McGill University Math 354: Honors Analysis 3 Fall 2012 Solutions to selected problems. x y

Size: px
Start display at page:

Download "McGill University Math 354: Honors Analysis 3 Fall 2012 Solutions to selected problems. x y"

Transcription

1 McGill Uiversity Math 354: Hoors Aalysis 3 Fall 212 Assigmet 3 Solutios to selected problems Problem 1. Lipschitz uctios. Let M K be the set o all uctios cotiuous uctios o [, 1] satisyig a Lipschitz coditio with costat K >, i.e. such that For M K, deie the orm by Prove that (x) (y) K x, y [, 1]. (1) sup x [,1] (x) + sup x,y [,1] (x) (y). i) deies the orm o M K, i.e. c c ad that + g + g. ii) Coclude that d(, g) : g deies a distace o M K. iii) (extra credit) M K is closed, ad that it is the closure o the set o all dieretiable uctios o [, 1] satisyig (t) K. iv) The set M K M K is ot closed. v) (ot or credit). What do you thi is the closure o M? Solutio: For (i), we remar that liearity is obvious rom the deiitio o. The triagle iequality ollows rom the triagle iequality or the sup-orm, ad rom taig the supremum over x y i the ollowig iequality: (x) + g(x) (y) g(y) (x) (y) + g(x) g(y). The statemet o (ii) ollows by a stadard argumet o how a orm deies a distace. For (iii), assume that { (x)} is a sequece i Lip K, ad that (x) (x) as, i.e. that or ay ɛ > there exists N such that or all N, we have (x) (x) (y) + (y) sup (x) (x) + sup < ɛ. (2) x x y Cosider the expressio (x) (y) /. It ollows rom (2) that or N, we have (x) (y) ɛ + (x) (y) ɛ + K. Sice ɛ was arbitrary, we see that Lip K (ad is thereore automatically cotiuous), hece Lip K is closed. Next, i C 1 ([, 1]) with sup t (t) K, the Lip K by the itermediate value theorem, sice (y) (x) (y x) (θ), where θ [x, y]. Next, assume that (x) satisies (1) ad cosider the Berstei polyomial B (, x). It suices to show that

2 Lemma. I Lip K, the B (, x) Lip K. Ideed, B is clearly dieretiable or all, approximates uiormly as by a result proved i class. Also, the iequality B (, y) B (, x) K y x implies that B (, x) K (i we assume that B (, y) > K or some y [, 1], we shall get a cotradictio with Lipschitz iequality by choosig x close eough to y ad applyig the itermediate value theorem). Proo o the Lemma. (Tae rom the ote Lipschitz costats or the Berstei polyomials o a Lipschitz cotiuous uctio by B. Brow, D. Elliott ad D. Paget, Joural o Approximatio Theory 49, , 1987). Let x < y 1. The B (, y) j j j ( ) (1 y) j j ( ) (1 y) j j j ( ) j (x + (y x)) j ( j ) [ j!x (y x) j (1 y) j!(j )!( j)! ( ] j )x (y x) j ( ) j (3) Ater chagig the order o summatio ad writig + l j, (3) becomes B (, y) l We ext write a similar idetity or B (, x): B (, x) ( ) x ( ) x l!x (y x) l (1 y) l!(l)!( l)! ( ) ((y x) + (1 y)) ( ( ) + l ) [ ( ] )(y x) l (1 y) l l l!x (y x) l (1 y) l!l!( l)! ( ) (4) (5) Subtractig (5) rom (4) we id that B (, y) B (, x) By the Lipschitz coditio, l ( ) + l so it ollows rom the precedig iequality that B (, y) B (, x) K!x (y x) l (1 y) l!l!( l)! l ( ) ( ) l K, [ ( + l!x (y x) l (1 y) l!l!( l)! ) ( ) l. ( )]

3 The last expressio is equal to K K!(y x) l l!( l)! l ( l l ( l ) (y x) l ( l K B ((z) z, y x) K(y x), ) [ l ( ] l )x (1 y) l ) (x + 1 y) l where i the last lie we have used the idetity B ((z) z, x) x. That idetity ollows easily rom the secod combiatorial idetity give i the hadout about Berstei approximatio theorem. Summarizig, we have show that B (, y) B (, x) K(y x), which iishes the proo o the Lemma, as well as the part (iii) o Problem 1. For items (iv) ad (v), we ote that approximatio is uderstood i terms o the d (uiorm) distace. Thus, it ollows rom Berstei approximatio theorem that the closure o K Lip K is the whole C([, 1]). Ideed, or ay cotiuous uctio o [, 1], there exists N such that or ay > N, sup x (x) B (, x) ɛ. Now, we claim that B (, x) K Lip K. Ideed, (d/dx)b (, x) is cotiuous ad thus taes a maximum value, say K. As discussed beore, that shows that B (, x) Lip K. To see that ot every cotiuous uctio lies i K Lip K, cosider the uctio (x) x. The (d/dx)(x) 1/(2 x) goes to iiity as x. It is also easy to see that or ay K >, there exist < x < y < 1 such that y x 1 > K. y x y + x This happes i x ad y are small eough. Thus, x / Lip K or ay K, ad hece it is ot cotaied i their uio. Problem 2. Fredholm equatio. Use the ixed poit theorem to prove the existece ad uiqueess o the solutio to homogeeous Fredholm equatio (x) λ K(x, y)(y)dy. Here K(x, y) is a cotiuous uctio o [, 1] 2 satisyig K(x, y) M is called the erel o the equatio. Cosider the mappig o C([, 1]) ito itsel give by (A)(x) λ K(x, y)(y)dy. Let d d be the usual maximum distace betwee uctios. Prove that i) Prove that d(a, Ag) λm d(, g). ii) Coclude that A has a uique ixed poit i C([, 1]) or λ < 1/M, e.g. there exists a uique C([, 1]) such that A. iii) Prove that is a solutio o the Fredholm equatio.

