Real Analysis Problems

Real Analysis Problems Cristian E. Gutiérrez September 14, 29 1

1 CONTINUITY 1 Continuity Problem 1.1 Let r n be the sequence of rational numbers and Prove that f(x) = 1. f is continuous on the irrationals. 2. f is discontinuous on the rationals. 3. Calculate f(x) dx. {n:r n<x} 1 2 n. Hint: for (3) set A(x) = {n : r n < x} so f(x) = n=1 χ A(x)(n) 1 ; use Fubini. 2n Problem 1.2 Let f n (x) = sin x + 4 n 2 π 2 on [, + ). Prove that 1. f n is equicontinuous on [, + ). 2. f n is uniformly bounded. 3. f n pointwise on [, + ). 4. There is no subsequence of f n that converges to uniformly. 5. Compare with Arzelà-Ascoli. Problem 1.3 Prove that the class of Lipschitz functions f in [a, b] with Lipschitz constant K and f(a) = is a compact set in C([a, b]). Problem 1.4 Let B be the unit ball in C([a, b]). Define for f C([a, b]) T f(x) = b a ( ) x 2 + e x2 +y f(y) dy. Prove that T (B) is relatively compact in C[a, b]. 2

1 CONTINUITY Problem 1.5 Let g n : R R be the continuous function that is zero outside the interval [, 2/n], g n (1/n) = 1, and g n is linear on (, 1/n) and (1/n, 2/n). Prove that g n pointwise in R but the convergence is not uniform on any interval containing. Let r k be the rational numbers and define f n (t) = 2 k g n (t r k ). k=1 Prove that 1. f n is continuous on R. 2. f n pointwise in R. 3. f n does not converge uniformly on any interval of R. Problem 1.6 Let f C(R) and f n (x) = 1 n ( x + k ). Show that f n con- n verges uniformly on every finite interval. n 1 f k= Problem 1.7 Consider the following statements: (a) f is a continuous function a.e. on [, 1] (b) there exists g continuous on [, 1] such that g = f a.e. Show that (a) does not imply (b), and (b) does not imply (a). Problem 1.8 Show that if f and g are absolutely continuous functions in [a, b] and f (x) = g (x) a.e., then f(x) g(x) = constant, for each x [a, b]. Problem 1.9 If f is absolutely continuous in [, 1], then f 2 is absolutely continuous in [, 1]. Problem 1.1 Let < α 1. A function f C α ([, 1]), f is Hölder continuous of order α, if there exists K such that f(x) f(y) K x y α for all x, y [, 1]. Prove that 1. C α [, 1] C β ([, 1]) for < β α 1. 2. If f is Hölder continuous of order one, that is, f is Lipschitz, then f is absolutely continuous on [, 1]. 3

1 CONTINUITY 3. Let a >. Define g(x) = x a sin(x a ) for x and f() =. 4. Prove that g / BV [, 1]. Prove that g C α ([, 1]) with α = a/(1 + a), and g / C β ([, 1]) for β > a/(1 + a). Problem 1.11 Consider the space BV [a, b] of bounded variation functions in [a, b] with the norm k f BV = sup f(α i ) f(α i 1 ) + f(a), Γ i=1 where the supremum is taken over all partitions Γ = {α,..., α k } of [a, b]. Prove that 1. (BV [a, b], ) is a Banach space; 2. the set of absolutely continuous functions in [a, b] is a closed subspace of BV [a, b]. Problem 1.12 Show that the series ( 1) k k + x converges for each x R and the sum is a Lipschitz function. k=1 Problem 1.13 Let f : [a, b] R be continuous and {x k } [a, b] a Cauchy sequence. Prove that f(x k ) is a Cauchy sequence. Problem 1.14 Let f : R R be such that f(x) f(y) M x y α for some M >, α > 1 and for all x, y R. Prove that f is constant. Problem 1.15 Let f(x) = x 2 sin(1/x 2 ) for x [ 1, 1], x, and f() =. Show that f is differentiable on [ 1, 1] but f is unbounded on [ 1, 1]. Problem 1.16 Suppose f n f uniformly in Ω open, and {x n } Ω with x n x Ω. Prove that f n (x n ) f(x). Problem 1.17 If x n is a sequence such that x n+1 x n 1/2 n, then {x n } is a Cauchy sequence. 4

