rudin实分析与复分析答案

��������� � chapter one 3. Prove that if f is a real function on a measurable space X such that {x : f(x) ≥ r} is measurable for every rational r, then f is measurable. [proof]:For each real number α, there exists an descending sequence {rn} of rational numbers such that lim n→∞ rn = α. Moreover,we have (α, +∞) = ∞⋃ n=1 [rn, +∞). Hence, f−1((α, +∞)) = ∞⋃ n=1 f−1([rn, +∞)). Since sets f−1([rn, +∞)) are measurable for each n, the set f −1((α, +∞)) is also measurable. Then f is measurable. 4. Let {an} and {bn} be sequences in [−∞, +∞], and prove the following assertions: (a) lim sup n→∞ (−an) = − lim inf n→∞ an. (b) lim sup n→∞ (an + bn) ≤ lim sup n→∞ an + lim sup n→∞ bn. provided none of the sums is of the form ∞−∞. (c) If an ≤ bn for all n, then lim inf n→∞ an ≤ lim inf n→∞ bn. Show by an example that strict inequality can hold in (b). [proof]: (a) Since sup k≥n (−ak) = − inf k≥n ak, n = 1, 2, · · · . Therefore, let n →∞, it obtains lim n→∞ sup k≥n (−ak) = − lim n→∞ inf k≥n ak. 1 By the definations of the upper and the lower limits, that is lim sup n→∞ (−an) = − lim inf n→∞ an. (b) Since sup k≥n (ak + bk) ≤ sup k≥n ak + sup k≥n bk, n = 1, 2, · · · . Hence lim n→∞ sup k≥n (ak + bk) ≤ lim n→∞ [sup k≥n ak + sup k≥n bk] = lim n→∞ sup k≥n ak + lim n→∞ sup k≥n bk. By the definations of the upper and the lower limits, that is lim sup n→∞ (an + bn) ≤ lim sup n→∞ an + lim sup n→∞ bn. example: we define an = (−1) n, bn = (−1) n+1, n = 1, 2, · · · . Then we have an + bn = 0, n = 1, 2, · · · . But lim sup n→∞ an = lim sup n→∞ bn = 1. (c)Because an ≤ bn for all n, then we have inf k≥n (ak) ≤ inf k≥n bk, n = 1, 2, · · · . By the definations of the lower limits, it follows lim inf n→∞ an ≤ lim inf n→∞ bn. 5. (a)Supose f : X → [−∞, +∞] and g : X → [−∞, +∞] are measurable. Prove that the sets {x : f(x) < g(x)}, {f(x) = g(x)} are measurable. (b)Prove that the set of points at which a sequence of measurable real- valued functions converges (to a finite limit) is measurable. 2 [proof]: (a)Since for each rational r, the set {x : f(x) < r < g(x)} = {x : f(x) < r} ∩ {x : r < g(x)} is measurable. And {x : f(x) < g(x)} = ⋃ r∈Q {x : f(x) < r < g(x)} Therefore the set {x : f(x) < g(x)} is measurable. Also beacause |f − g| is measurable function and the set {x : f(x) = g(x)} = ∞⋂ n=1 {x : |f(x)− g(x)| < 1 n } is also measurable. (b) Let {fn(x)} be the sequence of measurable real-valued functions, A be the set of points at which the sequence of measurable real-valued functions {fn(x)} converges (to a finite limit). Hence A = ∞⋂ r=1 ∞⋃ k=1 ∞⋂ n,m≥k {x : |fn(x)− fm(x)| < 1 r }. Therefore, A is measurable. 