4 Solutio: For (i), we id that A(x) Ag(x) λ 1 K(x, y)((xy) g(y))dy λk(x, y) (y) g(y) dy. The last expressio is λ M d(, g)dy λ Md(, g). For (ii), we remar that it ollows rom (i) that A is a cotractio mappig provided λ < 1/M. The existece o a uique ixed poit ollows rom stadard results. Item (iii) ollows rom the deiitio o A. Problem 3. Relative topology. Let X be a metric space, ad let Y be a subset o X (with the iduced distace). Prove that a set B is ope i Y i ad oly i B Y A, where A is ope i X. Solutio: Suppose B Y is ope i Y. So, or every y B there exists a positive umber r y such that U Y (y, r y ) : {z Y : d(y, z) < r y } B. Let A y B U X (y, r y ). Clearly, A is a ope subset o X (it is a uio o ope balls). Also, A Y y B (U X (y, r y ) Y ) y B U Y (y, r y ) B, sice U Y (y, r y ) B. For the coverse, let V be ope i X, ad let y V Y. The U X (y, t) V or some t >. But U Y (y, t) U X (y, t) Y V Y. Thus, V Y is ope i Y, QED. Problem 4. Let X X, where X is ope or all. Suppose that the restrictio X is cotiuous or all ; prove that is cotiuous o X. Solutio: Let : X Y be our map, ad let V Y be ope. To prove cotiuity, we have to show that 1 (V ) : {x X : (x) V } is ope. Now, 1 (V ) 1 (V ) A. But the latter set is the preimage o V uder the restrictio A, ad hece is ope sice A is cotiuous or all. Thus, 1 (V ) is ope as a uio o ope sets. A dieret proo uses sequeces. Let x y X. We wat to show that (x ) (y). Sice ope sets A m cover the whole space, we have y A m or some m. Sice A m is ope, we have B(y, r) A m or some r >. It ollows that x B(y, r) or large eough, so x A m. Sice Am is cotiuous, we have (x ) (y), QED. Problem 5. Cosider C([a, b]), the vector space o all cotiuous uctios o [a, b], equipped with the usual orm p, 1 p. Cosider a map Φ : C([a, b]) C([a, b]) deied by Φ() 2. For what values o p is this map cotiuous? Please justiy careully your aswer. Solutio: We have Φ( + h) Φ() 2h + h 2. First let p. Let C([a, b]), deote M, ad choose < ɛ < 1/3. Let h < mi(ɛ/(3m), ɛ). The 2h + h 2 2Mɛ 3M + ɛ2 ɛ(2/3 + ɛ) < ɛ, hece Φ is cotiuous at, hece Φ is cotiuous or p. Cosider ext 1 p <, ad assume WLOG that [a, b] [, 1]. Let (x), the Φ( + h) Φ() h 2. We would lie to choose h C([a, b]) such that h p is small, but h 2 p is large to prove that Φ is ot cotiuous at.

5 Let h(x) δ 1/p or x [, 1/], h(x) or x [2/, 1], ad let h(x) be liear or x [1/, 2/]. The it is easy to show that δ p (h(x)) p dx 2δ p, hece δ h p δ 2 1/p, ad the expressio goes to as δ. O the other had, it is also easy to show that δ 2p (h(x)) 2p dx 2δ 2p, hece δ 1/(2p) h 2 p δ (2) 1/(2p), ad the expressio diverges as, showig that Φ is ot cotiuous at. Problem 6. Let M be a bouded subset i C([, 1]). Prove that the set o uctios F (x) x (t)dt, M (6) has compact closure (i the space o cotiuous uctios with the uiorm distace d ). Solutio: Let F be a amily o uctios deied by (6). By Arzela-Ascoli Theorem, it suices to show that F is bouded ad equicotiuous. We irst remar, that sice M is bouded i C([, 1]) (with d, or uiorm, distace), there exists C > such that (t) < C or all t [, 1] ad or all M. It ollows that or ay x [, 1] ad or ay M, x x F (x) (t)dt (t) dt Cx C, so F is bouded i C([, 1]). To prove that F is equicotiuous, we ix ɛ >, ad let δ ɛ/c. Suppose < δ. Assume (without loss o geerality) that x < y. The or ay F F, y y F (y) F (x) (t)dt (t) dt < δ C ɛ, x so F is equicotiuous. This iishes the proo. Problem 7. Let X be a compact metric space with a coutable base, ad let A : X X be a map satisyig d(ax, Ay) < d(x, y) or all x, y X. Prove that A has a uique ixed poit i X. Solutio: Cosider the uctio (x) d(x, Ax). We irst show that is cotiuous. Ideed, i d(y, x) < ɛ, the d(ay, Ax) < ɛ as well sice A is cotractig, thereore d(x, Ax) d(y, Ay) < 2ɛ. Sice ɛ was arbitrary, cotiuity ollows. A cotiuous uctio o a compact set attais its miimum, say at a poit y X. Suppose that the miimum is positive, i.e. that Ay y. The d(a 2 y, Ay) < d(ay, y) sice A is cotractig, which cotradicts the miimality. Thereore, d(a, Ay) ad so y is a ixed poit. Uiqueess ollows rom the cotractig property o A i the usual way. Problem 8 (extra credit). Give a example o a o-compact but complete metric space X ad a map A : X X as i Problem 7 such that A does t have a ixed poit. Solutio: Cosider the example rom Problem 4, Assigmet 2: X N with d(m, ) 1+1/(m+ ). That distace deies discrete topology i X, so X is certaily complete (ay Cauchy sequece is evetually costat). Cosider ay icreasig map o N N, or example A() The A decreases the distace. Ideed, 1 + 1/(m + ) > 1 + 1/(m ), or all m >. It is also clear that A has o ixed poits. x

6 Problem 9 (extra credit). Let C([, 1]). Prove that or ay ɛ > ad N N there exists a uctio g C([, 1]) such that d 1 (, g) < ɛ ad g 2 > N. Solutio. The idea is based o a observatio that a uctio x α, 1/2 α < 1 is itegrable but ot square-itegrable o the iterval [, 1]. So, we ix 1/2 α < 1. We also let M : max x (x). Next, give ɛ > we ca choose < δ such that x α dx (2δ)1 α 1 α < ɛ/3, as well as (x) dx < ɛ/3. We also let M : max x (x). I additio, we require δ 1 α < ɛ/3 ad δ < /epsilo/3m We costruct g(x) as ollows: or 2δ x 1, we let (x) g(x). For delta x 2δ, we let x (1 + t)δ, t 1, ad deie g(x) (1 t)δ α + t(2δ) (i.e. we iterpolate liearly betwee ad g. O the iterval [, δ], we remar that as η, we have δ η x 2α dx, so we ca choose η > so that δ η x 2α dx > N 2. We ially let g(x) x α or x [η, δ], ad g(x) η α or x [, η]. We eed to veriy that g has the required properties. For the irst property we remar that (x) g(x) dx (x) g(x) dx (x) dx + g(x) dx The irst itegral i the right-had side is less tha ɛ/3 by the choice o δ. The secod itegral is less tha δ x α dx + δ max(δ α, M) ɛ 3 + max(δ1 α, δm) < 2ɛ 3, also by the choice o δ. Addig the two estimates, we id that (x) g(x) dx < ɛ. For the secod iequality, we id that g(x) 2 dx δ g(x) 2 dx δ η x 2α dx > N 2, so g 2 > N ad the secod requiremet is satisied. Problem 1. Tube Lemma. Let X be a metric space, ad let Y be a compact metric space. Cosider the product space X Y. I V is a ope set o X Y cotaiig the slice {x } Y o X Y, the V cotais some tube W Y about {x } Y, where W is a eighborhood o x i X. Give a example showig that the Tube Lemma does ot hold i Y is ot compact. Solutio. Let ρ be the distace o X ad σ the distace o Y. We deie the maximum distace d o X Y by d((x 1, y 1 ), (x 2, y 2 )) max(ρ(x 1, x 2 ), σ(y 1, y 2 )). (7) This deies the d distace i case R 2 R R. It easy to see that ope balls or the metric d have the orm U V, where U is a ope ball i X, ad V is a ope ball i Y (ad similarly or closed balls). It is also easy to see that the topology deied by the distace d max(ρ, sigma) is equivalet to topologies deied by d p : (ρ p + σ p ) 1/p, just lie or R 2, (i.e. ope ad closed sets coicide or all distaces), so we ca mae our calculatio usig the distace d without loss o geerality.

