Mathematics II, course 2013-2014 Juan Pablo Rinc´on–Zapatero October 24, 2013 Summary: The course has four parts that we describe below. (I) ”Topology in Rn” is a brief review of the main concepts and properties of the Euclidean space, including open and closed sets, compact sets, sequences, limits, continuity and differen- tiability of functions, and separation of convex sets; (II) ”Metric Spaces” generalize topological concepts of the Euclidean space to arbitrary sets endowed with a metric, with the objective of understanding Theorem of Banach for contraction mappings; (III) ”Parametric optimization problems” studies parametric continuity and parametric mono- tonicity of solutions of optimization problems. (IV) ”Fixed point theorems” analyzes the theorems of Brower and Kakutani for functions and correspondences, respectively and the Theorem of Banach for contraction mappings in complete metric spaces. In parts (II), (III) and (IV), several applications to economic problems are given, as the existence of Nash equilibria in noncooperative games, consumer theory, or the existence of equilibria in a pure exchange economy. Bibliography: • Apostol, T.M. Mathematical Analysis, Second Edition, Addison-Wesley, 1974. • Berge, C. Espaces Topologiques. Fonctions Multivoques, Deuxime dition, Dunod, 1966. • Rudin, W. Principles of Mathematical Analysis, Third Edition, McGraw-Hill, 1976. • ———–. Real and Complex Analysis, Third Edition, McGraw-Hill, 1987. • Ok, F.A. Real Analysis with Economic Applications, Princeton University Press, 2007. • Sundaram, R.K. A First Course in Optimization Theory, Cambridge University Press, 2005. • Sydsaeter, Hammond, Seierstad, Strom. Further Mathematics for Economic Analysis, Second Edition, Prentice Hall, 2008. 1
Transcript
Mathematics II, course 2013-2014
Juan Pablo Rincon–Zapatero
October 24, 2013
Summary:The course has four parts that we describe below.(I) ”Topology in Rn” is a brief review of the main concepts and properties of the Euclideanspace, including open and closed sets, compact sets, sequences, limits, continuity and differen-tiability of functions, and separation of convex sets;(II) ”Metric Spaces” generalize topological concepts of the Euclidean space to arbitrary setsendowed with a metric, with the objective of understanding Theorem of Banach for contractionmappings;(III) ”Parametric optimization problems” studies parametric continuity and parametric mono-tonicity of solutions of optimization problems.(IV) ”Fixed point theorems” analyzes the theorems of Brower and Kakutani for functions andcorrespondences, respectively and the Theorem of Banach for contraction mappings in completemetric spaces.
In parts (II), (III) and (IV), several applications to economic problems are given, as the existenceof Nash equilibria in noncooperative games, consumer theory, or the existence of equilibria ina pure exchange economy.
Bibliography:
• Apostol, T.M. Mathematical Analysis, Second Edition, Addison-Wesley, 1974.
We start with some concepts that are probably very familiar to you. If x, y ∈ Rm, the innerproduct of the two vectors is
x · y =m∑i=1
xiyi.
The main properties of the inner product are: i) x · x ≥ 0 and x · x = 0 if and only if x = 0;ii) x · y = y · x; iii) (λx) · y = λ(x · y), λ ∈ R; iv) (x + y) · z = x · y + y · z.
We define the Euclidean norm of x by
‖x‖ = (x · x)1/2.
Lemma 1.1 (Cauchy–Schwartz inequality). If x, y ∈ Rm, then |x · y| ≤ ‖x‖‖y‖.
Proof. Let a = ‖x‖2, b = ‖y‖2, and c = x ·y. If b = 0, there is nothing to prove, so we supposethat b > 0. Then, for any λ ∈ R
0 ≤n∑i=1
(xi − λyi)2 = a− 2λc+ λ2b.
Letting λ = c/b, we obtain 0 ≤ a− c2/b and so c2 ≤ ab.
Lemma 1.2. The Euclidean norm satisfies
1. ‖x‖ ≥ 0;
2. If ‖x‖ = 0, then x = 0;
3. If λ ∈ R, then ‖λx‖ = λ‖x‖;
4. (The triangle inequality) ‖x + y‖ ≤ ‖x‖+ ‖y‖.
Proof. The triangle inequality is a consequence of the Cauchy–Schwartz inequality
If we set x = a− b and y = b− c, then the triangle inequality becomes
‖a− c‖ ≤ ‖a− b‖+ ‖b− c‖
that means that the length of one side of a triangle is less than or equal to the sum of thelengths of the other two sides.
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1.2 Sequences and limits
Definition 1.3. A sequence x1,x2, . . . tends in the Euclidean norm of Rm to a limit x as ntends to infinity
xn → x as n→∞,
if, given any ε > 0, there exists an n0(ε) such that
‖xn − x‖ < ε for all n ≥ n0(ε).
Lemma 1.4. We work in Rm with the Euclidean norm.
1. The limit is unique;
2. If xn → x and n(1) < n(2) < · · · < n(j) < · · · , then xn(j) → x as j →∞.
3. If xn → x and yn → y, then xn + yn → x + y;
4. If xn → x and λn, λ ∈ R with λn → λ, then λnxn → λx.
Next theorem establishes that any bounded sequence has a convergent subsequence.
Theorem 1.5 (Bolzano–Weierstrass). If xn ∈ Rm and there exists a K such that ‖xn‖ ≤ Kfor all n, then we can find n(1) < n(2) < · · · < n(j) < · · · and x ∈ Rm such that xn(j) → x asj →∞.
1.3 Open, closed and bounded sets
Definition 1.6. Let x ∈ Rm and r > 0. The open ball of center x and radius r is the set
B(x, r) = y ∈ Rm : ‖x− y‖ < r.
The closed ball of center x and radius r is the set
B(x, r) = y ∈ Rm : ‖x− y‖ ≤ r.
Definition 1.7. A set U ⊆ Rm is open if, whenever x ∈ U , there exists an r > 0 such thatB(x, r) ⊆ U .
Thus every point of an open set is surrounded by a ball consisting only of points of the set.
Definition 1.8. The set N is a neighborhood of the point x if we can find an r > 0 such thatB(x, r) ⊆ N .
Now we identify sets that are closed under the operation of taking limits.
Definition 1.9. A set F ∈ Rm is closed if whenever xn ∈ F for each n and xn → x as n→∞,then x ∈ F .
Lemma 1.10. A subset U of Rm is open if and only if its complement Rm\U is closed.
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Proof.Necessity: Suppose that U is open and let xn ∈ Rm\U with xn → x as n → ∞. Arguing bycontradiction, if x /∈ Rm\U , then x ∈ U ; since U is open, B(x, ε) ⊆ U for some ε > 0. Asxn → x, there exists N such that ‖xN − x‖ < ε, and so xN ∈ B(x, ε) ⊆ U , contradicting thestatement that xn ∈ Rm\U , arriving to a contradiction. Thus, x ∈ Rm\U and Rm\U is closed.Sufficiency: Suppose that Rm\U is closed. Arguing by contradiction, if U is not open, thenthere must exist an a ∈ U such that, given any ε > 0, there exists y /∈ U with ‖y − a‖ < ε.In particular, we can find xn ∈ Rm\U such that ‖xn − a‖ < 1/n. This means that xn → a asn→∞, and so, since Rm\U is closed, a ∈ Rm\U , hence a ∈ U ∩ Rm\U , which is absurd.
Lemma 1.11. Let x ∈ Rm and r > 0.
