NAME: MATH 172: Lebesgue Integration and Fourier Analysis (winter 2012) Final exam. Wednesday, March 21, time: 2.5h

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1 NAME: SOLUTION problem # points max MATH 172: Lebesgue Integration and Fourier Analysis (winter 2012 Final exam Wednesday, March 21, 2012 time: 2.5h Please answer the questions below using the given space. If you need extra space, you can use the back of the page or an additional sheet of paper. The given space, however, should be large enough for your answer. Note that answers do not require complete sentences. As long as we can see that you know how to solve the problem and structure your proof, you will receive all points. You don t need to reprove anything that has been covered in class or on the problem sets. Please turn in your scrap paper as well, as it might contain valuable attempts. It can only improve your score! Before starting your exam, please write your name in the field above! 1

2 (1 (15 points Decide whether the following statements are true or false. Justify your answer briefly (i.e. either give a short proof or provide a counterexample. (a Let K R be the standard Cantor ternary set (i.e. K = K i and each K i+1 arises from K i by removing exactly one third in the center of each interval. Then there is a sequence of numbers x 1, x 2,... R such that [0, 1] (K + x i. TRUE FALSE 2 For any x R we have λ(k + x = λ(k = 0, i.e. K + x is a nullset. = (K + x i is a nullset if [0, 1] (K + x i, then ( 1 = λ([0, 1] λ (K + x i = 0. Contradiction! (b If A (0, 1 is an open set, then for every ε > 0 there is a special polygon P F such that A P and λ(p λ(a < ε. TRUE FALSE Let K [0, 1] be a Cantor ternary set with λ(k > 0 and consider A = [0, 1] \ K. Since K has empty interior, we have A = [0, 1]. if A P, then (0, 1 P = λ(p 1 > λ(a. This contradicts the claim for ε = λ(k = 1 λ(a.

3 (c Let A R n, A L be a Lebesgue set which is contained in a ball of radius r > 0 (i.e. A B(0, r. Then for every ε > 0 there are open sets B 1, B 2 B(0, r such that A B 1 and B(0, r \ A B 2 and such that λ(b 1 + λ(b 2 < λ(b(0, r + ε. TRUE FALSE 3 By the Theorem on Approximation there are F A G such that λ(g \ K < ε. Set B 1 = G B(0, r, B 2 = B(0, r \ F. Then B 1 B 2 = B(0, r and B 1 B 2 G \ K and hence λ(b 1 + λ(b 2 = λ(b 1 B 2 + λ(b 1 B 2 < λ(b(0, r + ε.

4 (2 (20 points (a Let X be a space and M a σ-algebra on X. Consider a sequence f 1, f 2,... : X R of M-measurable functions. Show that the following set is measurable with respect to M (i.e. show that B M. B = {x X : all but finitely many of the values f 1 (x, f 2 (x,... are the same} (Note: a sequence of numbers a 1, a 2,... R has the property that all but finitely many of its elements are the same, if there is a finite set I N such that a k = a j for all k, j N \ I. 4 Then g k : X R, g k (x = f k+1 (x f k (x are M-measurable. = A k = {x X : f k+1 (x = f k (x} = g 1 k (0 M. B = {x X : f k (x, f k+1 (x,... is constant} = A j M. k=1 k=1 j=k (b Let X be a space and M a σ-algebra on X. Consider two measures λ, µ on M such that λ(a µ(a for all A M. Show that for any M-measurable function f : X [0, which only takes non-negative values, we have f dλ f dµ. If s : X [0, is simple and non-negative, then m s = a i χ Ai a i 0, A i M. m m s dλ = a i λ(a i a i µ(a i = s dµ. Hence { } f dλ = sup s dλ : s S, s f { } sup s dµ : s S, s f = f dλ.

5 5 (c Show that there is a sequence f 1, f 2,... L 1 (R such that f j f k 1 = f j f k dλ = 1 whenever j k. Set f k = 1 2 χ [k,k+1]. (d Consider a space X, a σ-algebra M on X and a measure λ on M. Let f : X R be an M-measurable and integrable function such that f dλ = 0. Show that for any t > 0 λ ( f 1( [t, 1 2t f dλ. 0 = f dλ = f + dλ f dλ = f + dλ = f dλ f dλ = f + dλ + f dλ = 2 f + dλ Since tχ f 1 ([t, f + we have tλ(f 1 ([t, f + dλ = 1 2 f dλ.

