Final Exam Selected Solutions APPM 5450 Spring 2015 Applied Analysis 2 Date: Thursday, May

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1 Fial Exam Selected Solutios APPM 5450 Sprig 2015 Applied Aalysis 2 Date: Thursday, May Istructor: Dr. Becker Your ame: If the mathematical field is ot specified, you may assume it is R or C at your coveiece. The symbol H deotes a arbitrary Hilbert space. Your proofs may use ay major result discussed i class (if you are usure, please ask). Total poits possible: 104. N.B. Ulike the homeworks, the grades may be curved. Poits are ot distributed accordig to difficulty. Whe justifyig your aswer, do as little work as possible, e.g., avoid explicitly evaluatig itegrals if ot ecessary. Partial credit is possible. For problems 1 ad 2, PLEASE WRITE DIRECTLY ON THIS SHEET Problem 1: (20 pts) Defiitios. State the followig defiitios ad/or theorems. You may skip oe defiitio (please clearly mark which oe should ot be graded). 2 poits each. (1) Defie the Sobolev space H s (T) for s > 0. (2) Sobolev embeddig theorem, ay versio Solutio: H k (R) C(R) for k > 1/2 (careful, you should really specify R sice for R d you eed k > d/2). Do t cofuse this with Riema-Lebesgue! It s ot i C 0 (R). (3) Baach-Alouglu theorem, ay variat (4) Briefly defie the Fourier trasform o S(R) (the space of Schwartz fuctios). Solutio: Via the stadard itegral defiitio. (5) Briefly defie the Fourier trasform o S(R). Solutio: I a weak sese, e.g., for T S(R) defie T s.t. for all ϕ S, T, ϕ = T, ϕ. (6) Briefly defie the Fourier trasform o L 2 (R). Solutio: The stadard aswer is that it is defied by usig the desity of S ad extedig this via the BLT theorem. Valid alteratives: it is a diagoal operator with respect to the Hermite fuctios, mappig coefficiets (c ) =0,1,... to (( i) c ) =0,1,.... If you defie it via the itegral form, -1 poits (the itegral form is valid oly if f L 1 or f S). Note that your aswer should ever cotai ay sums, oly itegrals. Remember: for L 2 (R), use the Fourier trasform ad itegrals, while for L 2 (T), use Fourier series ad sums. (7) What does it mea for (ϕ ) S to coverge to a limit ϕ? Solutio: It meas for all α, β multi-idices, the ϕ ϕ α,β 0. The pseudo-orm is defied i eq. (11.3). (8) Lebesgue domiated covergece theorem (9) Fubii s theorem (10) Defie what it meas for a fuctio f : X Y to be measurable with respect to measure spaces (X, A) ad (Y, B). (11) I the above defiitio, if Y = R ad B is the Borel σ-algebra geerated by the stadard topology o R, what is a simplified defiitio of a measurable fuctio? Solutio: That for all c R, the set {x f(x) < c} is measurable. (Equivaletly, sets like {x f(x) c}, etc., are also valid). Problem 2: (30 pts) Mark true/false (or yes/o). No justificatio eeded. H deotes a Hilbert space. 2 poits each. Solutio: Most studets did very well o this, but almost o oe got questio 6 correct. (1) Let C, D H. If C = D, is C = D? Solutio: False. This is oly true if D is closed. For example, let H = R 2 ad D = {qe 1 q Q} where e 1 is the first caoical uit vector. The C = spa(e 2 ) so C = spa(e 1 ) = D D. 1

2 (2) If µ(x) <, the L p (X) L q (X) if p q. Solutio: False. (3) If µ(x) <, the L p (X) L q (X) if p q. Solutio: True. (4) If µ(x) =, the L p (X) L q (X) if p q. Solutio: False. (5) If µ(x) =, the L p (X) L q (X) if p q. Solutio: False. (6) Is the Heaviside fuctio H(x) = χ (0, ) (x) weakly differetiable? Solutio: No, sice if it were, it would violate the Sobolev embeddig theorem, sice weakly differetiable implies it is i H 1 [sorry for overloadig the H symbol] but it is clearly ot cotiuous. See example i the book for more discussio. It has a