1 CONTINUITY Problem 1.18 Let a < b, c < d. Define ax sin 2 1 + bx x cos2 1, for x > x f(x) =, for x = cx sin 2 1 + dx x cos2 1, for x >. x Calculate D f, D + f, D f, D + f at x =. Problem 1.19 Let f be a continuous function in [-1,2]. Given x 1, and n 1 define the sequence of functions f n (x) = n 2 x+ 1 n x 1 n f(t) dt. Show that f n is continuous in [,1] and f n converges uniformly to f in [,1]. Problem 1.2 Let f n : R R be a uniformly bounded sequence of functions. Show that for each countable subset S R there exists a subsequence of f n which converges in S. Hint: select the subsequence by using a diagonal process Problem 1.21 Let f be a continuous function in [, 1] such that f is absolutely continuous in [, ɛ] for every ɛ, < ɛ < 1. Show that f is absolutely continuous in [, 1]. Problem 1.22 Let f n be absolutely continuous functions in [a, b], f n (a) =. Suppose f n is a Cauchy sequence in L 1 ([a, b]). Show that there exists f absolutely continuous in [a, b] such that f n f uniformly in [a, b]. Problem 1.23 Let f n (x) = cos(n x) on R. Prove that there is no subsequence f nk converging uniformly in R. Problem 1.24 Let f : R n R be a bounded function. If E R n, then the oscillation of f over E is defined by osc E f = sup f inf f, E E and for x R n define ω f (x) = lim δ osc Bδ (x)f. Prove 1. If E F, then osc E f osc F f. 2. Prove that ω f is well defined and ω f (x) = inf δ> osc Bδ (x)f. 3. For each α > the set {x : ω f (x) < α} is open. 4. The set of points of discontinuity of f is an F σ. 5

2 SEMICONTINUOUS FUNCTIONS 2 Semicontinuous functions Definition 2.1 Let Ω R n be open. The function f : Ω R is lower (upper) semicontinuous if f(z) lim inf x z f(x) = lim δ inf x z <δ f(x) (f(z) lim sup x z f(x) = lim δ sup x z <δ f(x)). Problem 2.2 Prove that the following statements are equivalent: 1. f is lower semicontinuous; 2. f 1 (c, + ) is open for each c R; 3. f 1 (, c] is closed for each c R; 4. If x k z, then f(z) sup k f(x k ); 5. For each z Ω and for each ɛ > there exists a neighborhood V of z such that f(x) f(z) ɛ for each x V. Problem 2.3 Let g : Ω R and define Prove that f is lower semicontinuous. f(x) = lim inf z x g(z). Problem 2.4 Prove that 1. if E R n, then χ E is lower semicontinuous if and only if E is open. 2. the function if x is irrational f(x) = 1 if x = m/n with n N, m Z, and m/n irreducible n is upper semicontinuous. Problem 2.5 Let {f α } α J be a family of lower semicontinuous function in Ω. Prove that f(x) = sup α J f α (x) is lower semicontinuous. Problem 2.6 Prove that the function defined in Problem 1.1 is lower semicontinuous in R. 6

