Real Analysis with Economic Applications - Hints References doc

b By part a and the Bolzano-Weierstrass Theorem, every real Cauchy sequence has a convergent subsequence.. You can easily prove this by using Theorem C.2 and imitating the way we proved

Hints for Selected Exercises

Hints For Selected Exercises of Chapter A

Exercise 7 Let me show that if R is transitive, then xPRy and yRz implies xPRz

SincePR⊆ R,it is plain thatxRzholds in this case Moreover, ifzRxholds as well, then

yRx, forR is transitive andyRz.But this contradictsxPRy

Exercise 13 (c) Suppose c (S) = ∅ (which is possible only if |S| ≥ 3) Take any

x1 ∈ S Since c (S) = ∅, there is an x2 ∈ S\{x1} with x2 x1 Similarly, there is

an x3 ∈ S\{x1, x2} with x3 x2 Continuing this way, I find S = {x1, , x|S|} with

x|S| · · · x1.Now find a contradiction to being acyclic.

Exercise 14 Apply Sziplrajn’s Theorem to the transitive closure of the relation ∗:=

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 havef (x) ={y ∈ X : y /∈ f(y)}for somex∈ X

Exercise 29 Show that inf S =− sup{−s ∈ R : s ∈ S}

Exercise 30 Consider first the case where a ≥ 0 Apply Proposition 6.(b) (twice) to find a , b ∈ Q such that 0 ≤ a < a < b < b Now x ∈ (a , b ) iff 0 < b −ax−a < 1 while

1

√

2 ∈ (0, 1)\Q (Why?)

Exercise 36. |ym− a| ≤ zm− xm+|xm− a| for eachm

Exercise 38 For every m∈ N, there exists an xm ∈ S withxm+ m1 > 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 allk, k , l ∈ N.Use (i) and (ii) and a 3ε argument to establish that(yk)is Cauchy So,

by the previous exercise,yk→ x for somex∈ R.Now use the inequality

Exercise 45 (a) This follows from the fact that|xk+1+ xk+2+· · ·| =

x −Skxi





for all k ∈ N (By the way, the converse of (a) is true too — just verify that(Sm

xi)is a real Cauchy sequence (Exercise 40).)

Exercise 48 (a) Try the sequence ((−1)m)

(b) Suppose that S∞

xi = x ∈ R Then sm → x,so, for any −∞ < y < x, there is

an M ∈ N withsm ≥ y for all m≥ M.But then

1 m

Exercise 53 Use the Bolzano-Weierstrass Theorem.

Exercise 54 Take anyy∈ (f(a), f(b)),and defineS := {x ∈ [a, b] : f(x) < y}.Show thatf (sup S) = y

Exercise 56 (a) Defineg ∈ C[a, b]byg(t) := f (t)− f(a) +t−a

b−a

(f (b)− f(a)), and apply Rolle’s Theorem tog

Exercise 57 Consider the case α = 1, and write Ub

f > ra(f ) −ε4 and

] b a

g +4ε > Ra(g) ≥

] b a

g > ra(g) −ε4 (Find one dissection forf that does the job, and another for g.Then combine these dissec- tions by )Now verify thatRa(f + g)≤ Ra(f ) + Ra(g)andra(f + g)≥ ra(f ) + ra(g)

Put it all together to writeRa(f +g)−ra(f +g) < εand

Exercise 68 See Example K.9.

Hints For Selected Exercises of Chapter B

Exercise 2 As simple as the claim is, you still have to invoke the Axiom of Choice to prove it Let’s see Take any setSwith|S| = ∞ I wish to find an injectiong fromNinto

Exercise 5 (b) There is a bijection between [0, 1] and N∞ You can prove this by

“butterfly hunting.” (If you get stuck, sneak peek at the hint I gave for Exercise 15 below.) Exercise 7 Let me do this in the case whereAis countably infinite Take any countably 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), ifb∈ C

b, otherwise This is a bijection, is it not?

Exercise 8 Use Proposition 3.

Exercise 9 Its concavity ensures thatf is right-differentiable (Exercise A.68), and that

f+ is decreasing (Example K.9) Now show thatx∈ Df iff xis a point of discontinuity of

f+, and use Exercise 8.

Exercise 10 (c) Recall Exercise A.20.

Exercise 13 LetX := 2S and consider f ∈ XS defined byf (x) :={y ∈ S : x y}

Exercise 14 Adapt the proof of Proposition 5.

Exercise 15 (b) You know that 2N ∼card {0, 1}∞, so it is enough to prove that

[0, 1] ∼card {0, 1}∞ You can do this by adopting the trick we used to prove the countability of R Construct the bijection f from [0, 1] onto {0, 1}∞ as follows For any given x∈ [0, 1], the first term of f (x) is0 ifx∈ [0,12], and 1ifx ∈ (12, 1] In the former case, the second term off (x)is 0ifx∈ [0,14]and 1ifx∈ (14,12] In the latter case, Exercise 16 Recall Exercise 2.

un-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 and assign to each such interval I one (if any)y ∈ X such that u(y) belongs to I This procedure yields a countable setA inX.Define nextB :={(x, y) ∈ (X\A)2 : x y and

x z y for no z ∈ A}, and check that B is countable (The key is to observe that if

(x, y) ∈ B, thenx w y for no w ∈ X.) Now verify that Y := A∪ {t ∈ X : either

(t, x)or(x, t) is contained inB for somex∈ X}works as desired.

1 − u(x), if x ∈ Q +

−1 − u(x), otherwise ?

Hints For Selected Exercises of Chapter C

Exercise 1 The answer to the first question is: No!

Exercise 7 Suppose we can Then a nonempty set S in this space is closed iff it is finite But the entire spaceNmust be closed, so we have a contradiction Thus the answer

to the question is: No!

Exercise 9. bdR(Q) = R.