distributioal derivative but ot a weak derivative, ad this is why we say that sometimes T f is ot the same as T f. (7) Is it possible that i l 1 (N), weak covergece always implies strog covergece? Solutio: Yes, ad i fact it is true (Schur 1924). If 1 < p <, this is ot possible sice l p is reflexive, so the Baach- Alouglu theorem says the uit ball is weakly compact, ad the uit ball i a ifiite dimesioal space is ever strogly compact. For p = 1, the space is ot reflexive so this is ot precluded (this is what the questio is gettig at). Note that we discussed this o problem 4b i HW 14 from last semester. (8) L p (R) is separable for all 1 p. Solutio: False, sice this is ot true for p =. (9) L p ([0, 1]) is separable for all 1 p. Solutio: False, sice this is ot true for p =. (10) C([0, 1]) is dese i L ([0, 1]). Solutio: False, sice that would imply it is separable. (11) C c (R) is dese i L p (R) for 1 p <. Solutio: True. (12) C c (R) is complete with respect to the uiform orm. Solutio: False, for example a Cauchy sequece may coverge to e x2. (13) Let f L 2 (T) ad defie the partial sum f N = N = N f e where (e ) is the Fourier basis. Does f f L 2 0? Solutio: Yes. (14) If f, g L 1 (R), is fg L 1 (R)? Solutio: No. (15) If f, g L 1 (R), is f g L 1 (R)? Solutio: Yes, follows from Fubii s theorem. Problem 3: (12 pts) Short respose. 2 poits each. For examples of fuctios, do t forget to specify their domai. (1) The space L 2 (T) cotais periodic fuctios, but the fuctios are ot defied poitwise because they are really equivalece classes. If they are ot defied poitwise, how ca they be periodic? Briefly discuss. Solutio: Because cotiuous fuctios are defied poitwise, we ca easily defie C(T). The defie L 2 (T) as the closure of C(T) uder the L 2 orm. (2) O I = [ π, π], f(x) = x is very smooth, i.e., f C (I), whereas g(x) = x is ot a smooth, i.e., g C(I) \ C 1 (I). Do you expect the Fourier coefficiets of f to decrease faster tha those of g? Is this true? Solutio: If we cosider the periodic extesio, the f is o loger eve cotiuous, so f / C(T), whereas g C(T), so we actually expect the Fourier coefficiets of g to decrease faster, which is what we observe. (3) Give a example of a operator that is positive but ot coercive. Solutio: If your example is of a liear operator, the the space is ecessarily ifiite dimesioal. Take l 2 (N) for example, ad defie T (x) = (1/x ) for x = (x ) l 2. More geerally, use ay diagoal operator (i terms of a orthoormal basis, or just o l 2 coordiate-wise) such that A(x) = (a x ) ad a > 0 but a 0. Aother example is example 9.5, usig (T f)(x) = x f(x) o L 2 ([0, 1]). Sice x [0, 1], we see f, T f 0 always, but this ca be arbitrarily small as the support of f clusters aroud 0. This questio was missed by a lot of studets. Oe of the issues is positive vs. o-egative. We say a operator T is positive or positive defiite if x, T x > 0 for all x, ad it is o-egative or positive semi-defiite if we just have x, T x 0. 2

3 (4) Give a example of fuctios (f ) that coverge poitwise a.e. to f but lim f f. Solutio: May possibilities, e.g., i L 1, take boxes supported o [0, 1/] with height so f = 0 ad lim f = 1 0 = 0. This questio was aswered correctly by most studets. (5) Give a example of a fuctio that is Lebesgue itegrable but ot Riema itegrable. Solutio: The idicator fuctio of the ratioal umbers. Note that if a fuctio is very ice ad o-egative (so o cacellatios happe), the the itegrals will be the same. So it s ot the case that, say, f(x) = x 1/2 over [0, ) is Lebesgue itegral but ot Riema itegral (it s either). Rather, you eed somethig fuy happeig, such as the idicator of the ratioals. For