3 MEASURE 3 Measure Problem 3.1 Let E j be a sequence of sets in R n and E = lim sup j E j = k=1 j=k E j. Show that lim sup j χ Ej (x) = χ E (x). Problem 3.2 Consider the set of rational numbers Q = {q j } j=1. Prove that R \ (q j 1j, q 2 j + 1j ). 2 j=1 Problem 3.3 Construct a countable family of closed intervals contained in [, 1] such that the union covers [, 1] but there is no finite subcovering. Problem 3.4 Let E be a set in R n. Show that there exists a sequence G i of open sets, G 1 G 2 E such that G i = E e. i=1 Problem 3.5 If E [, 1], E = and f(x) = x 3, then f(e) =. Problem 3.6 True or false, justify your answer. 1. The class of Lebesgue measurable sets has cardinality 2 c. 2. Every perfect set has positive measure. 3. Every bounded function is measurable. Problem 3.7 Let µ be an exterior measure and let A n be a sequence of sets such that µ (A n ) < +. n=1 Show that µ (lim A n ) =. ( lim A n = n=1 j=n A j) Problem 3.8 Let f : [, 1] [, 1] be the Cantor function, and let µ be the Lebesgue-Stieltjes measure generated by f, i.e. µ((a, b]) = f(b) f(a), for (a, b] [, 1]. Show that µ is singular with respect to Lebesgue measure. 7

3 MEASURE Problem 3.9 Let Ω be an uncountable set. The class of sets defined by F = {A Ω : A is countable or A c is countable} is a σ-algebra. Define the measures: µ(a) = + if A is infinite, µ(a) = #(A) if A is finite; ν(a) = if A is countable and ν(a) = 1 if A is uncountable. Show that ν µ but an integral representation of the form ν(a) = f dµ is not valid. Why A this does not contradict the Radon-Nikodym theorem? Problem 3.1 Show that given δ, < δ < 1, there exists a set E δ [, 1] which is perfect, nowhere dense and E δ = 1 δ. Hint: the construction is similar to the construction of the Cantor set, except that at the k th stage each interval removed has length δ3 k. Problem 3.11 Let {I 1,, I N } be a finite family of intervals in R such that Q [, 1] N j=1i j. Prove that N j=1 I j 1. Is this true if the family of intervals is infinite? Problem 3.12 Let {E j } j=1 be a sequence of measurable sets in R n such that E j E i = for j i. Prove that E j = j=1 E j. j=1 Problem 3.13 Let A, B R n such that A is measurable. If A B = then A B e = A + B e. Hint: given ɛ > there exists F closed such that F A and A \ F < ɛ. Problem 3.14 Let {E j } j=1 be a sequence of measurable sets in R n. Prove that 1. lim inf j E j and lim sup j E j are measurable. 2. lim inf j E j lim inf j E j. 3. If j k E j < for some k then lim sup j E j lim sup j E j. Problem 3.15 Let A R measurable with A <. Show that the function f(x) = (, x) A is nondecreasing, bounded and uniformly continuous in R. 8

3 MEASURE Problem 3.16 Show that Q is not a set of type G δ. Construct a set G of type G δ such that Q G and G =. Problem 3.17 Let E be a measurable set in R with positive measure. We say that x R is a point of positive measure with respect to E if E I > for each open interval I containing x. Let E + = {x R : x is of positive measure with respect to E}. Prove that 1. E + is perfect. 2. E \ E + =. Problem 3.18 Let E be the set of numbers in [, 1] whose binary expansion has in all the even places. Show that E =. Problem 3.19 Let f k be measurable and f k f a.e. in R n. Prove that there exists a sequence of measurable sets {E j } j=1 such that R n \ j=1e j = and f k f uniformly on each E j. Problem 3.2 Let E R n be a measurable set with positive measure, and let D R n be a countable dense set. Prove that R n \ qk D(q k + E) =. Problem 3.21 An ellipsoid in R n with center at x is a set of the form E = {x R n : A(x x ), x x 1}, where A is an n n positive definite symmetric matrix and, denotes the Euclidean inner product. Prove that ω n E = (det A), 1/2 where ω n is the volume of the unit ball. Hint: formula of change of variables. Problem 3.22 Let f : R R be a continuously differentiable function. Define Φ : R 2 R 2 by Φ(x, y) = (x + f(x + y), y f(x + y)). Prove that Φ(E) = E for each measurable set E R 2. Hint: formula of change of variables. 9

