Real Analysis with Economic Applications - Hints References doc

Hints for Selected E xercises Hints For Selected Exercises of Chapter A Exercise 7.LetmeshowthatifR is transitive, then xP R y and yRz implies xP R z. Since P R ⊆ R, it is plain that xRz holdsinthiscase.Moreover,ifzRx holds as well, then yRx,forR is transitive and yRz . But this contradicts xP R y. Exercise 13. (c)Supposec  (S)=∅ (which is possible only if |S| ≥ 3). Ta ke any x 1 ∈ S. Since c  (S)=∅, there is an x 2 ∈ S\{x 1 } with x 2  x 1 . Similarly, there is an x 3 ∈ S\{x 1 ,x 2 } with x 3  x 2 . Continuing this way, I find S = {x 1 , , x |S| } with x |S|  ··· x 1 . Now find a con tradiction to  being acyclic. Exercise 14. Apply Sziplrajn’s Theorem to the transitiv e closure of the relation  ∗ :=  ∪ ({x ∗ }×Y ). Exercise 16.(e) inf A = W A and sup A = W {B ∈ X : V A ⊆ B} for any class A ⊆ X . Exercise 18.Define the equivalence relation ∼ on X by x ∼ y iff f(x)=f(y), let Z := X/ ∼ , and let g be the associated quotient map. Exercise 20.If f was such a surjection, we would have f(x)={y ∈ X : y/∈ f(y)} for some x ∈ X. Exercise 29.Showthatinf S = − sup{−s ∈ R : s ∈ S}. Exercise 30.Considerfirst the case w here a ≥ 0. Apply Proposition 6.(b) (twice) to find a  ,b  ∈ Q su ch that 0 ≤ a<a  <b  <b.Now x ∈ (a  ,b  ) iff 0 < x−a  b  −a  < 1 while 1 √ 2 ∈ (0, 1)\Q. (Wh y?) Exercise 36. |y m − a| ≤ z m − x m + |x m − a| for each m. Exercise 38. For every m ∈ N, there exists an x m ∈ S with x m + 1 m > sup S. Exercise 40.(b) By part (a) and the Bolzano-Weierstrass Theorem, every real Cauchy sequence has a convergent subsequence. Exercise 41. Note that |y k − y k  | ≤ |y k − x kl | + |x kl − x k  l | + |x k  l − y k  | for all k, k  ,l ∈ N. Use (i) and (ii) and a 3ε argument to establish that (y k ) is Cauchy. So, by the previous exercise, y k → x for some x ∈ R. Now use the inequality |x kl − x| ≤ |x kl − y k | + |y k − x| ,k,l=1, 2, to conclude that x kl → x. Exercise 43. How about  1, −1, 1 2 , 1, −1, 1 2 ,  and  −1, 1, 1 2 , −1, 1, 1 2 ,  ? 563 Exercise 45.(a) This follows from the fact that |x k+1 + x k+2 + ···|=    x − S k x i    for all k ∈ N. (By the wa y, the con verse of (a) is true too — just verify that ( S m x i ) is a real Cauchy sequence (Exercise 40).) Exercise 48.(a) Try the sequence ((−1) m ). (b)Supposethat S ∞ x i = x ∈ R. Then s m → x, so, for any −∞ <y<x, there i s an M ∈ N with s m ≥ y for all m ≥ M. But then 1 m # M S i=1 s i + m S i=M+1 s i $ ≥ 1 m M S i=1 s i + m−M m y, m ≥ M, so letting m →∞, and recalling that y is arbitrarily chosen in (−∞,x), I find lim inf 1 m S m s i ≥ x. Now argue similarly for lim sup 1 m S m s i . Exercise 53. Use the Bolzano-Weierstrass Theorem. Exercise 54.Takeany y ∈ (f(a),f(b)), and define S := {x ∈ [a, b]:f(x) <y}. Show that f(sup S)=y. Exercise 56.