SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL (1 + µ(f n )) f(x) =. But we don t need the exact bound.) Set


 Thomas Gray
 2 years ago
 Views:
Transcription
1 SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL 28 Nottion: N {, 2, 3,...}. (Tht is, N.. Let (X, M be mesurble spce with σfinite positive mesure µ. Prove tht there is finite positive mesure ν on (X, M such tht µ ν nd ν µ. (Recll tht µ ν mens tht µ is bsolutely continuous with respect to ν, tht is, for ny E M, ν(e implies µ(e. Solution. If µ is finite, tke ν µ. Hence we ssume tht µ(x. By the σfiniteness ssumption, we cn write X n F n for mesurble sets F n such tht µ(f n < for ech n N. For ny mesurble set A M, define µ(a F n ν(a 2 n ( + µ(f n. n One checks tht ν is mesure by writing ν(a f dµ with A f(x 2 n ( + µ(f n whenever x F n. (The function f cn lso be written s ( f 2 n χ Fn. ( + µ(f n n Alterntively, show directly tht ν is mesure: clerly ν(, nd countble dditivity of ν follows from countble dditivity of µ. Clerly ν(x <. If µ(a, then µ(a F n for ech n N, whence ν(a. On the other hnd, if ν(a, then µ(a F n for ech n N. This implies tht µ(a by countble dditivity. 2. Let (X, µ be mesure spce with µ(x <. For f L 2 (X, µ, prove tht f log( f is in L (X, µ. (Tke y log(y when y. If (X, µ (R, m, is it still the cse tht f L 2 implies f log( f L (X, µ? Solution. First consider the cse µ(x <. Since y y log(y is continuous on (, ] nd lim y + y log(y exists (it is zero, there is constnt M such tht y log(y M for ll y [, ]. (In fct, methods of elementry clculus show tht for ll y (, ], we hve y log(y e, so we cn tke M e. But we don t need the exct bound. Set E { x X : f(x }. For x E, log ( f(x < f(x. Hence f(x log ( f(x < f(x 2. Therefore, using f L 2 (X, µ t the lst step, f(x log ( f(x dµ f(x log ( f(x dµ + f(x log ( f(x dµ X E E X X\E f 2 dµ + M dµ X\E f 2 dµ + Mµ(X \ E X f 2 dµ + Mµ(X <. If (X, µ (R, m, the sttement fils. Here is counterexmple. (A counterexmple, with proof, must be present in correct solution. Set { x 3/4 x f(x x <. Then R f 2 dm x 3/2 dx 2,
2 2 SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL 28 but f log( f dm x 3/4 log(x 3/4 dx 3 4 R x 3/4 log(x dx > 3 4 e x 3/4 dx. Alternte solution to the second prt (sketch. For ny α ( 2, ], the function { x α x f(x x < is counterexmple. There re mny others. 3. Let f : [, ] R be continuous function. ( Prove tht lim (b Prove tht lim x n f(x dx. (3 points. (n + x n f(x dx f(. (7 points Solution to 3.(. For n N, set g n (x x n f(x for ll x [, ]. Then lim g n (x for lmost ll x [, ] (in fct, for ll x [, ] except possibly x, f bounded nd hence is integrble on [, ], nd g n (x f(x for ll x [, ] nd ll n N. The Lebesgue Dominted Convergence Theorem therefore implies tht lim x n f(x dx ( lim g n(x dx. Alternte solution to 3.(. Let >. We find N N such tht n N implies xn f(x dx <. Set M sup f(x nd α x [,] 2M +. Without loss of generlity, is smll enough tht α >. Choose N N so lrge tht α N < 2M +. Let n N stisfy n N. Then α x n f(x dx x n f(x dx x n f(x dx + x n f(x dx α ( α Mα n + M( α < Mα N + M( α < M α 2M + α n M dx + ( + M α 2M + M dx <. Solution to 3.(b. Using the chnge of vribles y x n+, so x y /(n+ nd dy (n + x n dx, we get (n + x n f(x dx f ( y /(n+ dy. Set M sup x [,] f(x. For n N, set g n (x f ( y /(n+ for ll y [, ]. Then lim g n (y f( for ll y [, ] except y, the constnt function M is integrble on [, ], nd g n (y M for ll y [, ] nd ll n N. The Lebesgue Dominted Convergence Theorem therefore implies tht lim g n (x dx We cn lso prove this by direct estimtes. f( dx f(.
