FINAL EXAMINATION IN FOUNDATION OF ANALYSIS (TMA4225)

Transcription

1 Norwegia Uiversity of Sciece ad Techology Departmet of Mathematical Scieces Page of 7 Cotact durig exam: Eugeia Maliikova (735) FINAL EXAMINATION IN FOUNDATION OF ANALYSIS (TMA4225) Moday, December 4th 26 Time: 9: 3: Grades to be aouced: December 25th, 26 S. Problem Let f ad g be bouded fuctios o [a, b]. a) Prove that if both f ad g are Riema itegrable the fg is Riema itegrable. b) Prove that if both f ad g are absolutely cotiuous the fg is absolutely cotiuous. a) Let S f (S g ) be the set of the poits at which f (g) is discotiuous. We kow that a bouded fuctio is Riema itegrable if ad oly if it is cotiuous almost everywhere. Thus S f = S g =. Further let S fg be the set of the poits x [a, b] at which fg is discotiuous. If f ad g are cotiuous at some poit x, the fg is cotiuous at x. Therefor S fg S f S g. We have S fg S f + S g = ad the fuctio fg is Riema itegrable sice it is cotiuous almost everywhere.

2 Page 2 of 7 Aother solutio was give i Problem set (problem 3 c)). b) First, sice f ad g are absolutely cotiuous o [a, b], they are uiformly cotiuous o [a, b]. We wat to check that for ay ɛ > there exists δ such that (fg)(b ) (fg)(a ) < ɛ wheever (b a ) < δ. Let M = max{ f, g }, where f = max x [a,b] f(x). We have f(b)g(b) f(a)g(a) f(b) g(b) g(a) + g(a) f(b) f(a) M( g(b) g(a) + f(b) f(a) ). Now, suppose that ɛ >, sice f ad g are absolutely cotiuous, we ca fid δ such that f(b ) f(a ) < ɛ 2M ad g(b ) g(a ) < ɛ 2M, wheever (b a ) < δ. The for ay system of itervals (a, b ) with total legth less tha δ, f(b )g(b ) f(a )g(a ) M( g(b ) g(a ) + f(b ) f(a ) ) < ɛ. So fg AC([a, b]). This part of the problem is from Problem set (problem 3). Problem 2 a) Let E,...E be measurable subsets of [, ]. Suppose that each poit x [, ] belogs to at least k of these sets. Prove that max E k. b) Let E be a measurable subset of R. Use the Lebesgue differetiatio theorem to show that for a.e. x E E (x h, x + h) =. h 2h a) Let f(x) = = χ E (x), where χ E is the characteristic fuctio of E (i.e. the fuctio that equals at each poit of E ad zero at other poits).

3 Page 3 of 7 Each poit of [, ] belogs to at least k of the sets meas that f(x) k for ay x [, ]. For each E = χ E dm. We have E = = The max E = E k. = [,] χ E dm = [,] This part of problem 2 is from Problem set 6 (problem ). f(x)dm k. b) We apply the Lebesgue Differetiatio theorem to the characteristic fuctio of the set E. If E = + the χ E is ot itegrable but it is locally itegrable, we may apply the theorem for each fuctio χ E [,]. It implies that for a. e. x R the it below exists ad h I particular, for a. e. x E, E (x h, x + h) h 2h χ E (t) χ E (x) dm(t) =. 2h (x h,x+h) = χ E (t) = χ E (x) =. h 2h (x h,x+h) Problem 3 Let f be a o-egative measurable fuctio o a measure space (X, A, µ). Let λ f (t) = µ{f > t} be its distributio fuctio. a) Show that f L (µ) if ad oly if = λ f() + = λ 2 f ( ) < +. b) Show that if f L 2 (µ) the t t 2 λ f (t) =. a)we have fdµ = λ f dm. λ f dm = λ f dm + X λ f dm = (,+ ) /( ) =2 / =2 f(t)dm(t) + λ f (t)dm(t),