1. The open ball B(x, r) is open.
2. The closed ball B(x, r) is closed.
Lemma 1.12. Consider the collection τ of open sets in Rm.
1. ∅ ∈ τ , Rm ∈ τ ;
2. If Uα ∈ τ for all α ∈ A, then⋃α∈A Uα ∈ τ ;
3. If U1, U2, . . . , Un ∈ τ , then⋂nj=1 Uj ∈ τ .
Lemma 1.13. Consider the collection F of closed sets in Rm.
1. ∅ ∈ F , Rm ∈ F ;
2. If Fα ∈ F for all α ∈ A, then⋂α∈A Fα ∈ F ;
3. If F1, F2, . . . , Fn ∈ F , then⋃nj=1 Fj ∈ F .
Example 1.14. The intersection of open sets need not be open. In R we have
∞⋂j=1
(−j−1, 1) = [0, 1).
To show this, let x belongs to the intersection set at the left of the inequality. Then, −j−1 <x < 1 for every j, so 0 ≤ x < 1. Conversely, if 0 ≤ x < 1, then −j−1 < x < 1 for any j andthen x ∈
⋂∞j=1(−j−1, 1).
Definition 1.15. An open cover of a set E in Rm is a collection Uα of open subsets of Rmsuch that E ⊆ ∪αUα.
Definition 1.16. A subset K of Rm is said to be compact if every open cover of K contains afinite subcover.
This means that if Uα is an open cover of K, then there are finitely many indices α1, . . . , αnsuch that
K ⊆ Uα1 ∪ · · · ∪ Uαn .
Theorem 1.17 (The Heine–Borel Theorem ). A subset K in Rm is compact if and only if itis closed and bounded1.
1I quote here the following poem of Conway:
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Theorem 1.18.
1. If K is a compact subset in Rm, then every sequence in K has a subsequence convergingto a point of K;
2. If K is a subset of Rm such that every sequence in K has a subsequence converging to anelement of K, then K is a compact set.
Proof. 1. If xn ∈ K, then it is bounded and so, by the Theorem of Bolzano–Weierstrass,has a convergent subsequence xn(j) → x ∈ K since K is closed.
2. We argue by contradiction, supposing first that K is not closed. Then, we can findxn ∈ K and x /∈ K such that xn → x as n→∞. Since any subsequence of a convergentsequence converges to the same limit, no subsequence of the xn can converge to a pointof K: Second, if K is not bounded we can find xn ∈ K such that ‖xn‖ > n. If x is anypoint in Rm, then the inequality
‖xn − x‖ ≥ ‖xn‖ − ‖x‖ > n− ‖x‖
shows that no subsequence of the xn can converge.
1.4 Completeness of Rm
One of the most useful criteria of convergence of sequences is Cauchy’s Principle of Convergence,which allows us to prove the convergence of a sequence in Rm without any knowledge aboutthe limit.
Definition 1.19. A sequence xn defined in Rm is a Cauchy sequence if for every ε > 0 thereexists an integer n0(ε) such that |xp − xq| < ε for all p, q ≥ n0(ε).
The followings lemma holds in any metric space, not only in Rm with the Euclidean metric.
Lemma 1.20. Any convergent sequence in Rm is a Cauchy sequence.
Proof. Suppose that xn → x as n→∞. Let ε > 0. We can find n0(ε) such that
‖xn − x‖ < ε
2∀n ≥ n0(ε).
Then‖xp − xq‖ ≤ ‖xp − x‖+ ‖xq − x‖ < ε
2+ε
2= ε, ∀p, q 6= n0(ε).
The following result, however, is not true in arbitrary metric spaces.
If K is closed and bounded, say Heine–Borel,And also Euclidean, then we can tellThat, if it we smotherWith a large open cover,There’s a finite refinement as well!
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Lemma 1.21. Any Cauchy sequence in Rm converges.
Proof. It is easy to see that a Cauchy sequence is bounded. By definition, given any ε > 0,there exists n0(ε) such that ‖xp − xq‖ < ε for all p, q ≥ n0(ε). In particular, if n ≥ n0(1), wehave
‖xn‖ ≤ ‖xn − xn0(1)‖+ ‖xn0(1)‖ < 1 + ‖xn0(1)‖.
Hence ‖xn‖ ≤ max1≤r≤n0(1) ‖xr‖ + 1. Then, by the Bolzano–Weierstrass theorem, there isa subsequence xn(j) → x as j → ∞. Thus, given any ε > 0, one can find J(ε) such that‖xn(j) − x‖ < ε for all j ≥ J(ε). Take now n ≥ n0(ε/2), j ≥ J(ε/2) and n(j) > n0(ε/2). Then
Theorem 1.22 (Cauchy’s Principle of Convergence). A sequence in Rm converges if and onlyif it is a Cauchy sequence.
1.5 Continuous functions
Definition 1.23. Let D ⊆ Rm. We say that a function f : D → Rp is continuous at x ∈ D if,given ε > 0, there exists δ(ε,x) > 0 such that, whenever y ∈ D and ‖x− y‖ < δ(ε,x), we have
‖f(x)− f(y)‖ < ε.
If f is continuous at every x ∈ D, we say that f is a continuous function on D.
Lemma 1.24. Let D ⊆ Rm and f : D → Rp. Let xn,x ∈ D such that xn → x as n → ∞.Then f is continuous at x if and only if f(xn)→ f(x) as n→∞.
Proof. If f is continuous at x, for every ε > 0 there exists δ(ε,x) > 0 such that for any ysatisfying ‖x − y‖ < δ(ε,x), we have ‖f(x) − f(y)‖ < ε. But if xn converges to x as n → ∞,there exists an integer n0(δ(ε,x)) such that n ≥ n0 implies ‖x− xn‖ < δ(ε,x) and hence that‖f(x)− f(xn)‖ < ε. Hence the sequence f(xn) converges to f(x) as n→∞.
To prove the sufficient condition, suppose that it is satisfied but that f is not continuous atx. Then there exists ε > 0 such that for any δ > 0 we can find y satisfying ‖x − y‖ < δ and‖f(x) − f(y)‖ ≥ ε. Let δn → 0 as n → ∞ and xn with ‖x − xn‖ < δ and ‖f(x) − f(xn)‖ ≥ ε.The sequence xn converges to x as n → ∞, hence the sequence f(xn) converges to f(x) asn → ∞. But this is in contradiction with the inequality ‖f(x) − f(xn)‖ ≥ ε for all n. Theassumption that f is not continuous at x is thus false.
Another way of looking at continuity, which is crucial to define continuity in topologicalspaces without a metric, is given in the following result.
Lemma 1.25. The function f : Rm → Rp is continuous if and only if f−1(U) is open wheneverU is an open set in Rp.
Theorem 1.26 (Weierstrass’ Theorem). Let K be a compact subset of Rm and f : K → Rp acontinuous function. Then f(K) is closed and bounded.
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Proof. By Theorem 1.18 we need only prove that any sequence in f(K) has a subsequenceconverging to a limit in f(K). Let yn ∈ f(K). By definition, ∃xn ∈ K with f(xn) = yn.Since K is closed and bounded, there is n(j) → ∞ and x ∈ K such that xn(j) → x, thusyn(j) = f(xn(j))→ f(x) ∈ f(K) since f is continuous.
Theorem 1.27 (Extreme Value Theorem). Let K be a compact subset of Rm and f : K → Ra continuous function. Then there exists x1,x2 ∈ K such that
f(x2) ≤ f(x) ≤ f(x1)
for all x ∈ K.