Exercise 10 Observe that intX(S)∩ Y is an open subset of Y that is contained in

S∩ Y.Conversely, ifY is open inX,thenintY(S∩ Y ), being open inY,must be open in

X.Since this set is obviously contained inS∩ Y, we have

intY(S ∩ Y ) ⊆ int X (S ∩ Y ) = int X (S) ∩ int X (Y ) = intX(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 xm in N1

m ,X(x)∩ S for each m ∈ N

What islim xm?

(c) ⇒(a) Use Proposition 1.

Exercise 19 Pick any x∈ X,and use the countability ofX to find an r > 0such that

d(x, y) = r for ally ∈ X and d(x, y) > r for somey ∈ X Isn’t {Nr,X(x), X\Nr,X(x)}

a partition ofX?

Exercise 22 For any f, g ∈ C[0, 1]withf|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, so sinceX is connected, we haveU (y)∪ L (x) = X.Now to derive a contradiction, assume that z and w are two points 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 Sincez w,we must then also have w∈ U (y) Now show thatO := L (z)∩ L (w) = L (z) ∩ L (w) Conclude from this thatO must be clopen in X,soO = X by Proposition 2 Butz /∈ O

Exercise 28 (a) Try the open cover {N1

m ,X(x) : x∈ X} for eachm∈ N

Exercise 29 (b) Give a counter-example by using a suitable discrete space.

Exercise 30 Take any f ∈ X and let Sm :={x ∈ T : |f(x)| ≥ m} for each m ∈ N

Now, eachSm is a compact inT,andS1 ⊇ S2 ⊇ · · ·.Then{Sm}has the finite intersection property, while, iff was not bounded, we would have W∞

Si =∅

Exercise 35 The first statement easily follows from Theorem 2 and the definition of

d∞ and D∞ Use the “baby” Arzelà-Ascoli Theorem to prove the second one.

Exercise 40. ∞ = B(N) In the case of C[0, 1], proceed just like I did in Example 11.[5].

Exercise 41 Use Theorem 2.

Exercise 42 (a) Recall Exercise A.61.

(b) (i) Watch the metric! Watch the metric! (ii) What is the domain of d1 ?

Exercise 43 (c) Let(fm) be a Cauchy sequence inC1[0, 1].Then(fm)is also Cauchy

in C[0, 1] so d∞(fm, f ) → 0 for some f ∈ C[0, 1] Similarly, d∞(fm, g) → 0 for some

g ∈ C[0, 1].Now 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 nonconvergent Cauchy sequence (xm)in X,and define Sm := {xk : k = m, m + 1, }

for eachm∈ 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 + ex)on R

Exercise 51 I wish to show that Φis a contraction First, observe that for anym∈ 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(Φm(x), Φ(x)) : x ∈ X} → 0, there exists an M ∈ N

such that d(Φ(z), Φm(z)) < ε2 for all m ≥ M and all z ∈ X Thus, letting K :=sup{Km : m = 1, 2, },I get

d(Φ(x), Φ(y)) ≤ ε + d(Φ M (x), Φ M (y)) ≤ ε + Kd(x, y) for all x, y ∈ X.Since K < 1by hypothesis, it follows thatΦis a contraction.

The claim would not be true if all we had was lim d(Φm(x), Φ(x)) = 0for allx ∈ X

For instance, consider the case whereX = Rand Φm := (1− m1)idR, m = 1, 2, Exercise 52 Metrize Rn byd∞

Exercise 53 Recall Exercise 48.

Is this map monotonic on [a, b]?What’s its minimum/maximum?)

Exercise 63 Completeness is not invariant under the equivalence of metrics.

Exercise 65 No Consider, for instance, the case in which (Xi, di) is R and Oi :=(−1i,1i), i = 1, 2, .

Exercise 68. c0 is denseR∞ relative to the product metric.

Exercise 69 Exercise 66 shows that any open set in X∞(Xi, di) can be written as a union of sets of the form of XmOi× Xm+1× Xm+1× · · ·,whereOi is an open subset ofXi,

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 > 0such thatNε,X(x)⊆ X\A,so

d(x, A)≥ ε

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)| < 2Kε and |g(x) − g(y)| < 2Kε whenever y∈ Nδ,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 caseX = R

and f = g =idX

Exercise 15 Use Theorem C.2 to show that if (xm) is a sequence inX that converges

tox,then the set{x, x1, x2, }is compact inX

Exercise 16 (b) Ris 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 if” part If S =∅, the claim is trivial So take any nonempty setS ⊆ X, and let y∈ f(clX(S)) Then there exists anx∈ clX(S) such that

y = f (x).Clearly, there must exist an(xm)∈ S∞such that xm → x(Exercise C.14) By continuity off, we then findf (xm)→ f(x).So f (x)∈ clX(f (S))

Exercise 24. Q is countable and R\Qis not So, by the Intermediate Value Theorem ?

Exercise 25 Punch a hole in Rn Now do the same inR.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 aδx > 0such that dY(f (x), f (y)) < ε2

for all y ∈ X with dX(x, y) < δx Now use the compactness of X to find finitely many

sub-S? Can you extendϕtoRn?

Exercise 31 Define g ∈ R[a,b] by g(t) := f (t)− αt, and show that g must have a minimum on[a, b]

Exercise 37 (a) Since u is strictly increasing, {α ∈ R+ : x (α, , α)} is bounded from above, so itssupbelongs toR.Since uis lower semicontinuous, this set is closed, and hence contains itssup

Exercise 40 (a) The idea is to show that in this casef must be upper semicontinuous everywhere Let’s use Proposition 4 to this end Take anyx∈ X,and let(xm)be any real sequence withxm → x.Then

lim sup f (xm) = lim sup f ((xm− x + x∗) + (x − x∗))

= lim sup f (xm− x + x∗) + f (x − x∗).

Now define(ym) := (xm− x + x∗),and notice thatym → x∗.So?