7. Suppose that fn : X → [0, +∞] is measurable for n = 1, 2, 3, · · · , f1 ≥ f2 ≥ f3 ≥ · · · ≥ 0, fn(x) → f(x) as n → ∞,for every x ∈ Xand f1 ∈ L 1(µ). Prove that then lim n→∞ ∫ X fndµ = ∫ X fdµ. and show that this conclusion does not follow if the condition ”f1 ∈ L 1(µ)”is omitted. [proof]: Define gn = f1−fn, n = 1, 2, · · ·, then gn is increasing measurable on X for n = 1, 2, 3, · · ·. Moreover, gn(x) → f1(x)− f(x) as n →∞, for every x ∈ X. By the Lebesgue’s monotone convergence theorem, we have lim n→∞ ∫ X gndµ = ∫ X (f1 − f)dµ. Since f1 ∈ L 1(µ), it shows lim n→∞ ∫ X fndµ = ∫ X fdµ. 3 [counterexample]: Let X = (−∞, +∞), and we define fn = χEn, En = [n, +∞), n = 1, 2, · · · . 8. Suppose that E is measurable subset of measure space (X, µ) with µ(E) > 0; µ(X − E) > 0, then we can prove that the strict inequality in the Fatou’s lemma can hold. 10. Suppose that µ(X) < ∞, {fn} is a sequence of bounded complex measurable functions on X ,and fn → f uniformly on X. Proved that lim n→∞ ∫ X fndµ = ∫ X fdµ, and show that the hypothesis ”µ(X) < ∞”cannot be omitted. [proof]: There exist Mn > 0, n = 1, 2, · · · such that |fn(x)| ≤ Mn for every x ∈ X, n = 1, 2, · · ·. Since fn → f uniformly on X, there exists M > 0 such that |fn(x)| ≤ M, |f(x)| ≤ M on X, n = 1, 2, · · · . By the Lebesgue’s dominated convergence theorem, we have lim n→∞ ∫ X fndµ = ∫ X fdµ. [counterexample]: Let X = (−∞, +∞), and we define fn = 1 n on X, n = 1, 2, · · · . 12. Suppose f ∈ L1(µ). Proved that to each ε > 0 there exists a δ > 0 such that ∫ E |f |dµ < ε whenever µ(E) < δ. [proof]: Since ∫ X |f |dµ = sup ∫ X sdµ, the supremum being taken over all simple measurable functions s such that 0 ≤ s ≤ |f |. Hence, for each ε > 0, there exists a simple measurable function s with 0 ≤ s ≤ |f | such that∫ X |f |dµ ≤ ∫ X sdµ + ε 2 . Suppose that M > 0, satisfying 0 ≤ s(x) ≤ M on X, and let δ = ε 2M , we have∫ E |f |dµ ≤ ∫ E sdµ + ε 2 ≤ Mm(E) + ε 2 < ε when m(E) < δ. 4 ��������� � chapter two 1. Let {fn} be a sequence of real nonnegative functions on R 1, and con- sider the following four statements: (a)If f1 and f2 are upper semicontinuous, then f1 + f2 is upper semicon- tinuous. (b)If f1 and f2 are lower semicontinuous, then f1 + f2 is lower semicon- tinuous. (c)If each {fn} is upper semicontinuous, then ∞∑ 1 fn is upper semicontin- uous. (d)If each {fn} is lower semicontinuous, then ∞∑ 1 fn is lower semicontinu- ous. Show that tree of these are true and that one is false. What happens if the word ”nonnegative” is omitted? Is the truth of the statements affected if R 1 is replaced by a general topological space? [proof]: (a) Because for every real number α, {x : f1(x) + f2(x) < α} = ⋃ r∈Q [{x : f1(x) < r} ∩ {x : f2(x) < α− r}] and f1 and f2 are upper semicontinuous, therefore f1 + f2 is upper semicon- tinuous. (b) Because for every real number α, {x : f1(x) + f2(x) > α} = ⋃ r∈Q [{x : f1(x) > r} ∩ {x : f2(x) > α− r}] and f1 and f2 are upper semicontinuous, therefore f1 + f2 is upper semicon- tinuous. (c) This conclusion is false. counterexample: for each n ∈ N, define fn = { 1; x = rn 0; others 1 which {rn} are the all rationals on R 1. Then {fn} is a sequence of real non- negative upper semicontinuous functions on R1, moreover we have f(x) = ∞∑ 1 fn(x) = χQ. But f(x) is not upper semicontinuous. (d) According to the conclusion of (b), it is easy to obtain. If the word ”nonnegative” is omitted, (a) and (b) are still true but (c) and (d) is not. The truth of the statements is not affected if R1 is replaced by a general topological space. 2. Let f be an arbitrary complex function on R1, and define φ(x, δ) = sup{|f(s)− f(t)| : s, t ∈ (x− δ, x + δ)} φ(x) = inf{φ(x, δ) : δ > 0} Prove that φ is upper semicontinuous, that f is continuous at a point x if and only if φ(x) = 0, and hence that the set of points of continuity of an arbitrary complex function is a Gδ. [proof]: It is obvious that φ(x) = inf{φ(x, 1 k ) : k ∈ N} ≥ 0 and φ(x, 1 k ) ≥ φ(x, 1 k + 1 ), k ∈ N For any α > 0, we will prove that the set {x : φ(x) < α} is open. Suppose that φ(x) < α, then there exists a N > 0 such that φ(x, 1 k ) < α, k > N. Take y ∈ U(x, 1 2k ), since U(y, 1 2k ) ⊂ U(x, 1 k ) and then φ(y) ≤ φ(y, 1 2k ) ≤ φ(x, 1 k ) < α So we have y ∈ {x : φ(x) < α}. Therefore, U(x, 1 2k ) ⊂ {x : φ(x) < α} 2 Thus, it show that {x : φ(x) < α} is open. Then it follows that φ is upper semicontinuous. In the following, we first suppose that φ(x) = 0. for any � > 0, it exists δ > 0 such that φ(x, δ) < � And so for any y ∈ (x− δ, x + δ) we have |f(x)− f(y)| ≤ φ(x, δ) < � That is f is continuous at x. Next, provided that f is continuous at x. Then for any � > 0 it exists k ∈ N satifying |f(x)− f(y)| ≤ � 2 , ∀y ∈ (x− 1 k , x + 1 k ) and so φ(x, 1 k ) ≤ �. It shows φ(x) = inf{φ(x, 1 k ) : k ∈ N} = 0 Finally, since {x : φ(x) = 0} = ∞⋂ n=1 {x : φ(x) < 1 n } then it shows that the set of points of continuity of an arbitrary complex function is a Gδ. 3. Let X be a metric space, with metric ρ. For any nonempty E ⊂ X, define ρE(x) = inf{ρ(x, y) : y ∈ E} Show that ρE is a uniformly continuous function on X. If A and B are disjoint nonempty closed subsets of X, examine the relevance of the function f(x) = ρA(x) ρA(x) + ρB(x) to Urysohn’s lemma. [Proof]: Since for any x1, x2 ∈ X and y ∈ E, we have ρ(x1, y) ≤ ρ(x1, x2) + ρ(x2, y) Hence ρE(x1) ≤ ρ(x1, x2) + ρE(x2) 3 and so |ρE(x1)− ρE(x2)| ≤ ρ(x1, x2) Therefore, it shows that ρE is a uniformly continuous function on X. In the following, suppose that K ⊂ V ⊂ X, K is compact subset, and V is open subset. Define f(x) = ρV c(x) ρV c(x) + ρK(x) then 0 ≤ f ≤ 1 and f is continuous on X. Moreover, K ≺ f ≺ V . 