Proof. Since f(K) 6= ∅ is bounded, it has a supremum, M = sup f(K). Since f(K) is closed,M ∈ f(K), that is to say, M = f(x1) for some x1 ∈ K. We obtain x2 similarly.
That is, a real–valued continuous function on a closed bounded set attains global minimumand maximum on that set.
1.6 Convex and concave functions
Definition 1.28. A subset C ⊆ Rm is convex if λx + (1− λ)y ∈ C for all x, y ∈ C and each0 ≤ λ ≤ 1. A function f : C −→ R is said to be convex provided
f(λx + (1− λ)y) ≤ λf(x) + (1− λ)f(y)
for all x, y ∈ C and each 0 ≤ λ ≤ 1.
The function f is concave if −f is convex.
Example 1.29 (Young’s inequality). For any 1 < p, q <∞,1
p+
1
q= 1
ab ≤ ap
p+bq
q(a, b > 0).
The function x 7→ ex is convex, thus using the definition of convexity we have
ab = eln a+ln b = e1p
ln ap+ 1q
ln bq ≤ 1
peln ap +
1
qeln bq =
ap
p+bq
q.
Theorem 1.30. Suppose f : Rm −→ R is convex. Then for each x ∈ Rm there exists p ∈ Rmsuch that the inequality
f(y) ≥ f(x) + p · (y − x) (1)
holds for all y ∈ Rm.
Definition 1.31. The vector p in (1) above is called a subgradient of f at x. The set of allsubgradients at x is called the subdifferential of f at x and this set is denoted ∂f(x).
Example 1.32.
∂(|x|) =
−1, if x < 0;
[−1, 1], if x = 0;
1, if x > 0.
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Remark 1.33. (i) The mapping y 7→ f(x) + p · (y− x) determines the supporting hyperplaneof f at x. Inequality (1) says the graph of f lies above each supporting hyperplane. If f isdifferentiable at x, then p = ∇f(x).
(ii) If f is of class C2, then f is convex (concave) if and only if the Hessian matrix Hf(x)is positive (negative) semidefinite or positive (negative) definite. This means
m∑i,j=1
fxixj(x)hihj ≥ 0(≤ 0) ∀x,h ∈ Rm.
Example 1.34. The function f(K,L) = KαL1−α with 0 < α < 1 is concave in R+×R+. It iscontinuous in R+ × R+ and of class C2 in R++ × R++. The derivatives are:
)The first principal minor have negative sign, and the second is the determinant, which is 0, thusthe quadratic form is negative semidefinite for every (K,L) ∈ R++×R++, thus f is concave inthis region. Since it is continuous in R+ × R+, we can conclude that f is also concave here.
Theorem 1.35 (Jensen’s Inequality I). Let f : C ⊆ Rm −→ R be convex. Then
f
(n∑j=1
λjxj
)≤
n∑j=1
λjf(xj)
for all x1, . . . ,xn ∈ C, for all λ1, . . . , λn ≥ 0 and∑n
j=1 λj = 1.
Example 1.36. Using Jensen’s inequality one can show that for any strictly positive real num-bers x1, . . . , xn the geometric mean is below the arithmetic mean:
(x1 · · ·xn)1/n ≤ x1 + · · ·+ xnn
.
For, let f(x) = − lnx, which is convex in R++ and let λj = 1/n. Then
− ln
(x1 + · · ·+ xn
n
)≤ − 1
n(lnx1 + · · ·+ lnxn) = − ln (x1 · · ·xn)1/n
and taking exponentials we are done.
Remark 1.37. Suppose X is a random variable with a discrete distribution with finitely manypairwise distinct x1, . . . , xn. The expectation of X is
EX =n∑j=1
xjProbX = xj.
Jensen’s inequality implies that for any convex function f : R −→ R
Ef(X) ≥ f(EX).
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There exists useful generalizations. Here we present the following: Let λ be the Lebesguemeasure on Rm and let u : U −→ R be a measurable function such that the Lebesgue integral∫U|u(x)| dx is finite, where U is an open and bounded subset of Rm.
Theorem 1.38 (Jensen’s Inequality II). Let f : R −→ R be a convex function. Then
f
(1
λ(U)
∫U
u(x) dx
)≤ 1
λ(U)
∫U
f(u(x)) dx.
Proof. Since f is convex, for each x ∈ R there exists p ∈ R such that
f(y) ≥ f(x) + p(y − x) for all y ∈ R.
Let
x =1
λ(U)
∫U
u(x) dx, y = u(x).
Then
f(u(x)) ≥ f
(1
λ(U)
∫U
u(x) dx
)+ p
(u(x)− 1
λ(U)
∫U
u(x) dx
).
Integrating with respect to x over U we get the desired inequality, since∫U
(u(x)− 1
λ(U)
∫U
u(x) dx
)dx =
∫U
u(x) dx− λ(U)
λ(U)
∫U
u(x) dx = 0
1.7 Separation Theorems
Separation theorems refer to the possibility of separating disjoint sets in Rm by means of a linearvariety. In general, this is not possible for arbitrary sets, unless they are convex. Separationtheorems useful in establishing optimality conditions without differentiability requirements inconvex programming. They are also fundamental in welfare economics.
Given a nonzero vector u in Rm and a scalar c, the set H = x ∈ Rm : u · x = c is ahyperplane in Rm, with u as a normal vector. The associated closed half-spaces are
H+ = x : u · x ≥ c, H− = x : u · x ≤ c.
Given two subsets A and B of Rm, we say that H separates A and B if for instance A ⊆ H+
and B ⊆ H−. In other words, A and B can be separated by a hyperplane if there exists unonzero and a scalar c such that
u · x ≤ c ≤ u · y for all x ∈ A and all y ∈ B.
If both inequalities are strict, then the hyperplane strictly separates A and B.
Theorem 1.39. Let A be a closed, convex set of Rm and let y be a point in Rm such thaty /∈ A. Then there exists a nonzero vector u ∈ Rm and a scalar α such that A and y arestrictly separated by the hyperplane H = x ∈ Rm : u · x = c.
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Given a set A in Rm, we say that a point x is in the boundary of A if for any ε > 0,A ∩ B(x, ε) 6= ∅ and Ac ∩ B(x, ε) 6= ∅. We will write in this case x ∈ ∂A and say that x is aboundary point of A. Note that boundary points are never interior points. A hyperplane His said to be a support (or supporting hyperplane) for a convex set A at a point y ∈ ∂A if Hseparates A and y. The existence of a supporting hyperplane to a convex set is assured inthe following result.
Theorem 1.40. Let A be a convex set in Rm and suppose that y is not an interior point of A.Then there exists a nonzero vector u ∈ Rm and a scalar α such that A and y are separatedby the hyperplane H = x ∈ Rm : u · x = c.
Theorem 1.41. Let A and B be two disjoint nonempty convex sets in Rm. Then there existsa nonzero vector u ∈ Rm and a scalar α such that A and B are separated by the hyperplaneH = x ∈ Rm : u · x = c.
Consider the following generalization of the above theorem. Let us denote by A the set oninterior points of the subset A.
Theorem 1.42. Let A and B two nonempty convex sets in Rm such that A is nonempty andA and B are disjoints. Then there exists a nonzero vector u ∈ Rm and a scalar α such that Aand B are separated by the hyperplane H = x ∈ Rm : u · x = c.