the opposite directio (see ext problem) you eed cacellatio. (6) Give a example of a fuctio that has a improper Riema itegral but is ot Lebesgue itegrable. Solutio: The sic fuctio o R o R. The Lebesgue itegral does ot allow cacellatios the same way a improper Riema itegral does. Note that if si(x)/x is ot itegrable o [0, ), the either is si(1/x)/x o [0, ) sice they are the same after chagig variables x 1/x. The former does ot decay fast eough at while the latter oscillates too much ear 0. You ca make lots of examples by exploitig cacellatio ad usig coditioally coverget series. For example, o [0, ), let i(x) be the fuctio that returs the greatest iteger less tha x, e.g., i(3.7) = 3. The the fuctio f(x) = 1/x ( 1) i(x) is modulated ad has cacellatios, so it has a improper Riema itegral but ot a Lebesgue oe; the same coclusio holds for the fuctio f(x) = i= 1 ( 1) χ [,+1]. I particular, for both fuctios, we exploit the fact that we kow from udergrad aalysis that the sum =1 1 ( 1) is coditioally coverget. Also ote that a lot of the itegrals for Fourier coefficiets that we write are ot well-defied i the Lebesgue sese, but we write them that way because it is short-had otatio for the Fourier trasform (which we defie usig desity argumets if f L 2 \ L 1 ). Most of these itegrals are welldefied i the improper Riema sese, so these would make examples. For example, f(x) = 1/x o all of R, because the cotributios from x > 0 ad x < 0 cacel each-other out. This questio was missed by a lot of studets. Problem 4: (6 pts) Covergece. 3 poits each. (1) Let f (x) = e iπx L 2 ([0, 1]). Does f coverge strogly, or weakly, ad if so, what is the limit? Justify your aswer. Solutio: It coverges weakly to 0 because the dual space is also L 2 ([0, 1]) ad this is a subset of L 1 ([0, 1]) so the Riema-Lebesgue lemma applies, ad the decay property of the Fourier trasform implies the weak covergece. It does ot coverge strogly because f (x) 2 = 1 so f = 1. [Note that via Baach-Alouglu, we ca immediately ifer that it has a weakly coverget subsequece, ad you could try to use this fact to prove that the whole sequece must coverge.] Or, ot the equivalece with Fourier series ad use Bessel s iequality. Or ote that it is orthoormal ad (via Bessel s iequality) we kow all orthoormal sequeces caot coverge strogly but do coverge weakly. Quite a few studets did ot correctly calculate f, which should be 1. Recall that e ix = 1 for all x. Rubric: 2 poits for right aswer without correct justificatio, 0 to 1 poit for wrog aswers depedig o argumets. (2) Let f (x) = e iπx χ [,] L 2 (R). Does f coverge strogly, or weakly, ad if so, what is the limit? Justify your aswer. Solutio: The Riema-Lebesgue lemma o loger applies sice the dual space is L 2 (R) ad this is ot a subset of L 1 (R). I fact, we do ot have weak covergece because a ecessary coditio is that f is bouded, ad it is ot (sice f 2 = 2). Of course (f ) does ot coverge strogly either. Problem 5: (6 pts) Covergece ad Itegrals. Let (f ) ad f be itegrable fuctios o [1, ) such that f f a.e. Give a short proof or a couter-example for the followig statemets: (3 poits each) 3