7 The poit (x, y) is a iterior poit o V or all y Y, hace there exist r r(y) > such that the ball U X (x, r(y)) U Y (y, r(y)) cetered at (x, y) is cotaied i V. Call the correspodig balls U X (y) ad U Y (y). The balls {U X (y) U Y (y)} y Y orm a ope cover o {x } Y. Sice {x } Y is isometric to Y, it is compact. Accordigly, there exist iitely may y Y, say y 1, y 2..., y such that j1 U X(y j ) U Y (y j ) cover {x } Y. Let r mi 1 j {r(y j )}. The we ca let W U(x, r) ad the coclusio will hold. For the couterexample i case o ocompact Y, let X Y R, x (so that {x } Y is the y-axis), ad cosider the ope set V be {(x, y) : y < 1/ x, or y.} Problem 11. Let B deote the set o all sequeces (x ) such that lim x. Cosider l 1 as a subset o l. Prove that the closure o l 1 i l is equal to B. Problem 12. a) Let A X be coected, ad let {A α } α I be a amily o coected subsets o X. Show that A A α or all α I, the A ( α I A α ) is coected. b) Let X ad Y be coected metric spaces. Show that X Y is coected. Solutio. a) Let B A ( α I A α ). Suppose or cotradictio that B is ot coected. The by Lemma we ca assume that B C D, where C ad D are disjoit ope subsets o X that have oempty itersectio with B. By a result proved i class, we ow that A A α is coected or all α. The A A α has to lie etirely i C or etirely i D, otherwise they will separate A A α. Thus A A α C (say). But this must the hold or all α, so B C ad D B. Cotradictio iishes the proo. (b) Give x X, cosider the map : Y X Y give by (y) (x, y). The map is cotiuous ad Y is coected, so {x } Y is coected or every Y. Next, ix y Y. We similarly id that X {y } is coected. I Problem 5, let A X {y }, ad let A x {x} Y, where the idex α is replaced by x X. Now, A A x (x, y ). It ollows rom Problem 5 that (X {y }) ( x X {x} Y ) is coected. But the above set is just X Y, so the proo is iished.

McGill University Math 354: Honors Analysis 3 Fall 2012 Solutions to selected problems

McGill University Math 354: Honors Analysis 3 Fall 2012 Solutions to selected problems McGill Uiversity Math 354: Hoors Aalysis 3 Fall 212 Assigmet 3 Solutios to selected problems Problem 1. Lipschitz fuctios. Let Lip K be the set of all fuctios cotiuous fuctios o [, 1] satisfyig a Lipschitz

More information

Lecture Notes for Analysis Class

Lecture Notes for Analysis Class Lecture Notes for Aalysis Class Topological Spaces A topology for a set X is a collectio T of subsets of X such that: (a) X ad the empty set are i T (b) Uios of elemets of T are i T (c) Fiite itersectios

More information

Sequences and Series of Functions

Sequences and Series of Functions Chapter 6 Sequeces ad Series of Fuctios 6.1. Covergece of a Sequece of Fuctios Poitwise Covergece. Defiitio 6.1. Let, for each N, fuctio f : A R be defied. If, for each x A, the sequece (f (x)) coverges

More information

Solutions to home assignments (sketches)

Solutions to home assignments (sketches) Matematiska Istitutioe Peter Kumli 26th May 2004 TMA401 Fuctioal Aalysis MAN670 Applied Fuctioal Aalysis 4th quarter 2003/2004 All documet cocerig the course ca be foud o the course home page: http://www.math.chalmers.se/math/grudutb/cth/tma401/

More information

Math Solutions to homework 6

Math Solutions to homework 6 Math 175 - Solutios to homework 6 Cédric De Groote November 16, 2017 Problem 1 (8.11 i the book): Let K be a compact Hermitia operator o a Hilbert space H ad let the kerel of K be {0}. Show that there

More information

University of Colorado Denver Dept. Math. & Stat. Sciences Applied Analysis Preliminary Exam 13 January 2012, 10:00 am 2:00 pm. Good luck!

University of Colorado Denver Dept. Math. & Stat. Sciences Applied Analysis Preliminary Exam 13 January 2012, 10:00 am 2:00 pm. Good luck! Uiversity of Colorado Dever Dept. Math. & Stat. Scieces Applied Aalysis Prelimiary Exam 13 Jauary 01, 10:00 am :00 pm Name: The proctor will let you read the followig coditios before the exam begis, ad

More information

f n (x) f m (x) < ɛ/3 for all x A. By continuity of f n and f m we can find δ > 0 such that d(x, x 0 ) < δ implies that

f n (x) f m (x) < ɛ/3 for all x A. By continuity of f n and f m we can find δ > 0 such that d(x, x 0 ) < δ implies that Lecture 15 We have see that a sequece of cotiuous fuctios which is uiformly coverget produces a limit fuctio which is also cotiuous. We shall stregthe this result ow. Theorem 1 Let f : X R or (C) be a

More information

Chapter 3 Inner Product Spaces. Hilbert Spaces

Chapter 3 Inner Product Spaces. Hilbert Spaces Chapter 3 Ier Product Spaces. Hilbert Spaces 3. Ier Product Spaces. Hilbert Spaces 3.- Defiitio. A ier product space is a vector space X with a ier product defied o X. A Hilbert space is a complete ier

More information

Math 508 Exam 2 Jerry L. Kazdan December 9, :00 10:20

Math 508 Exam 2 Jerry L. Kazdan December 9, :00 10:20 Math 58 Eam 2 Jerry L. Kazda December 9, 24 9: :2 Directios This eam has three parts. Part A has 8 True/False questio (2 poits each so total 6 poits), Part B has 5 shorter problems (6 poits each, so 3