4. Examine the proof of the Riesz theorem and prove the following two statements: (a) If E1 ⊂ V1 and E2 ⊂ V2, where V1 and V2 are disjoint open sets, then µ(E1 ∪ E2) = µ(E1) + µ(E2), even if E1 and E2 are not in M. (b) If E ∈ MF , then E = N ∪ K1 ∪ K2 ∪ · · ·, where {Ki} is a disjoint countable collection of compact sets and µ(N) = 0. [proof]: (a) By the proof of the Riesz theorem, there exists a Gδ set G such that E1 ∪ E2 ⊂ G and µ(E1 ∪ E2) = µ(G). Set G1 = V1 ∩ G, G2 = V2 ∩ G, then E1 ⊂ G1, E2 ⊂ G2 and G1 ∩ G2 = ∅. Therefore µ(E1)+µ(E2) ≤ µ(G1)+µ(G2) = µ(G1∪G2) = µ(G∩(V1∪V2)) ≤ µ(G) = µ(E1∪E2). And since µ(E1 ∪ E2) ≤ µ(E1) + µ(E2), It shows µ(E1 ∪ E2) = µ(E1) + µ(E2). (b) Since E ∈ MF , there exist a sequence of disjoint compact subsets {Kn} such that Kn ⊂ E − n−1⋃ i=0 Ki, m(E − n−1⋃ i=0 Ki) < m(Kn) + 1 n , n = 1, 2, · · · , which we order K0 = ∅. 4 Set K = ∞⋃ n=1 Kn. Since m(E) < +∞, we obtain that m(E − n⋃ i=1 Ki) < 1 n , n = 1, 2, · · · . Whence we have m(E −K) ≤ m(E − n⋃ i=1 Ki) < 1 n , n = 1, 2, · · · . Set N = E −K, then it shows that m(N) = 0 and E = K ∪N . 5. Let E be Cantor’s familiar ”middle thirds” set. Show that m(E) = 0. even though E and R1 have the same cardinality. [proof]: According to the construction of Cantor’s set, it shows that the set E is closed set and m(E) = m([0, 1])− ∞∑ n=1 2n−1 3n = 0. 11. Let µ be a regular Borel measure on a compact Hausdorff space X; assume µ(X) = 1. prove that there is a compact set K ⊂ X such that µ(K) = 1 but µ(H) < 1 for every proper compact subset H of K. [Proof]: Define the set Ω = {Kα : Kα ⊂ X, Kα is compact set with µ(kα) = 1} then X ∈ Ω. Order K = ⋂ Kα∈Ω Kα Thus K is compact subset. Assume that V ⊂ X is an open set with K ⊂ V . Since X is compact, V c is also compact. Again since V c ⊂ ⋃ Kα∈Ω Kcα then there exist finite sets Kα1 , · · · , Kαn such that V c ⊂ n⋃ k=1 Kcα k 5 Because µ(X) = 1 = µ(Kα), µ(Kcα k ) = 0, k = 1, 2, · · · , n and so µ(V c) = 0. We therefore have µ(V ) = 1. Since µ is regular, hence it shows that µ(K) = 1. By the construction of K, it follows that µ(H) < 1 for any proper compact subset of K. If U ⊂ X is open set with µ(U) = 0, it follows µ(U c) = 1. U c is compact subset since X is compact, and so K ⊂ U c by the construction of K. Hence, U ⊂ Kc. It thus shows that Kc is the largest open set in X whose measure is 0. 