3 SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL 28 3 Alternte solution to 3.(b. Let >. Set M sup x [,] f(x. Choose δ (, sufficiently smll tht f(x f( < 3 for x [ δ, ]. Choose N N so lrge tht ( δ N+ < 3M +. Let n N stisfy n N. Then Now nd (n + x n f(x dx δ δ δ (n + x n f(x dx + (n + x n f( dx + δ δ (n + x n f(x dx (n + x n [f(x f(] dx + δ (n + x n f(x dx. (n + x n f( dx f( f([ ( δ n+ ] f( ( δ n+ f( < 3, δ (n + x n [f(x f(] dx 3 δ (n + x n dx 3, (n + x n f(x dx < M( δn+ < 3. Putting the lst three inequlities together gives (n + x n f(x dx f( < Let C([, ] be the spce of continuous functions on [, ]. Prove tht ( /2 f f 2 dm [,] defines norm on C([, ]. Is C([, ] Bnch spce with respect to this norm? Justify your nswer. We give direct proof first. Solution. We check tht the formul given defines norm. It is obvious tht f for ll f C([, ] nd tht. It is cler tht αf α f for ll f C([, ] nd ll α C. Next, suppose tht f C([, ] nd f. Then f lmost everywhere with respect to m. For every nonempty open set U [, ], we hve m(u >. So W {x [, : f(x } contins no nonempty open set. Since f is continuous, W. Thus f is the zero element of C([, ]. Finlly, let f, g C([, ]; we clim tht f + g f + g. We my ssume f + g. We hve, using Hölder s inequlity t the fourth step, f + g 2 f + g 2 dm f + g ( f + g dm f + g f dm + f + g g dm [,] [,] ( /2 ( /2 ( /2 ( f + g 2 dm f dm 2 + f + g 2 dm g 2 dm [,] [,] [,] [,] f + g ( f + g. The clim follows by dividing by f + g. However, C([, ] with this norm is not Bnch spce, since C([, ] is not complete with respect this norm. There re mny exmples one cn construct. For exmple, for n N define f n C([, to be the piecewise liner function x [, 2 ] f n (x 2 n ( x 2 [,] 2n x ( 2 2n, 2 + 2n 2n, ]. x [ 2 [,] /2
4 4 SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL 28 We clim tht (f n n N is Cuchy sequence. Let >. Choose N N with N > 4/ 2. Let m, n N stisfy m, n N. Then f m (x f n (x for x [, ]\ [ 2 2N, 2 + ] 2N nd fm (x f n (x f m (x+f n (x 2 for x [ 2 2N, 2 + ] 2N. Therefore ([ f m f n 2 4m 2 2N, 2 + ] 4 2N N < 2. The clim is proved. We finish the proof by showing tht lim f n does not exist. Suppose tht f C([, ] nd lim f n f. Since f n is rel for ll n N, it is esy to see tht f n Re(f f n f. Since limits re unique if they exist, f must be rel. Choose δ > such tht ( x 2 < δ implies f(x f 2 < 4. First suppose tht f( 2 2. For every n > 2/δ, if x ( 2 + δ 4, 2 + ] δ 2, then f(x > 4 while f n(x. So ( [ /2+δ/2 /2 ( ( ] 2 /2 δ δ f n f f n f dm 2 /2+δ/ This contrdicts lim f n f. Otherwise, f ( 2 < 2. Then similr rgument on [ 2 δ 2, 2 ] δ 4 shows f n f > δ/8 for n > 2/δ. Agin, we hve contrdicted lim f n f. This completes the solution. It is not enough to just sy f n χ [, 2 ] in the norm given, nd tht χ [, 2 ] C([, ]. It isn t even enough to prove tht there is no function in C([, ] which is equl lmost everywhere to χ [, 2 ]. One must show tht (f n n N does not converge in C([, ]. (Note, though, tht this pproch does work in the lternte solution, becuse the lternte solution explicitly uses the embedding of C([, ] in L 2 ([, ]. An lternte proof of the tringle inequlity cn be derived from the fct tht the norm defined on C([, ] comes from sclr product. Alternte solution. There is n obvious liner mp I : C([, ] L 2 ([, ]. We clim tht I is injective. Suppose I(f. Then f lmost everywhere with respect to m. For every nonempty open set U [, ], we hve m(u >. So W {x [, : f(x } contins no nonempty open set. Since f is continuous, W. Thus f is the zero element of C([, ]. Since I is injective nd liner, the formul f I(f L2 ([,] is norm on C([, ]. We know tht the rnge of I is dense in L 2 ([, ]. We clim tht it is not ll of L 2 ([, ]. Given this, C([, ] is nonclosed subspce of L 2 ([, ], nd therefore not complete; this will finish the solution. Set f χ [, 2 ]. If f were in the rnge of I, then there would be g C([, ] such tht f g lmost everywhere. But this is clerly impossible. 5. Let X, Y, nd Z be Bnch spces. Let S : Y Z be bounded injective liner opertor nd let T : X Y be liner opertor. Suppose S T : X Z is bounded. Prove tht T is bounded. (Hint: use the Closed Grph Theorem. Solution. We prove tht T hs closed grph nd then pply the Closed Grph Theorem to conclude tht T is bounded. To show tht T hs closed grph, it is sufficient to prove tht whenever (ξ n n N is sequence in X such tht lim ξ n nd lim T ξ n η, then η. Since S is bounded, we hve lim S(T ξ n Sη. Since S T is bounded, we hve lim (S T ξ n. Hence Sη, nd injectivity of S implies tht η. 6. Let m be Lebesgue mesure on [, 28]. For f L 2 ([, 28], let T f : L 2 ([, 28] C be the liner functionl T f (g fg dm [, 28] for g L 2 ([, 28]. Suppose (f n n N is sequence of functions in L 2 ([, 28] such tht f n L2 ([, 28] for ech n N. Prove tht if (f n n N converges to zero lmost everywhere, then for ech g L 2 ([, 28], lim T f n (g. (In other words, prove tht (f n n N converges wekly to zero in L 2 ([, 28].