More information

HOMEWORK #10 SOLUTIONS

HOMEWORK #10 SOLUTIONS Math 33 - Aalysis I Sprig 29 HOMEWORK # SOLUTIONS () Prove that the fuctio f(x) = x 3 is (Riema) itegrable o [, ] ad show that x 3 dx = 4. (Without usig formulae for itegratio that you leart i previous

More information

Assignment 5: Solutions

Assignment 5: Solutions McGill Uiversity Departmet of Mathematics ad Statistics MATH 54 Aalysis, Fall 05 Assigmet 5: Solutios. Let y be a ubouded sequece of positive umbers satisfyig y + > y for all N. Let x be aother sequece

More information

ABOUT CHAOS AND SENSITIVITY IN TOPOLOGICAL DYNAMICS

ABOUT CHAOS AND SENSITIVITY IN TOPOLOGICAL DYNAMICS ABOUT CHAOS AND SENSITIVITY IN TOPOLOGICAL DYNAMICS EDUARD KONTOROVICH Abstract. I this work we uify ad geeralize some results about chaos ad sesitivity. Date: March 1, 005. 1 1. Symbolic Dyamics Defiitio

More information

A. Appendix to: The Generalization Error of Dictionary Learning with Moreau Envelopes

A. Appendix to: The Generalization Error of Dictionary Learning with Moreau Envelopes A. Appedix to: A.. roo o Lemma roo. For a proo o (9 ad ( see Corollary i (Georgogiais, 06. The cotiuity o e h ollows rom Theorem.5 i (Rockaellar & Wets, 009. From the calculus rules o the geeralized subgradiets

More information

(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3

(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3 MATH 337 Sequeces Dr. Neal, WKU Let X be a metric space with distace fuctio d. We shall defie the geeral cocept of sequece ad limit i a metric space, the apply the results i particular to some special

More information

HOMEWORK #4 - MA 504

HOMEWORK #4 - MA 504 HOMEWORK #4 - MA 504 PAULINHO TCHATCHATCHA Chapter 2, problem 19. (a) If A ad B are disjoit closed sets i some metric space X, prove that they are separated. (b) Prove the same for disjoit ope set. (c)

More information

Real Variables II Homework Set #5

Real Variables II Homework Set #5 Real Variables II Homework Set #5 Name: Due Friday /0 by 4pm (at GOS-4) Istructios: () Attach this page to the frot of your homework assigmet you tur i (or write each problem before your solutio). () Please

More information

Theorem 3. A subset S of a topological space X is compact if and only if every open cover of S by open sets in X has a finite subcover.

Theorem 3. A subset S of a topological space X is compact if and only if every open cover of S by open sets in X has a finite subcover. Compactess Defiitio 1. A cover or a coverig of a topological space X is a family C of subsets of X whose uio is X. A subcover of a cover C is a subfamily of C which is a cover of X. A ope cover of X is

More information

Where do eigenvalues/eigenvectors/eigenfunctions come from, and why are they important anyway?

Where do eigenvalues/eigenvectors/eigenfunctions come from, and why are they important anyway? Where do eigevalues/eigevectors/eigeuctios come rom, ad why are they importat ayway? I. Bacgroud (rom Ordiary Dieretial Equatios} Cosider the simplest example o a harmoic oscillator (thi o a vibratig strig)

More information

Math 220B Final Exam Solutions March 18, 2002

Math 220B Final Exam Solutions March 18, 2002 Math 0B Fial Exam Solutios March 18, 00 1. (1 poits) (a) (6 poits) Fid the Gree s fuctio for the tilted half-plae {(x 1, x ) R : x 1 + x > 0}. For x (x 1, x ), y (y 1, y ), express your Gree s fuctio G(x,

More information

Singular Continuous Measures by Michael Pejic 5/14/10

Singular Continuous Measures by Michael Pejic 5/14/10 Sigular Cotiuous Measures by Michael Peic 5/4/0 Prelimiaries Give a set X, a σ-algebra o X is a collectio of subsets of X that cotais X ad ad is closed uder complemetatio ad coutable uios hece, coutable

More information

Topologie. Musterlösungen

Topologie. Musterlösungen Fakultät für Mathematik Sommersemester 2018 Marius Hiemisch Topologie Musterlösuge Aufgabe (Beispiel 1.2.h aus Vorlesug). Es sei X eie Mege ud R Abb(X, R) eie Uteralgebra, d.h. {kostate Abbilduge} R ud

More information

ANSWERS TO MIDTERM EXAM # 2

ANSWERS TO MIDTERM EXAM # 2 MATH 03, FALL 003 ANSWERS TO MIDTERM EXAM # PENN STATE UNIVERSITY Problem 1 (18 pts). State ad prove the Itermediate Value Theorem. Solutio See class otes or Theorem 5.6.1 from our textbook. Problem (18

More information

STAT 516 Answers Homework 6 April 2, 2008 Solutions by Mark Daniel Ward PROBLEMS

STAT 516 Answers Homework 6 April 2, 2008 Solutions by Mark Daniel Ward PROBLEMS STAT 56 Aswers Homework 6 April 2, 28 Solutios by Mark Daiel Ward PROBLEMS Chapter 6 Problems 2a. The mass p(, correspods to either o the irst two balls beig white, so p(, 8 7 4/39. The mass p(, correspods

More information

Product measures, Tonelli s and Fubini s theorems For use in MAT3400/4400, autumn 2014 Nadia S. Larsen. Version of 13 October 2014.

Product measures, Tonelli s and Fubini s theorems For use in MAT3400/4400, autumn 2014 Nadia S. Larsen. Version of 13 October 2014. Product measures, Toelli s ad Fubii s theorems For use i MAT3400/4400, autum 2014 Nadia S. Larse Versio of 13 October 2014. 1. Costructio of the product measure The purpose of these otes is to preset the

More information

MATH 413 FINAL EXAM. f(x) f(y) M x y. x + 1 n

MATH 413 FINAL EXAM. f(x) f(y) M x y. x + 1 n MATH 43 FINAL EXAM Math 43 fial exam, 3 May 28. The exam starts at 9: am ad you have 5 miutes. No textbooks or calculators may be used durig the exam. This exam is prited o both sides of the paper. Good

More information

A Proof of Birkhoff s Ergodic Theorem

A Proof of Birkhoff s Ergodic Theorem A Proof of Birkhoff s Ergodic Theorem Joseph Hora September 2, 205 Itroductio I Fall 203, I was learig the basics of ergodic theory, ad I came across this theorem. Oe of my supervisors, Athoy Quas, showed