12. Show that every compact subset of R1 is the support of a Borel measure. [Proof]: Assume that µ is the Lebesgue measure on R1 and K is a com- pact subset of R1. Define φ(f) = ∫ K fdµ; f ∈ Cc(R 1). Then φ is a positive linear functional on Cc(R 1). By Riesz representation theorem, there exists a σ-algebra M in R1 which contains all Borel sets in R1, and there exists a unique positive measure σ on M such that φ(f) = ∫ R1 fdσ, f ∈ Cc(R 1). By theorem2.14, it shows σ(K) = inf{φ(f) : K ≺ f} = µ(K) σ(X) = sup{φ(f) : f ≺ X} ≤ µ(K) and so σ(X) = σ(K) = µ(K) < +∞. Therefore, it follows for any measurable set A ⊂ X, σ(A) = σ(A ∩K). That is, K is the support of σ. 14. Let f be a real-valued Lebesgue measurable function on Rk. Prove that there exist Borel functions g and h such that g(x) = h(x) a.e.[m] and g(x) ≤ f(x) ≤ h(x) for every x ∈ Rk. 6 [Proof]: Firstly, assume that 1 > f ≥ 0. By the proof of Theorem 2.24 in text, we have f(x) = ∞∑ n=1 tn(x), ∀x ∈ R k and 2ntn is the characteristic function of a Lebesgue measurable set Tn ⊂ R k(n ∈ N). According to the construction of Lebesgue measurable set, there exist Borel measurable sets Fn, En(n ∈ N) satisfying Fn ⊂ Tn ⊂ En; m(En − Fn) = 0; n ∈ N. Define g = ∞∑ n=1 2−nχFn; h = ∞∑ n=1 2−nχEn then g, h are Borel measurable functions on Rk and g(x) ≤ f(x) ≤ h(x). Moreover, g(x) = h(x) a.e.[m]. Secondly, it is easy that the conclusion is correct for 0 ≤ f < M , and hence it is also for bounded real-valued Lebesgue measurable function on Rk. Finally, if f be a real-valued Lebesgue measurable function on Rk and if Bn = {x : |f(x)| ≤ n}, n = 1, 2, · · · then we have χBn(x)f(x) → f(x), as n →∞. According to the conclusion above, there exist Borel functions gn and hn such that gn(x) = hn(x) a.e.[m] and gn(x) ≤ χBn(x)f(x) ≤ hn(x) for every x ∈ R k. Take g = lim sup gn; h = lim inf hn then g(x) = h(x) a.e.[m] and g(x) ≤ f(x) ≤ h(x) for every x ∈ Rk. 15. It is easy to guess the limits of∫ n 0 (1− x n )ne x 2 dx and ∫ n 0 (1 + x n )ne−2xdx as n →∞. Prove that your guesses are correct. [proof]: Define fn(x) = χ[0,n](1− x n )ne x 2 ; gn(x) = χ[0,n](1 + x n )ne−2x; n ∈ N 7 Then it is easy that |fn(x)| ≤ e −x 2 , and fn(x) → e −x 2 ; |gn(x)| ≤ e −x, and fn(x) → e −x. By the Lebesgue’s Dominated Convergence Theorem, it shows lim n→∞ ∫ n 0 (1− x n )ne x 2 dx = lim n→∞ ∫ ∞ 0 fn(x)dx = ∫ ∞ 0 e− x 2 dx = 2 lim n→∞ ∫ n 0 (1 + x n )ne−2xdx = lim n→∞ ∫ ∞ 0 gn(x)dx = ∫ ∞ 0 e−xdx = 1 17. Define the distance between points (x1, y1) and (x2, y2) in the plane to be |y1 − y2| if x1 = x2; 1 + |y1 − y2| if x1 6= x2. show that this is indeed a metric, and that the resulting metric space X is locally compact. If f ∈ Cc(X), let x1, x2, · · · , xn be those values of x for which f(x, y) 6= 0 for at least one y (there are only finitely many such x!), and define Λf = n∑ j=1 ∫ +∞ −∞ f(xj, y)dy Let µ be the measure associated with this Λ by Theorem2.14. If E is the x-axis, show that µ(E) = ∞ although µ(K) = 0 for every compact K ⊂ E. [proof]:Order Pi = (xi, yi), i = 0, 1, 2, 3, · · · . d(P1, P2) = { |y1 − y2| if x1 = x2; 1 + |y1 − y2| if x1 6= x2 Since d(P1, P2) ≤ { |y1 − y3|+ |y2 − y3| if x1 = x2; 1 + |y1 − y3|+ |y2 − y3| if x1 6= x2 Therefore, d(P1, P2) ≤ d(P1, P3) + d(P2, P3). That is d is metric. Take X = (R2, d). Then Pn → P ⇔ ∃ N > 0, such that if n > N, xn = x; and|yn − y| → 0. 