Exercise 41 (a) f − f(0)satisfies Cauchy’s functional equation

(b) Show first that f (λx + (1− λ)y) = λf(x) + (1 − λ)f(y)for any0≤ x, y ≤ 1and

λ∈ Rsuch that0≤ λx + (1 − λ)y ≤ 1.(Note Section F.2.2 contains a generalization of this exercise.)

i 2 for eachm SinceSm 1

i 2 converges — this is why I’m so sure that

fm s are well-defined —S∞

i=m+1

1

i 2 → 0 asm→ ∞.Thus fm → f uniformly.

Exercise 50 Define αm := sup{|ϕm(x)| : x ∈ T } for each m and α := sup{|ϕ(x)| :

x ∈ T } (Is α finite?) Since d∞(ϕm, ϕ) → 0, there exists an M ∈ N such that

|ϕ(x) − ϕm(x)| < 1 for all x ∈ T and m ≥ M Then K := max{α1, , αM, α + 1}

should do it.

Exercise 55 Take T = Nand 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 necessarily closed in X,the closedness of these sets in

T does not imply their closedness in X.You should use uniform continuity.

Exercise 67 Write φ = (φ1, φ2, ), and define φ∗i ∈ RT by φ∗i(x) := inf{φi(w) +Kd(w, x)α : w∈ T }for eachi Now check that φ∗ = (φ∗1, φ∗2, ) does the job.

Exercise 68 (a) Recall the Homeomorphism Theorem (Section 3.1).

(b) LetX be[0, 1] with the discrete metric, andY be[0, 1] with the usual metric Exercise 69 Try the projection operator (Example 5).

Exercise 73 If f is a fixed point of Φ, then, for any 0 < t < 1, we must have

f (t) = f (tm 2

)for any m ∈ N.Then f|[0,1) = 0

Exercise 77 Let0denote then-vector of0s, and write βf(x)for(f1(x), , fn(x))for any x ∈ Bn

Exercise 2. Γ(X) = S.

Exercise 4 For the “only if” part, definef byf (y) := x wherex is any element ofX

withy∈ 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 pi = 0 for somei,use the definition of upper hemicontinuity directly By Example 2, this is the only problematic case.

Exercise 11 If (ym)∈ Γ(S)∞,then there exists an(xm)∈ S∞ such thatym∈ Γ(xm)

for each m Use Theorem C.2 to extract a subsequence of (xm) that converges to some

x∈ S,and then use Proposition 2 to get your hands on a suitable subsequence of(ym)

Exercise 12 (a) Supposef is closed Take anyy∈ Y and letObe an open subset ofX

withf−1(y)⊆ O.Notice thatY\f(X\O)is an open subset ofY that containsy.Moreover,

f−1(Y\f(X\O)) ⊆ O.Conversely, ifSis a closed subset ofXandy∈ Y \f(S),thenX\S

is open inX and f−1(y)⊆ X\S,so there exists aδ > 0such thatf−1(Nδ,Y(y))⊆ X\S

So?

Exercise 21 Use Proposition D.1 and triangle inequality (twice) to verify thatΓ has a closed graph Then apply Proposition 3.(a).

Exercise 22 Let X0 := X and Xi+1 := Γ(Xi) for all i∈ Z+ Recall Example C.8 to

be able to conclude that S :=W∞

Xi 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(xm)∈ X∞ and(ym)∈ R∞ withxm

→ xfor somex∈ X,and

0≤ ym ≤ ϕ(xm)for eachm.Sinceϕis continuous,(ϕ(xm))converges Deduce that(ym)

is a bounded sequence, and then use the Bolzano-Weierstrass Theorem and Proposition 2

to conclude that Γis upper hemicontinuous.

To show that Γ is lower hemicontinuous, take any (x, y) ∈ X × R and (xm) ∈ X∞

such that xm

→ xand 0≤ y ≤ ϕ(x).If y = ϕ(x), then defineym := ϕ(xm) for eachm

Ify < ϕ(x),then there exists an integerM with0≤ y ≤ ϕ(xm)for eachm ≥ M,so pick

ym from [0, ϕ(xm)] arbitrarily form = 1, , M, and letym := y for eachm≥ M.Either way,ym → yand ym ∈ Γ(xm) for eachm.Apply Proposition 4.

Exercise 28. Γ is not upper hemicontinuous at any x ∈ [0, 1] ∩ Q, but it is upper hemicontinuous at anyx∈ [0, 1]\Q To prove the latter fact, recall that[0, 1]∩ Qis 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

(xm) ∈ [0, 1]∞ with xm → x Let (zm) be a sequence in [0, 1] ∩ Q with zm → y and

(wm) a sequence in [0, 1]\Q with wm → y Now define ym := zm if xm is irrational and

ym := wm if xm is rational, m = 1, 2, Then (ym) ∈ [0, 1]∞ satisfies ym ∈ Γ(xm)for eachm, andym → y

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 υm ∈ N1

m ,R(υ)∩ u(T ) such that

Γ(υm)∩ O = ∅ Take any x ∈ Γ(υ) ∩ O, and show that u(x) > υ is impossible If

u(x) = υ, then use the monotonicity ofu

Exercise 32 (a) If a < 0 < b < 1,thendH([0, 1], [a, b]) = max{|a| , 1 − b} (b) Yes Exercise 35 Take any Cauchy sequence (Am) in c(Y ) and define Bm := clY(Am ∪

Am+1∪ · · ·)for eachm.Show thatB1 (hence eachBm)is compact By the Cantor-Fréchet Intersection Theorem,W∞

Bi ∈ c(Y ).Show thatAm →W∞

Bi

Exercise 36 Let F := {f1, , fk} for somek ∈ N Define the self-map Φon c(Y ) by

Φ(A) := f1(A)∪ · · · ∪ fk(A), and check thatΦis a contraction, the contraction coefficient

of which is smaller than the maximum of those of fi s Now apply Exercise 35 and the Banach Fixed Point Theorem.