1.8 Problems
We work in Rm with the Euclidean norm.
1. (a) Show that the limit is unique, that is, if xn → x and xn → y, then x = y
(b) If xn → x, then ‖xn − x‖ → 0 and ‖xn‖ → ‖x‖. However, ‖xn‖ → ‖x‖ does notimply xn → x.
(c) Suppose xn ∈ Rm, x ∈ Rm, λn ∈ R and λ ∈ R. If xn → x and λn → λ, thenλnxn → λx.
(d) If xn → x and yn → y, show that xn · yn → x · y.
Solution:
(a) Since xn → x and xn → y, for any ε > 0 there exists n0(ε) such that for anyn ≥ n0(ε)
‖xn − x‖ < ε/2 and ‖xn − y‖ < ε/2.
By way of contradiction, if x 6= y, then taking ε = ‖x−y‖ > 0 and using the triangleinequality we obtain
‖x− y‖ ≤ ‖xn − x‖+ ‖xn − y‖ < ε/2 + ε/2 = ‖x− y‖,
a contradiction. Thus, the limit is unique.
(b) The definition establishes that convergence of the sequence of vectors xn to x isequivalent to convergence of the sequence of real numbers ‖xn − x‖ to 0. To show
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the second part, observe that for arbitrary x, y, ‖x‖ ≤ ‖x − y‖ + ‖y‖ and ‖y‖ ≤‖x− y‖+ ‖x‖. Hence we have the two inequalities
‖x‖ − ‖y‖ ≤ ‖x− y‖ and ‖y‖ − ‖x‖ ≤ ‖x− y‖,
which means ∣∣‖x‖ − ‖y‖∣∣ ≤ ‖x− y‖.
Since xn converges to x, for any ε there exists n0 such that for any n ≥ n0 ‖xn−x‖ <ε, thus ∣∣‖xn‖ − ‖x‖∣∣ ≤ ‖xn − x‖ < ε,
hence ‖xn‖ → ‖x‖. (This implies that, as function, the norm ‖ · ‖ : Rm −→ R iscontinuous). However, if we consider the sequence (1,−1, 1,−1, . . .), |xn| → 1 butxn does not converge.
(c) Observe that by triangle inequality and homogeneity of the norm
Since the norm is continuous, taking limits we get ‖λnxn− λx‖ → 0, hence λnxn →λx.
(d) Observe that by triangle inequality and the Cauchy–Schwartz inequality
|xn · yn − x · y| ≤ |xn · (yn − y)|+ |y · (xn − x)|≤ ‖xn‖ ‖yn − y‖+ ‖y‖‖xn − x‖.
Since the norm is continuous, taking limits we get |xn · yn − x · y| → 0, hencexn · yn → x · y. This means that the inner product mapping is continuous.
2. Show that the open ball is open and that the closed ball is closed.
3. (a) Show that the arbitrary union of open sets is open.
(b) Show that the finite intersection of open sets is open.
(c) Show that arbitrary intersection of closed sets is closed.
(d) Show that the finite union of closed sets is closed.
4. Let U1, U2, . . . be open sets in R such that U1 ⊇ U2 ⊇ . . .. Show, by means of examplesthat
⋂∞j=1 Uj may be (a) open but not closed, (b) closed but not open, (c) open and closed
or (d) neither open nor closed.
Solution: For (a) let Uj = (0, 1), for (b) let Uj = (−1/j, 1/j), for (c) let Uj = (0, 1/j)and for (d) let Uj = (−1/j, 1). Then
⋂∞j=1 Uj is (a) (0, 1), (b) 0, (c) ∅ and (d) [0, 1),
respectively.
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5. If A is open in Rm and B is open in Rp, show that A×B is open in Rm × Rp.
6. Show that a function f : Rm −→ Rp is continuous if and only if f−1(U) is open wheneverU is an open set in Rp.
Solution: Recall that f−1(U) = x ∈ Rm : f(x) ∈ U.Necessity: Suppose f is continuous and U is open in Rp. Let x ∈ f−1(U), so f(x) ∈ Uand hence there exists ε > 0 such that
B(f(x), ε) ⊆ U.
Since f is continuous at x, there exists δ > 0 such that for any y ∈ B(x, δ)
‖f(x)− f(y)‖ < ε or, in other way, f(y) ∈ B(f(x), ε).
Thus, we have shownf(B(x, δ)) ⊆ B(f(x), ε) ⊆ U,
which meansB(x, δ) ⊆ f−1(U),
thus f−1(U) is open.
Sufficiency: Suppose that f−1(U) is open if U is open in Rp. Let x ∈ Rm and ε > 0.Since B(f(x), ε) is open, f−1(B(f(x), ε)) is open and x ∈ f−1(B(f(x), ε)), so there existsδ > 0 such that
B(x, δ) ⊆ f−1(B(f(x), ε)),
orf(B(x, δ)) ⊆ B(f(x), ε),
which means‖x− y‖ < δ ⇒ ‖f(x)− f(y)‖ < ε,
hence f is continuous.
7. (a) Let A = (0, 1). Show that the map f : A −→ R given by f(x) = 1/x is continuousbut f(A) is unbounded (the continuous image of a bounded set need not be bounded).
(b) Let A = [1,∞) and let f : A −→ R be given by f(x) = 1/x. Show that A is closedand f is continuous but f(A) is not closed (the continuous image of a closed set neednot be closed).
(c) Show that the projection functions π1, π2 : R2 −→ R defined by π1(x, y) = x andπ2(x, y) = y are continuous. Show that the set A = (x, 1/x) : x > 0 is closed inR2 but that π1(A) is not.
(d) Show, however, that if F is a closed set in R2 with π2(F ) bounded, then π1(F ) mustbe closed.
Solution:
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(a) f(A) = (0,∞).
(b) f(A) = (0, 1].
(c) If ‖(xn, yn)−(x, y)‖ → 0, then xn → x and yn → y, thus both projection functions arecontinuous. Now, if (xn, yn) ∈ A and ‖(xn, yn)− (x, y)‖ → 0, then yn = 1/xn → 1/xand yn → y, hence y = 1/x and (x, y) ∈ A, thus A is closed. However, π1(A) =(0,∞) is not closed since 1/n ∈ π1(A) and 1/n→ 0 /∈ π1(A) as n→∞.
(d) If xn ∈ π1(F ) and xn → x we can find yn ∈ π2(F ) such that (xn, yn) ∈ F . Sinceπ2(F ) is bounded, by the Bolzano–Weierstrass Theorem there exists a convergentsubsequence yn(j) of yn converging to some y as j → ∞. Since xn(j) → x and F isclosed, we have (x, y) ∈ F , so that x ∈ π1(F ).
8. Let A ⊆ Rm. A function f : A −→ R is said to be upper semicontinuous on A if, givenany x ∈ A and any ε > 0, we can find δ(ε,x) > 0 such that
f(y)− f(x) < ε for all y ∈ A with ‖y − x‖ < δ(ε,x).
(a) Give an example of a function f : R −→ R which is upper semicontinuous but notcontinuous.
(b) If A ⊆ Rm and f : A −→ R is such that both f and −f are upper semicontinuouson A, show that f is continuous on A.
(c) For a function f : A −→ R, the upper sections are defined as the subsets of Rmgiven by
Sλ(f) = x ∈ A : f(x) ≥ λ, where λ ∈ R.
Suppose that the set A is closed. Show that f is upper semicontinuous if and only ifSλ is closed for any λ.