8 and for any 0 < δ < 1, U(P0, δ) = {P ∈ X : d(P0, P ) < δ} = {P ∈ X : x = x0, and |y − y0| < δ} U(P0, δ) = {P ∈ X : d(P0, P ) ≤ δ} = {P ∈ X : x = x0, and |y − y0| ≤ δ} Moreover, U(P0, δ) is compact subset of X. Thus X is locally compact Haus- dorff space. In the following, we order the set for f ∈ Cc(X) Ωf = {x ∈ R : ∃ y ∈ R such that f(x, y) 6= 0} Let K be the support of f , K is compact. Then there exists finite P1, P2, · · · , Pn ∈ K for any 0 < δ < 1 such that K ⊂ n⋃ j=1 U(Pj, δ) and so Ωf = {x1, x2, · · · , xn}. Since Λf = n∑ j=1 ∫ +∞ −∞ f(xj, y)dy it is easy that Λ is a positive linear functional on Cc(X). There thus exist a measure µ associated with this Λ by Theorem2.14. Let E be the x-axis and 0 < δ < 1. Define V = ⋃ Pi∈E U(Pi, δ); where Pi = (xi, 0); it shows V is open and E ⊂ V . For any finite elements P1, P2, · · · , Pn ∈ E, the set F = n⋃ j=1 U(Pj, δ 2 ) is compact with F ⊂ V . By the Urysohn’s lemma, there exists f ∈ Cc(X) satisfying F ≺ f ≺ V . So we have µ(V ) ≥ Λf ≥ nδ. Hence, it shows µ(V ) = ∞ as n → ∞. Therefore, µ(E) = ∞. Assume that K ⊂ E is compact, then it is obvious that K is finite set. Take K = {P1, P2, · · · , Pm} 9 For any 0 < � < 1, define G = m⋃ j=1 U(Pj, �) then g is open and K ⊂ G. By the Urysohn’s lemma, there exists g ∈ Cc(X) satisfying K ≺ g ≺ G. Since µ(K) ≤ Λg ≤ 2� whence µ(K) = 0. 21. If X is compact and f : X → (−∞, +∞) is upper semicontinuous, prove that f attains its maximum at some point of X. [proof]: for every t ∈ f(X), define Et = {x ∈ X : f(x) < t} Assume that f does not attain its maximum at some point of X. Then it shows X = ⋃ t∈f(X) Et. Since f is continuous, Et is open sets. Again since X is compact, there exists finite ti(i = 1, 2, · · · , n) ∈ f(X) such that X = n⋃ i=1 Eti . Take t0 = max{t1, t2, · · · , tn} then we have f(x) < t0, ∀x ∈ X. But it is in contradiction with t0 ∈ f(X). It therefore is proved that f attains its maximum at some point of X. 10 ��������� � chapter three 1. Prove that the supremum of any collection of convex functions on (a, b) is convex on (a, b) and that pointwise limits of sequences of convex functions are convex. What can you say about upper and lower limits of sequences of convex functions? [Proof]: Suppose that α ≥ 0, β ≥ 0 and α + β = 1. Let {fi}i∈I be any collection of convex functions on (a, b). for any x, y ∈ (a, b), we have fi(αx+ βy) ≤ αfi(x) + βfi(y), ∀i ∈ I. It prove that sup i∈I fi(αx+ βy) ≤ α sup i∈I fi(x) + β sup i∈I fi(y). So sup i∈I fi is convex. Assume I = N and f is pointwise limits of sequences of convex functions {fi}. according to the properties of limit, we have lim i→∞ fi(αx+ βy) ≤ α lim i→∞ fi(x) + β lim i→∞ fi(y) That is f(αx+ βy) ≤ αf(x) + βf(y) and so f is convex. By the definition of upper limits and the conclusion above, it is easy that the upper limits of sequences of convex functions is convex. But the lower limits of sequences of convex functions is false. 