6 (3 (10 points Consider the Borel σ-algebra B on R n, let U R n be a subset which is not Borel measurable and set V = R n \ U. Let M be the σ-algebra which is generated by B and U (i.e. M is generated by the family N = B {U}. Show that for any A M there are sets B, C B such that A = (B U (C V. 6 Set M = {A M : there are B, C B such that A = (B U (C V }. Then N M : If A B, then A = (A U (A V M. U = (R n U ( V M. And M is a σ-algebra: M, since B (see above. If A i = (B i U (C i V M, then (( (( A i = B i U If A = (B U (C V M, then M M and hence M = M. A c = (B c U (C c V M. C i V M.

7 (4 (15 points Let M be a σ-algebra on a space X and λ a measure on M. Consider an M-measurable function f : X [0, which only takes non-negative values. (a Assume that 0 f(x 1 for all x X and that f dλ <. Show that f k dλ = λ(f 1 (1. 7 Observe that since 0 f (x 1, we have f k f for all k. Since f is integrable, we obtain by LDCT f k dλ = Here we used that for a [0, 1] f k dλ = ak = 0 if a < 1 1 if a = 1 χ f 1 (1 dλ = λ(f 1 (1. (b Assume that 0 f(x 1 for all x X. We don t make any assumptions on the integral of f! Show that ( f ( 1 k dλ = k f dλ = λ f 1 (0, 1]. Observe that f 1 1 f 1 2 f by LICT f 1 k dλ = f 1 k dλ = χ f 1 ((0,1] dλ = λ(f 1 ((0, 1]. Here we used that for a 0 a 1 k = 0 if a = 0 1 if a > 0

8 8 (c Assume that f(x 0 for all x X and that f dλ <. Show that ( f ( 1 k dλ = λ f 1 (0,. Set A = f 1 ([0, 1], B = f 1 ((1,. Then X = A B. By part (b f 1 k χa dλ = f A 1 k dλ = λ(a f 1 ((0, 1] = λ(f 1 ((0, 1]. Moreover, since f 1 k χ B f (because a 1, k 1 = a 1 k a we obtain by LDCT f 1 k χb dλ = f 1 k dλ = f 1 k χb dλ = f 1 k χa dλ + χ B dλ = λ(b = λ(f 1 ((1,. f 1 k χb dλ = λ(f 1 ((0, 1] + λ(f 1 ((1, = λ(f 1 ((0,.

9 (5 (10 points Let f : R [0, be an L-measurable and integrable function which only takes non-negative values. For any k N define f k : R R by f k (x = f(x + k and set h(x = inf f k (x. Show that h = 0 almost everywhere. 9 By Fatou s Lemma h dλ = inf f k dλ inf But h is non-negative and periodic: if l Z, then f k dλ = f dλ <. Hence h(x + l = inf > h dλ = f (x + l + k = inf f (x + k = h(x. l= hχ [l,l+1 dλ = hχ [0,1 dλ = 0. l= hχ [0,1 dλ. By the sum-formula above, this implies h dλ = 0 thus h = 0 a.e.

10 (6 (10 points Give an example for a B-measurable function f : R R such that f p < for all p (1, 2, but f p = whenever p = 1 or p [2, ]. Justify your answer. Recall that for 1 p < ( f p = 1/p f p dλ 10 and f = inf{m 0 : f M a.e.}. Set f 1 (x = 1 x χ (1,, f 2 (x = 1 x χ (0,1, f = f 1 + f 2. Then for any p [1, ] (since f 1, f 2 f f 1 p, f 2 p f p f 1 p + f 2 p. f p < if and only if both f 1 p, f 2 p <. Obviously f 2 =. For p < f 1 p p = and f 2 p p = x p dx < if and only if p > 1 x p/2 dx < if and only if p/2 < 1.

11 (7 (10 points Let f, g : (0, R be B-measurable and integrable functions. Define a function h : (0, (0, R by ( x h(x, y = f(xy g x y y. You may in the following assume without proof that h is B-measurable. (a Show that h is integrable. (b Show that h dλ = 1 ( ( f dλ g dλ. (0, (0, 2 (0, (0, 11 We first assume (a. Use the C 1 -diffeo Φ : (0, (0, (0, (0,, (x, y (xy, x y. Its differential is dφ = ( y x y 1 xy 2, so the Jacobian is det dφ = 2 x y. By the Transformation Law, since we obtain By Fubini (0, (0, = (0, h(x, y = 1 2 f (Φ 1(x, yg(φ 2 (x, y det dφ(x, y, (0, (0, h dλ = 1 2 f (xg(y dxdy = (0, ( f (x dx g(y dy = (0, (0, (0, f (xg(y dxdy. ( f (xg(y dx dy (0, ( ( f (x dx (0, g(y dy (0,. (a follows the same way by replacing h with h (x, y = f (xy g ( x y x y resp. g with f resp. g. and f