More information

Ma 4121: Introduction to Lebesgue Integration Solutions to Homework Assignment 5

Ma 4121: Introduction to Lebesgue Integration Solutions to Homework Assignment 5 Ma 42: Itroductio to Lebesgue Itegratio Solutios to Homework Assigmet 5 Prof. Wickerhauser Due Thursday, April th, 23 Please retur your solutios to the istructor by the ed of class o the due date. You

More information

Math 220A Fall 2007 Homework #2. Will Garner A

Math 220A Fall 2007 Homework #2. Will Garner A Math 0A Fall 007 Homewor # Will Garer Pg 3 #: Show that {cis : a o-egative iteger} is dese i T = {z œ : z = }. For which values of q is {cis(q): a o-egative iteger} dese i T? To show that {cis : a o-egative

More information

n p (Ω). This means that the

n p (Ω). This means that the Sobolev s Iequality, Poicaré Iequality ad Compactess I. Sobolev iequality ad Sobolev Embeddig Theorems Theorem (Sobolev s embeddig theorem). Give the bouded, ope set R with 3 ad p

More information

Math 140A Elementary Analysis Homework Questions 3-1

Math 140A Elementary Analysis Homework Questions 3-1 Math 0A Elemetary Aalysis Homework Questios -.9 Limits Theorems for Sequeces Suppose that lim x =, lim y = 7 ad that all y are o-zero. Detarime the followig limits: (a) lim(x + y ) (b) lim y x y Let s

More information

REAL ANALYSIS II: PROBLEM SET 1 - SOLUTIONS

REAL ANALYSIS II: PROBLEM SET 1 - SOLUTIONS REAL ANALYSIS II: PROBLEM SET 1 - SOLUTIONS 18th Feb, 016 Defiitio (Lipschitz fuctio). A fuctio f : R R is said to be Lipschitz if there exists a positive real umber c such that for ay x, y i the domai

More information

FUNDAMENTALS OF REAL ANALYSIS by

FUNDAMENTALS OF REAL ANALYSIS by FUNDAMENTALS OF REAL ANALYSIS by Doğa Çömez Backgroud: All of Math 450/1 material. Namely: basic set theory, relatios ad PMI, structure of N, Z, Q ad R, basic properties of (cotiuous ad differetiable)

More information

Homework 2. Show that if h is a bounded sesquilinear form on the Hilbert spaces X and Y, then h has the representation

Homework 2. Show that if h is a bounded sesquilinear form on the Hilbert spaces X and Y, then h has the representation omework 2 1 Let X ad Y be ilbert spaces over C The a sesquiliear form h o X Y is a mappig h : X Y C such that for all x 1, x 2, x X, y 1, y 2, y Y ad all scalars α, β C we have (a) h(x 1 + x 2, y) h(x

More information

Math 61CM - Solutions to homework 3

Math 61CM - Solutions to homework 3 Math 6CM - Solutios to homework 3 Cédric De Groote October 2 th, 208 Problem : Let F be a field, m 0 a fixed oegative iteger ad let V = {a 0 + a x + + a m x m a 0,, a m F} be the vector space cosistig

More information

Measure and Measurable Functions

Measure and Measurable Functions 3 Measure ad Measurable Fuctios 3.1 Measure o a Arbitrary σ-algebra Recall from Chapter 2 that the set M of all Lebesgue measurable sets has the followig properties: R M, E M implies E c M, E M for N implies

More information

Sequences and Limits

Sequences and Limits Chapter Sequeces ad Limits Let { a } be a sequece of real or complex umbers A ecessary ad sufficiet coditio for the sequece to coverge is that for ay ɛ > 0 there exists a iteger N > 0 such that a p a q

More information

PRELIM PROBLEM SOLUTIONS

PRELIM PROBLEM SOLUTIONS PRELIM PROBLEM SOLUTIONS THE GRAD STUDENTS + KEN Cotets. Complex Aalysis Practice Problems 2. 2. Real Aalysis Practice Problems 2. 4 3. Algebra Practice Problems 2. 8. Complex Aalysis Practice Problems

More information

Solution. 1 Solutions of Homework 1. Sangchul Lee. October 27, Problem 1.1

Solution. 1 Solutions of Homework 1. Sangchul Lee. October 27, Problem 1.1 Solutio Sagchul Lee October 7, 017 1 Solutios of Homework 1 Problem 1.1 Let Ω,F,P) be a probability space. Show that if {A : N} F such that A := lim A exists, the PA) = lim PA ). Proof. Usig the cotiuity

More information

Math 299 Supplement: Real Analysis Nov 2013

Math 299 Supplement: Real Analysis Nov 2013 Math 299 Supplemet: Real Aalysis Nov 203 Algebra Axioms. I Real Aalysis, we work withi the axiomatic system of real umbers: the set R alog with the additio ad multiplicatio operatios +,, ad the iequality

More information

1 Introduction. 1.1 Notation and Terminology

1 Introduction. 1.1 Notation and Terminology 1 Itroductio You have already leared some cocepts of calculus such as limit of a sequece, limit, cotiuity, derivative, ad itegral of a fuctio etc. Real Aalysis studies them more rigorously usig a laguage

More information

Definition 4.2. (a) A sequence {x n } in a Banach space X is a basis for X if. unique scalars a n (x) such that x = n. a n (x) x n. (4.

Definition 4.2. (a) A sequence {x n } in a Banach space X is a basis for X if. unique scalars a n (x) such that x = n. a n (x) x n. (4. 4. BASES I BAACH SPACES 39 4. BASES I BAACH SPACES Sice a Baach space X is a vector space, it must possess a Hamel, or vector space, basis, i.e., a subset {x γ } γ Γ whose fiite liear spa is all of X ad

More information

Math 104: Homework 2 solutions

Math 104: Homework 2 solutions Math 04: Homework solutios. A (0, ): Sice this is a ope iterval, the miimum is udefied, ad sice the set is ot bouded above, the maximum is also udefied. if A 0 ad sup A. B { m + : m, N}: This set does

More information

Final Solutions. 1. (25pts) Define the following terms. Be as precise as you can.

Final Solutions. 1. (25pts) Define the following terms. Be as precise as you can. Mathematics H104 A. Ogus Fall, 004 Fial Solutios 1. (5ts) Defie the followig terms. Be as recise as you ca. (a) (3ts) A ucoutable set. A ucoutable set is a set which ca ot be ut ito bijectio with a fiite

More information

6.3 Testing Series With Positive Terms

6.3 Testing Series With Positive Terms 6.3. TESTING SERIES WITH POSITIVE TERMS 307 6.3 Testig Series With Positive Terms 6.3. Review of what is kow up to ow I theory, testig a series a i for covergece amouts to fidig the i= sequece of partial

More information

2.1. The Algebraic and Order Properties of R Definition. A binary operation on a set F is a function B : F F! F.

2.1. The Algebraic and Order Properties of R Definition. A binary operation on a set F is a function B : F F! F. CHAPTER 2 The Real Numbers 2.. The Algebraic ad Order Properties of R Defiitio. A biary operatio o a set F is a fuctio B : F F! F. For the biary operatios of + ad, we replace B(a, b) by a + b ad a b, respectively.