3 MEASURE Problem 3.23 Let E [, 1] such that there exists δ > satisfying E [a, b] δ (b a) for all [a, b] [, 1]. Prove that E = 1. Problem 3.24 We say that the sets A, B R n are congruent if A = z + B for some z R n. Let E R n be measurable such that < E < +. Suppose that there exists a sequence of disjoint sets {E i } i=1 such that for all i, j, E i and E j are congruent, and E = j=1e j. Prove that all the E j s are non measurable. 1

4 MEASURABLE FUNCTIONS 4 Measurable functions Let E be a measurable set and f n : E R a sequence of measurable functions. (A) f n f pointwise in E if f n (x) f(x) for all x E. (B) f n f pointwise almost everywhere in E if f n (x) f(x) for almost all x E. (C) f n f almost uniformly in E if for each ɛ > there exists a measurable set F E such that E \ F ɛ and f n f uniformly in F. (D) f n f in measure in E if for each ɛ >, lim {x E : f n (x) f(x) ɛ} =. (E) f n is a Cauchy sequence in measure in E if for each ɛ, δ >, there exists N such that {x E : f n (x) f m (x) ɛ} δ for all n, m N. f : E R is a measurable function if f 1 ((, a)) is a Lebesgue measurable set for all a R. f : E R is a Borel measurable function if f 1 ((, a)) is a Borel set for all a R. Problem 4.1 If f n f almost uniformly in E, then f n f in measure in E. Problem 4.2 If f n f in measure and f n g in measure, then f = g a.e. Problem 4.3 If f n is a Cauchy sequence in measure, then there exists a subsequence f nk that is a Cauchy sequence almost uniformly. Problem 4.4 If f n is a Cauchy sequence in measure, then there exists f measurable such that f n f in measure. Problem 4.5 Give an example of a sequence f n that converges in measure but does not converge a.e. Problem 4.6 If E < and f n f a.e. in E, then f n f in measure. Show that this is false if E =. Problem 4.7 If f n f in measure, then there exists a subsequence f nk f nk f a.e. such that 11

4 MEASURABLE FUNCTIONS Problem 4.8 If f n f in measure, then f n is a Cauchy sequence in measure. Problem 4.9 Let f, f k : E R be measurable functions. Prove that 1. If f k f in measure, then any subsequence f kj contains a subsequence f kjm f a.e. as m,. 2. Suppose E <. If any subsequence f kj contains a subsequence f kjm f a.e. as m, then f k f in measure. Hint: for (2) suppose by contradiction that f k f in measure. Then there exists ɛ > such that {x E : f k (x) f(x) ɛ }. Hence there is an increasing sequence k j such that {x E : f kj (x) f(x) ɛ } r >, for some r > and for all j. By hypothesis there is a subsequence f kjm f a.e. as m. Now use Egorov to get a contradiction. Problem 4.1 Let f : E R be a measurable function. Prove that if B R is a Borel set then f 1 (B) is measurable. Hint: consider A = {A R : f 1 (A) is measurable} and show that A is a σ-algebra that contains the open sets of R. Problem 4.11 If f : E R is a measurable function and g : R R is Borel measurable, then g f is measurable. 12