Exercise 42 Modify the argument given in the last paragraph of the proof of Maximum Theorem.

Exercise 43 Take any (θm) ∈ Θ∞ with θm → θ for some θ ∈ Θ, and suppose that

ϕ∗(θ) < lim sup ϕ∗(θm).Then there exist an ε > 0and a strictly increasing (mk)∈ N∞

such that, for some xm k ∈ Γ(θmk), k = 1, 2, , and K ∈ N, we have ϕ∗(θ) + ε ≤ϕ(xm k, θmk) for all k ≥ K (Why?) Now use the upper hemicontinuity of Γ to extract a subsequence of(xmk)that converges to a point in Γ(θ),and then derive a contradiction by using the upper semicontinuity of ϕ

Exercise 44 (a) Let α := 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 thatΓis upper hemicontinuous It is in fact lower hemicontinuous as well To see this, take any υ ∈ [α, β], x ∈ Γ(υ), and any sequence

(υm)∈ [α, β]∞ with υm → υ.LetI :={m : υm ≤ υ} andJ :={m : υm > υ} Define

S := T ∩ {λx + (1 − λ)(1, , 1) : λ ∈ [0, 1]}.

Clearly,β ≥ υm ≥ υfor allm ∈ J.By the Intermediate Value Theorem, for eachm ∈ J,

there exists an 0 ≤ λm ≤ 1 such that u(λmx + (1− λm)(1, , 1)) = υm Now define

(xm) ∈ T∞ as xm := x if m ∈ I, and as xm := λmx + (1− λm)(1, , 1) otherwise Then xm ∈ Γ(υm) for each m and xm → x (Why?) Conclusion: Γ is a continuous correspondence The stage is now set for applying the Maximum Theorem.

(b) eu(p, u∗(p, ι)) ≤ ι follows from definitions To show that < cannot hold here, you will need the continuity of eu and strict monotonicity of u∗

Exercise 53 (a) Let me formulate the problem in terms of capital accumulation In that case, lettingγ := 1− α,the problem is to choose a real sequence (xm)in order to

Maximize S∞

i=0

δi(pf (xi) − (x i+1 − γx i )) such that γxm ≤ x m+1 ≤ x m + θ, m = 0, 1,

It is easily seen thatf must have a unique positive fixed point, denote it byx.¯ Clearly, the firm will never operate with an input level that exceeds x,¯ so the solution of the problem must belong toX∞, whereX := [0, ¯x] Define 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(xm)∈ X∞ in order to

Maximize S∞

i=0

δiϕ(x i , x i+1 ) such that x m+1 ∈ Γ(x m ), m = 0, 1,

(b) The optimal policy correspondence P forD(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 property,

P (x) ∈ arg max{pf(x) − (y − γx) + δ(pf(y) −(P2(x) − γy)) + δ2V (P2(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 onexthat satisfies this, no?

Exercise 57 Study the correspondence Γ : T ⇒ Rn withΓ(x) := {y ∈ Rn : (x, y) ∈

S},whereT := {x ∈ Rn : (x, y)∈ S for somey∈ S}

Exercise 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 the claim is elementary To prove the “≥” part, define

Γ1 : Y ⇒ X by Γ1(y) := arg max f (·, y) and Γ2 : X ⇒ Y by Γ2(x) := arg min f (x,·)

Now define the self-correspondenceΓonX×Y byΓ(x, y) := Γ1(y)×Γ2(x).Use Kakutani’s Fixed Point Theorem to find an (x∗, y∗) such that (x∗, y∗) ∈ Γ(x∗, y∗), and check that

maxx∈Xminy∈Y f (x, y)≥ f(x∗, y∗)≥ miny∈Y maxx∈Xf (x, y)

Exercise 60 Let T := {x ∈ X : x ∈ Γ(x)}, and note that T =∅ Define g := f|T,

and show that g is a self-map on T.Moreover, if xis a fixed point of g,then it is a fixed point of bothf and Γ

Exercise 65 (a) Let Oy :={x ∈ X : y ∈ Ψ(x)}, y ∈ Rn.Then {Oy : y ∈ Rn

} is an open cover ofX, so since X is compact, there are finitely many y1, , yk

∈ Rn such that

{O(yi) : i = 1, , k}covers X.Now proceed as in the proof of the Approximate Selection Lemma.

(b) Apply part (a) and Brouwer’s Fixed Point Theorem.

Exercise 66 (b) If the claim was false, then, by Carathéodory’s Theorem, we could find

(xm), (zm), (ym,i)∈ X∞ and (λm,i)∈ [0, 1]∞, i = 1, , n + 1 such thatSn+1

for allm∈ N.Use the sequential compactness of X,and the convex-valuedness and closed graph property ofΓto derive a contradiction.

Hints For Selected Exercises of Chapter F

Exercise 2. X ={0} has to be the case.

Exercise 3 Sure. ((0, 1),⊕),where x⊕ y := x + y − 1

2,is such a group.

Exercise 4 (c) Suppose there are y, z ∈ X such that y ∈ Y \Z and z ∈ Z\Y If

(Y ∪ Z, +)is a group, then, y + z ∈ Y ∪ Z,that is, eithery + z ∈ Y orz ∈ Y ∪ Z But

ify + z ∈ Y,thenz =−y + y + z ∈ Y,contradiction.