5 SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL 28 5 Solution. To simplify the nottion, set I [, 28] nd bbrevite L2 ([, 28] to. Let >. Since g 2 L (I, there exists δ > such tht for ny mesurble set A I with m(a < δ, g 2 dm < 2 4. Set For n N define A E n { x I : f n (x > } 2 28( g +. nd T n E k. Define Z n T n. Then f n lmost everywhere implies m(z. Since m(i <, there is N N such tht for ll n N, m(t n < δ. Suppose tht n N. Then, using f n nd m(t n < δ t the second step, ( /2 ( /2 f n g dm f n 2 dm g 2 dm < T n T n T n 2. Also, using f n on I \ T n t the second step, ( /2 ( f n g dm f n 2 dm g 2 dm I\T n I\T n I\T n Therefore m(i \ T n /2 ( I I kn /2 /2 g 2 dm m(i /2 g < 2. T fn (g f n g dm < We cn lso use Egoroff s Theorem. The rgument involving the sets E n in the first solution is prt of the proof of Egoroff s Theorem, but one sees from tht solution tht we don t need the full strength of Egoroff s Theorem. Alternte solution. To simplify the nottion, set I [, 28] nd bbrevite L2 ([, 28] to. Let >. Choose δ > s in the first solution. Use Egoroff s Theorem to find subset B I such tht m(i \ B < δ nd f n uniformly on B. Choose N N such tht for ll n N, sup f n (x < x B 2 28( g +. For n N, we hve T fn (g f n g dm I I\B f n g dm + f n g dm. B The first integrl on the right is less thn 2 by the rgument used in the first solution for T n f n g dm, nd the second integrl on the right is less thn 2 by the rgument used in the first solution for I\T n f n g dm. So T fn (g <. 7. Find ll entire functions f such tht f(z z 5/2 for ll z C. Solution. We show tht the only such function is the constnt function f(z for ll z C. First, clerly f(. Suppose f is not identiclly zero. Then there re n entire function h with h( nd k N such tht f(z z k h(z for ll z C. We clim tht k 3. To prove the clim, choose δ > such tht h(z h( < h( /2 for ll z C with z < δ. Then for z < δ, we hve h(z > h( /2, so z k 2 zk h(z h( 2 z 5/2 h(. For this to be true for ll z C with z < δ, we must hve k 5/2. Since k N, it follows tht k 3.
6 6 SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL 28 Given the clim, whenever z, h(z f(z z k f(z. z 5/2 Hence h is bounded entire function, nd h must be constnt by Liouville s theorem. So f(z h(z k for ll z C. Then h( z 5/2 k for ll z C \ {}. Since k 3, this contrdicts h(. Hence f is identiclly zero. Alternte solution. We show tht the only such function is the constnt function f(z for ll z C. Set g(z z 2 f(z for z C \ {}. Then g(z z /2 for z C \ {}, so lim z g(z, Therefore g hs removble singulrity t, tht is, there is n entire function h such tht h(z g(z for ll z C \ {}. Moreover, h( lim z g(z. Therefore there is n entire function k such tht h(z zk(z for ll z C. Tht is, f(z z 3 k(z for ll z C. Now z 3 k(z z 5/2 for ll z C, so k(z z /2 for ll z C \ {}. Therefore lim z k(z. Since k is entire, Liouville s theorem implies tht k(z for ll z C. Therefore f(z for ll z C. 8. Let n, nd let f(z z n + n z n + + z + be monic polynomil with complex coefficients. Prove tht mx z f(z. Solution. Set g(z + n z + n 2 z z n for z C. Then g( nd g(z z n f(/z for z. Suppose mx z f(z <. Then mx z g(z <. Hence g hs locl mximum inside the open unit disk D. This forces g to be constnt. Hence g. This contrdicts mx z g(z <. Alternte solution. Suppose mx z f(z <. Then for z we hve z n (z n f(z < z n. So Rouché s Theorem sys tht z z n nd z z n f(z hve the sme number of zeros in D {z C: z < }, counting multiplicity. But z z n hs n zeros in D, nd z z n f(z, being polynomil of degree t most n, hs t most n zeros in the whole complex plne. This contrdiction shows tht mx z f(z. 9. Suppose f L ((, (using Lebesgue mesure. Prove tht F (z defines holomorphic function on {z C: Im(z > }. f(te itz dt Solution. Set Ω {z C: Im(z > }. We prove tht if Ω then exists, by showing tht it is equl to F (z F ( lim z z f(tite it dt. It suffices to let (z n n N be ny sequence in Ω \ {} such tht lim z n, nd prove tht F (z n F ( ( lim z n z n We will use the Lebesgue Dominted Convergence Theorem. For n N define g n : (, C by for t (,. Then g n (t eitzn e it z n F (z n F ( z n f(tite it dt. f(tg n (t dt for ll n N. Since for ech t (, the function z e itz is holomorphic function with derivtive z ite itz, we hve lim g n (t ite it for ll t (,. So lim f(tg n (t f(tite it for ll t (,.
7 SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL 28 7 (2 For z C \ {} nd t (,, we estimte: e itz z (it k z k k! t t k z k t k! k k k t k z k (k! te t z. Set b Im(. Then b >. Choose N N so lrge tht n N implies z n < b/2. For such n, (2 implies (3 e it(zn z n tebt/2, so (4 g n (t e it e it(zn z n te bt/2. The function t te bt/2 is bounded on (,, so t f(t te bt/2 is integrble on (,. By (4, we hve f(tg n (t f(t te bt/2 for ll n N nd ll t (,, nd we lredy sw tht lim f(tg n (t f(tite it for ll t (,, so ( follows from the Lebesgue Dominted Convergence Theorem. This completes the solution. Alternte solution. This solution differs only in the method used to get the estimte (3. As in the first solution, set b Im( >, nd choose N N so lrge tht n N implies z n < b/2. For such n, we hve Im(z n > b/2, so Im(z n Im(z n Im( > b 2. Further let t (,. Then, if α [, ], we hve exp(itα(zn exp ( Re(itα(zn exp ( tαim(z n < exp ( tαb/2 e bt/2. Setting h t (z e itz for z C, we then hve e it(z n ht (z h t ( z n sup The estimte (3 follows. α [,] h t (α(z n z n sup it exp(itα(zn t zn e bt/2. α [,] One cn lso solve this problem by combining Fubini s Theorem nd Morer s Theorem. This is in principle net wy to do it. Unfortuntely, one must prove tht F is continuous in order to use Morer s Theorem. Second lternte solution. Set Ω {z C: Im(z > }. We first clim tht F is continuous on Ω. To prove the clim, let z Ω, nd let (z n n N be ny sequence in Ω such tht lim z n z. For n N nd t (,, set g n (t f(te itzn. Then lim g n (t f(te itz for ll t (,. For ll n N nd t (,, using Im(z n > t the lst step, we hve e itzn exp(re(itz n exp( tim(z n. Therefore g n (t f(t. Since f is integrble on (,, the Lebesgue Dominted Convergence Theorem implies tht F (z n f(te itzn dt f(te itz dt F (z s n. The clim is proved. Now let be ny tringle in Ω. We show tht F (z dz, by using Fubini s Theorem nd the fct tht the integrnd in the definition of F is holomorphic s function of z. Assume tht is defined on [, b]. Then, by definition, b F (z dz F ((s (s ds. Define function g : [, b] (, C by g(s, t f(t exp(it(s (s. The function (s, t exp(it(s (s is continuous except on the product of finite subset of [, b] with (,. Therefore it is mesurble. The function (s, t f(t is clerly mesurble. Therefore g is the product of
8 8 SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL 28 two mesurble functions, hence mesurble. So g is mesurble. (Checking tht g is mesurble is n essentil prt of the solution. We cn now use Fubini s Theorem for nonnegtive functions to clculte b ( g d(m m f(t exp(it(s (s b b dt ds f (s ds f (s ds, [,b] (, which is finite. Therefore Fubini s Theorem for integrble functions cn be pplied, to get b ( F (z dz g(s, t dt ds ( b f(t exp(it(s (s ds dt ( f(t exp(itz dz dt. Since Ω is convex, Cuchy s Theorem implies tht the inner integrl on the right is zero for ll t (,. Therefore F (z dz. Since is n rbitrry tringle in Ω, Morer s Theorem now implies tht F is holomorphic.