(a)Define g ∈ C[a, b] by g(t):=f(t) − f(a)+  t−a b−a  (f(b) − f(a)), and apply Rolle’s Theorem to g. Exercise 57. Consider the case α =1, and write U b a f := U b a f(t)dt and U b a g := U b a g(t)dt. Take any ε > 0, and show that there exists an a ∈ D[a, b] with ] b a f + ε 4 >R a (f) ≥ ] b a f>r a (f) − ε 4 and ] b a g + ε 4 >R a (g) ≥ ] b a g>r a (g) − ε 4 . (Find one dissection for f that does the job, and another for g. Then combine these dissec- tions by .) Now verify that R a (f + g) ≤ R a (f)+R a (g) and r a (f + g) ≥ r a (f)+r a (g). PutitalltogethertowriteR a (f +g)−r a (f +g) < ε and    R a (f + g) − ( U b a f + U b a g)    < ε. Exercise 61.Letmeshowthatf(a) > 0 would yield a contradiction, and you do the rest. Suppose f(a) > 0. Then,bycontinuityoff, we have c := inf{x ∈ (a, b]:f(x) > f(a) 2 } >a.(Yes?) So, by Exercise 58, U b a f(t)dt = U c a f(t)dt+ U b c f(t)dt ≥ f(a) 2 (c−a) > 0, contradiction. Exercise 68. See Example K.9. Hints For Selected Exercises of Chapter B Exercise 2. As simple a s the claim is, you s till have to invoke th e Axiom of C h oice to prove it. Let’s see. Take any set S with |S| = ∞. Iwishtofind an injection g from N into S. Let A := 2 S \{∅}. By the Axiom of hoice, there exists a map f : A → V A such that f(A) ∈ A for each A ∈ A. Now define g : N → S, recursively, as follo ws, g(1) := f(S), g(2) := f(S\{g(1)}),g(3) := f(S\{g(1),g(2)}), and so on. 564 Exercise 5.(b) There is a bijection bet ween [0, 1] and N ∞ .Youcanprovethisby “butterfly hunting.” (If y ou get stuc k, sneak peek at the hin t I gav e for Exercise 15 belo w.) Exercise 7. Let me do this in the case where A is countably infinite. Take any coun tably infinite subset C of B (Exercise 2 ). Since both C and A ∪ C are countably infinite, there exists a bijection g from C onto A ∪ C. (Why?) Now define f : B → A ∪ B by f(b):=  g(b), if b ∈ C b, otherwise . This is a bijection, is it not? Exercise 8. Use Proposition 3. Exercise 9. Its concavity ensures that f is right-differentiable (Exercise A.68), and that f  + is decreasing (Example K.9). Now show that x ∈ D f iff x is a point of discontinuity of f  + , and use Exercise 8. Exercise 10.(c) Recall Exercise A.20. Exercise 13.Let X := 2 S and consider f ∈ X S defined by f(x):={y ∈ S : x  y}. Exercise 14. Adapt the proof of Proposition 5. Exercise 15.(b)Youknowthat 2 N ∼ card {0, 1} ∞ , so i t is enough to pro ve that [0, 1] ∼ card {0, 1} ∞ . You can do this by adopting the trick we used to prove the un- countability of R. Construct the bijection f from [0, 1] onto {0, 1} ∞ as follo ws. For any given x ∈ [0, 1], the first term of f(x) is 0 if x ∈ [0, 1 2 ], and 1 if x ∈ ( 1 2 , 1]. In the former case, the second term of f(x) is 0 if x ∈ [0, 1 4 ] and 1 if x ∈ ( 1 4 , 1 2 ]. In the latter case, Exercise 16. Recall Exercise 2. Exercise 22.False. Exercise 24. Adapt the proof of Proposition 9 for the “if” part. For the converse, let u represent , and consider the set of all closed intervals with distinct rational points andassigntoeachsuchinterval