Exercise 12 Scalar multiplication does not play a role here; recall Exercise 4.(c) Exercise 15 (a) R2.(b) R2

\{(0, 0)}

Exercise 17 Letm ≥ n+ 2,and pick anymvectorsx1, , xm inS Since there can be

at mostn + 1affinely independent vectors in Rn, {x1, , xm

}is affinely dependent That

is, there is a nonzero m-vector(α1, , αm) such that Sm

αixi = 0 and Sm

αi = 0.Let

I :={i : αi > 0} and J := {i : αi < 0}.Then A := {xi : i ∈ I} and B :={xi : i∈ J}

are equal to the task Indeed, whereβi := αi/S

i∈Iαi for eachi∈ I,we haveS

i∈Iβixi

∈co(A)∩ co(B)

Exercise 18 (a) The claim is obviously true if|S| = n + 1.Suppose that it is also true when|S| = m,for an arbitrarily picked integerm≥ n + 1.We wish to show that it would then hold in the case|S| = m + 1as well Indeed, if|S| = m + 1,the induction hypothesis implies that there is at least one vectorxi inW

(S\{Si}),for eachi = 1, , m + 1.We are done if xi = xj for somei = j Otherwise, the cardinality of the set T :={x1, , xm+1

}

exceeds n + 2, so by Radon’s Lemma, there exist disjoints sets A and B in T such that

co(A)∩ co(B) = ∅.Now show that ifx∈ co(A)∩ co(B), thenx∈W

Exercise 28. KL(−x) = KL(x)for all x∈ X

Exercise 30 Start with a basis for null(L) (which exists by Theorem 1), and then extend this to a basis for X by using Theorem 1 If S is the set of vectors you added in this extension, thenL(S) is a basis forL(X)

Exercise 31 (b) Here is how to go from (i) to (iii) Take anyα∈ R\{0}andx, x ∈ X

Suppose z ∈ Γ(αx) Then, by linearity of Γ, α1z ∈ Γ(x), so z ∈ αΓ(z) Thus, Γ(αx) ⊆αΓ(x).Now take any z ∈ Γ(x + x ).Then, for any w∈ Γ(−x ),we have

z + w ∈ Γ(x + x ) + w ⊆ Γ(x + x ) + Γ(−x ) ⊆ Γ(x)

sinceΓis linear Thus z ∈ Γ(x) − w ⊆ Γ(x) + Γ(x ),since, by the first part of the proof,

−w ∈ Γ(x )

(c) Ify∈ Γ(x),then, for any z ∈ Γ(x), we have z = y + z− y ∈ y + Γ(x) − Γ(x) ⊆

y + Γ(0)

(e) We have P (Γ(0)) = {0}, soP ◦ Γ is single-valued by part (a) Take any x ∈ X

and lety := P (Γ(x)).Then there is a z ∈ Γ(x)such thaty = P (z).We havez− P (z) ∈null(P )(sinceP is idempotent), soz−P (z) ∈ Γ(0)which meansz−P (z)+Γ(0) = Γ(0)

by part (c) But then, using part (c) again,Γ(x)− y = z + Γ(0) − y = z − P (z) + Γ(0) =Γ(0).Since 0∈ Γ(0),it follows that y∈ Γ(x)

Exercise 32 (a) InR2 a carefully selectedS with|S| = 4 would do the job.

(b) Take any λ ∈ R and x, y ∈ S, and suppose z := λx + (1− λ)y ∈ S If λ < 0,

Exercise 34 The problem is to show that injectivity and surjectivity alone are able

to entail the invertibility of L Let S := {x1, , xm

} be a basis for X, m ∈ N, and let

yi := L(xi) for each i In both cases, the idea is to define L−1 as “the” linear operator

in L(X, X) with L−1(yi) = xi If {y1, , ym

} is a basis for X, then we are fine For in that case there is a unique F ∈ L(X, X) with F (yi) = xi, and this F is the inverse of

L.(A linear operator is determined completely by its actions on the basis, right?) Well, if

L is injective, then L(S) is linearly independent in X, and if L is surjective, L(S) spans

L(X) = X.(Why?) So, in either case, L(S) is a basis forX

Exercise 35 (b) Let A be a basis for X and B a basis for Y Suppose |A| ≤ |B|, and let Σ be the set of all injections in BA For any σ ∈ Σ, define Lσ ∈ L(X, Y ) by

Lσ(x) := σ(x) for allx∈ A.(IsLσ well-defined?) Now show that{Lσ : σ ∈ Σ}is a basis forL(X, Y )while |Σ| = |A| |B| A similar argument works for the case|A| > |B|as well Exercise 37. 0∈ H/ is needed only for the existence part.

Exercise 52 Define xi := v(Ai∪ {i}) − v(Ai)whereA1 =∅and Ai :={1, , i − 1}

for eachi≥ 2.Show thatx∈ core(v)

Exercise 53 This is much easier than it looks Use P (∅, v) = 0 and the Principle of Mathematical Induction to show that there is a unique potential function To prove the second claim, fix any (N, v) ∈ G, and use Proposition 8 to write v = S

A∈N αvAuA for someαv

Hints For Selected Exercises of Chapter G

Exercise 4 Let H = {x : L(x) = α}, where L ∈ L(X, R) and α ∈ R Use the convexity ofS and the Intermediate Value Theorem to show that if L(x) > α > L(y)for

somex, y ∈ S,thenL(z) = α for somez ∈ S.

Exercise 8 (e) Let Γ(x0) = {z}, and let y, y ∈ Γ(x) for some x ∈ X Let x :=2x0 − x (which belongs to the domain of Γ since X is a linear space), and observe that

1

2{y, y } + 12Γ(x )⊆ {z}by convexity ofΓ.Deduce that y = y

Exercise 9 (b)Cis infinite dimensional, because{(1, 0, 0, ), (0, 1, 0, ), }is a basis forc0.Moreover, the sequences(xm) := (1,−1, 0, 0, )and(ym) := (−2, 1, 0, 0, )belong

toC,but 12(xm) + 12(ym) /∈ C ∪ {(0, 0, )}

(c) For any vector in c0 , there exist two vectors inC such that that vector lies at the midpoint of those two vectors.

Exercise 12 The question is ifco(C) is a cone (Yes it is!)