5 INTEGRATION 5 Integration Problem 5.1 Let f : [a, b] R be integrable and nonnegative. Prove that ( b 2 ( b 2 ( b 2 f(x) cos x dx) + f(x) sin x dx) f(x) dx). a a Hint: write f(x) = f(x) f(x) cos x and use Schwartz on the left hand side. Problem 5.2 Show that lim Hint: see Makarov, prob. 6.21, p.17. x 2 1 + + x 2 n dx = 2/3. [,1] x n 1 + + x n Problem 5.3 Let f 1 : [, M] R be a bounded function. Define f n+1 (x) = x f n (t) dt, n = 1, 2,. Prove that the series n=2 f n(x) is uniformly convergent on [, M] and the sum is a continuous function on [, M]. Problem 5.4 Let f : R n R be integrable. Prove that 1. {x R n : f(x) = } =. 2. For each ɛ >, {x R n : f(x) ɛ} <. 3. For each ɛ > there exists a compact set K such that f(x) dx < ɛ. R n \K 4. For each ɛ > there exist M R and A R n measurable such that f(x) M in A and f(x) dx < ɛ. R n \A 5. For each ɛ > there exist δ > such that if A R n is measurable and such that A < δ then f(x) dx < ɛ. A Problem 5.5 Let f : E R be measurable with E <. Then f is integrable if and only if m {x Rn : f(x) m} <. Hint: use Abel s summation by parts formula, a where A k = k j=1 a j. N a k b k = A N b N + k=1 N 1 k=1 A k (b k b k 1 ), 13

5 INTEGRATION Problem 5.6 Let f, f k : R n R be integrable functions such that f k f a.e. Then R n f k (x) f(x) dx if and only if R n f k (x) dx R n f(x) dx. Problem 5.7 Let P [, 1] be a perfect nowhere dense set, i.e., ( P ) =, with positive measure. Show that χ P (x) is not Riemann integrable on [, 1] but it is Lebesgue integrable on [, 1]. Problem 5.8 Let f : R n R be integrable and ɛ >. Prove that 1. There exists g : R n R simple and integrable such that R n f(x) g(x) dx < ɛ. 2. There exists h : R n R continuous with compact support such that R n f(x) h(x) dx < ɛ. Problem 5.9 (Fatou in measure) If f k and f k f in measure in E then f(x) dx lim inf f k (x) dx. k E Hint: let A k = f E k(x) dx and A = lim inf k A k ; there exists a subsequence A km A as m. Since f km f in measure, there exists a subsequence f kmj f a.e. as j. Now apply Fatou. E Problem 5.1 (Monotone convergence in measure) If f k, f k f k+1, and f k f in measure in E then f(x) dx = lim f k (x) dx. k E E Problem 5.11 (Lebesgue in measure) If f k g a.e. with g integrable in E, and f k f in measure in E then f(x) dx = lim f k (x) dx. k E E Problem 5.12 Let f, f k : R n R be measurable functions such that f k f a.e. and there exists g integrable such that f k (x) g(x) a.e. for all k. Prove that f k f almost uniformly. 14

5 INTEGRATION Problem 5.13 Let E < and let X denote the class of measurable functions in E. If f, g X define f(x) g(x) d(f, g) = 1 + f(x) g(x) dx. E Show that (X, d) is a metric space and f k f in (X, d) if and only if f k f in measure in E. Problem 5.14 Let C [, 1] be the Cantor set and f : [, 1] R given by {, if x C f(x) = p, if x I [, 1] P, I interval such that l(i) = 3 p Show that f is measurable and f(x) dx = 3. Problem 5.15 Prove that 1. x sin x 1 + (nx) dx = o( 1 ), if a > 1. a n dx 2. lim (1 + x n )n x = 1 1/n Problem 5.16 Calculate lim 1. f n (x) = 2. f n (x) = log(x + n) e x cos x n nx log x 1 + n 2 x 2 3. f n (x) = n x 1 + n 2 x 2 4. f n (x) = nx 1 + n 2 x 2 5. f n (x) = n3/2 x 1 + n 2 x 2 f n (x)dx where: 6. f n (x) = np x r log x, r >, p < min{2, 1 + r} 1 + n 2 x2 15