I one (if any) y ∈ X su ch that u(y) belongs to I.This procedure yiel ds a countable set A in X. Define next B := {(x, y) ∈ (X\A) 2 : x  y and x  z  y for no z ∈ A}, and chec k that B is coun table. (The k ey is to observe that if (x, y ) ∈ B, then x  w  y for no w ∈ X.) Now verify that Y := A ∪ {t ∈ X : eit her (t, x) or (x, t) is contained in B for some x ∈ X} worksasdesired. Exercise 29. No. Exercise 32. How about U := {u, v}, where u(x):=  x x+1 , if x ∈ Q + −x x+1 , otherwise and v(x):=  1 − u(x), if x ∈ Q + −1 − u(x), otherwise ? Hints For Selected Exercises of Chapter C Exercise 1. The answer to the first q uestion is: No! 565 Exercise 7. Suppose we can. Then a nonempty set S in this space is closed iff it is finite. But the en tire space N must be closed, so we have a contradiction. Thus the answer to the question is: No! Exercise 9. bd R (Q)=R. Exercise 10.Observethat in t X (S) ∩ Y is an open subset of Y that is contained in S ∩ Y. Conversely, if Y is open in X, then int Y (S ∩ Y ), being open in Y, must be open in X. Since this set is ob viously con tained in S ∩ Y, we have in t Y (S ∩ Y ) ⊆ int X (S ∩ Y )=int X (S) ∩ int X (Y )=int X (S) ∩ Y. Exercise 12. Try a suitable indiscreet space. Exercise 14. (a) ⇒ (b) The answer is hidden in Example 3.[4]. (b) ⇒ (c) In this case there exists at least one x m in N 1 m ,X (x) ∩ S for each m ∈ N. What is lim x m ? (c) ⇒ (a) Use Proposition 1. Exercise 19. Pick any x ∈ X, and use the countability of X to find an r>0 suc h that d(x, y) = r for all y ∈ X and d(x, y) >rfor s ome y ∈ X.Isn’t{N r,X (x),X\N r,X (x)} a partition of X? Exercise 22. For any f,g ∈ C[0, 1] with f| Q = g| Q , we have f = g. Exercise 25. Take any x, y ∈ X with x  y. Show that U  (y) ∪ L  (x) is clopen, s o since X is connected, we have U  (y) ∪ L  (x)=X. Now to derive a contradiction, assume that z and w are two p oints in X with z  w. Then either z ∈ U  (y) or z ∈ L  (x). Suppose former is the case; the latter case is analyzed similarly. Since z  w, we must then also have w ∈ U  (y). NowshowthatO := L  (z) ∩ L  (w)=L  (z) ∩ L  (w). Conclude from this that O must be clopen in X, so O = X by Proposition 2. But z/∈ O. Exercise 28. (a)Trytheopencover{N 1 m ,X (x):x ∈ X} for each m ∈ N. Exercise 29. (b) Give a counter-example by using a suitable discrete space. Exercise 30. Take any f ∈ X and let S m := {x ∈ T : |f(x)| ≥ m} for each m ∈ N. Now, each S m is a compact in T, and S 1 ⊇ S 2 ⊇ ···. Then {S m } has the finite intersection property, while, if f was not bounded, we would have W ∞ S i = ∅. Exercise 35. The first statement easily follo ws from Theorem 2 and the definition of d ∞ and D ∞ . Use the “bab y” Arzelà-Ascoli Theorem to prove the second one. Exercise 40.  ∞ = B(N). In the case o f C[0, 1], proceed just like I did in Example 11.[5]. Exercise 41.UseTheorem2. Exercise 42. (a) Recall Exercise A.61. (b) (i) Watc h the metric! Watch the metric! (ii) What is the domain of d 1 ? 