Exercise 13 The hardest part is to show that (ii) implies (iii) To show this, take any

x0 ∈ B and letY :=aff (B) − x0.ThenY is a linear subspace of X such thatx0 ∈ X\Y

Take any basisS forY,and show that there is a basis T forX withS∪ {x0} ⊆ T.For all

x ∈ X,there exists a unique λx : T → R such that x =S

z∈T λx(z)z,where, of course,

λx(z) = 0 for only finitely manyz ∈ T.(Why?) Define L∈ L(X, R)by L(x) := λx(x0);

that should do the job.

Exercise 14 Recall that span(X+) = X+− X+ (Exercise 10).

Exercise 20 (a) Let n := dim(aff (S)) The case n = 0 is trivial, so assume n ≥ 1,

and pick anyx1 ∈ S.Thenspan(S − x1)is an n-dimensional linear space, so it contains n

linearly independent vectors, sayx2

− x1, , xn+1

− x1.(This means that{x1, , xn+1

}is affinely independent in X.)Now show thatSn+1 1

n+1xi

∈ ri(co{x1, , xn+1

}) ⊆ ri(S)

(b) Use part (a) and Observation 1.

Exercise 22 No Evaluate al-int(S) and al-int(al-int(S))for the set S considered at the end of Example 4, for instance.

Exercise 25. al-intC[0,1](C[0, 1]+) ={f ∈ C[0, 1] : f 0}

Exercise 26 Let x ∈ al-intX(S), and pick any y ∈ X By definition, there exist

αx+y > 0 and βx−y > 0 such that x + αy = (1− α)x + α(x + y) ∈ S and x− βy =(1− β)x + β(x − y) ∈ S for all 0 ≤ α ≤ αx+y and 0 ≤ β ≤ βx−y Thus, defining

εy := min{αx+y, βx−y}establishes the “only if” part of the claim To prove the converse, assume that x satisfies the stated property, pick any y ∈ X, and use the property with respectx− y

Exercise 27 (a) Observe that (1− α)x ∈ al-intX((1− α)S)for any 0≤ α < 1, and use Exercise 26.

(b) That aff (S) = X is immediate from Observation 1 Let T := al-intX(S), and

to derive a contradiction, suppose aff (T ) ⊂ X Then, by Observation 1, al-intX(T ) =∅

Now use part (a) and the convexity of S to show that we cannot haveal-intX(T ) =∅

Exercise 30 (a) R\Q is algebraically closed in R,but Q is not algebraically open in

R

(b) Ifx is inS\al-intX(S), then, by convexity of S,there must exist ay∈ X\S such that(1− α)x + αy ∈ X\S for all 0 < α≤ 1.

Exercise 33 Letx∈W

{al-cl(A) : A ∈ A}and pick anyy∈W

{ri(A) : A ∈ A}.Now use the Claim I proved in Example 10 to getx∈ al-cl(W

Exercise 42 The argument underlying the Hahn-Banach Extension Theorem 1 will

do fine here Take any positive L ∈ L(Y, R) and z ∈ X\Y, and try to find a positive

K ∈ L(Z, R), whereZ := span(Y ∪ {z}) To this end, define A :={y ∈ Y : z X y}

and B :={y ∈ Y : y X z}, and check that these are nonempty sets Show that there is

an α∈ R withsup L(A)≤ α ≤ inf L(B) This number will do the job of αI used in the proof of Hahn-Banach Extension Theorem 1.

Exercise 44 Pick any x∈ S.DefineT := S− x, y := z − x,and Y := span(S − x)

Then al-intY(T ) =∅ and y∈ Y \T.First apply Proposition 1 and then Corollary 2 Exercise 45 False, even inR2!

Exercise 46 Separate{(x, t) : t ≥ ϕ(x)} and {(x, t) : t ≤ ψ(x)}

Exercise 59 The first possibility can be written asSm

w1juj+Sm

wj2(−uj) = v for somew1, w2

∈ Rm

+.Now use Farkas’ Lemma.

Exercise 60 Define the (n + 1)× nmatrix B:= [bij], where

Exercise 61 (a) Follow through the argument given before the statement of the result That is, show that A− B is a closed set, and then apply the argument outlined in the second paragraph of Proposition 4.

(b) Show that clRn(A− B)is a closed and convex set, and use Proposition 4.

Exercise 62 Use the Minkowski Separating Hyperplane Theorem to find a nonzero

k ∈ N, and suppose the claim is true for dim(S) ≤ k − 1 Now consider the case where

dim(S) = k Take any x ∈ S If x ∈ al-bdR n(S), then there exists a hyperplane H

that supportsS atx But then dim(H ∩ S) ≤ k − 1 (yes?), so the induction hypothesis

applies If, on the other hand, x ∈ al-intR n(S), take any y ∈ S\{x} (which exists since

dim(S) > 0) and observe that the line {λx + (1 − λ)y : λ ∈ R} intersects S in exactly two points in al—bdRn(S) Use what you learned in the previous case to express each of these points as convex combinations of the extreme points ofS

Exercise 73 By Example F.9, there is aw∈ Rn such thatL(y) = wy.ButL(y) = ωy

for all y∈ Y and ω ∈ w + Y⊥

Exercise 76 By the Projection Theorem, any x∈ Rn can be written asx = yx+ zx

for some unique(yx, zx)∈ Y × Y⊥ DefineP byP (x) := yx

Hints For Selected Exercises of Chapter H

Exercise 1 (a) Pick any s, t ∈ S with s t, and any r ∈ al-intY(S) Show that there is an 0 < α < 1 small enough so that s∗ := (1− α)r + αs ∈ al-intY(S) and

t∗ := (1−α)r+αt ∈ al-intY(S).Now take anyσ∈ Y and use weak continuity and the fact thats∗ ∈ al-intY(S)to find aθσ > 0small enough so thatS (1− θ)s∗+ θ(σ + t∗) t∗

for all 0≤ θ ≤ θσ.Conclusion: (1− θ)(s∗− t∗) + θσ ∈ Afor all 0≤ θ ≤ θσ

(b) span(C) ⊆ span(S) = {λ(s − t) : λ > 0 ands, t∈ S}

Exercise 8 This is a showcase for the Minkowski Separating Hyperplane Theorem Exercise 9 Use Bewley’s Expected Utility Theorem.