5 INTEGRATION 7. f n (x) = 8. f n (x) = 1 + nx (1 + x) n ( 1 + x ) n x sin( ) in the interval (, ). n n Problem 5.17 Prove the following identities. 1. lim n 1/β x 1/β (1 x) n dx x = β e uβ du if β >. n ( 2. lim 1 x n x n) (α 1) dx = 3. lim α 4. 5. 6. x 1/3 1 x log e x x (α 1) dx if α >. n e nx 2 dx = ( lim n e nx 2 ) dx if α >. α ( ) 1 dx = 9 x sin t e t x dt = x n 1 n 2 + 1 n=1 log x sin x dx = n=1 n=1 1 (3n + 4) 2. for 1 < x < 1. ( 1) n (2n)(2n)!. Problem 5.18 If f is Riemann integrable on [a, b] and f(x) = for x [a, b] Q, then b f(x) dx =. a Problem 5.19 Show that f(x) = { 1 if x = 1/n otherwise is Riemann integrable on [, 1]. Find the value of f(x) dx. Problem 5.2 Let f : [, 1] R be Riemann integrable. Prove that 1 lim n n f(k/n) = k=1 f(x) dx. 16

5 INTEGRATION Problem 5.21 Let f : [, 1] R be continuous, and f(x) for each x (, 1). Suppose that f(x) 2 = 2 for each x [, 1]. Prove that f(x) x. x f(t) dt Problem 5.22 Let f be a measurable function on [, 1] and let A = {x [, 1] : f(x) Z}. Prove that A is measurable and as n. [cos (π f(x))] 2n dx A, Problem 5.23 Find all the values of p and q such that the integral 1 dxdy x 2p + y2q converges. x 2 +y 2 1 Problem 5.24 Let f 1,, f k be continuous real valued functions on the interval [a, b]. Show that the set {f 1,, f k } is linearly dependent on [a, b] if and only if the k k matrix with entries has determinant zero. f i, f j = b a f i (x) f j (x) dx Problem 5.25 Let f : [, + ) R be continuously differentiable with compact support in [, + ); and < a < b <. Prove that f(b x) f(a x) x dx = f() ln(b/a). Problem 5.26 Find all the values of p such that the integral converges. π/2 x yp e sin x dx dy 17

5 INTEGRATION Problem 5.27 Recall that the convolution of two integrable functions f and g is defined by f g(x) = f(x y) g(y) dy. R n Let f such that R n f(x) dx = A < 1. Define the sequence f k = f f where the convolution is performed k times. Prove that all the f k s are integrable and f k in L 1 (R n ). Problem 5.28 Let f L 1 (, + ) be nonnegative. Prove that 1 n n x f(x) dx, as n. Problem 5.29 Let f in R, set g(x) = n= f(x + n). Show that if g L(R) then f = a.e. Problem 5.3 Given f L 2 ([, 1]) define Kf(x) = 1 x 4/3 x f(t) dt. Show that Kf 1 C f 2 where C is a constant independent of f. Hint: use Cauchy-Schwartz. Problem 5.31 Let A be a measurable subset of [, 2π], show that cos nx dx, as n. A Problem 5.32 Let f be a non-negative measurable function in [,1]. Show that the limit exists if and only if lim f(x) n dx {x [, 1] : f(x) > 1} =. sin x Problem 5.33 Prove that lim dx =. 1 + n x2 18

5 INTEGRATION Problem 5.34 (Variant of Fatou) Let g n L 1 (E) and g n g in L 1 (E). Suppose f n are measurable functions on E and f n g n a.e. for each n. Prove that lim sup f n lim sup f n. E E Problem 5.35 (Variant of Lebesgue) Let g n L 1 (E) and g n g in L 1 (E). Suppose f n are measurable functions on E such that f n f a.e. or f n f in measure, and f n g n a.e. for each n. Prove that f n f, as n, and consequently lim f n = E E E f. Problem 5.36 Let E <. Prove that f n f in measure if and only if { f n f 1} as n. E Problem 5.37 Let f L 1 (, 1) and suppose that lim x 1 f(x) = A <. Show that lim n x n f(x) dx = A. 19