Phil Wertheimer UMD Math Qualifying Exam Solutions Analysis  January, 2015
Problem 1 Let m denote the Lebesgue mesure restricted to the compct intervl [, b]. () Prove tht function f defined on the compct intervl [, b] is Lipschitz if nd only if there is constct c nd function
More informationThe Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More informationUNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3
UNIFORM CONVERGENCE Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 Suppose f n : Ω R or f n : Ω C is sequence of rel or complex functions, nd f n f s n in some sense. Furthermore,
More informationf p dm = exp Use the Dominated Convergence Theorem to complete the exercise. ( d φ(tx))f(x) dx. Ψ (t) =
M38C Prctice for the finl Let f L ([, ]) Prove tht ( /p f dm) p = exp p log f dm where, by definition, exp( ) = To simplify the problem, you my ssume log f L ([, ]) Hint: rewrite the left hnd side in form
More informationW. We shall do so one by one, starting with I 1, and we shall do it greedily, trying
Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering)
More informationNotes on length and conformal metrics
Notes on length nd conforml metrics We recll how to mesure the Eucliden distnce of n rc in the plne. Let α : [, b] R 2 be smooth (C ) rc. Tht is α(t) (x(t), y(t)) where x(t) nd y(t) re smooth rel vlued
More informationMath Solutions to homework 1
Mth 75  Solutions to homework Cédric De Groote October 5, 07 Problem, prt : This problem explores the reltionship between norms nd inner products Let X be rel vector spce ) Suppose tht is norm on X tht
More informationProblem Set 4: Solutions Math 201A: Fall 2016
Problem Set 4: s Mth 20A: Fll 206 Problem. Let f : X Y be onetoone, onto mp between metric spces X, Y. () If f is continuous nd X is compct, prove tht f is homeomorphism. Does this result remin true
More informationHomework 4. (1) If f R[a, b], show that f 3 R[a, b]. If f + (x) = max{f(x), 0}, is f + R[a, b]? Justify your answer.
Homework 4 (1) If f R[, b], show tht f 3 R[, b]. If f + (x) = mx{f(x), 0}, is f + R[, b]? Justify your nswer. (2) Let f be continuous function on [, b] tht is strictly positive except finitely mny points
More informationAnalytical Methods Exam: Preparatory Exercises
Anlyticl Methods Exm: Preprtory Exercises Question. Wht does it men tht (X, F, µ) is mesure spce? Show tht µ is monotone, tht is: if E F re mesurble sets then µ(e) µ(f). Question. Discuss if ech of the
More informationA PROOF OF THE FUNDAMENTAL THEOREM OF CALCULUS USING HAUSDORFF MEASURES
INROADS Rel Anlysis Exchnge Vol. 26(1), 2000/2001, pp. 381 390 Constntin Volintiru, Deprtment of Mthemtics, University of Buchrest, Buchrest, Romni. emil: cosv@mt.cs.unibuc.ro A PROOF OF THE FUNDAMENTAL
More informationFunctional Analysis I Solutions to Exercises. James C. Robinson
Functionl Anlysis I Solutions to Exercises Jmes C. Robinson Contents 1 Exmples I pge 1 2 Exmples II 5 3 Exmples III 9 4 Exmples IV 15 iii 1 Exmples I 1. Suppose tht v α j e j nd v m β k f k. with α j,
More informationProperties of the Riemann Integral
Properties of the Riemnn Integrl Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University Februry 15, 2018 Outline 1 Some Infimum nd Supremum Properties 2
More informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 200910 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some
More informationFUNDAMENTALS OF REAL ANALYSIS by. III.1. Measurable functions. f 1 (
FUNDAMNTALS OF RAL ANALYSIS by Doğn Çömez III. MASURABL FUNCTIONS AND LBSGU INTGRAL III.. Mesurble functions Hving the Lebesgue mesure define, in this chpter, we will identify the collection of functions
More informationTheoretical foundations of Gaussian quadrature
Theoreticl foundtions of Gussin qudrture 1 Inner product vector spce Definition 1. A vector spce (or liner spce) is set V = {u, v, w,...} in which the following two opertions re defined: (A) Addition of
More informationAdvanced Calculus: MATH 410 Uniform Convergence of Functions Professor David Levermore 11 December 2015
Advnced Clculus: MATH 410 Uniform Convergence of Functions Professor Dvid Levermore 11 December 2015 12. Sequences of Functions We now explore two notions of wht it mens for sequence of functions {f n
More informationFundamental Theorem of Calculus for Lebesgue Integration
Fundmentl Theorem of Clculus for Lebesgue Integrtion J. J. Kolih The existing proofs of the Fundmentl theorem of clculus for Lebesgue integrtion typiclly rely either on the Vitli Crthéodory theorem on
More information7.2 Riemann Integrable Functions