12 (8 For this problem it is recommended that you use the Fourier transform. Recall that for any f L 1 (R n f(ξ = f(xe ix ξ dx. R n And the Fourier Inversion Theorem states that if f, f L 1 (R n, then f (x = (2π n f( x. In particular, if f is of Schwartz class (i.e. f S, then also f S and the inversion formula ( holds. (10 points Show that for any h : R n R, h S, there is a function f C 2 (R n such that f f = h. The Fourier Transform of both sides is n 2 f iξ k x f n = 2 f i 2 x f = f f = f f = ĥ i 2 (assuming that f S and n f (ξ = 2 f (ξ = x 2 k=1 k n k=1 iξ k f x k (ξ = n ξk f 2 (ξ = ξ 2 f (ξ. ξ 2 f f = ĥ = f (ξ = ĥ(ξ ξ To define f, we want to take the Fourier Transform of the right hand side, to obtain f. Observe first that for any k, l = 1,..., n ĥ(ξ ξ 2 + 1, ĥ(ξ ξ k ξ 2 + 1, ĥ(ξ ξ k ξ l ξ ĥ(ξ k=1 = ĥ(ξ ξ 2 + 1, ξ ĥ(ξ k ξ 2 + 1, ξ ĥ(ξ kξ l ξ L1 (R n. the Fourier Transform of ĥ(ξ exists and is C 2 (actually, it s even in S and setting ξ 2 +1 ( ĥ(ξ f (x = (2π n ( x ξ we obtain ( ξ 2 ( f (x f (x = (2π n ĥ(ξ ĥ(ξ ( x+(2π n ( x = (2π n ĥ( x = h(x. ξ ξ Here we used that whenever g(ξ, ξ k g(ξ, ξ k ξ l g(ξ L 1 (R n, then n 2ĝ(ξ n ĝ = = i ξ k g(ξ n = ξ xk 2 k 2 x g(ξ = ξ 2 g(ξ. k k=1 k=1 k=1 12 (

13 (9 (10 points Let f : R R be an L-measurable and integrable function and define the function F : R R by fχ [0,t] dλ if t 0 F (t = fχ [t,0] dλ if t < 0 13 Show that for any compactly supported smooth function g Cc fg dλ = F g dλ (R we have where g denotes the derivative of g. You can use without proof the fact that F is continuous. First solution Let ε > 0. By the denseness of C c such that f f 1 < ε. Set f χ [0,t] dλ if t 0 F (t = f χ [t,0] dλ if t < 0 F is smooth and (F (x = f (x. (R in L 1 (R, we can find an f C (R = t 0 f (x dx. Since g Cc (R, we know that g, g 1 < C <. f f χ [0,t] dλ if t 0 F (t F (t f f dλ < ε. f f χ [t,0] dλ if t < 0 f g dλ + F g dλ f g dλ + F g dλ + (f f g dλ + (F F g dλ. The first term is zero by the Fundamental Theorem of Calculus. The second an third terms is bounded by f f g dλ < Cε, F F g dλ ε g dλ Cε. Letting ε 0, yields the result. f g dλ + F g dλ < 2Cε.

14 14 First solution We compute F (tg (t dt = ( [0, f (xχ [0,t] (x dx g (t dt (,0 ( f (xχ [t,0] (x dx g (t dt By Fubini (observe that (x, t f (xχ [0,t] (xg (t and (x, t f (xχ [t,0] (xg (t are integrable, since g has compact support and f is integrable, this is equal to ( ( f (xg (tχ [0,t] (x dt dx f (xg (tχ [t,0] (x dt dx [0, (,0 If t 0, then χ [0,t] (x = χ [x, (tχ [0, (x and if t < 0, then χ [t,0] (x = χ (,x] (tχ (,0] (x. the term above is equal to ( f (xχ [0, (x g (tχ [x, (t dt dx [0, ( f (xχ (,0] (x g (tχ (,x] (t dt dx (,0 By the Fundamental Theorem of Calculus if x 0, then g (tχ [x, (t dt = g (t dt = g(x and if x 0, then [0, (,0 g (tχ (,x] (t dt = ( is equal to f (xχ [0, (xg(x dx x x g (t dt = g(x f (xχ (,0] g(x0 dx = f (xg(x dx. (