6 SOLUTIONS CONTINUITY 6 Solutions Continuity 1.1 Let y < x with y irrational. Write f(x) f(y) = {n:y r n<x} 1/2 n. Given ɛ > there exists N such that n N 1/2n < ɛ. Let δ = min{ y r k : 1 k N}. If x y < δ and n is such that y r n < x, then n > N. Because if n N and y r n < x, then r n y < x y < δ which is impossible by definition of δ. If y Q, then y = r m for some m and so f(x) f(y) = {n:y r n<x} 1/2n 1/2 m for any x > y. We have 1.2 Write f(x) dx = = n=1 χ A(x) (n) 1 2 dx = 1 n n=1 1 2 n χ (rn,+ )(x) dx = 2 n n=1 n=1 χ A(x) (n) dx 1 2 n [, 1] (r n, + ). sin x + 4 n 2 π 2 sin y + 4 n 2 π 2 = cos 1 ξ + 4 n 2 π 2 2 (x y), ξ + 4 n 2 π2 with some ξ between x + 4 n 2 π 2 and y + 4 n 2 π 2. To show the pointwise convergence pick y =. For (3), f n (z) = 1 when z + 4 n 2 π 2 = (k + 1/2)π. 1.3 The class is bounded in C[a, b] and equicontinuous. 1.4 If f(x) 1 on [a, b], then T (f) is bounded. Show that T (f) are equicontinuous writing T (f)(x) T (f)(z) = b a ( x 2 + z 2 )f(y) dy + b a (e x2 e z2 ) e y f(y) dy. 1.5 The continuity follows from the uniform convergence of the series. Given ɛ > there exists N such that k=n+1 2 k < ɛ. Write f n (t) = N 2 k g n (t r k ) + 2 k g n (t r k ) = A n (t) + B n (t). k=1 k=n+1 We have supp g n ( r k ) [t 1/2n, t]. Let E = {r 1, r 2,, r N }. Suppose first that t / E and let δ = dist(t, E). If 1/2n < δ, then r k [t 1/2n, t] for all 2

6 SOLUTIONS CONTINUITY 1 k N and so A n (t) = for n > 1/2δ. If on the other hand t E, then lim sup A n (t) N k=1 2 k lim sup g n (t r k ) =. Let (a, b) be an interval. There exists r m (a, b) and so r m + 1/n (a, b) for all n sufficiently large. Therefore sup x (a,b) f n (x) f n (r m + 1/n) 2 m g n (1/n) = 2 m for all n sufficiently large. 1.6 f n (x) is a Riemann sum and converges to x+1 f(t) dt as n. Let (a, b) x be a finite interval. Write f n (x) x+1 x n 1 f(t) dt = k= x+(k+1)/n x+k/n (f(x + k/n) f(t)) dt. Since f is uniformly continuous on [a, b + 1] we have that f(x + k/n) f(t) < ɛ for all x [a, b], x + k/n t x + (k + 1)/n, k n 1, and for all n sufficiently large. Therefore x+1 sup f n(x) f(t) dt < ɛ for all n sufficiently large. x [a,b] 1.7 (a) (b) f(x) = 1/x; (b) (a) g(x) = 1, f(x) = 1 for x Q and f(x) = for x R \ Q. x 1.8 By the absolute continuity f(x) f(a) = x f (t) dt and g(x) g(a) = a a g (t) dt, for all a x b, so f(x) g(x) = f(a) g(a) for all a x b. x 1.9 f(x) 2 f(y) 2 = (f(x) + f(y))(f(x) f(y)) and f is bounded. 1.1 Let h >. We have g(x + h) g(x) g(x + h) g() + g(x) g() (x+h) a +x a 2(x+h) a. Also g (x) 2a/x for x. By the mean value theorem, g(x + h) g(x) C a min{(x + h) a, h/x}. Next either x a+1 h or x a+1 < h, and then conclude g C α ([, 1]). To show the negative statement notice that g(x) = at x = ξ = (nπ) 1/a with n = 1, 2,. Let δ > small, and ξ = ((n + δ)π) 1/a. Suppose by contradiction that g(ξ) g(ξ ) K ξ ξ β. We have g (x) = a x xa sin x a cos x a a x ( cos x a x a sin x a ) a x ( cos x a x a ). From the mean value theorem g(ξ) g(ξ ) = g( ξ) ξ ξ, with ((n+1)δ) 1/a ξ (nδ) 1/a. Choosing δ > small, we get that g ( ξ) C δ n 1/a for all n large. Also notice that ξ ξ n (1+a)/a for n large. Combining the orders of magnitude we obtain a contradiction unless β a/(1 + a). 1.11 Let {f n } be a Cauchy sequence. Given x (a, b) take the partition {a, x, b}. We have f n (x) f m (x) (f n f m )(x) (f n f m )(a)) + (f n f m )(a) f n f m BV, as n, m. Thus, for each x [a, b], {f n (x)} is a Cauchy sequence. Let f(x) = lim f n (x). Let Γ = {x,, x N } be a partition of [a, b]. Given ɛ > there exists 21