(d) If K ⊆ Rm is compact and f : K −→ R is upper semicontinuous, show that f has aglobal maximum on K.
(e) Is it necessarily true that an upper semicontinuous function is bounded below? Is itnecessarily true that, if f is bounded below, it attains a global minimum?
(f) A function f such that −f is upper semicontinuous is called lower semicontinuous.Show that a lower semicontinuous functions attains global minimum on compact sets.
Solution:
(a) The function f(x) = 0 for x 6= 0 and f(0) = 1 is upper semicontinuous but notcontinuous.
(b) If f is upper semicontinuous on A, then given any x ∈ A and any ε > 0, we can findδ(ε,x) > 0 such that
f(y)− f(x) < ε for all y ∈ A with ‖y − x‖ < δ(ε,x).
Thus, since −f is also upper semicontinuous
−(f)(y)− (−f)(x) < ε for all y ∈ A with ‖y − x‖ < δ(ε,x).
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Thus we have the two inequalities
f(y)− f(x) < ε and − (f(y)− f(x)) < ε,
that means
|f(y)− f(x)| < ε for all y ∈ A with ‖y − x‖ < δ(ε,x),
hence f is continuous on A.
(c) Necessity: Note that Sλ(f) = A∩x ∈ Rm : f(x) ≥ λ. If we prove that F = x ∈Rm : f(x) ≥ λ is closed, we are done since we have an intersection of two closedsets. Let x /∈ F . Then ε = λ− f(x) > 0 and since f is upper semicontinuous, thereexists δ > 0 such that ‖y − x‖ < δ implies f(y) < f(x) + ε = λ, thus y /∈ F . Wehave shown that B(x, δ) ⊆ Rm\F , hence Rm\F is open and F is closed.
Sufficiency: Let x ∈ A and let ε > 0. Define λ = f(x) + ε. Obviously, x ∈Rm\Sλ(f), which is open by assumption. Thus, there exists δ > 0 such thatB(x, δ) ⊆ Rm\Sλ(f), that is, for ‖y − x‖ < δ we have f(y) < λ = f(x) + ε,which proves that f is upper semicontinuous.
(d) Let us show that f is bounded above. Reasoning by contradiction, if f is notbounded above, then we can find xn ∈ K such that f(xn) ≥ n. Since K is closedand bounded, we can extract a convergent subsequence xn(j) → x ∈ K as j → ∞.Let ε = 1 and choose j0 such that for any j ≥ j0
‖xn(j) − x‖ < δ(1,x).
Since f is upper semicontinuous,
f(xn(j)) < f(x) + 1,
thus we get the contradiction n(j) ≤ f(x) + 1 for any j, which is absurd, sincen(j) → ∞ as j → ∞. Thus, f is bounded above. Let M be the supremum of theset f(K), that is,
M = supx∈K
f(x),
and let us show that there is some x1 ∈ K such that it is attained, M = f(x1).Consider a sequence xn ∈ K approaching the supremum, f(xn) → M as n → ∞.Since K is closed and bounded, there is a subsequence xn(j) converging to somex1 ∈ K. Since f is upper semicontinuous, by a reasoning similar as above, forε = 1/j it is possible to select j0 such that for any j ≥ j0 one has
f(xn(j)) < f(x1) + 1/j.
Taking limits as j →∞ we have
M ≤ f(x1).
Hence, M = f(x1).
(e) The function f : [0, 1] −→ R given by f(x) = −1/x for x 6= 0, f(0) = 0 is uppersemicontinuous but not bounded below.
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(f) The function f : [0, 1] −→ R given by f(x) = x for x 6= 0, f(0) = 1 is uppersemicontinuous and bounded below, but it does not attain a global minimum.
(g) Since −f is upper semicontinuous, it attains a global maximum on any compact setK ⊆ Rm, hence there exists x1 ∈ K such that (−f)(x) ≤ (−f)(x1) for all x ∈ K,thus f(x) ≥ f(x1) for all x ∈ K, which means that x1 is a global minimum of f .
9. Let A ⊆ Rm be closed and unbounded and let f : A −→ R be an upper semicontinuousfunction such for any sequence xn ∈ A with ‖xn‖ → ∞ as n→∞, it is possible to extracta subsequence xn(j) such that
f(xn(j))→ −∞ as j →∞.
Show that f attains a global maximum on A.
Solution: Let us show that the upper sections Sλ(f) are bounded. If not, then thereis some sequence xn ∈ Sλ(f) such that ‖xn‖ → ∞. It is possible then to extract asubsequence such that f(xn(j)) ≤ −j. Then −j ≥ λ for any j, which is absurd. Hence,the upper sections are bounded. On the other hand, we know from a previous problemthat Sλ(f) is closed since f is upper semicontinuous, thus for any λ the set Sλ(f) is closedand bounded. Fix any λ0 for which Sλ0(f) 6= ∅. Let x0 ∈ Sλ0(f) a global maximum of fin the set Sλ0(f). We have
supf(x) : x ∈ A = supf(x) : x ∈ Sλ0(f) = f(x0),
hence f attains its global maximum on A.
10. Let A ⊆ Rm be closed and let b /∈ A. Show that there is y ∈ A such that
‖b− y‖ ≤ ‖b− x‖ ∀x ∈ A,
that is, there is some point y ∈ A which distance to b is minimum among all points ofA.
Solution: Consider the function g(x) = ‖b − x‖. It is continuous since the norm iscontinuous. If A is bounded, then g attains a global minimum y ∈ A by Weierstrass’Theorem. If A is unbounded, then there is some sequence xn ∈ A such that ‖xn‖ → ∞,hence
g(xn) = ‖xn − b‖ ≥ ‖xn‖ − ‖b‖ → ∞ as n→∞.
According to the previous problem, there is y ∈ A minimizing g on A and we are done.
11. (a) Give an example of a continuous and bounded function that does not attain maximumand minimum on a bounded set.
(b) Give an example of a continuous and bounded function that does not attain maximumand minimum on a bounded set.
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(c) Give an example of a function which is neither upper semicontinuous nor lowersemicontinuous in the interval [0, 1].
Solution:
(a) f(x) = x and A = (0, 1).
(b) f(x) = arctan x and A = R.
(c) f(x) = x if x 6= 0, 1, and f(0) = f(1) = 1/2.
1.9 Homework
We work in the Euclidean space Rm.
1. Show that the open ball is open and that the closed ball is closed.
2. (a) Show that the arbitrary union of open sets is open.
(b) Show that the finite intersection of open sets is open.
(c) Show that arbitrary intersection of closed sets is closed.
(d) Show that the finite union of closed sets is closed.
3. If A is open in Rm and B is open in Rp, show that A×B is open in Rm × Rp.
4. Show that if F ⊆ K with F closed and K compact, then F is compact.
5. Which of the following statements are true and which are false? Give proofs or counterex-amples.
(a) If A is an open (closed) subset of Rm+1, then
x ∈ Rm : (0,x) ∈ A
is open (closed) in Rm.
(b) If A is an open (closed) subset of Rm, then
(0,x) ∈ Rm+1 : x ∈ A
is open (closed) in Rm+1.
6. Let A ⊆ Rm and let f : A −→ Rp be a function. We call
G = (x, f(x)) : x ∈ A
the graph of f .
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(a) Show that, if G is bounded, then A is.
(b) Show that it is possible to have G closed and f continuous but A not closed.