4 (1) If f f uiformly, the lim f 1 = f. 1 Solutio: No. Take f = 2 χ [2,2 +1 ] so f = 1 ad it coverges to 0 uiformly, but lim f = 1 0 = 0. Or take f to be 0 outside [1, + 1] ad liear iside [1, + 1] with f(1) = 2/ ad f( + 1) = 0, so that f 1 (x) dx = 1 but yet it coverges uiformly to 0. This does ot violate the uiform covergece theorem sice that theorem requires a bouded domai of itegratio. (2) If (f ) is mootoe decreasig, the lim if f 1 = f. 1 Solutio: Yes. Use Fatou s lemma (the limif should be a big hit to use Fatou). Specifically, to apply Fatou we eed o-egative fuctios. Sice f f a.e., we have that f f is o-egative a.e.. We ca work directly with f istead of f f sice we ca just subtract off f sice it is itegrable so this is fiite. Thus f = lim f = lim if f lim if We also have f 1 f is o-egative a.e. for all, ad similarly ca subtract off f 1, so aother applicatio of Fatou gives f = lim f = lim if f lim if f ad takig the egative out gives f lim if f. Combiig the two iequalities gives the equality. We ca actually quickly prove a stroger statemet: sice f f ad f f 1 L 1, we ca apply the MCT, ad coclude that i fact f = lim f. These two problems were take from part of a questio from a old prelim test, ad studet scores o both these problems were low (oly 3 studets gave correct or early-correct aswers to the 2d questio). f Problem 6: (6pts) Spectral theory. Let A B(H). 3 poits each. (1) Prove λ σ r (A) implies λ σ p (A ). Solutio: Let T = A λi so T = A λi. The H = ra(t ) ker(t ) so λ σ r (A) implies H ra(t ) ad hece there is a o-trivial elemet i ker(t ) which implies λ σ p (A ). This questio is fudametal. See Prop i the book. (2) If A = A, prove σ r (A) =. Solutio: If A is self-adjoit, the spectrum is real. By the previous questio, we have λ σ r (A) implies λ = λ σ p (A ) = σ p (A), but σ r (A) σ p (A) = by defiitio, hece we caot have λ σ r (A). Problem 7: (8 pts) Fourier trasform. 2 poits each. (1) Let f(x) = e iωx for ω R be a fuctio o R. Which of the followig spaces does f live i: S(R), S (R), L 1 (R), L 2 (R) (combiatios allowed, e.g., oe or all )? Solutio: It is i S (R) ad oe of the others. If we chose I = [0, 1] istead of I = R, for example, the it would be i all of S(I), S (I), L 1 (I), L 2 (I). def (2) Let F be the Fourier trasform. What is f = F(f) for f as above? Solutio: A acceptable aswer is δ ω (i.e. δ( ω)), though 2πδ ω is more precise. Note that the review sheet metioed two key Fourier trasform pairs: the sic fuctio ad box fuctio, ad the Gaussia fuctio (ad a scaled versio of itself). This δ ω ad e iωx is aother key pair you should memorize forever. 4

5 For ituitio, use eq (11.33) which states that δ(x) = 1/ 2 e ikx dk, though this is ot meat i a precise sese sice this is ot Lebesgue itegrable. Workig with the itegral themselves is messy sice you eed to use a ultraviolet cutoff like Prop It is better to use Prop which says ê iωx ϕ = τ ω ϕ for ϕ S, ad therefore this also holds for S sice it is defied weakly so all properties of the Fourier trasform o S carry over to S. The just use the fact that we kow δ = 1/ 2 (example 11.31) ad the shift this. For referece, to use eq (11.33) iformally, we do (for ϕ S arbitrary) f, ϕ = f, ϕ = f(x) ϕ(x) dx ( ) = f(x) e ixy ϕ(y) dy dx ( ) = ϕ(y) e ixy f(x) dx dy ( ) = ϕ(y) e ix(ω y) dx dy = ϕ(y) 2δ(ω y) dy = 2 δ(ω ), ϕ = 2 δ( ω), ϕ. (Why is this iformal? Because we have chaged the order of itegratio eve though Fubii s theorem does t apply, ad we are usig eq (11.33) which is ot literally true.) (3) Let ϕ S(R). What is δ, ϕ? Solutio: It is ϕ (0). See Example i the book. (4) Let ϕ S(R). What is δ ϕ? Solutio: First, observe that we are lookig for the aswer to be a fuctio, ot a scalar. The equatio above Example i the book defies the covolutio as δ ϕ = δ, Rτ x ϕ where R is the reversal operatio ad τ x the shift-by-τ, so if we have ϕ(y) the (τ x ϕ)(y) = ϕ(y x) ad (Rτ x ϕ)(y) = ϕ(x y). The actio of the distributio is δ, ψ = δ, ψ = ψ (0) (see previous questio). If ψ = Rτ x ϕ the via the chai rule, ψ (y) = ϕ (x y) thus (δ ϕ)(x) = ϕ (x), so the aswer is ϕ. This example is very importat. Whe you covolve a fuctio with the delta fuctio, you get back the