More information

Real Numbers R ) - LUB(B) may or may not belong to B. (Ex; B= { y: y = 1 x, - Note that A B LUB( A) LUB( B)

Real Numbers R ) - LUB(B) may or may not belong to B. (Ex; B= { y: y = 1 x, - Note that A B LUB( A) LUB( B) Real Numbers The least upper boud - Let B be ay subset of R B is bouded above if there is a k R such that x k for all x B - A real umber, k R is a uique least upper boud of B, ie k = LUB(B), if () k is

More information

Part A, for both Section 200 and Section 501

Part A, for both Section 200 and Section 501 Istructios Please write your solutios o your ow paper. These problems should be treated as essay questios. A problem that says give a example or determie requires a supportig explaatio. I all problems,

More information

lim za n n = z lim a n n.

lim za n n = z lim a n n. Lecture 6 Sequeces ad Series Defiitio 1 By a sequece i a set A, we mea a mappig f : N A. It is customary to deote a sequece f by {s } where, s := f(). A sequece {z } of (complex) umbers is said to be coverget

More information

Metric Space Properties

Metric Space Properties Metric Space Properties Math 40 Fial Project Preseted by: Michael Brow, Alex Cordova, ad Alyssa Sachez We have already poited out ad will recogize throughout this book the importace of compact sets. All

More information

Archimedes - numbers for counting, otherwise lengths, areas, etc. Kepler - geometry for planetary motion

Archimedes - numbers for counting, otherwise lengths, areas, etc. Kepler - geometry for planetary motion Topics i Aalysis 3460:589 Summer 007 Itroductio Ree descartes - aalysis (breaig dow) ad sythesis Sciece as models of ature : explaatory, parsimoious, predictive Most predictios require umerical values,

More information

CS537. Numerical Analysis and Computing

CS537. Numerical Analysis and Computing CS57 Numerical Aalysis ad Computig Lecture Locatig Roots o Equatios Proessor Ju Zhag Departmet o Computer Sciece Uiversity o Ketucky Leigto KY 456-6 Jauary 9 9 What is the Root May physical system ca be

More information

Analytic Continuation

Analytic Continuation Aalytic Cotiuatio The stadard example of this is give by Example Let h (z) = 1 + z + z 2 + z 3 +... kow to coverge oly for z < 1. I fact h (z) = 1/ (1 z) for such z. Yet H (z) = 1/ (1 z) is defied for

More information

On the behavior at infinity of an integrable function

On the behavior at infinity of an integrable function O the behavior at ifiity of a itegrable fuctio Emmauel Lesige To cite this versio: Emmauel Lesige. O the behavior at ifiity of a itegrable fuctio. The America Mathematical Mothly, 200, 7 (2), pp.75-8.

More information

1 lim. f(x) sin(nx)dx = 0. n sin(nx)dx

1 lim. f(x) sin(nx)dx = 0. n sin(nx)dx Problem A. Calculate ta(.) to 4 decimal places. Solutio: The power series for si(x)/ cos(x) is x + x 3 /3 + (2/5)x 5 +. Puttig x =. gives ta(.) =.3. Problem 2A. Let f : R R be a cotiuous fuctio. Show that

More information

Axioms of Measure Theory

Axioms of Measure Theory MATH 532 Axioms of Measure Theory Dr. Neal, WKU I. The Space Throughout the course, we shall let X deote a geeric o-empty set. I geeral, we shall ot assume that ay algebraic structure exists o X so that

More information

Math 128A: Homework 1 Solutions

Math 128A: Homework 1 Solutions Math 8A: Homework Solutios Due: Jue. Determie the limits of the followig sequeces as. a) a = +. lim a + = lim =. b) a = + ). c) a = si4 +6) +. lim a = lim = lim + ) [ + ) ] = [ e ] = e 6. Observe that

More information

5. Matrix exponentials and Von Neumann s theorem The matrix exponential. For an n n matrix X we define

5. Matrix exponentials and Von Neumann s theorem The matrix exponential. For an n n matrix X we define 5. Matrix expoetials ad Vo Neuma s theorem 5.1. The matrix expoetial. For a matrix X we defie e X = exp X = I + X + X2 2! +... = 0 X!. We assume that the etries are complex so that exp is well defied o

More information

A) is empty. B) is a finite set. C) can be a countably infinite set. D) can be an uncountable set.

A) is empty. B) is a finite set. C) can be a countably infinite set. D) can be an uncountable set. M.A./M.Sc. (Mathematics) Etrace Examiatio 016-17 Max Time: hours Max Marks: 150 Istructios: There are 50 questios. Every questio has four choices of which exactly oe is correct. For correct aswer, 3 marks

More information

Common Coupled Fixed Point of Mappings Satisfying Rational Inequalities in Ordered Complex Valued Generalized Metric Spaces

Common Coupled Fixed Point of Mappings Satisfying Rational Inequalities in Ordered Complex Valued Generalized Metric Spaces IOSR Joural of Mathematics (IOSR-JM) e-issn: 78-578, p-issn:319-765x Volume 10, Issue 3 Ver II (May-Ju 014), PP 69-77 Commo Coupled Fixed Poit of Mappigs Satisfyig Ratioal Iequalities i Ordered Complex

More information

1+x 1 + α+x. x = 2(α x2 ) 1+x

1+x 1 + α+x. x = 2(α x2 ) 1+x Math 2030 Homework 6 Solutios # [Problem 5] For coveiece we let α lim sup a ad β lim sup b. Without loss of geerality let us assume that α β. If α the by assumptio β < so i this case α + β. By Theorem

More information

Differentiable Convex Functions

Differentiable Convex Functions Differetiable Covex Fuctios The followig picture motivates Theorem 11. f ( x) f ( x) f '( x)( x x) ˆx x 1 Theorem 11 : Let f : R R be differetiable. The, f is covex o the covex set C R if, ad oly if for

More information

MIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS

MIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS MIDTERM 3 CALCULUS MATH 300 FALL 08 Moday, December 3, 08 5:5 PM to 6:45 PM Name PRACTICE EXAM S Please aswer all of the questios, ad show your work. You must explai your aswers to get credit. You will

More information

The Boolean Ring of Intervals

The Boolean Ring of Intervals MATH 532 Lebesgue Measure Dr. Neal, WKU We ow shall apply the results obtaied about outer measure to the legth measure o the real lie. Throughout, our space X will be the set of real umbers R. Whe ecessary,

More information

CS321. Numerical Analysis and Computing

CS321. Numerical Analysis and Computing CS Numerical Aalysis ad Computig Lecture Locatig Roots o Equatios Proessor Ju Zhag Departmet o Computer Sciece Uiversity o Ketucky Leigto KY 456-6 September 8 5 What is the Root May physical system ca

More information

MAS111 Convergence and Continuity

MAS111 Convergence and Continuity MAS Covergece ad Cotiuity Key Objectives At the ed of the course, studets should kow the followig topics ad be able to apply the basic priciples ad theorems therei to solvig various problems cocerig covergece

More information

If a subset E of R contains no open interval, is it of zero measure? For instance, is the set of irrationals in [0, 1] is of measure zero?

