) and THREE of the four questions in SECTION
More informationLEBESGUE INTEGRATION. Introduction
LEBESGUE INTEGATION EYE SJAMAA Supplementary notes Math 414, Spring 25 Introduction The following heuristic argument is at the basis of the denition of the Lebesgue integral. This argument will be imprecise,
More information17 The functional equation
18.785 Number theory I Fall 16 Lecture #17 11/8/16 17 The functional equation In the previous lecture we proved that the iemann zeta function ζ(s) has an Euler product and an analytic continuation to the
More informationMath 118B Solutions. Charles Martin. March 6, d i (x i, y i ) + d i (y i, z i ) = d(x, y) + d(y, z). i=1
Math 8B Solutions Charles Martin March 6, Homework Problems. Let (X i, d i ), i n, be finitely many metric spaces. Construct a metric on the product space X = X X n. Proof. Denote points in X as x = (x,
More informationMath 127C, Spring 2006 Final Exam Solutions. x 2 ), g(y 1, y 2 ) = ( y 1 y 2, y1 2 + y2) 2. (g f) (0) = g (f(0))f (0).
Math 27C, Spring 26 Final Exam Solutions. Define f : R 2 R 2 and g : R 2 R 2 by f(x, x 2 (sin x 2 x, e x x 2, g(y, y 2 ( y y 2, y 2 + y2 2. Use the chain rule to compute the matrix of (g f (,. By the chain
More informationMath 328 Course Notes
Math 328 Course Notes Ian Robertson March 3, 2006 3 Properties of C[0, 1]: Sup-norm and Completeness In this chapter we are going to examine the vector space of all continuous functions defined on the
More informationconsists of two disjoint copies of X n, each scaled down by 1,
Homework 4 Solutions, Real Analysis I, Fall, 200. (4) Let be a topological space and M be a σ-algebra on which contains all Borel sets. Let m, µ be two positive measures on M. Assume there is a constant
More informationSolutions Final Exam May. 14, 2014
Solutions Final Exam May. 14, 2014 1. (a) (10 points) State the formal definition of a Cauchy sequence of real numbers. A sequence, {a n } n N, of real numbers, is Cauchy if and only if for every ɛ > 0,
More informationRiemann integral and volume are generalized to unbounded functions and sets. is an admissible set, and its volume is a Riemann integral, 1l E,
Tel Aviv University, 26 Analysis-III 9 9 Improper integral 9a Introduction....................... 9 9b Positive integrands................... 9c Special functions gamma and beta......... 4 9d Change of
More informationSolution to Exercise 3
Spring 207 MATH502 Real Analysis II Solution to Eercise 3 () For a, b > 0, set a sin( b ), 0 < f() = 0, = 0. nπ+π/2 Show that f is in BV [0, ] iff a > b. ( ) b Solution. Put n = for n 0. We claim that
More informationMath 115 ( ) Yum-Tong Siu 1. Derivation of the Poisson Kernel by Fourier Series and Convolution
Math 5 (006-007 Yum-Tong Siu. Derivation of the Poisson Kernel by Fourier Series and Convolution We are going to give a second derivation of the Poisson kernel by using Fourier series and convolution.
More informationSolutions to Tutorial 11 (Week 12)
THE UIVERSITY OF SYDEY SCHOOL OF MATHEMATICS AD STATISTICS Solutions to Tutorial 11 (Week 12) MATH3969: Measure Theory and Fourier Analysis (Advanced) Semester 2, 2017 Web Page: http://sydney.edu.au/science/maths/u/ug/sm/math3969/
More informationCHAPTER VI APPLICATIONS TO ANALYSIS
CHAPTER VI APPLICATIONS TO ANALYSIS We include in this chapter several subjects from classical analysis to which the notions of functional analysis can be applied. Some of these subjects are essential
More information3. (a) What is a simple function? What is an integrable function? How is f dµ defined? Define it first
Math 632/6321: Theory of Functions of a Real Variable Sample Preinary Exam Questions 1. Let (, M, µ) be a measure space. (a) Prove that if µ() < and if 1 p < q
More informationProduct measures, Tonelli s and Fubini s theorems For use in MAT4410, autumn 2017 Nadia S. Larsen. 17 November 2017.