7.2 Riemnn Integrble Functions Theorem 1. If f : [, b] R is step function, then f R[, b]. Theorem 2. If f : [, b] R is continuous on [, b], then f R[, b]. Theorem 3. If f : [, b] R is bounded nd continuous
More informationEntrance Exam, Real Analysis September 1, 2009 Solve exactly 6 out of the 8 problems. Compute the following and justify your computation: lim
1. Let n be positive integers. ntrnce xm, Rel Anlysis September 1, 29 Solve exctly 6 out of the 8 problems. Sketch the grph of the function f(x): f(x) = lim e x2n. Compute the following nd justify your
More informationThe Banach algebra of functions of bounded variation and the pointwise Helly selection theorem
The Bnch lgebr of functions of bounded vrition nd the pointwise Helly selection theorem Jordn Bell jordn.bell@gmil.com Deprtment of Mthemtics, University of Toronto Jnury, 015 1 BV [, b] Let < b. For f
More informationMATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1
MATH34032: Green s Functions, Integrl Equtions nd the Clculus of Vritions 1 Section 1 Function spces nd opertors Here we gives some brief detils nd definitions, prticulrly relting to opertors. For further
More informationf(x)dx . Show that there 1, 0 < x 1 does not exist a differentiable function g : [ 1, 1] R such that g (x) = f(x) for all
3 Definite Integrl 3.1 Introduction In school one comes cross the definition of the integrl of rel vlued function defined on closed nd bounded intervl [, b] between the limits nd b, i.e., f(x)dx s the
More information2 Fundamentals of Functional Analysis
Fchgruppe Angewndte Anlysis und Numerik Dr. Mrtin Gutting 22. October 2015 2 Fundmentls of Functionl Anlysis This short introduction to the bsics of functionl nlysis shll give n overview of the results
More informationODE: Existence and Uniqueness of a Solution
Mth 22 Fll 213 Jerry Kzdn ODE: Existence nd Uniqueness of Solution The Fundmentl Theorem of Clculus tells us how to solve the ordinry differentil eqution (ODE) du = f(t) dt with initil condition u() =
More informationPresentation Problems 5
Presenttion Problems 5 21355 A For these problems, ssume ll sets re subsets of R unless otherwise specified. 1. Let P nd Q be prtitions of [, b] such tht P Q. Then U(f, P ) U(f, Q) nd L(f, P ) L(f, Q).
More informationLecture 3. Limits of Functions and Continuity
Lecture 3 Limits of Functions nd Continuity Audrey Terrs April 26, 21 1 Limits of Functions Notes I m skipping the lst section of Chpter 6 of Lng; the section bout open nd closed sets We cn probbly live
More informationA HELLY THEOREM FOR FUNCTIONS WITH VALUES IN METRIC SPACES. 1. Introduction
Ttr Mt. Mth. Publ. 44 (29), 159 168 DOI: 1.2478/v1127956z t m Mthemticl Publictions A HELLY THEOREM FOR FUNCTIONS WITH VALUES IN METRIC SPACES Miloslv Duchoň Peter Mličký ABSTRACT. We present Helly
More informationMath 61CM  Solutions to homework 9
Mth 61CM  Solutions to homework 9 Cédric De Groote November 30 th, 2018 Problem 1: Recll tht the left limit of function f t point c is defined s follows: lim f(x) = l x c if for ny > 0 there exists δ
More informationMATH 174A: PROBLEM SET 5. Suggested Solution
MATH 174A: PROBLEM SET 5 Suggested Solution Problem 1. Suppose tht I [, b] is n intervl. Let f 1 b f() d for f C(I; R) (i.e. f is continuous relvlued function on I), nd let L 1 (I) denote the completion
More informationAdvanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004
Advnced Clculus: MATH 410 Notes on Integrls nd Integrbility Professor Dvid Levermore 17 October 2004 1. Definite Integrls In this section we revisit the definite integrl tht you were introduced to when
More information4402 Geometry/Topology: Differentiable Manifolds Northwestern University Solutions of Practice Problems for Final Exam
4402 Geometry/Topology: Differentible Mnifolds Northwestern University Solutions of Prctice Problems for Finl Exm 1) Using the cnonicl covering of RP n by {U α } 0 α n, where U α = {[x 0 : : x n ] RP
More informationII. Integration and Cauchy s Theorem
MTH6111 Complex Anlysis 200910 Lecture Notes c Shun Bullett QMUL 2009 II. Integrtion nd Cuchy s Theorem 1. Pths nd integrtion Wrning Different uthors hve different definitions for terms like pth nd curve.
More informationBest Approximation in the 2norm
Jim Lmbers MAT 77 Fll Semester 111 Lecture 1 Notes These notes correspond to Sections 9. nd 9.3 in the text. Best Approximtion in the norm Suppose tht we wish to obtin function f n (x) tht is liner combintion
More informationACM 105: Applied Real and Functional Analysis. Solutions to Homework # 2.