566 Exercise 43. (c)Let(f m ) be a Cauchy sequence in C 1 [0, 1]. Then (f m ) is also Cauchy in C[0, 1] so d ∞ (f m ,f) → 0 for some f ∈ C[0, 1]. Similarly, d ∞ (f  m ,g) → 0 for some g ∈ C[0, 1]. No w use the result you proved in part (b). Exercise 44. To prove the “if” part (which is a tad harder than the “only if” part), pick a noncon vergent Cauch y sequence (x m ) in X, and define S m := {x k : k = m, m +1, } for each m ∈ N. Exercise 46. (a) Recall Exercise 41. Exercise 47. The argument is analogous to that I gave to prove Theorem 3. Exercise 50. (a) First draw a graph and then try the map x → 1+ln(1+e x ) on R. Exercise 51. IwishtoshowthatΦ is a contraction. First, observe that for any m ∈ N and x, y ∈ X, d(Φ(x), Φ(y)) ≤ d(Φ(x), Φ m (x)) + d(Φ m (x), Φ m (y )) + d(Φ m (y), Φ(y)). Now pick any ε > 0. Since sup{d(Φ m (x), Φ(x)) : x ∈ X} → 0, there exists an M ∈ N such that d(Φ(z), Φ m (z)) < ε 2 for all m ≥ M and all z ∈ X.Thus,lettingK := sup{K m : m =1, 2, }, Iget d(Φ(x), Φ(y)) ≤ ε + d(Φ M (x), Φ M (y )) ≤ ε + Kd(x, y) for all x, y ∈ X. Since K<1 by hypothesis, it follows that Φ is a contraction. The claim would not be true if all we had was lim d(Φ m (x), Φ(x)) = 0 for all x ∈ X. For instance, consider the case where X = R and Φ m := (1 − 1 m )id R ,m=1, 2, Exercise 52. Met rize R n by d ∞ . Exercise 53. Recall Exercise 48. Exercise 56. ln  1+b 1+a  ≤ b−a for all 0 ≤ a ≤ b. (Study the map t → (t−a)−ln  1+t 1+a  . Is this map monotonic on [a, b]? What’s its minimum/maxim um ?) Exercise 63. Completeness is not invariant under the equivalence of metrics. Exercise 65. No. Consider, for instance, the case in wh ich (X i ,d i ) is R and O i := (− 1 i , 1 i ),i=1, 2, . Exercise 68. c 0 is dense R ∞ relative to the product metric. Exercise 69. Exercise 66 shows that any open set in X ∞ (X i ,d i ) can be written as a unionofsetsoftheformof X m O i × X m+1 ×X m+1 ×···, where O i is an open subset of X i , i =1, , m. Apply this characterization to the definition of connectedness, and see what you get. Hints For Selected Exercises of Chapter D Exercise 2 .(a)Ifx ∈ X\A, then there exists an ε > 0 such that N ε,X (x) ⊆ X\A, so d(x, A) ≥ ε. 567 Exercise 4.Yes. Exercise 7. (c)Yes. Exercise 8. For any ε > 0, there exists a δ > 0 such that, for all x ∈ X, we have |f(x) − f(y)| < ε 2K and |g(x) − g(y)| < ε 2K whenever y ∈ N δ,X (x). Then |f(x)g(x) − f(y)g(y )| ≤ |f(x)||g(x) − g(y)| + |g(y)||f(x) − f (y)| < ε. Without boundedness, the claim would not be true. For instance, consider the case X = R and f = g = id X . Exercise 15. UseTheoremC.2toshowthatif(x m ) is a sequence in X that converges to x, then the set {x, x 1 ,x 2 , } is compact in X. Exercise 16. (b) R is complete but (0, 1) is not. (c) Because it is nonexpansive. Exercise 19. This is just Exercise 9 in disguise. Exercise 20. (a) Here is the “only i f” part. If S = ∅, the claim is trivial. So