(c) Show that, if A is closed and f is continuous, then G is closed.
(d) Show that G and A may be closed but f discontinuous.
7. Give an example of a subset S of R and a function f : S −→ R continuous and strictlymonotonous increasing in S such that its inverse, f−1, is not continuous on f(S). (Hint:look for a suitable S)
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2 Parametric Optimization Problems
2.1 Continuity of correspondences
Definition 2.1. A correspondence Γ : X ⊆ Rm ⇒ Rp is a mapping
Γ : X ⊆ Rm −→ 2Rp
,
where 2Rp
is the collection of all subsets of Rp, including the empty set ∅.
Definition 2.2. We say that Γ is upper hemicontinuous (uhc) at x0 ∈ X iff for every open setU ⊆ Rp such that Γ(x0) ⊆ U , there is an open set N ⊆ Rm such that x0 ∈ N and for everyx ∈ X ∩N , Γ(x) ⊆ U .
Remark 2.3 (Not uhc). If Γ is not uhc at x0, then there is an open set U ⊇ Γ(x0), a sequencexn ∈ X ∩B(x0,
1n) and a sequence yn ∈ Γ(xn) for which yn /∈ U for all n ≥ 1.
Definition 2.4. We say that Γ is lower hemicontinuous (lhc) at x0 ∈ X iff for every open setU ⊆ Rp such that Γ(x0) ∩ U 6= ∅, there is an open set N ⊆ Rm such that x0 ∈ N and for everyx ∈ X ∩N , Γ(x) ∩ U 6= ∅.
Remark 2.5 (Not lhc). If Γ is not lhc at x0, then there is y0 ∈ Γ(x0), an open set U 3 y0 anda sequence xn ∈ X ∩B(x0,
1n) such that for any element y ∈ Γ(xn), y /∈ U for all n ≥ 1.
Proposition 2.6. If Γ is uhc or lhc at x0 ∈ X and single valued in B(x0, r) ∩ X for somer > 0, then it is a continuous mapping at x0.
Proof. We can consider that a Γ is a mapping in B(x0, r)∩X. Let U be an open set such thaty0 = Γ(x0) ∈ U . Then, if Γ is uhc at x0, then there is an open set N with x0 ∈ N and Γ(x) ⊆ Ufor any x ∈ X ∩ N . Take N ⊆ B(x0, r), so that Γ is singled valued. Then, Γ(x) ∈ U for anyx ∈ X ∩N , so that X ∩N ⊆ Γ−1(U) and since X ∩N is open in X, we have shown that Γ iscontinuous at x0. The argument for the lhc case is the same, as Γ(x) ∩ U 6= ∅ and Γ(x) ⊆ Uare identical conditions for a mapping.
Reciprocally, if f : X −→ Rp is a continuous mapping, then Γ defines as Γ(x) = f(x) isboth usc and lhc.
Definition 2.7. We say that the correspondence Γ is continuous at x0 ∈ X iff it is borh uhcand lhc at x0.
Definition 2.8. We say that the correspondence Γ is uhc, lhc or continuous at X iff it is uhc,lhc or continuous at every x ∈ X, respectively.
Exercise 2.9. Study if Γ : R+ ⇒ R+ is uhc/lhc in the following cases.
1. Γ(x) = [0, x].
2. Γ(x) = [0, x).
3. Γ(x) = (0, x].
4. Γ(x) = (0, x) if x > 0, Γ(0) = 0.
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1. Γ is uhc: Let U be open with [0, x0] ⊆ U . Then there is δ > 0 such that (−δ, x0 + δ) ⊆ U .Let N = (x0 − δ/2, x0 + δ/2). Since f(x) = x is increasing, Γ(x) ⊆ [0, x0 + δ/2) for anyx ∈ R+ ∩N , thus Γ(R+ ∩N) ⊆ [0, x0 + δ/2) ⊆ (−δ, x0 + δ) ⊆ U , thus Γ is uhc.
Γ is lhc: Let U be open with [0, x0] ∩ U 6= ∅. Then there is some point y0 ∈ (0, x0) ∩ U .Let δ = x0 − y0 > 0 and consider N = (x0 − δ/2, x0 + δ/2). Let us check that for anyx ∈ N one has y0 ∈ Γ(x) = [0, x], so that Γ(x) ∩ U 6= ∅. The smallest of these intervalscontain [0, x0 − δ/2) and x0 − δ/2 = x0/2 + y0/2 > y0, thus y0 ∈ Γ(x) ∩ U for all x ∈ Nand Γ is lhc.
2. Γ is not uhc: it suffice to show an U open with [0, x0) ⊆ U for some x0 and Γ(R+∩N) * Ufor any open set N containing x0. Let U = (−1, x0), so that Γ(x0) ⊆ U and U is open.For any N neighborhood of x0, there is some δ > 0 such that (x0 − δ, x0 + δ) ⊆ N . Butx+ δ/2 ∈ (x0 − δ, x0 + δ) and Γ(x0 + δ/2) = [0, x0 + δ/2) * U .
Γ is lhc: the same proof as above.
3. Γ is uhc: for any x0 > 0, and any U ⊇ (0, x0], there is some δ > 0 such that (0, x0] ⊆(0, x0 + δ) ⊆ U . Take N = (x0 − δ/2, x0 + δ/2). Then for any x ∈ R+ ∩ N one hasΓ(x) ⊆ (0, x0 + δ/2) ⊆ (0, x0 + δ) ⊆ U , thus Γ is uhc.
Γ is lhc: the same proof as above.
4. Γ is not uhc at x0 > 0: the same proof as above.
Γ is lhc at x0 > 0: the same proof as above.
Γ is both uhc and lhc at 0.
Part 1 of the above exercise is a particular case of the following result.
Exercise 2.10. Let Γ(x) = [g(x), f(x)], where functions g, f : R −→ R are continuous andg(x) ≤ f(x) for all x. Show that Γ is a continuous set–valued map.
Proof. Γ is uhc: Let U be open with [g(x0), f(x0)] ⊆ U . There exists ε > 0 such that [g(x0)−ε, f(x0) + ε] ⊆ U . Given that g (f) is continuous, there exists δ′ > 0 (δ′′ > 0) such that ifx ∈ (x0 − δ′, x0 + δ′), (x ∈ (x0 − δ′, x0 + δ′)), then g(x) > g(x0)− ε (f(x) < f(x0) + ε). Takingδ = minδ′, δ′′ we have g(x0)− ε < g(x) ≤ f(x) < f(x0) + ε, thus [g(x), f(x)] ⊆ U .
Γ is lhc: Let U be open with [g(x0), f(x0)] ∩ U 6= ∅. There is some y0 ∈ U ∩ (g(x0), f(x0))because U is open. Let ε > 0 such that ε ≤ y0 − g(x0) and ε ≤ f(x0)− y0. Then, since both gand f are continuous, there exists δ > 0 such that for any x ∈ (x0 − δ, x0 + δ)
thus y0 ∈ (g(x), f(x)), and [g(x), f(x)] ∩ U 6= ∅, hence Γ is lhc.
Exercise 2.11. Show that Γ(x) = [−|x|, |x|] with x ∈ R is continuous.
Proof. The function x 7→ |x| is continuous and apply the result above.
Definition 2.12. The domain of Γ is the set domΓ = x ∈ X : Γ(x) 6= ∅. The image of theset A ⊆ domΓ by the correspondence Γ is the set
Γ(A) =⋃x∈A
Γ(x).