6 SOLUTIONS CONTINUITY n such that N (f n f m )(x i ) (f n f m )(x i 1 )) < ɛ i=1 for each n, m n. Letting n yields N (f f m )(x i ) (f f m )(x i 1 )) < ɛ i=1 for each m n. Hence f f m BV < ɛ for m n. Let f n be absolutely continuous such that f n f BV. Given ɛ > there exists N such that k i=1 (f n f)(x i ) (f n f)(x i 1 ) < ɛ for each n N and any partition Γ = {x,, x k }. Since f N is absolutely continuous, there exists δ > such that if (a 1, b 1 ),, (a r, b r ) is any collection of disjoint intervals of [a, b] with r i=1 (b i a i ) < δ, then r i=1 f N(b i ) f N (a i ) < ɛ. If Γ = {a, a 1, b 1, a 2, b 2,, a r, b r, b}, then r k f(b i ) f(a i ) (f f N )(b i ) (f f N )(a i ) + i=1 i=1 f f N BV + ɛ < 2 ɛ. k f N (b i ) f N (a i ) i=1 22

7 SOLUTIONS SEMICONTINUITY 7 Solutions Semicontinuity 2.6 Suppose y Q, y = r m. Then f(x) 1 + f(y) for all x > y, so 2m for any δ >. We now estimate inf f(x) 1 + f(y) > f(y) y<x<y+δ 2m inf f(x). If x y, then y δ<x y f(x) = f(y) {n:x r n<y} 1/2 n. Suppose y δ < x y, then {n : x r n < y} {n : y δ < r n < y} and so f(x) f(y) 1/2 n. {n:y δ<r n<y} We claim that for each N 1 there exists δ > such that if r j (r m δ, r m ) then j N. Suppose by contradiction that this is false. Then there exists N 1 such that for each δ = 1/k there exists a j(k) such that r j(k) (r m 1/k, r m ) and j(k) < N. Hence j(k) takes only a finite number of values between 1 and N, and therefore there exists a subsequence k i such that k i as i with j(k i ) = s and r s (r m 1/k i, r m ) for all i. Letting i we get r m r s < r m, a contradiction. Now given ɛ >, there exists N 1 such that n N 1/2n < ɛ, and by the claim there exists δ > such that {n : y δ < r n < y} {n : n N}. Consequently, 1/2 n < ɛ, {n:y δ <r n<y} 1/2 n n N and so f(x) f(y) ɛ for y δ < x y and < δ δ. Thus, for all < δ δ, and so lim inf f(x) f(y) ɛ y δ<x y inf δ y δ<x y f(x) f(y) ɛ for any ɛ >. 23