Exercise 11 (b) A variational preference need not satisfy theC-Independence Axiom∗

Exercise 17 ConsiderLfirst onA,and then use the Hahn-Banach Extension Theorem 1.

Exercise 25 This axiom, SYM, C.INV and IIA imply Axiom PO.

Exercise 27 (b) Roth (1980) argues strongly that the “right” solution here is(1

2,1

2, 0)

(But Aumann (1985) disagrees.)

Exercise 28 This requires you to find a clever way of using Kakutani’s Fixed Point Theorem.

Hints For Selected Exercises of Chapter I

Exercise 8 For (b), use the continuity of vector addition to show that A + x is open for anyx.Now take the union over allx inB

Exercise 9. Q∞ doesn’t work Why?

Exercise 12 Let X be a metric linear space andSx :={λx : λ ∈ R} for eachx∈ X

A0∪A1∪···,and show thatcl[0,1](A) = [0, 1]whileλΓ(x)+(1−λ)Γ(y) ⊆ Γ(λx+(1−λ)y)

for any x, y ∈ S and λ∈ A.But then, for any0 < α < 1,there exists a (λm)∈ A∞ such that λm → αand Gr(Γ) is λm -convex for each m Now use the closed graph property of

Γand the continuity of scalar multiplication and vector addition.

Exercise 14 For each m ∈ N, definefm ∈ C1[0, 1] by fm(t) := m1 sin mt Check that

d∞(fm, 0) → 0 whereas d∞(fm, 0) = 1 for each m Conclusion: D is a discontinuous linear operator.

Exercise 20.x→ |L(x)|is a continuous seminorm on X when Lis continuous.

Exercise 21 (a) The “only if” part follows from the definition of continuity To see the “if” part, pick any K > 0 with |L(x)| ≤ K for all x ∈ O Take any (xm) ∈ X∞

Exercise 22 (a) Take any ε > 0 and x ∈ X Then there exists a δ > 0 with

Γ(Nδ,X(0))⊆ Nε,Y(Γ(0)) So, recalling Exercise F.31.(b),

Now consider the sequence(y− zm)

Exercise 23 (a) We haveΓ(x) = L(x) + Γ(0)for someL∈ L(X, Y ).(Recall Exercise F.31.(c).)

(b) Recall Exercise F.31.(e).

Exercise 26 (a) Take anyx∈ O.Given thatOis open and convex, we can find ay∈ O

and a0 < λ < 1such thatx0 = λx+(1−λ)y.(Why?) ThenλΓ(x)+(1−λ)Γ(y) ⊆ Γ(x0),

so if I choose anyz ∈ Γ(y),I get Γ(x)⊆ λ1(Γ(x0)− (1 − λ)z)

Exercise 30 You know that cone(S)must be algebraically closed from Theorem G.1 Exercise 33 Note that xi+ θNε,X(0) = θNε,X(xi)and imitate the argument I gave to prove Lemma C.1.

Exercise 38 (b) The closed unit ball BX is a bounded set, obviously Is idX(BX)

compact whenX is infinite dimensional?

Exercise 39 (a) The image of a relatively compact set must be relatively compact under

L

Exercise 40 Start as in the hint I gave above for Exercise 13 — 12 was not really playing

an essential role there.

Exercise 45 Observe that, by Proposition 11, it is enough to show that co(S) ⊆al-intX(co(S)) to prove the second claim Let x ∈ co(S) Then x can be written as a convex combination of finitely many members of S Let’s assume that two members of S

are enough for this (since this is only a hint), that is,x = λy + (1− λ)z for somey, z ∈ S

and 0≤ λ ≤ 1 Now take anyw ∈ X Since y, z ∈ S and S is open, there exists a small enoughα∗ > 0such that(1− α∗)y + α∗w∈ S and (1− α∗)z + α∗w∈ S.(Yes?) Thus (1 − α∗)x + α∗w = λ ((1 − α∗)y + α∗w) + (1 − λ) ((1 − α∗)z + α∗w) ∈ co(S) Sinceco(S)is convex, the line segment between(1− α∗)x + α∗wandxis contained within

co(S),so(1− α∗)x + α∗w∈ co(S)for all 0≤ α ≤ α∗

Exercise 47 (c) Let me show that clX(S) ⊆ clX(intX(S)) Let y ∈ clX(S) Then there is a(ym)∈ S∞ with ym → y Pick any x∈ intX(S) and use Lemma 3 to conclude

Exercise 48 The answer to the question is: No.

Exercise 50 Either recall the proofs of Propositions G.2 and Theorem G.2, or use these results along with those of Section 5.2.

Exercise 51 Recall Example G.12.

Hints For Selected Exercises of Chapter J

Exercise 2 False Recall(R2, d1/2)

Exercise 6 Fix any x, y ∈ X and β ≥ α > 0, and show that claimed inequality is equivalent to the following: β

are Lipschitz continuous, then so is ϕ|A∪B To prove this, take arbitrary points x and y

in A and B, respectively, and find (z, w) ∈ A × B with z = x and w = y Now define

T :=co{z, w}, and notice thatϕ|T is concave Deduce from this that

ϕ(y)−ϕ(x) x−y ≤ ϕ(x)−ϕ(z)x−z ≤ K A and ϕ(x)−ϕ(y)x−y ≤ ϕ(y)−ϕ(w)y−w ≤ K B , whereKA and KB are the Lipschitz constants ofϕ|A and ϕ|B,respectively LettingK :=max{KA, KB}, therefore, we find|ϕ(x) − ϕ(y)| ≤ K x − y

Exercise 21 Recall Corollary I.1.