2. If φ is convex on (a, b) and if ψ is convex and nondecreasing on the range of φ, prove that ψ ◦ φ is convex on (a, b). For φ > 0, show that the convexity of lnφ implies the convexity of φ , but not vice versa. [Proof]: Assume that x, y ∈ (a, b);α, β ≥ 0and α + β = 1. Then ψ ◦ φ(αx+ βy) ≤ ψ(αφ(x) + βφ(y)) ≤ αψ ◦ φ(x) + βψ ◦ φ(y). 1 and so ψ ◦ φ is convex on (a, b). Since et is nondecreasing convex on R and φ = eln φ, it shows from the conclusion above that the convexity of lnφ implies the convexity of φ for φ > 0. [counterexample]: φ(x) = x2 is convex on (0,∞) but lnφ(x) = 2 lnx is not convex. Moreover, for φ > 0, it can show that the convexity of logc φ(c > 1) implies the convexity of φ. 3. Assume that φ is a continuous real function on (a, b) such that φ( x + y 2 ) ≤ 1 2 φ(x) + 1 2 φ(y) for all x and y ∈ (a, b). Prove that φ is convex. [Proof]: According to the definition of convex function,we only need to prove the case 0 < λ < 1 2 . Take E = { k 2n | k = 1, 2, · · · , 2n − 1; n ∈ N}. For n = 1, it is obvious that φ( x + y 2 ) ≤ 1 2 φ(x) + 1 2 φ(y) for all x and y ∈ (a, b). For n = 2, suppose that λk = k 4 , k = 1. φ(λ1x+ (1− λ1)y) = φ( 1 2 (x+y 2 + y)) ≤ 1 2 φ(x+y 2 ) + 1 2 φ(y) ≤ 1 2 [1 2 φ(x) + 1 2 φ(y)] + 1 2 φ(y) = 1 4 φ(x) + 3 4 φ(y) = λ1φ(x) + (1− λ1)φ(y) Assume that the case λ = k 2n−1 , k = 1, 2, · · · , 2n−1 − 1 is correct. For any λ ∈ E with λ = k 2n ; 1 ≤ k ≤ 2n−1, n ≥ 2, and for all x and y ∈ (a, b), it shows that (i) φ(λx+ (1− λ)y) = φ( x+ y 2 ) ≤ 1 2 φ(x) + 1 2 φ(y) = λφ(x) + (1− λ)φ(y) 2 when k = 2n−1. (ii)for 1 ≤ k < 2n−1, λx + (1− λy) = k 2n x + (1− k 2n )y = 1 2 ( k 2n−1 x+ (1− k 2n−1 )y + y) Since k 2n−1 x+ (1− k 2n−1 )y ∈ (a, b) and so φ(λx + (1− λ)y) ≤ 1 2 φ( k 2n−1 x + (1− k 2n−1 )y) + 1 2 φ(y) ≤ k 2n φ(x) + 1 2 (1− k 2n−1 )φ(y) + 1 2 φ(y) = λφ(x) + (1− λ)φ(y) For any 0 < λ < 1, there exists a sequence {λn} ⊂ E with lim n→∞ λn = λ. Thus φ(λx+ (1− λ)y) = lim n→∞ φ(λnx + (1− λn)y) ≤ lim n→∞ (λnφ(x) + (1− λn)φ(y)) = λφ(x) + (1− λ)φ(y) Therefore, it shows that φ is convex on (a, b). 4. Suppose f is a complex measurable function on X, µ is a positive measure on X, and φ(p) = ∫ X |f |pdµ = ‖f‖pp (0 < p <∞). Let E = {p : φ(p) <∞}, Assume ‖f‖∞ > 0. (a) If r < p < s, r ∈ E, and s ∈ E, Prove that p ∈ E. (b) Prove that lnφ is convex in the interior of E and that φ is continuous on E. (c) By (a), E is connected. Is E necessarily open? losed?