fuctio see refereces o Gree s fuctio. Most studets missed this problem. Covolutios were ot emphasized i this class but they are fair game o the prelim, as they are metioed i our text ad they are cosidered part of udergrad aalysis. Problem 8: (3 pts) Give a example of a measurable space (X, µ) ad measurable sets (E ) such that E 1 E 2 E 3... ad ( ) lim µ(e ) µ E. Solutio: The ecessary igrediet i the couter-example is that some E has ifiite measure. For example, let X = R ad E = [, ) so E =. Th ifiite measure igrediet is ecessary sice o problem 2 from homework 11, we proved that if ay E has fiite measure, the we do have equality. Problem 9: (13 pts) Bouded liear operators. (1)* (1 pt) Let X be a Baach space. If ϕ(x) = ϕ(y) for all ϕ X, prove x = y. 5

6 Solutio: This is problem 5.6 from the book ad was problem 1 i HW 10 from last semester. You first defie a fuctioal ψ o the subspace spaed by x (wlog let y = 0 sice ϕ is liear) such that ψ(x) = x (uless x = 0), the use Hah-Baach to exted this to a fuctioal o the whole space. Therefore if ϕ(x) = 0 for all fuctios ϕ, we have ψ(x) = 0 but also ψ(x) = x if x 0, so this is a cotradictio. (2) (2 pts) A closed operator T : X Y, X ad Y ormed liear spaces, is such that if for (x ) X, x x ad T (x ) y, the T (x) = y. Explai how this differs from a cotiuous operator, ad state whether closed operators are cotiuous, or vice-versa, or either. Solutio: This is weaker tha beig cotiuous sice we pre-suppose that T (x ) coverges. For cotiuous (i.e., sequetially cotiuous), the fact that T (x ) coverges (if x coverges) is a cosequece, ot a coditio. Hece cotiuous implies closed, but ot vice-versa. (3) (0 pts, fact) Let H be a Hilbert space, ad let A : H H ad B : H H be operators (ot ecessarily liear or bouded) with the property that Ax, y = x, By for all x, y H. The you ca prove A (ad hece B) must be liear operators. (4) (3 pts) Uder the same assumptios as part (3), prove A (ad hece B) must be bouded as well. Hit: you may use the followig corollary of the ope-mappig theorem kow as the Closed Graph Theorem : if X ad Y are Baach ad A : X Y is liear, the A is closed iff it is bouded. Solutio: We wish to show A is closed, so we ca apply the theorem ad coclude it is also bouded. Let x x ad A(x ) y, ad we seek to prove y = A(x). The for all x H, y, z = lim Ax, z = lim x, Bz = x, Bz = Ax, z sice the ier product is cotiuous. Usig part (a), we coclude y = Ax as required. (5) (4 pts) Let H 1 (T) be the Sobolev space o the torus. 2 poits each: i. Defie the weak derivative (deote the operator by D). ii. Is D : H 1 (T) L 2 (T) L 2 (T) bouded? Briefly justify Solutio: We ca defie it weakly such that if f H 1, the for all ϕ C 1 (T), T, ϕ = T, ϕ. Or, defie it o the Fourier coefficiets ad map f to (i) f. It is ot bouded because as we see via the Fourier coefficiet defiitio, it icreases the magitude of coefficiets. (6) (3 pts) I chapter 10, which we did ot cover, the book defies a cocept called the formal adjoit of the weak derivative D. Ca this be the same cocept of adjoit that we discussed this semester? Please discuss why or why ot. Solutio: We just argued i 5(ii) that D is ot bouded, but we proved i 4 that ay operator with a adjoit mus be bouded. Therefore D caot have a adjoit i the sese we defied. The formal adjoit is differet sice it chages the domai. Basically, a liear operator i the sese we have discussed i class must be defied o the whole domai, but you could restrict it. This is related to how we view H 1 : is ca be a subset of L 2 (that is, usig the L 2 orm), which is ot complete, or it ca be its ow Hilbert space with the H 1 orm (i which case it is complete). 6