take any nonempty set S ⊆ X, and let y ∈ f(cl X (S)). Then there exists an x ∈ cl X (S) such that y = f(x). Clearly,theremustexistan(x m ) ∈ S ∞ such that x m → x (Exercise C.14). By continuity of f, we then find f(x m ) → f(x). So f(x) ∈ cl X (f(S)). Exer cise 24. Q is countable and R\Q is not. So, by the Intermediate Value Theorem ? Exercise 25. Punch a hole in R n . No w do the same in R. Do you get similar spaces? Exercise 27. You can easily prove this by using Theorem C.2 and imitating the way we proved Proposition A.11. Let me suggest a more direct argument here. Take any ε > 0. Since f is continuous, for any x ∈ X there exists a δ x > 0 such that d Y (f(x),f(y)) < ε 2 for all y ∈ X with d X (x, y) < δ x . Now use the compactness of X to find finitely many x 1 , , x m ∈ X with X = V {N 1 2 δ x i ,X (x i ):i =1, ,m}. Let δ := 1 2 min{δ x 1 , , δ x m }, and observe that d X (x, y) < δ implies d X (x i ,y) < δ x i for some i ∈ {1, , m} with d X (x, x i ) < 1 2 δ x i . (Good old triangle inequality!) But d Y (f(x),f(y)) ≤ d Y (f(x),f(x i ))+ d Y (f(x i ),f(y)). So? Exercise 30. Suppose there exists a sequence (x m ) ∈ T ∞ without a convergent sub- sequence. It is okay to assume that all terms of this sequence are distinct. Define S := {x 1 ,x 2 , }. Show that S is clo sed and define ϕ ∈ R S by ϕ(x i ):=i. Is ϕ contin uous on S? Can you extend ϕ to R n ? Exercise 31. Define g ∈ R [a,b] by g(t):=f(t) − αt, and sho w that g must have a minim um on [a, b]. Exercise 37. (a)Sinceu is strictly increasing, {α ∈ R + : x  (α, , α)} is bounded from above, so its sup belongs to R. Since u is lower semicon tinuous, this set is closed, and hence contains its sup . 568 Exercise 40. (a) The idea is to show that in this case f must be upper semicontinuous everywhere. Let’s use Proposition 4 to this end. Tak e any x ∈ X, and let (x m ) be any real sequence with x m → x. Then lim sup f(x m ) = lim sup f((x m − x + x ∗ )+(x − x ∗ )) = lim sup f(x m − x + x ∗ )+f(x − x ∗ ). Now define (y m ):=(x m − x + x ∗ ), and notice that y m → x ∗ . So? Exercise 41. (a) f − f(0) satisfies Cauchy’s functional equation. (b)Showfirst that f(λx +(1− λ)y)=λf(x)+(1− λ)f(y) for any 0 ≤ x, y ≤ 1 and λ ∈ R such that 0 ≤ λx +(1− λ)y ≤ 1. (Note. Section F.2.2 contains a generalization of this exercise.) Exercise 48. Define f m ∈ C[−1, 1] by f m (t):= S m t i i 2 , m =1, 2, and show that d ∞ (f,f m ) ≤ S ∞ i=m+1 1 i 2 for each m. Since S m 1 i 2 converges — this is why I’m so sure that f m sarewell-defined — S ∞ i=m+1 1 i 2 → 0 as m →∞. Thus f m → f uniformly. Exercise 50. Define α m := sup{|ϕ m (x)| : x ∈ T} for each m and α := sup{|ϕ(x)| : x ∈ T }. (Is α finite?) Since d ∞ (ϕ m , ϕ) → 0, there exists an M ∈ N such that |ϕ(x) − ϕ m (x)| < 1 for all x ∈ T and m ≥ M.ThenK := max{α 1 , , α M , α +1} should do it. Exercise 55. Take T = N and recall that  ∞ is not separable. Exercise 58. Imitate the argument given