Product measures, Tonelli s and Fubini s theorems For use in MAT4410, autumn 017 Nadia S. Larsen 17 November 017. 1. Construction of the product measure The purpose of these notes is to prove the main
More informationMATH 6337 Second Midterm April 1, 2014
You can use your book and notes. No laptop or wireless devices allowed. Write clearly and try to make your arguments as linear and simple as possible. The complete solution of one exercise will be considered
More informationAdvanced Analysis Qualifying Examination Department of Mathematics and Statistics University of Massachusetts. Tuesday, January 16th, 2018
NAME: Advanced Analysis Qualifying Examination Department of Mathematics and Statistics University of Massachusetts Tuesday, January 16th, 2018 Instructions 1. This exam consists of eight (8) problems
More informationSingular Integrals. 1 Calderon-Zygmund decomposition
Singular Integrals Analysis III Calderon-Zygmund decomposition Let f be an integrable function f dx 0, f = g + b with g Cα almost everywhere, with b
More informationMath 172 HW 1 Solutions
Math 172 HW 1 Solutions Joey Zou April 15, 2017 Problem 1: Prove that the Cantor set C constructed in the text is totally disconnected and perfect. In other words, given two distinct points x, y C, there
More informationMATH34032 Mid-term Test 10.00am 10.50am, 26th March 2010 Answer all six question [20% of the total mark for this course]
MATH3432: Green s Functions, Integral Equations and the Calculus of Variations 1 MATH3432 Mid-term Test 1.am 1.5am, 26th March 21 Answer all six question [2% of the total mark for this course] Qu.1 (a)
More informationTopics in Harmonic Analysis Lecture 6: Pseudodifferential calculus and almost orthogonality
Topics in Harmonic Analysis Lecture 6: Pseudodifferential calculus and almost orthogonality Po-Lam Yung The Chinese University of Hong Kong Introduction While multiplier operators are very useful in studying
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 645-4A (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 informationTOOLS FROM HARMONIC ANALYSIS
TOOLS FROM HARMONIC ANALYSIS BRADLY STADIE Abstract. The Fourier transform can be thought of as a map that decomposes a function into oscillatory functions. In this paper, we will apply this decomposition
More information2. Function spaces and approximation
2.1 2. Function spaces and approximation 2.1. The space of test functions. Notation and prerequisites are collected in Appendix A. Let Ω be an open subset of R n. The space C0 (Ω), consisting of the C
More informationProblem set 5, Real Analysis I, Spring, otherwise. (a) Verify that f is integrable. Solution: Compute since f is even, 1 x (log 1/ x ) 2 dx 1
Problem set 5, Real Analysis I, Spring, 25. (5) Consider the function on R defined by f(x) { x (log / x ) 2 if x /2, otherwise. (a) Verify that f is integrable. Solution: Compute since f is even, R f /2
More informationThe Heine-Borel and Arzela-Ascoli Theorems
The Heine-Borel and Arzela-Ascoli Theorems David Jekel October 29, 2016 This paper explains two important results about compactness, the Heine- Borel theorem and the Arzela-Ascoli theorem. We prove them
More informationThe Hilbert transform
The Hilbert transform Definition and properties ecall the distribution pv(, defined by pv(/(ϕ := lim ɛ ɛ ϕ( d. The Hilbert transform is defined via the convolution with pv(/, namely (Hf( := π lim f( t
More informationReal Variables # 10 : Hilbert Spaces II
randon ehring Real Variables # 0 : Hilbert Spaces II Exercise 20 For any sequence {f n } in H with f n = for all n, there exists f H and a subsequence {f nk } such that for all g H, one has lim (f n k,
More informationOutline of Fourier Series: Math 201B
Outline of Fourier Series: Math 201B February 24, 2011 1 Functions and convolutions 1.1 Periodic functions Periodic functions. Let = R/(2πZ) denote the circle, or onedimensional torus. A function f : C
More informationMEASURE AND INTEGRATION: LECTURE 18
MEASURE AND INTEGRATION: LECTURE 18 Fubini s theorem Notation. Let l and m be positive integers, and n = l + m. Write as the Cartesian product = +. We will write points in as z ; x ; y ; z = (x, y). If
More informationg(x) = P (y) Proof. This is true for n = 0. Assume by the inductive hypothesis that g (n) (0) = 0 for some n. Compute g (n) (h) g (n) (0)
Mollifiers and Smooth Functions We say a function f from C is C (or simply smooth) if all its derivatives to every order exist at every point of. For f : C, we say f is C if all partial derivatives to
More information4. Product measure spaces and the Lebesgue integral in R n.