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The graph of Γ is the set
Ω = (x, y) : x ∈ domΓ, y ∈ Γ(x) ⊆ X × Γ(X).
Definition 2.13. We say that Γ has open, closed, bounded, compact or convex values iff forevery x ∈ X the sets Γ(x) are open, closed, bounded, compact or convex, respectively.
Definition 2.14. We say that Γ has the closed graph property at x0 ∈ X iff for any sequence(xn) in X converging to x0 and for any sequence (yn), yn ∈ Γ(xn), converging to some y0, onehas y0 ∈ Γ(x0).
Definition 2.15. We say that Γ has the closed graph property iff it has the closed graph propertyat every x ∈ X.
This means that Ω is relatively closed in X × Rp, so that Ω is closed in Rm × Rp if X isclosed.
Proposition 2.16. If Γ has the closed graph property at x0 and it is locally bounded at x0 ∈ X(there exists an open set N with x0 ∈ N and Γ(X ∩N) is a bounded set), then Γ is uhc at x0.
Proof. Suppose N as in the theorem, and let B(x0, r) ⊆ N . We argue by contradiction,supposing that Γ is not uhc at x0. Then, according to Remark 2.3 there is an open set U ⊇Γ(x0), a sequence xn ∈ X ∩B(x0,
1n) and a sequence yn ∈ Γ(xn) for which yn /∈ U for all n ≥ 1.
However, yn ∈ Γ(xn) ⊆ Γ(B(x0,1n)) ⊆ Γ(B(x0, 1)) for all n ≥ 1. Since Γ is locally bounded,
Γ(B(x0, 1)) is a bounded set. By the Bolzano–Weierstrass Theorem, there is a subsequenceyn(j) that converges to some y0. But Γ has the closed graph property at x0, thus given thatxn → x0, y0 ∈ Γ(x0). On the other hand, yn(j) ∈ Rp\U that is closed, hence y0 ∈ Rp\U ory0 /∈ U and hence y0 /∈ Γ(x0), a contradiction.
Corollary 2.17 (Compact graph uhc test). If the graph of Γ is compact, then Γ is uhc.
Exercise 2.18. Study whether the correspondence
Γ(x) =
[0, 1/x], x > 0;
0, x = 0.
is uhc/lhc or has the closed graph property. Is it closed, bounded, compact, convex–valued?
Proof. For points x > 0, Γ is cns by Example 2.10. At x = 0 is not uhc, but it is lhc. SinceX = R+ is closed in R, to have the closed graph property is equivalent to the graph of Γ to beclosed in R2. Obviously, the graph is not closed (the vertical segment y = 0 is no in the graph),hence Γ has no the closed graph property. It is closed, bounded, compact, convex–valued.
Exercise 2.19. Study whether the correspondence
Γ(x) =
[0, 1/x], x > 0;
R+, x = 0.
is uhc/lhc or has the closed graph property. Is it closed, bounded, compact, convex–valued? Isthe graph of Γ convex?
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Proof. For points x > 0, Γ is cns by Example 2.10. At x = 0 is both uhc, and lhc. It is uhcsince for any U open with Γ(0) ⊆ U we must have Γ(x) ⊆ U for any x, since Γ(x) is compactvalued. It is lhc at x = 0 since for any open set U with Γ(0)∩U 6= ∅, let y0 + ε ∈ Γ(0)∩U . LetN = (0, δ) with 0 < δ < 1/(y0 + ε). Then, for 0 < x < δ, 1/x > 1/δ > y0 + ε, thus y0 ∈ Γ(x)for any x ∈ R+ ∩N . It has the closed graph property, since the graph of Γ is closed in R2.
It is closed and convex valued, and bounded valued for x > 0. The graph of Ω is not convex,since the points (1, 1), (2, 1
2) are in Ω, but 1
2(1, 1) + 1
2(2, 1
2) = (3
2, 3
4) /∈ Ω, since 3
4 (3/2)−1 =
23.
Proposition 2.20. If f : X×Y ⊆ Rm×Rp −→ R is continuous and Y is closed and bounded,then the correspondence Γ defined as
Γ(x) = y ∈ Y : f(x, y) ≤ 0
is uhc in its domain.
Proof. Notice that the set F = (x, y) ∈ X × Y : f(x, y) ≤ 0 is closed since f is continuousand F = f−1((−∞, 0]) is the inverse image of the closed set2 (−∞, 0], Hence, the graph of Γ
Ω = (x, y) : x ∈ X, f(x, y) ≤ 0 = F ∩ (X × Y )
is relatively closed in X × Y and given that Y is closed, F is relatively closed in X ×Rp. ThusΓ has the closed graph property. Since Γ(x) ⊆ Y for all x, it is also locally bounded, thus it isuhc by Proposition 2.16.
Example 2.21. Let X = [0, 1] and Y = [0, 2] and let f : X × Y −→ [0, 2] be the continuousfunction defined as
f(x, y) =
x+ y − 1, if 0 ≤ y ≤ 1;
xy, if 1 ≤ y ≤ 2.
The associated correspondence Γ in Proposition 2.20 is uhc according to the above result.Its graph is
6
-0 1
1
2
X
Y
@@@@@@
Ω
This example shows that Γ is not in general lhc.
2Recall that f is continuous if and only if inverse images of closed (open) sets are closed (open).
22
Proposition 2.22. If f : X ⊆ Rm −→ R is continuous and Y is compact, then the correspon-dence Γ defined as
Γ(x) = y ∈ Y : y ≤ f(x)
is continuous in its domain.
2.2 The Theorem of the Maximum
Given the function f : X × Y ⊆ Rm × Rp −→ R and the correspondence Γ : X ⇒ Y considerthe problem
v(x) = supy∈Γ(x)
f(x, y).
Function v is the value function and the set valued map
Γ∗(x) = y ∈ Y : v(x) = f(x, y)
is the optimal correspondence.
Proposition 2.23. Suppose that Γ is uhc with compact values and that K ⊆ X is compact.Then, Γ(K) = ∪x∈KΓ(x) is compact.
Proof.Γ(K) is bounded: If not, there is xn ∈ K and yn ∈ Γ(xn) such that ‖yn‖ ≥ n for all n. SinceK is compact, there is a subsequence xn(j) → x0 ∈ K. Let U open and bounded such thatΓ(x0) ⊆ U (this is possible since Γ(x0) is bounded). Since Γ is uhc, Γ(xn(j)) ⊆ U for any j,thus for some M , n(j) ≤ ‖yn(j)‖ ≤M for any j, which is absurd. Thus, Γ(K) is bounded.
Γ(K) is closed: Let yn ∈ Γ(K) such that yn → y0. There are xn ∈ K such that yn ∈ Γ(xn) anda subsequence xn(j) → x0 ∈ K. Let U open and bounded such that Γ(x0) ⊆ U (this is possiblesince Γ(x0) is bounded). Since Γ is uhc, Γ(xn(j)) ⊆ U for any j, thus yn(j) is a bounded sequence.By the Bolzano–Weierstrass Theorem, yn(j) → y′0. Suppose y′0 /∈ Γ(x0). Then, since Γ(x0) iscompact, there is some open set W such that y′0 ∈ W and U ∩W = ∅. Hence yn(j) ∈ Rp\Wthat is closed, thus the limit y′0 /∈ W , a contradiction. Hence y′0 ∈ Γ(x0) and since the limit isunique, y0 ∈ Γ(x0), thus y0 ∈ Γ(K) and Γ(K) is therefore closed.