in the proof of the Arzelà-Ascoli Theorem. Exercise 62. This is just Exercise 30 (with a touch of the Tietze Extension Theorem). Exercise 66. (a)Since T is not n ecessarily closed in X, the closedness of these sets in T does not imply their closedness in X. You should use uniform continuit y. Exercise 67. Write φ =(φ 1 , φ 2 , ), and define φ ∗ i ∈ R T by φ ∗ i (x):=inf{φ i (w)+ Kd(w,x) α : w ∈ T } for each i. Now check that φ ∗ =(φ ∗ 1 , φ ∗ 2 , ) does the job. Exercise 68. (a) Recall the Homeomorphism Theorem (Section 3.1). (b)Let X be [0, 1] with the discrete metric, and Y be [0, 1] with the usual metric. Exercise 69. Try the projection operator (Example 5). Exercise 73. If f is a fixed poin t of Φ, then, for any 0 <t<1, we must have f(t)=f(t m 2 ) for any m ∈ N. Then f| [0,1) =0. Exercise 77. Let 0 denote the n-v ector of 0s, and write β f (x) for (f 1 (x), , f n (x)) for any x ∈ B n α . If β f (x) = 0 for all x ∈ B n α , then I can define the map Φ : B n α → R n by Φ(x):= −α d 2 (β f (x),0) β f (x). Would a fixed point of Φ satisfy the given boundary condition? Does Φ have a fixed point? Hints For Selected Exercises of Chapter E Exercise 1. σ(R + )=[0, π 2 ) ∪ { π 2 , 5π 2 , 9π 2 , }. 569 Exercise 2. Γ(X)=S. Exercise 4. For the “only if” part, define f by f(y):=x where x is any element of X with y ∈ Γ(x). Is f well-defined? Exercise 8. How about Γ(0) = [−1, 1] and Γ(t):={0} for all 0 <t≤ 1? Exercise 10. Yes. To verify its upper hemicontinuity at any (p, ι) with p i =0for some i, use the definition of upper hemicontinuity directly. By Example 2, this is the only problematic case. Exercise 11. If (y m ) ∈ Γ(S) ∞ , then there exists an (x m ) ∈ S ∞ such that y m ∈ Γ(x m ) for each m. Use Theorem C.2 to extract a subsequence of (x m ) that converges to some x ∈ S, and then use Prop osition 2 to get your hands on a suitable subsequence of (y m ). Exercise 12. (a)Supposef is closed. Tak e any y ∈ Y and let O be an open subset of X with f −1 (y) ⊆ O. Notice that Y \f(X\O) is an open subset of Y that contains y. Moreo ver, f −1 (Y \f(X\O)) ⊆ O. Conv ersely, if S is a closed subset of X and y ∈ Y \f(S), then X\S is open in X and f −1 (y) ⊆ X\S, so there exists a δ > 0 such that f −1 (N δ,Y (y)) ⊆ X\S. So? Exercise 21. Use Proposition D.1 and triangle inequality (twice) to verify that Γ has a closed graph. Then apply Proposition 3.(a). Exercise 22. Let X 0 := X and X i+1 := Γ(X i ) for all i ∈ Z + . Recall Example C.8 to be able to conclude that S := W ∞ X i is a nonempty compact set. S is a fixed set of Γ. Exercise 24. The answer to the last question here is: No. Exercise 27. Take any (x m ) ∈ X ∞ and (y m ) ∈ R ∞ with x m → x for some x ∈ X, and 0 ≤ y m ≤ ϕ(x m ) for each m. Since ϕ is continuous, (ϕ(x m )) con verges. Deduce that (y m ) is a bounded sequence, and then use the Bolzano-Weierstrass Theorem and Proposition 2 to conclude that Γ is upper hemicontinuous. To show