Exercise 22 Let (xk) be a sequence (of sequences) in c0, and suppose

2 for all k ≥ K, and letM := MK.Clearly,|xm| < εfor all m≥ M

And yes, c is also a closed subspace of ∞

Exercise 28 Recall Dirichlet’s Rearrangement Theorem.

Exercise 31 (a) Pick any z ∈ X\Y and let β := d · (z, Y ) > 0 For any (fixed)

0 < α < 1,there must exist a y∈ Y withβ ≤ z − y ≤ βα Definex := z−y1 (z− y)

Exercise 32 Yes No.

Exercise 34 Pick any open convex set O1 with 0 ∈ O1 ⊆ N1,X(0), and say d1 :=diam(O1) Findk1 ∈ N and x1(1), , xk1(1) ∈ S such that S ⊆ Vk 1

clX(Nd 1 ,X(xi(1)))

Now pick any open convex setO2 with0∈ O2 ⊆ Nd 1 ,X(0),and sayd2 :=diam(O2).Find

k2 ∈ N and x1(2), , xk2(2)∈ S such that S ⊆Vk 2

clX(Nd 2 ,X(xi(2))) Proceed this way inductively to obtain a sequence(Om)of open convex sets such that, for eachm = 2, 3, ,

(i)0∈ Om ⊆ Nd m −1 ,X(0);and (ii) there exist akm ∈ Nand x1(m), , xk m(m)∈ S with

S ⊆ Vk mclX(Nd m ,X(xi(m))) (Here dm := diam(Om) for each m.) Now let S(m) :=co{x1(m), , xkm(m)}and define the self-correspondenceΓm on S(m)by

Γm(x) := (Γ(x) + clX(Ndm,X(0))) ∩ S(m),

m = 1, 2, Proceed as in the proof of the Glicksberg-Fan Fixed Point Theorem.

Exercise 36 Recall how I proved Corollary E.2.

Exercise 38 (c) Since clX(Φ(S))is compact, there exists a subsequence of(Φ(y(m)))

that converges inclX(Φ(S)) The limit of this subsequence must be a fixed point ofΦ

Exercise 41 Let S be any nonempty closed, bounded, and convex subset of X such that 0 ∈ intX(S) (What if intX(S) = ∅? What if 0 ∈ int/ X(S)?) Define the retraction

r ∈ SX by r(x) :=

x, ifx∈ S

1

ϕS(x)x, otherwise , where ϕS is the Minkowski functional of S

Proceed as in the proof of Proposition D.10.

Exercise 42 (b) For any f ∈ F, if x ∈ Fix(f) and g ∈ F, then g(x) = g(f (x)) =

f (g(x)),sog(x)∈ Fix(f).That is, g is self-map on Fix(f ), for any f, g∈ F

Exercise 46 Take an arbitrary k > 0, and let Ck stand for the class of all Lipschitz continuous real functionsgon[0, 1]such thatg(0) = 0andkis greater than or equal to the Lipschitz constant ofg Check thatCk is a convex set, and use the Arzelà-Ascoli Theorem

to prove that Ck is a compact subset of C[0, 1] Now define the continuous self-mapΦ on

Ck by Φ(f )(x) := H(x, f (h(x))) Check that Φ(Ck) ⊆ Ck for anyk > 1−Kqp , and apply the Schauder Fixed Point Theorem 1.

Exercise 47 Use Krasnoselski˘ı’s Theorem.

Exercise 58 (This is quite similar to Example 6.[1].) Letting M := x∗ , we have

|L(x)| ≤ M x for anyx∈ X,by the Cauchy-Schwarz Inequality (of Exercise 12).

Exercise 60 For anyx∈ X,we have K(L(x)) Z ≤ K ∗ L(x) Y ≤ K ∗ L ∗ x

Exercise 63 (b) We have f (x,·) ∗ ≤ |||f||| x for anyf ∈ bX,Y and x∈ X

(c) Fix an arbitrarily small ε > 0.Since(Lm)is Cauchy, there is anM ∈ Nsuch that

|Lk(x)− Ll(x)| < εfor allx∈ Xandk, l≥ M.But thenε ≥ liml→∞|Lk(x)− Ll(x)| =

|Lk(x)− L(x)| for all x∈ X and k ≥ M

Exercise 70 Let H be the set of all closed halfspaces that contain S Since W

H is closed and convex, thatcoX(S)⊆W

His immediate Conversely, ifx /∈ coX(S),then, by Theorem 1, there exist an α∈ R and a nonzero L∈ X∗ such that inf L(coX(S))≥ α >L(x).So x /∈ L−1([α,∞)) ∈ H.(Could you use Proposition G.2 here?)

Exercise 82 LetY :=span{x} and defineL∈ Y∗ by L(λx) := λ x for any λ∈ R

Now use the Hahn-Banach Extension Theorem 2.

Exercise 83 We have L0 ∗ = L0(x)

Exercise 84 By Exercise 82, x ≤ sup{|L(x)| : L ∈ X∗ and L ∗ ≤ 1}.Now invoke the definition of · ∗

Exercise 86 There must exist az∈ X withd · (z, Y ) > 0.LetZ :=span(Y ∪{z}) =

Y +{λz : λ ∈ R}.DefineL0 ∈ RZ byL0(y + λz) := λd · (z, Y )for ally ∈ Y andλ∈ R,

and check that L0 ∈ Z∗ (In fact, L0 ∗ = 1.) Now apply the Hahn-Banach Extension Theorem 2.

Exercise 89 Suppose there is a subspace Y of X and an L0 ∈ Y∗ such thatL1|Y =

L0 = L2|Y and L1 ∗ = L0 ∗ = L2 ∗ for two distinctL1, L2 ∈ X∗.IfL := 12L1+12L2 , thenL∈ X∗ and L ∗ = L0 ∗.Show that this contradicts the rotundity of X∗