If a subset E of R contains no open interval, is it of zero measure? For instance, is the set of irrationals in [0, 1] is of measure zero? 2 Lebesgue Measure I Chapter 1 we defied the cocept of a set of measure zero, ad we have observed that every coutable set is of measure zero. Here are some atural questios: If a subset E of R cotais a

More information

Properties of Fuzzy Length on Fuzzy Set

Properties of Fuzzy Length on Fuzzy Set Ope Access Library Joural 206, Volume 3, e3068 ISSN Olie: 2333-972 ISSN Prit: 2333-9705 Properties of Fuzzy Legth o Fuzzy Set Jehad R Kider, Jaafar Imra Mousa Departmet of Mathematics ad Computer Applicatios,

More information

Dupuy Complex Analysis Spring 2016 Homework 02

Dupuy Complex Analysis Spring 2016 Homework 02 Dupuy Complex Aalysis Sprig 206 Homework 02. (CUNY, Fall 2005) Let D be the closed uit disc. Let g be a sequece of aalytic fuctios covergig uiformly to f o D. (a) Show that g coverges. Solutio We have

More information

page Suppose that S 0, 1 1, 2.

page Suppose that S 0, 1 1, 2. page 10 1. Suppose that S 0, 1 1,. a. What is the set of iterior poits of S? The set of iterior poits of S is 0, 1 1,. b. Give that U is the set of iterior poits of S, evaluate U. 0, 1 1, 0, 1 1, S. The

More information

PRACTICE FINAL/STUDY GUIDE SOLUTIONS

PRACTICE FINAL/STUDY GUIDE SOLUTIONS Last edited December 9, 03 at 4:33pm) Feel free to sed me ay feedback, icludig commets, typos, ad mathematical errors Problem Give the precise meaig of the followig statemets i) a f) L ii) a + f) L iii)

More information

The Wasserstein distances

The Wasserstein distances The Wasserstei distaces March 20, 2011 This documet presets the proof of the mai results we proved o Wasserstei distaces themselves (ad ot o curves i the Wasserstei space). I particular, triagle iequality

More information

The Discrete-Time Fourier Transform (DTFT)

The Discrete-Time Fourier Transform (DTFT) EEL: Discrete-Time Sigals ad Systems The Discrete-Time Fourier Trasorm (DTFT) The Discrete-Time Fourier Trasorm (DTFT). Itroductio I these otes, we itroduce the discrete-time Fourier trasorm (DTFT) ad

More information

New Characterization of Topological Transitivity

New Characterization of Topological Transitivity ew Characterizatio o Topological Trasitivity Hussei J Abdul Hussei Departmet o Mathematics ad Computer Applicatios, College o Sciece, Uiversity o Al Muthaa, Al Muthaa, Iraq Abstract Let be a dyamical system,

More information

Supplemental Material: Proofs

Supplemental Material: Proofs Proof to Theorem Supplemetal Material: Proofs Proof. Let be the miimal umber of traiig items to esure a uique solutio θ. First cosider the case. It happes if ad oly if θ ad Rak(A) d, which is a special

More information

MATH 112: HOMEWORK 6 SOLUTIONS. Problem 1: Rudin, Chapter 3, Problem s k < s k < 2 + s k+1

MATH 112: HOMEWORK 6 SOLUTIONS. Problem 1: Rudin, Chapter 3, Problem s k < s k < 2 + s k+1 MATH 2: HOMEWORK 6 SOLUTIONS CA PRO JIRADILOK Problem. If s = 2, ad Problem : Rudi, Chapter 3, Problem 3. s + = 2 + s ( =, 2, 3,... ), prove that {s } coverges, ad that s < 2 for =, 2, 3,.... Proof. The

More information

Chapter 2. Periodic points of toral. automorphisms. 2.1 General introduction

Chapter 2. Periodic points of toral. automorphisms. 2.1 General introduction Chapter 2 Periodic poits of toral automorphisms 2.1 Geeral itroductio The automorphisms of the two-dimesioal torus are rich mathematical objects possessig iterestig geometric, algebraic, topological ad

More information

Abstract Vector Spaces. Abstract Vector Spaces

Abstract Vector Spaces. Abstract Vector Spaces Astract Vector Spaces The process of astractio is critical i egieerig! Physical Device Data Storage Vector Space MRI machie Optical receiver 0 0 1 0 1 0 0 1 Icreasig astractio 6.1 Astract Vector Spaces

More information

A REMARK ON A PROBLEM OF KLEE

A REMARK ON A PROBLEM OF KLEE C O L L O Q U I U M M A T H E M A T I C U M VOL. 71 1996 NO. 1 A REMARK ON A PROBLEM OF KLEE BY N. J. K A L T O N (COLUMBIA, MISSOURI) AND N. T. P E C K (URBANA, ILLINOIS) This paper treats a property

More information

Taylor Polynomials and Approximations - Classwork

Taylor Polynomials and Approximations - Classwork Taylor Polyomials ad Approimatios - Classwork Suppose you were asked to id si 37 o. You have o calculator other tha oe that ca do simple additio, subtractio, multiplicatio, or divisio. Fareched\ Not really.