that Γ is low er hemicontinuous, tak e any (x, y) ∈ X × R and (x m ) ∈ X ∞ such that x m → x and 0 ≤ y ≤ ϕ(x). If y = ϕ(x), then define y m := ϕ(x m ) for each m. If y<ϕ(x), then there exists an integer M with 0 ≤ y ≤ ϕ(x m ) for each m ≥ M, so pick y m from [0, ϕ(x m )] arbitrarily for m =1, , M, and let y m := y for each m ≥ M. Either way, y m → y and y m ∈ Γ(x m ) for each m. Apply Proposition 4. Exercise 28. Γ is not upper hemicontinuous at any x ∈ [0, 1] ∩ Q, but it is upper hemicontinuous at any x ∈ [0, 1]\Q. To prov e the latter fact, recall that [0, 1] ∩ Q is dense in [0, 1], so there is one and only one open subset of [0, 1] that contains [0, 1] ∩ Q. Γ is lower hemicontinuous. To see this, take any x ∈ [0, 1],y ∈ Γ(x), and any (x m ) ∈ [0, 1] ∞ with x m → x. Let (z m ) be a sequence in [0, 1] ∩ Q with z m → y and (w m ) asequencein[0, 1]\Q with w m → y. Now define y m := z m if x m is irrational and y m := w m if x m is ra tional, m =1, 2, .Then(y m ) ∈ [0, 1] ∞ satisfies y m ∈ Γ(x m ) for each m, and y m → y. 570 Exercise 29. Upper hemicontinuity is just the closed graph property here. For the lower hemicontinuity at any υ ∈ u(T ), suppose there exists an open subset O of T such that Γ(υ) ∩ O = ∅, but for any m ∈ N there exists an υ m ∈ N 1 m ,R (υ) ∩ u(T ) suc h that Γ(υ m ) ∩ O = ∅. Take any x ∈ Γ(υ) ∩ O, and show that u(x) > υ is impossible. If u(x)=υ, then use the monotonicity of u. Exercise 32.(a)Ifa<0 <b<1, then d H ([0, 1], [a, b]) = max{|a| , 1 − b}. (b)Yes. Exer cise 35. Take any Cauchy sequence (A m ) in c(Y ) and define B m := cl Y (A m ∪ A m+1 ∪···) for each m. Show that B 1 (hence each B m ) is compact. By the Cantor-Fréc het Intersection Theorem, W ∞ B i ∈ c(Y ). Show that A m → W ∞ B i . Exer cise 36. Let F := {f 1 , ,f k } for some k ∈ N. Define the self-map Φ on c(Y ) by Φ(A):=f 1 (A) ∪ ···∪ f k (A),andcheckthatΦ is a contraction, the contraction coefficient of whic h is smaller than the maximum of those of f i s. Now apply Exercise 35 and the Banach Fixed Poin t Theorem. Exercise 42. Modify the argument given in the last paragraph of the proof of Maximum Theorem. Exercise 43. Take any (θ m ) ∈ Θ ∞ with θ m → θ for some θ ∈ Θ, and suppose that ϕ ∗ (θ) < lim sup ϕ ∗ (θ m ). Then there exist an ε > 0 and a strictly in creasing (m k ) ∈ N ∞ such that, for some x m k ∈ Γ(θ m k ),k=1, 2, , and K ∈ N, we have ϕ ∗ (θ)+ε ≤ ϕ(x m k , θ m k ) for all k ≥ K. (Why?) Now use the upper hemicontinuity of Γ to extract a subsequence of (x m k ) that conv erges to a point in Γ(θ), and then derive a contradiction by using the upper semicontinuity of ϕ. Exercise 44.