4 Page 4 of 7 the equality above follows from the mootoe covergece theorem (we ca itegrate a series of o-egative fuctios term by term). The fuctio λ f is o-icreasig, we have λ f () =2 =2 λ f (t)dm(t) λ f ( ) = =2 λ f (). To estimate the itegral over [/, /( )] we use agai that fuctio is o-icreasig ad estimate the legth of this iterval whe 2. The = 2 2 λ f ( ) = = = ( ), 2( ) 2 ( ) 2, 2 =2 ( ) 2( ) λ 2 f =2 /( ) / f(t)dm(t) =2 2 2 λ f (/). Assume first that f L (µ). The λ f dm < + ad the iequalities above imply λ f () < +, =2 = 2 2 λ f ( ) < +. The λ() < + ad λ f () + = = 2 λ f( ) < +. To prove the opposite implicatio, assume that the sum above is fiite, it meas that two series coverge: λ f () <, λ f( 2 ) < +. = Usig the estimates we got above, we have = λ f (t)dm(t) < +, The λ f (t)dm(t) < + ad f L (µ). λ f (t)dm(t) < +. b) Usig the formula g(f(x))dµ(x) = X g (t)λ f (t)dm(t)

5 Page 5 of 7 with g(t) = t 2 ad measurable o-egative fuctio f, we get f 2 dµ = 2tλ f (t)dm(t). X Assume that f L 2 (µ). The 2tλ f (t)dm(t) < +. We claim that I fact we have, = / /(+) t t 2tλ f (t)dm(t) = 2sλ f (s)dm(s) =. () 2tλ f (t)dm(t) <. Sice the series coverges, the tail of the series turs to zero ad Thus () follows. / 2tλ f (t)dm(t) =. The fuctio λ f is o-icreasig ad t 2sλ f (s)dm(s) λ f (t) t 2sds = t 2 λ f (t). The expressio i the left-had side of the iequality above turs to zero whe t goes to zero. Thus t t 2 λ f (t) =. Problem 4 Let f(x) = si x x. a) Show that for ay b > b b e xy si xdxdy. b) Use the idetity to calculate b e xy si xdx = + y e by cos b + ye by si b 2 + y 2 b b f(x)dx.

6 Page 6 of 7 (Justify your calculatios.) a) First, e xy dy = /x for x >, thus f(x) = si x x b b = si xe xy dy. We get si xe xy dydx. We wat to iterchage the order of itegratio usig the Fubii theorem. So we check first that the fuctio is itegrable over [, b] (, + ). Usig the estimate si xe xy xe xy, we get b si xe xy dydx Thus, chagig the order of itegratio, b b x b e xy dydx = b si xe xy dxdy. x dx = b < +. x b) Applyig the result of part a) ad the give expressio for the itegral, we get b ( ) + y e by cos b + ye by si b dy = 2 + y 2 g b (y)dy. We wat to fid the it of the itegral above, whe b goes to. We cosider fuctios g (y), where is positive iteger. First, if y > the g (y) = e y cos + ye y si + y 2 + y e y + y 2 2e y. It implies that for each y > ( ) g (y) = + y e y cos + ye y si = 2 + y 2 + y. 2 Moreover, sice, we have g (y) + 2e y for ay y >. The fuctio + 2e y +y 2 +y 2 is itegrable over (, + ) ad usig the domiated covergece theorem, we get g (y)dy = + y 2 dy = π 2.

7 Page 7 of 7 To explai why it is eough to cosider iteger b, we remark that f(x). For ay b > x let = [b] be the iteger part of b, i.e. b < +, the b f(x)dx b f(x)dx dx x <. Fially, b b, Z f(t)dt = π 2. Remark: Oe may also apply the domiated covergece theorem for the family of fuctios g b. I preferred to use the sequece g sice we formulated the theorem for the sequeces i the course.