ACM 05: Applied Rel nd Functionl Anlysis. Solutions to Homework # 2. Andy Greenberg, Alexei Novikov Problem. RiemnnLebesgue Theorem. Theorem (G.F.B. Riemnn, H.L. Lebesgue). If f is n integrble function
More informationPROBLEMS AND NOTES: UNIFORM CONVERGENCE AND POLYNOMIAL APPROXIMATION
PROBLEMS AND NOTES: UNIFORM CONVERGENCE AND POLYNOMIAL APPROXIMATION SAMEER CHAVAN Abstrct. These re the lecture notes prepred for the prticipnts of IST to be conducted t BP, Pune from 3rd to 15th November,
More informationSTUDY GUIDE FOR BASIC EXAM
STUDY GUIDE FOR BASIC EXAM BRYON ARAGAM This is prtil list of theorems tht frequently show up on the bsic exm. In mny cses, you my be sked to directly prove one of these theorems or these vrints. There
More informationRegulated functions and the regulated integral
Regulted functions nd the regulted integrl Jordn Bell jordn.bell@gmil.com Deprtment of Mthemtics University of Toronto April 3 2014 1 Regulted functions nd step functions Let = [ b] nd let X be normed
More informationAppendix to Notes 8 (a)
Appendix to Notes 8 () 13 Comprison of the Riemnn nd Lebesgue integrls. Recll Let f : [, b] R be bounded. Let D be prtition of [, b] such tht Let D = { = x 0 < x 1
More information1. On some properties of definite integrals. We prove
This short collection of notes is intended to complement the textbook Anlisi Mtemtic 2 by Crl Mdern, published by Città Studi Editore, [M]. We refer to [M] for nottion nd the logicl stremline of the rguments.
More informationThis is a short summary of Lebesgue integration theory, which will be used in the course.
3 Chpter 0 ntegrtion theory This is short summry of Lebesgue integrtion theory, which will be used in the course. Fct 0.1. Some subsets (= delmängder E R = (, re mesurble (= mätbr in the Lebesgue sense,
More informationExam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH1105 Instructor: Attila Máté 1
Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixedpoint itertion to converge when solving the eqution
More information2 Definitions and Basic Properties of Extended Riemann Stieltjes Integrals
2 Definitions nd Bsic Properties of Extended Riemnn Stieltjes Integrls 2.1 Regulted nd Intervl Functions Regulted functions Let X be Bnch spce, nd let J be nonempty intervl in R, which my be bounded or
More informationMAA 4212 Improper Integrals
Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly welldefined, is too restrictive for mny purposes; there re functions which
More informationANALYSIS QUALIFYING EXAM FALL 2016: SOLUTIONS. = lim. F n
ANALYSIS QUALIFYING EXAM FALL 206: SOLUTIONS Problem. Let m be Lebesgue measure on R. For a subset E R and r (0, ), define E r = { x R: dist(x, E) < r}. Let E R be compact. Prove that m(e) = lim m(e /n).
More informationThe HenstockKurzweil integral
fculteit Wiskunde en Ntuurwetenschppen The HenstockKurzweil integrl Bchelorthesis Mthemtics June 2014 Student: E. vn Dijk First supervisor: Dr. A.E. Sterk Second supervisor: Prof. dr. A. vn der Schft
More informationEuler, Ioachimescu and the trapezium rule. G.J.O. Jameson (Math. Gazette 96 (2012), )
Euler, Iochimescu nd the trpezium rule G.J.O. Jmeson (Mth. Gzette 96 (0), 36 4) The following results were estblished in recent Gzette rticle [, Theorems, 3, 4]. Given > 0 nd 0 < s
More informationMORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)
MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give
More information0.1 Properties of regulated functions and their Integrals.
MA244 Anlysis III Solutions. Sheet 2. NB. THESE ARE SKELETON SOLUTIONS, USE WISELY!. Properties of regulted functions nd their Integrls.. (Q.) Pick ny ɛ >. As f, g re regulted, there exist φ, ψ S[, b]:
More informationMath 426: Probability Final Exam Practice
Mth 46: Probbility Finl Exm Prctice. Computtionl problems 4. Let T k (n) denote the number of prtitions of the set {,..., n} into k nonempty subsets, where k n. Argue tht T k (n) kt k (n ) + T k (n ) by
More informationMain topics for the First Midterm
Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 23, Sections 4.14.8, nd Sections 5.15.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the
More informationGoals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite
Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite
More informationAbstract inner product spaces
WEEK 4 Abstrct inner product spces Definition An inner product spce is vector spce V over the rel field R equipped with rule for multiplying vectors, such tht the product of two vectors is sclr, nd the
More informationLecture 19: Continuous Least Squares Approximation
Lecture 19: Continuous Lest Squres Approximtion 33 Continuous lest squres pproximtion We begn 31 with the problem of pproximting some f C[, b] with polynomil p P n t the discrete points x, x 1,, x m for
More informationu(t)dt + i a f(t)dt f(t) dt b f(t) dt (2) With this preliminary step in place, we are ready to define integration on a general curve in C.
Lecture 4 Complex Integrtion MATHGA 2451.001 Complex Vriles 1 Construction 1.1 Integrting complex function over curve in C A nturl wy to construct the integrl of complex function over curve in the complex
More informationT b a(f) [f ] +. P b a(f) = Conclude that if f is in AC then it is the difference of two monotone absolutely continuous functions.