4 M. M. PELOSO 4. Product measure spaces and the Lebesgue integral in R n. Our current goal is to define the Lebesgue measure on the higher-dimensional eucledean space R n, and to reduce the computations
More informationContinuous Functions on Metric Spaces
Continuous Functions on Metric Spaces Math 201A, Fall 2016 1 Continuous functions Definition 1. Let (X, d X ) and (Y, d Y ) be metric spaces. A function f : X Y is continuous at a X if for every ɛ > 0
More informationA List of Problems in Real Analysis
A List of Problems in Real Analysis W.Yessen & T.Ma December 3, 218 This document was first created by Will Yessen, who was a graduate student at UCI. Timmy Ma, who was also a graduate student at UCI,
More informationFall f(x)g(x) dx. The starting place for the theory of Fourier series is that the family of functions {e inx } n= is orthonormal, that is
18.103 Fall 2013 1. Fourier Series, Part 1. We will consider several function spaces during our study of Fourier series. When we talk about L p ((, π)), it will be convenient to include the factor 1/ in
More information2) Let X be a compact space. Prove that the space C(X) of continuous real-valued 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 informationFunctional Analysis. Franck Sueur Metric spaces Definitions Completeness Compactness Separability...
Functional Analysis Franck Sueur 2018-2019 Contents 1 Metric spaces 1 1.1 Definitions........................................ 1 1.2 Completeness...................................... 3 1.3 Compactness......................................
More informationJUHA KINNUNEN. Harmonic Analysis
JUHA KINNUNEN Harmonic Analysis Department of Mathematics and Systems Analysis, Aalto University 27 Contents Calderón-Zygmund decomposition. Dyadic subcubes of a cube.........................2 Dyadic cubes
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 informationSOLUTIONS TO ASSIGNMENT 2 - MATH 355. with c > 3. m(n c ) < δ. f(t) t. g(x)dx =
SOLUTIONS TO ASSIGNMENT 2 - MATH 355 Problem. ecall ha, B n {ω [, ] : S n (ω) > nɛ n }, and S n (ω) N {ω [, ] : lim }, n n m(b n ) 3 n 2 ɛ 4. We wan o show ha m(n c ). Le δ >. We can pick ɛ 4 n c n wih
More informationMATH 51H Section 4. October 16, Recall what it means for a function between metric spaces to be continuous:
MATH 51H Section 4 October 16, 2015 1 Continuity Recall what it means for a function between metric spaces to be continuous: Definition. Let (X, d X ), (Y, d Y ) be metric spaces. A function f : X Y is
More informationModule 2: Reflecting on One s Problems
MATH55 Module : Reflecting on One s Problems Main Math concepts: Translations, Reflections, Graphs of Equations, Symmetry Auxiliary ideas: Working with quadratics, Mobius maps, Calculus, Inverses I. Transformations
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 informationMATH 202B - Problem Set 5
MATH 202B - Problem Set 5 Walid Krichene (23265217) March 6, 2013 (5.1) Show that there exists a continuous function F : [0, 1] R which is monotonic on no interval of positive length. proof We know there
More informationVISCOSITY SOLUTIONS. We follow Han and Lin, Elliptic Partial Differential Equations, 5.
VISCOSITY SOLUTIONS PETER HINTZ We follow Han and Lin, Elliptic Partial Differential Equations, 5. 1. Motivation Throughout, we will assume that Ω R n is a bounded and connected domain and that a ij C(Ω)
More information2. Metric Spaces. 2.1 Definitions etc.
