Real Analysis April f (t) dt. Show that g(x) 0 as x.

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1 eal Analysis April Part A. Let f L ( and let g(x x f (t dt. Show that g(x as x. Let {x n } be a sequence in such that x n as n, and define h n (t : χ [xn, f (t. Now, given ε >, there is some N such that hn (t χ [xn, f (t < ε. Thus, lim h n(t. Also, notice that hn (t f (t, so since f L (, by the Lebesgue n Dominated Convergence Theorem, we have lim g(x lim g(x n x n lim f (t dt n x n lim χ n [xn, f (t dt lim h n (t dt n lim h n(t dt n dt, as required.

2 . Prove or disprove: If f : [, ] [, ] is nondecreasing and continuous, then f is absolutely continuous. False. Counterexample: Cantor function Let C be the Cantor set. That is, let { } C x [, ] : x n a n3 n, where each a n or { 3 m ( 3k x [, ] \ 3 m, 3k } 3 { m m k } x [, ] \ open middle thirds. Notice that µ(c, since C is totally disconnected, where µ is the Lebesgue measure. Define the Cantor function f : [, ] [, ] as a n n, if x Cas defined above f (x n p k, for some p, k Z, if x [, ] \ C so that if x [, ] \ C, then f (x is equal to the value of f at the endpoints of the open middle third interval that x is in. Notice: f ( and f (, (i.e. f goes from to as x goes from to. Also, f is continuous everywhere. Now, since µ(c and f (x for every x C, f is differentiable almost everywhere with f. Thus, f is nondecreasing, also. So f satisfies the hypotheses. However, f (x dx dx f ( f (. So, by the contrapositive of the Fundamental Theorem of Calculus, f cannot be absolutely continuous. 3. Let E be a closed Lebesgue measurable subset of [, ]. Prove or disprove: a. If E c is dense, then E has measure. b. If E has measure, then E c is dense. a. False. Counterexample: Let E c Q (,. Notice: E c is dense in [, ] since the closure of E c is [, ] and since Q is dense in. Also, notice that E is closed since E c is open. But, µ(e c µ(q, so, µ(e µ([, ] µ(e c,if µ is the Lebesgue measure. b. True. Proof by contrapositive: If E c is not dense, then there is some ( δ > such that (y, y δ is not in E c, for some y E c. Thus, (y, y δ E and µ(e µ (y, y δ δ >.

3 4. Let f : be absolutely continuous, and assume that f L 4 ([, ] and that f (. Find all values of α so that for all such f. lim x x α f (x Since f is absolutely continuous on, it is also absolutely continuous on [, x], for any x, but let s only consider x [, ]. Also, f exists almost everywhere and is integrable on [, x], and, by the Fundamental Theorem of Calculus, f (t dt f (x f ( f (x. ( /4 Also, note that f L 4 ([, ] implies that f (t dt 4 <. Now, lim x x α f (x lim x α f (x x x f (t dt lim x x α lim x x α x f (t dt x lim x x α f (t on [, x] lim x x α 4/3 f (t 4 ( x lim x x α lim x x α x 3 4 ( lim x 3 4 α x ( x By Hölder s Inequality 3/4 ( x 4/3 dt f (t /4 4 dt f (t /4 4 dt χ [,x] f (t 4 dt Now, let {x n } be a sequence of positive real numbers in [, ] such that x n as n. Define g n (x : χ [,xn ] f (x 4. Then, g n (x as n, andg n (x f (x 4, which is integrable, so by the Dominated Convergence Theorem, we have the following: lim χ [,x] f (t 4 dt lim g n (t dt lim g x n n(t dt dt. n /4, provided x 3 4 α Thus, lim x x α f (x x n, it must be that 3 4 α. ( 4 lim x 3 α n g n (t dt n /4 n. So, since Therefore, if α 3, then lim 4 x x α f (x, as required. 3