More information

Convergence of random variables. (telegram style notes) P.J.C. Spreij

Convergence of random variables. (telegram style notes) P.J.C. Spreij Covergece of radom variables (telegram style otes).j.c. Spreij this versio: September 6, 2005 Itroductio As we kow, radom variables are by defiitio measurable fuctios o some uderlyig measurable space

More information

Advanced Real Analysis

Advanced Real Analysis McGill Uiversity December 26 Faculty of Sciece Fial Exam Advaced Real Aalysis Math 564 December 9, 26 Time: 2PM - 5PM Examier: Dr. J. Galkowski Associate Examier: Prof. D. Jakobso INSTRUCTIONS. Please

More information

Advanced Analysis. Min Yan Department of Mathematics Hong Kong University of Science and Technology

Advanced Analysis. Min Yan Department of Mathematics Hong Kong University of Science and Technology Advaced Aalysis Mi Ya Departmet of Mathematics Hog Kog Uiversity of Sciece ad Techology September 3, 009 Cotets Limit ad Cotiuity 7 Limit of Sequece 8 Defiitio 8 Property 3 3 Ifiity ad Ifiitesimal 8 4

More information

Empirical Processes: Glivenko Cantelli Theorems

Empirical Processes: Glivenko Cantelli Theorems Empirical Processes: Gliveko Catelli Theorems Mouliath Baerjee Jue 6, 200 Gliveko Catelli classes of fuctios The reader is referred to Chapter.6 of Weller s Torgo otes, Chapter??? of VDVW ad Chapter 8.3

More information

MA541 : Real Analysis. Tutorial and Practice Problems - 1 Hints and Solutions

MA541 : Real Analysis. Tutorial and Practice Problems - 1 Hints and Solutions MA54 : Real Aalysis Tutorial ad Practice Problems - Hits ad Solutios. Suppose that S is a oempty subset of real umbers that is bouded (i.e. bouded above as well as below). Prove that if S sup S. What ca

More information

Topics. Homework Problems. MATH 301 Introduction to Analysis Chapter Four Sequences. 1. Definition of convergence of sequences.

Topics. Homework Problems. MATH 301 Introduction to Analysis Chapter Four Sequences. 1. Definition of convergence of sequences. MATH 301 Itroductio to Aalysis Chapter Four Sequeces Topics 1. Defiitio of covergece of sequeces. 2. Fidig ad provig the limit of sequeces. 3. Bouded covergece theorem: Theorem 4.1.8. 4. Theorems 4.1.13

More information

Seunghee Ye Ma 8: Week 5 Oct 28

Seunghee Ye Ma 8: Week 5 Oct 28 Week 5 Summary I Sectio, we go over the Mea Value Theorem ad its applicatios. I Sectio 2, we will recap what we have covered so far this term. Topics Page Mea Value Theorem. Applicatios of the Mea Value

More information

ACO Comprehensive Exam 9 October 2007 Student code A. 1. Graph Theory

ACO Comprehensive Exam 9 October 2007 Student code A. 1. Graph Theory 1. Graph Theory Prove that there exist o simple plaar triagulatio T ad two distict adjacet vertices x, y V (T ) such that x ad y are the oly vertices of T of odd degree. Do ot use the Four-Color Theorem.

More information

Period Function of a Lienard Equation

Period Function of a Lienard Equation Joural of Mathematical Scieces (4) -5 Betty Joes & Sisters Publishig Period Fuctio of a Lieard Equatio Khalil T Al-Dosary Departmet of Mathematics, Uiversity of Sharjah, Sharjah 77, Uited Arab Emirates

More information

Functions of Bounded Variation and Rectifiable Curves

Functions of Bounded Variation and Rectifiable Curves Fuctios of Bouded Variatio ad Rectifiable Curves Fuctios of bouded variatio 6.1 Determie which of the follwoig fuctios are of bouded variatio o 0, 1. (a) fx x si1/x if x 0, f0 0. (b) fx x si1/x if x 0,

More information

Machine Learning Theory Tübingen University, WS 2016/2017 Lecture 12

Machine Learning Theory Tübingen University, WS 2016/2017 Lecture 12 Machie Learig Theory Tübige Uiversity, WS 06/07 Lecture Tolstikhi Ilya Abstract I this lecture we derive risk bouds for kerel methods. We will start by showig that Soft Margi kerel SVM correspods to miimizig

More information

Lecture 3 The Lebesgue Integral

Lecture 3 The Lebesgue Integral Lecture 3: The Lebesgue Itegral 1 of 14 Course: Theory of Probability I Term: Fall 2013 Istructor: Gorda Zitkovic Lecture 3 The Lebesgue Itegral The costructio of the itegral Uless expressly specified

More information

Chapter 3. Strong convergence. 3.1 Definition of almost sure convergence

Chapter 3. Strong convergence. 3.1 Definition of almost sure convergence Chapter 3 Strog covergece As poited out i the Chapter 2, there are multiple ways to defie the otio of covergece of a sequece of radom variables. That chapter defied covergece i probability, covergece i

More information

Complex Analysis Spring 2001 Homework I Solution

Complex Analysis Spring 2001 Homework I Solution Complex Aalysis Sprig 2001 Homework I Solutio 1. Coway, Chapter 1, sectio 3, problem 3. Describe the set of poits satisfyig the equatio z a z + a = 2c, where c > 0 ad a R. To begi, we see from the triagle

More information

b i u x i U a i j u x i u x j

b i u x i U a i j u x i u x j M ath 5 2 7 Fall 2 0 0 9 L ecture 1 9 N ov. 1 6, 2 0 0 9 ) S ecod- Order Elliptic Equatios: Weak S olutios 1. Defiitios. I this ad the followig two lectures we will study the boudary value problem Here

More information

Lecture 7: Properties of Random Samples

Lecture 7: Properties of Random Samples Lecture 7: Properties of Radom Samples 1 Cotiued From Last Class Theorem 1.1. Let X 1, X,...X be a radom sample from a populatio with mea µ ad variace σ

More information

FIR Filter Design: Part I

FIR Filter Design: Part I EEL3: Discrete-Time Sigals ad Systems FIR Filter Desig: Part I. Itroductio FIR Filter Desig: Part I I this set o otes, we cotiue our exploratio o the requecy respose o FIR ilters. First, we cosider some

More information

Lecture 2 Long paths in random graphs

Lecture 2 Long paths in random graphs Lecture Log paths i radom graphs 1 Itroductio I this lecture we treat the appearace of log paths ad cycles i sparse radom graphs. will wor with the probability space G(, p) of biomial radom graphs, aalogous

More information

Brief Review of Functions of Several Variables

Brief Review of Functions of Several Variables Brief Review of Fuctios of Several Variables Differetiatio Differetiatio Recall, a fuctio f : R R is differetiable at x R if ( ) ( ) lim f x f x 0 exists df ( x) Whe this limit exists we call it or f(

More information

Enumerative & Asymptotic Combinatorics

Enumerative & Asymptotic Combinatorics C50 Eumerative & Asymptotic Combiatorics Stirlig ad Lagrage Sprig 2003 This sectio of the otes cotais proofs of Stirlig s formula ad the Lagrage Iversio Formula. Stirlig s formula Theorem 1 (Stirlig s

More information

Solutions to Math 347 Practice Problems for the final

Solutions to Math 347 Practice Problems for the final Solutios to Math 347 Practice Problems for the fial 1) True or False: a) There exist itegers x,y such that 50x + 76y = 6. True: the gcd of 50 ad 76 is, ad 6 is a multiple of. b) The ifiimum of a set is

More information