Rel Vribles, Fll 2014 Problem set 5 Solution suggestions Exerise 1. Let f be bsolutely ontinuous on [, b] Show tht nd T b (f) P b (f) f (x) dx [f ] +. Conlude tht if f is in AC then it is the differene
More informationPolynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230
Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given
More informationMath 554 Integration
Mth 554 Integrtion Hndout #9 4/12/96 Defn. A collection of n + 1 distinct points of the intervl [, b] P := {x 0 = < x 1 < < x i 1 < x i < < b =: x n } is clled prtition of the intervl. In this cse, we
More informationFourier series. Preliminary material on inner products. Suppose V is vector space over C and (, )
Fourier series. Preliminry mteril on inner products. Suppose V is vector spce over C nd (, ) is Hermitin inner product on V. This mens, by definition, tht (, ) : V V C nd tht the following four conditions
More informationThe Bochner Integral and the Weak Property (N)
Int. Journl of Mth. Anlysis, Vol. 8, 2014, no. 19, 901906 HIKARI Ltd, www.mhikri.com http://dx.doi.org/10.12988/ijm.2014.4367 The Bochner Integrl nd the Wek Property (N) Besnik Bush Memetj University
More information1.3 The Lemma of DuBoisReymond
28 CHAPTER 1. INDIRECT METHODS 1.3 The Lemm of DuBoisReymond We needed extr regulrity to integrte by prts nd obtin the Euler Lgrnge eqution. The following result shows tht, t lest sometimes, the extr
More informationarxiv: v1 [math.ca] 11 Jul 2011
rxiv:1107.1996v1 [mth.ca] 11 Jul 2011 Existence nd computtion of Riemnn Stieltjes integrls through Riemnn integrls July, 2011 Rodrigo López Pouso Deprtmento de Análise Mtemátic Fcultde de Mtemátics, Universidde
More informationLecture 14: Quadrature
Lecture 14: Qudrture This lecture is concerned with the evlution of integrls fx)dx 1) over finite intervl [, b] The integrnd fx) is ssumed to be relvlues nd smooth The pproximtion of n integrl by numericl
More information7.2 The Definite Integral
7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where
More informationComplex integration. L3: Cauchy s Theory.
MM Vercelli. L3: Cuchy s Theory. Contents: Complex integrtion. The Cuchy s integrls theorems. Singulrities. The residue theorem. Evlution of definite integrls. Appendix: Fundmentl theorem of lgebr. Discussions
More informationReview of Riemann Integral
1 Review of Riemnn Integrl In this chpter we review the definition of Riemnn integrl of bounded function f : [, b] R, nd point out its limittions so s to be convinced of the necessity of more generl integrl.
More informationHomework 11. Andrew Ma November 30, sin x (1+x) (1+x)
Homewor Andrew M November 3, 4 Problem 9 Clim: Pf: + + d = d = sin b +b + sin (+) d sin (+) d using integrtion by prts. By pplying + d = lim b sin b +b + sin (+) d. Since limits to both sides, lim b sin
More informationn f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1
The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the
More informationTHE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS.
THE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrdlindeloftheorem/ This document is proof of the existenceuniqueness theorem
More informationCzechoslovak Mathematical Journal, 55 (130) (2005), , Abbotsford. 1. Introduction
Czechoslovk Mthemticl Journl, 55 (130) (2005), 933 940 ESTIMATES OF THE REMAINDER IN TAYLOR S THEOREM USING THE HENSTOCKKURZWEIL INTEGRAL, Abbotsford (Received Jnury 22, 2003) Abstrct. When relvlued
More informationLecture 1. Functional series. Pointwise and uniform convergence.
1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is
More informationLecture 3 ( ) (translated and slightly adapted from lecture notes by Martin Klazar)
Lecture 3 (5.3.2018) (trnslted nd slightly dpted from lecture notes by Mrtin Klzr) Riemnn integrl Now we define precisely the concept of the re, in prticulr, the re of figure U(, b, f) under the grph of
More informationINTRODUCTION TO INTEGRATION
INTRODUCTION TO INTEGRATION 5.1 Ares nd Distnces Assume f(x) 0 on the intervl [, b]. Let A be the re under the grph of f(x). b We will obtin n pproximtion of A in the following three steps. STEP 1: Divide
More informationDefinite integral. Mathematics FRDIS MENDELU
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the
More informationA product convergence theorem for Henstock Kurzweil integrals
A product convergence theorem for Henstock Kurzweil integrls Prsr Mohnty Erik Tlvil 1 Deprtment of Mthemticl nd Sttisticl Sciences University of Albert Edmonton AB Cnd T6G 2G1 pmohnty@mth.ulbert.c etlvil@mth.ulbert.c
More informationChapter 22. The Fundamental Theorem of Calculus
Version of 24.2.4 Chpter 22 The Fundmentl Theorem of Clculus In this chpter I ddress one of the most importnt properties of the Lebesgue integrl. Given n integrble function f : [,b] R, we cn form its indefinite
More informationOverview of Calculus I
Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,
More informationCHAPTER 4 MULTIPLE INTEGRALS
CHAPTE 4 MULTIPLE INTEGAL The objects of this chpter re fivefold. They re: (1 Discuss when sclrvlued functions f cn be integrted over closed rectngulr boxes in n ; simply put, f is integrble over iff
More informationMTH 5102 Linear Algebra Practice Exam 1  Solutions Feb. 9, 2016
Nme (Lst nme, First nme): MTH 502 Liner Algebr Prctice Exm  Solutions Feb 9, 206 Exm Instructions: You hve hour & 0 minutes to complete the exm There re totl of 6 problems You must show your work Prtil
More informationMATH 409 Advanced Calculus I Lecture 19: Riemann sums. Properties of integrals.