4 Part B 5. ecall that, if (X, ρ and (Y, σ are metric spaces, a function f : X Y is called Lipschitz continuous if there exists a constant λ such that σ ( f (x, f (y λρ(x, y for all x, y X. Let X be a compact metric space, and let Y. Show that the Lipschitz continuous functions are dense in the space of continuous functions with the uniform norm. Stone-Weierstrass Theorem: Suppose X is a compact Hausdorff space and A is a subalgebra of C(X,, which contains a non-zero constant function. Then A is dense in C(X, if, and only if, it separates points. In our case: A Lip(X, { f : X f is Lipschitz continuous}. We will need to show the following:. Lip(X, is a subalgebra of C(X, : (a Lip(X, C(X,. (b Lip(X, is a linear space. (c Lip(X, is closed under multiplication.. There is a non-zero constant function in Lip(X,. 3. Lip(X, separates points. 4. X is Hausdorff. Now, let s get to work.... (a Let x, y X and f Lip(X,, then f (x f (y λρ(x, y. Now, given ε >, there exists δ λ ε, such that ρ(x, x < δ. Then f (x f (x λρ(x, x < λδ ε. Thus, Lip(X, C(X,. (b Let a, b, f, g Lip(X,, x, y X, then ( a f bg (x ( a f bg (y a [ f (x f (y ] b [ g(x g(y ] a f (x f (y b g(x g(y a λ ρ(x, y b λ ρ(x, y ( a λ b λ ρ(x, y, Thus, (a f bg Lip(X,, so Lip(X, is a linear space. where ( a λ b λ. 4

5 (c Let x, y X, f, g Lip(X,. Since X is compact and f, g C(X,, there is some M such that f (t M and g(t M, for every t X. Then, f g(x f g(y f (xg(x f (yg(y f (xg(x [ f (yg(x f (yg(x ] f (yg(y g(x [ f (x f (y ] f (y [ g(x g(y ] M f (x f (y M g(x g(y M λ ρ(x, y M λ ρ(x, y M(λ λ ρ(x, y, where M(λ λ. Thus, f g Lip(X,, so Lip(X, is closed under multiplication.. Let x, y X, and let f c, c. Consider f (x f (y c c λρ(x, y, for λ. Thus, f Lip(X,. 3. Let x, y X such that x y, and suppose f (t ρ(t, q, for every t X, where q X. Then f (x f (y ρ(x, q ρ(y, q ρ(x, q ρ(q, y ρ(x, y. Thus, f Lip(X,. Also, for r X such that r q, we have that f (r ρ(r, q >, whereas f (q ρ(q, q. Thus, f separates points. 4. A topological space T is Hausdorff if for x, y T such that x y, there are disjoint open sets U, V T with x U and y V. Now, given that X is compact and is Hausdorff, let f Lip(X, and f (x f (y in. Then, f (x U and f (y V for open U, V such that U V. Since Lip(X, C(X, we know that f (U and f (V are both open in X. If t f (U f (V, then t f (U and t f (V, which implies that f (t U and f (t V, but U V. Thus, f (U f (V. Therefore, X is Hausdorff. Thus, by the Stone-Weierstrass Theorem, the Lipschitz continuous functions are dense in the space of continuous functions with the uniform norm, as required. 5

6 6. Use an appropriate Fourier series to compute n (n. Consider a function f : which is periodic with period one and which satisfies The Fourier series for f is c n c n c f (x χ (,/ (x χ /, (x for x < [ c n e πinx, where f (xe πinx dx e πinx dx e πinx πin e piin πin ] [ e πinx πin eπin πin e πinx dx ] for n. (e πin e πin ( πin cos(πn i sin(πn cos(πn i sin(πn πin cos(πn. πin ] So the Fourier series for f is dx [ [ ] x x ( cos(πn πin,n ( dx (. e πinx ( n πin,n e πinx Notice: if n is odd, then ( n (, but if n is even, then ( n (. So the Fourier series for f is: πi(n eπi(n x. 6

7 Now, by Parseval s Formula, f L ˆf l and so f ˆf L l Well, f L f (x dx and Thus, Therefore, ˆf l [ ] x dx [ ( ] x ( dx πi(n eπi(n x 4 π (n 4 π (n 4 π ( n 4 π ( n 4 π ( n 4 π ( ˆf l 8 π n (n n (n n (n n (n. n n ( from part (a. ( ( n ( (n (n (n f L (n π 8. 7

8 7. Assume that E is a compact Lebesgue measurable subset of, and let f (t cos(tx dx, t. Prove or disprove: a. f has compact support. b. f is infinitely differentiable. a. False. Counterexample: E Let E [, π], then E is a compact and Lebesgue measurable subset of. Then f (t π If t then f ( cos(tx dx t sin(tx π π dx π. x sin(πt, provided t. t Thus, f (t when t sin(πt, so sin(πt, which happens only when t Z. Now, the support of f is the closure of the set of all t such that f (t, is the closure of the complement of Z in, which is all of, which is NOT compact. b. True. One method (probably not the best method: Proof by induction. f (n (t E E ( n x n cos(tx dx if n is even ( n x n sin(tx dx if n is odd Using the Mean Value Theorem and the Lebesgue Dominated Convergence Thoerem (as in April 3 #6b and September #7b, first show true for base cases: n and n, then assuming true for the n th case, show true for the (n st and (n nd cases. 8

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