(a)Letα := u(0, , 0) and β := u(1, ,1).Define Γ :[α, β] ⇒ T by Γ(υ):=u −1 ([υ, β]). It is easy to check that Γ is upper hemicontinuous. It is in fact low er hemicontinuous as well. To see this, take any υ ∈ [α, β],x∈ Γ(υ), and any sequence (υ m ) ∈ [α, β] ∞ with υ m → υ. Let I := {m : υ m ≤ υ} and J := {m : υ m > υ}. Define S := T ∩ {λx +(1− λ)(1, , 1) : λ ∈ [0, 1]}. Clearly, β ≥ υ m ≥ υ for all m ∈ J. By the In termediate Value Theorem, for eac h m ∈ J, there exists an 0 ≤ λ m ≤ 1 such that u(λ m x +(1− λ m )(1, , 1)) = υ m .Nowdefine (x m ) ∈ T ∞ as x m := x if m ∈ I,andasx m := λ m x +(1− λ m )(1, , 1) otherwise. Then x m ∈ Γ(υ m ) for each m and x m → x. (W hy?) Conclusion: Γ is a con tin uous correspondence. The stage is now set for applying the Maximum Theorem. (b) e u (p, u ∗ (p, ι)) ≤ ι follo ws from definitions. To show that < cannot hold here, y ou will need the continuity of e u and strict monotonicity of u ∗ . Exercise 53. (a) Let me formulate the problem in terms of capital accumulation. In that case, letting γ := 1 − α, the problem is to choose a real sequence (x m ) in order to Maximize ∞ S i=0 δ i (pf(x i ) − (x i+1 − γx i )) such that γx m ≤ x m+1 ≤ x m + θ,m=0, 1, 571 It is easily seen that f must have a unique positive fixed point, denote it by ¯x. Clearly, the firm will nev er operate with an input lev el that exceeds ¯x, sothesolutionoftheproblem must belong to X ∞ , where X := [0, ¯x]. De fine Γ : X ⇒ X by Γ(x):=[γx, x + θ], and ϕ : Gr(Γ) → R by ϕ(a, b):=pf(a) − (b − γa). Then the problem of the firm can be written as choosing (x m ) ∈ X ∞ in order to Maximize ∞ S i=0 δ i ϕ(x i ,x i+1 ) such that x m+1 ∈ Γ(x m ),m=0, 1, (b) The optimal policy correspondence P for D(X, Γ,u,δ) satisfies P (x) = arg max{ϕ(x, y)+δV (y):γx ≤ y ≤ x + θ}, where V is the value function of the problem. It is single-valued, so I can treat is as a function. Then, using the one-deviation propert y, P (x) ∈ arg max{pf(x) −(y −γx)+δ(pf (y)−(P 2 (x)−γy)) + δ 2 V (P 2 (x)) : γx ≤ y ≤ x+θ}. By strict concavity of f, the solution must be interior, so the first-order-condition of the problem shows that −1+δpf  (P (x)) + δγ =0. But there is only one x that satisfies this, no? Exercise 57. Study the correspondence Γ : T ⇒ R n with Γ(x):={y ∈ R n :(x, y) ∈ S}, where T := {x ∈ R n :(x, y) ∈ S for some y ∈ S}. Exer cise 58. Study the correspondence Γ : X ⇒ X defined by Γ(x):={z ∈ X : ϕ(x, z ) ≥ max{ϕ(x, y):y ∈ X}}. Exercise 59. “≤” part of t he claim is elementary. To prove the “≥”part,define Γ 1 : Y ⇒ X by Γ 1 (y):=argmaxf(·,y) and Γ 2 : X ⇒ Y by Γ 2 (x):=argminf(x, ·). Now define the self-correspondence Γ on X×Y by Γ(x, y ):=Γ 1 (y) ×Γ 2 (x). Use Kakutani’s Fixed Poin t Theorem to find an (x ∗ ,y ∗ ) such that (x ∗ ,y ∗ ) ∈ Γ(x ∗ ,y ∗ ),andcheckthat max x∈X min y∈Y f(x, y) ≥ f(x ∗ ,y ∗ ) ≥ min y∈Y max x∈X f(x, y). Exer cise 60. Let T := {x ∈ X : x ∈ Γ(x)}, and note that T = ∅. Define g := f| T , and show that g is a s elf-map on T. Moreover, if x is a fixed point of g, then it is a fixed point of both f and Γ. Exercise 65. (a)LetO y := {x ∈ X : y ∈ Ψ(x)},y∈ R n . 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To verify its upper hemicontinuity at any (p, ι) with p i =0for some i,