MATH 409 Advnced Clculus I Lecture 19: Riemnn sums. Properties of integrls. Drboux sums Let P = {x 0,x 1,...,x n } be prtition of n intervl [,b], where x 0 = < x 1 < < x n = b. Let f : [,b] R be bounded
More informationarxiv: v1 [math.ca] 7 Mar 2012
rxiv:1203.1462v1 [mth.ca] 7 Mr 2012 A simple proof of the Fundmentl Theorem of Clculus for the Lebesgue integrl Mrch, 2012 Rodrigo López Pouso Deprtmento de Análise Mtemátic Fcultde de Mtemátics, Universidde
More informationConvex Sets and Functions
B Convex Sets nd Functions Definition B1 Let L, +, ) be rel liner spce nd let C be subset of L The set C is convex if, for ll x,y C nd ll [, 1], we hve 1 )x+y C In other words, every point on the line
More informationDefinite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function
More informationProperties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives
Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums  1 Riemnn
More informationLECTURE. INTEGRATION AND ANTIDERIVATIVE.
ANALYSIS FOR HIGH SCHOOL TEACHERS LECTURE. INTEGRATION AND ANTIDERIVATIVE. ROTHSCHILD CAESARIA COURSE, 2015/6 1. Integrtion Historiclly, it ws the problem of computing res nd volumes, tht triggered development
More informationChapter 4. Lebesgue Integration
4.2. Lebesgue Integrtion 1 Chpter 4. Lebesgue Integrtion Section 4.2. Lebesgue Integrtion Note. Simple functions ply the sme role to Lebesgue integrls s step functions ply to Riemnn integrtion. Definition.
More informationVariational Techniques for SturmLiouville Eigenvalue Problems
Vritionl Techniques for SturmLiouville Eigenvlue Problems Vlerie Cormni Deprtment of Mthemtics nd Sttistics University of Nebrsk, Lincoln Lincoln, NE 68588 Emil: vcormni@mth.unl.edu Rolf Ryhm Deprtment
More informationFundamental Theorem of Calculus
Fundmentl Theorem of Clculus Recll tht if f is nonnegtive nd continuous on [, ], then the re under its grph etween nd is the definite integrl A= f() d Now, for in the intervl [, ], let A() e the re under
More informationSection 5.1 #7, 10, 16, 21, 25; Section 5.2 #8, 9, 15, 20, 27, 30; Section 5.3 #4, 6, 9, 13, 16, 28, 31; Section 5.4 #7, 18, 21, 23, 25, 29, 40
Mth B Prof. Audrey Terrs HW # Solutions by Alex Eustis Due Tuesdy, Oct. 9 Section 5. #7,, 6,, 5; Section 5. #8, 9, 5,, 7, 3; Section 5.3 #4, 6, 9, 3, 6, 8, 3; Section 5.4 #7, 8,, 3, 5, 9, 4 5..7 Since
More informationPhysics 116C Solution of inhomogeneous ordinary differential equations using Green s functions
Physics 6C Solution of inhomogeneous ordinry differentil equtions using Green s functions Peter Young November 5, 29 Homogeneous Equtions We hve studied, especilly in long HW problem, second order liner
More informationLecture 1: Introduction to integration theory and bounded variation
Lecture 1: Introduction to integrtion theory nd bounded vrition Wht is this course bout? Integrtion theory. The first question you might hve is why there is nything you need to lern bout integrtion. You
More information1 i n x i x i 1. Note that kqk kp k. In addition, if P and Q are partition of [a, b], P Q is finer than both P and Q.
Chpter 6 Integrtion In this chpter we define the integrl. Intuitively, it should be the re under curve. Not surprisingly, fter mny exmples, counter exmples, exceptions, generliztions, the concept of the
More informationP 3 (x) = f(0) + f (0)x + f (0) 2. x 2 + f (0) . In the problem set, you are asked to show, in general, the n th order term is a n = f (n) (0)
1 Tylor polynomils In Section 3.5, we discussed how to pproximte function f(x) round point in terms of its first derivtive f (x) evluted t, tht is using the liner pproximtion f() + f ()(x ). We clled this
More informationConvergence of Fourier Series and Fejer s Theorem. Lee Ricketson
Convergence of Fourier Series nd Fejer s Theorem Lee Ricketson My, 006 Abstrct This pper will ddress the Fourier Series of functions with rbitrry period. We will derive forms of the Dirichlet nd Fejer
More informationIntegrals along Curves.
Integrls long Curves. 1. Pth integrls. Let : [, b] R n be continuous function nd let be the imge ([, b]) of. We refer to both nd s curve. If we need to distinguish between the two we cll the function the
More informationCalculus II: Integrations and Series
Clculus II: Integrtions nd Series August 7, 200 Integrls Suppose we hve generl function y = f(x) For simplicity, let f(x) > 0 nd f(x) continuous Denote F (x) = re under the grph of f in the intervl [,x]
More informationThe Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve.
Clculus Li Vs The Fundmentl Theorem of Clculus. The Totl Chnge Theorem nd the Are Under Curve. Recll the following fct from Clculus course. If continuous function f(x) represents the rte of chnge of F
More informationUniversitaireWiskundeCompetitie. Problem 2005/4A We have k=1. Show that for every q Q satisfying 0 < q < 1, there exists a finite subset K N so that
Problemen/UWC NAW 5/7 nr juni 006 47 Problemen/UWC UniversitireWiskundeCompetitie Edition 005/4 For Session 005/4 we received submissions from Peter Vndendriessche, Vldislv Frnk, Arne Smeets, Jn vn de
More informationHilbert Spaces. Chapter Inner product spaces
Chpter 4 Hilbert Spces 4.1 Inner product spces In the following we will discuss both complex nd rel vector spces. With L denoting either R or C we recll tht vector spce over L is set E equipped with ddition,
More informationNumerical Integration
Chpter 5 Numericl Integrtion Numericl integrtion is the study of how the numericl vlue of n integrl cn be found. Methods of function pproximtion discussed in Chpter??, i.e., function pproximtion vi the
More information