2. Metric Spaces 2.1 Definitions etc. The procedure in Section for regarding R as a topological space may be generalized to many other sets in which there is some kind of distance (formally, sets with
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 informationChapter 3: Baire category and open mapping theorems
MA3421 2016 17 Chapter 3: Baire category and open mapping theorems A number of the major results rely on completeness via the Baire category theorem. 3.1 The Baire category theorem 3.1.1 Definition. A
More information1.3.1 Definition and Basic Properties of Convolution
1.3 Convolution 15 1.3 Convolution Since L 1 (R) is a Banach space, we know that it has many useful properties. In particular the operations of addition and scalar multiplication are continuous. However,
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 informationWeighted norm inequalities for singular integral operators
Weighted norm inequalities for singular integral operators C. Pérez Journal of the London mathematical society 49 (994), 296 308. Departmento de Matemáticas Universidad Autónoma de Madrid 28049 Madrid,
More informationSobolev Spaces. Chapter 10
Chapter 1 Sobolev Spaces We now define spaces H 1,p (R n ), known as Sobolev spaces. For u to belong to H 1,p (R n ), we require that u L p (R n ) and that u have weak derivatives of first order in L p
More informationA review: The Laplacian and the d Alembertian. j=1
Chapter One A review: The Laplacian and the d Alembertian 1.1 THE LAPLACIAN One of the main goals of this course is to understand well the solution of wave equation both in Euclidean space and on manifolds
More informationGeometric intuition: from Hölder spaces to the Calderón-Zygmund estimate
Geometric intuition: from Hölder spaces to the Calderón-Zygmund estimate A survey of Lihe Wang s paper Michael Snarski December 5, 22 Contents Hölder spaces. Control on functions......................................2
More informationMetric Spaces and Topology
Chapter 2 Metric Spaces and Topology From an engineering perspective, the most important way to construct a topology on a set is to define the topology in terms of a metric on the set. This approach underlies
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 informationReview of Multi-Calculus (Study Guide for Spivak s CHAPTER ONE TO THREE)
Review of Multi-Calculus (Study Guide for Spivak s CHPTER ONE TO THREE) This material is for June 9 to 16 (Monday to Monday) Chapter I: Functions on R n Dot product and norm for vectors in R n : Let X
More informationNotes on the Lebesgue Integral by Francis J. Narcowich November, 2013
Notes on the Lebesgue Integral by Francis J. Narcowich November, 203 Introduction In the definition of the Riemann integral of a function f(x), the x-axis is partitioned and the integral is defined in
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 informationTopics in Harmonic Analysis Lecture 1: The Fourier transform
Topics in Harmonic Analysis Lecture 1: The Fourier transform Po-Lam Yung The Chinese University of Hong Kong Outline Fourier series on T: L 2 theory Convolutions The Dirichlet and Fejer kernels Pointwise
More informationMath 140A - Fall Final Exam
Math 140A - Fall 2014 - Final Exam Problem 1. Let {a n } n 1 be an increasing sequence of real numbers. (i) If {a n } has a bounded subsequence, show that {a n } is itself bounded. (ii) If {a n } has a
More informationAnalysis IV : Assignment 3 Solutions John Toth, Winter ,...). In particular for every fixed m N the sequence (u (n)
Analysis IV : Assignment 3 Solutions John Toth, Winter 203 Exercise (l 2 (Z), 2 ) is a complete and separable Hilbert space. Proof Let {u (n) } n N be a Cauchy sequence. Say u (n) = (..., u 2, (n) u (n),
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 informationChapter 8 Integral Operators
Chapter 8 Integral Operators In our development of metrics, norms, inner products, and operator theory in Chapters 1 7 we only tangentially considered topics that involved the use of Lebesgue measure,
More informationMATH 317 Fall 2016 Assignment 5
MATH 37 Fall 26 Assignment 5 6.3, 6.4. ( 6.3) etermine whether F(x, y) e x sin y îı + e x cos y ĵj is a conservative vector field. If it is, find a function f such that F f. enote F (P, Q). We have Q x
More information