Theorem 2.24 (Theorem of Berge or Theorem of the Maximum). Assume that v(x) < +∞for every x ∈ X.
1. If f is lsc and Γ is lhc, then v is lsc.
2. If f is usc and Γ is uhc with compact values, then v is usc.
3. If f is continuous and Γ is continuous with compact values, then v is continuous and Γ∗
is uhc with compact values.
Proof. 1. Let x0 ∈ X and ε > 0. By definition of v(x0), there exists y0 ∈ Γ(x0) such thatv(x0) ≤ f(x0, y0) + ε/2. Since f is lsc, there is δ > 0 such that
f(x0, y0) < f(x, y) +ε
2(2)
23
for any ‖(x, y) − (x0, y0)‖ < δ. Let U = B(y0, r). Since Γ is lhc, there is B(x0, r′) such
that Γ(x) ∩ U 6= ∅ for any x ∈ X ∩ B(x0, r′). Take r + r′ < δ. Then there is y ∈ Γ(x)
such that (2) holds, so that
f(x0, y0) < f(x, y) +ε
2≤ v(x) +
ε
2,
by the definition of v. Hence for any ‖x− x0‖ < r′
v(x0) ≤ f(x0, y0) + ε/2 < v(x) + ε
and v is lsc.
2. Let us show that the upper sections Sλ(v) are closed. Let xn ∈ Sλ(v), xn → x0. Since fis usc and Γ(xn) is compact for each n, there exists yn ∈ Γ(xn) such that
f(xn, yn) = v(xn) ≥ λ ∀n.
This means that (xn, yn) ∈ Sλ(f). Since xn → x, the set K = (xn) ∪ x0 is compact,thus by Proposition 2.23 Γ(K) is also compact. Hence there is a subsequence yn(j) →y0 ∈ Γ(K). But y0 ∈ Γ(x0), because otherwise it is possible to find an open set W suchthat y0 ∈ W and yn /∈ W by the uhc of Γ, reaching a contradiction with the fact thatthe set Rp\W is closed (see the proof of Proposition 2.23 above). Since Sλ(f) is closed,(xn(j), yn(j))→ (x0, y0) ∈ Sλ(f), thus f(x0, y0) ≥ λ. Since y0 ∈ Γ(x0) we have
v(x0) ≥ f(x0, y0) ≥ λ,
that is, x0 ∈ Sλ(v) and v is usc.
3. By parts (a) and (b) v is continuous. On the other hand
Γ∗(x) = y : v(x)− f(x, y) = 0 ∩ Γ(x)
is the intersection of a closed set (because both v and f are continuous) and a compactset (because Γ has compact values), thus Γ∗(x) is compact. Let us show that Γ∗ has theclosed graph property. Let (xn) → x0, (yn) → y0 and yn ∈ Γ(xn). Then y0 ∈ Γ(x0):otherwise, there is some open set W such that y0 ∈ W and yn /∈ W since Γ is uhc andΓ(x0) is compact (this argumentation has been used several times above). Let us showthat in fact y0 ∈ Γ∗(x0). Let z0 ∈ Γ(x0) arbitrary. Let the open set Uj = B(z0, 1/j). Sincexn → x0 and Γ is lhc, for each j there exists n(j) such that xn(j) → x0 and zn(j) ∈ Γ(xn(j))with zn(j) ∈ Uj, thus zn(j) → z0. Then
f(xn(j), zn(j)) ≤ f(xn(j), yn(j))
and taking limits, since f is continuous
f(x0, z0) ≤ f(x0, y0),
thus y0 ∈ Γ∗(x0) and Γ∗ has the closed graph property. Since Γ∗ is also locally bounded,then it is uhc by Proposition 2.16.
24
Theorem 2.25. Suppose that X ⊆ Rm is convex, f : X ×Rp −→ R is continuous and concaveand that Γ : X ⇒ Rp is a continuous set–valued mapping with convex graph Ω and with compactvalues. Let v be the value function v(x) = maxy∈Γ(x) f(x, y). Then we have
1. The value function v is concave and the optimal correspondence in convex–valued.
2. If f(x, ·) is strictly concave in y for every x ∈ X such that (x, y) ∈ Ω, then the optimalcorrespondence is single valued and it is continuous as a function.
Proof. We now that v is well defined and continuous and that the optimal correspondence Γ∗
exists and is uhc.
1. Let x1, x2 ∈ X and λ ∈ [0, 1]. We want to show
v(λx1 + (1− λ)x2) ≥ λv(x1) + (1− λ)v(x2).
Let yi ∈ Γ∗(xi), i = 1, 2, and let yλ = λy1 + (1 − λ)y2. Since Ω is convex we have (forxλ = λx1 + (1− λ)x2)
hence v is concave. moreover, it is well known that the maximum set of a concave functionis convex, thus given that for any x, f(x, y) is concave in y, we have Γ∗(x) is convex.
2. it is well known that the maximum set of a strictly concave function is empty or single–valued. Since Γ∗(x) is non empty, it is single valued and since it is uhc, it defines acontinuous function.
Example 2.26. Let the consumers’ problem
max√x1 + x2 s.t.: 0 ≤ 2x1 + 3x2 ≤ R.
What can you say about this problem without solving it explicitly?
Proof. The utility function u(x1, x2) =√x1 + x2 is continuous as it is the budget correspon-
dence Γ : R+ ⇒ R2+ defined by Γ(R) = (x1, x2) ∈ R2
+ : 0 ≤ 2x1 + 3x2 ≤ R. Thus, accordingto the Theorem of Berge, the value function or indirect utility function depends continuouslyon R. Since R1 < R2 implies Γ(R1) ⊆ Γ(R2), v(R1) ≤ v(R2), and since u is strictly increasingin both components, v is strictly increasing. Moreover, u is strictly concave in R2
+ and Ω isconvex (why?), thus v is concave and the demand correspondences, x∗1(R), x∗2(R), are singletonsand they are continuous functions of R.
25
2.3 Problems
1. For the following correspondences R ⇒ R, study whether they are continuous and withcompact or convex values. Study whether the graph is closed or convex.
(a) Γ1(x) = [0, 1) if x = 0; Γ1(x) = 0 otherwise.
(b) Γ2(x) = [0, 1] if x = 0; Γ1(x) = 0 otherwise.
(c) Γ3(x) = 0 if x = 0; Γ1(x) = [0, 1] otherwise.
Solution:
(a) It is continuous at any x 6= 0; at 0 it is uhc but not lhc. It has convex values butnot compact values, since Γ(0) is not compact. The graph is not closed, not convex.
(b) It is continuous at any x 6= 0; at 0 it is uhc but not lhc. It has convex and compactvalues. The graph is closed, not convex.
(c) It is continuous at any x 6= 0; at 0 it is lhc but not uhc. It has convex and compactvalues. The graph is not closed (the sequence (1/n, 1) is in the graph and convergesto (0, 1) as n→∞, which is not in the graph), not convex.
2. Let Γ : [0, 1] ⇒ [0, 1] be defined by
Γ(x) =
x
2
, if x ∈
[0,
1
2
];[x
2, x], if x ∈
(1
2, 1
].
Determine if Γ is uhc and/or lhc. Does it have compact and/or convex sections? Does ithave a convex graph?
Solution: It is uhc and lhc for x 6= 12. At 1
2is lhc but not uhc. It has compact and convex
values (or sections), but does not have a convex graph.
6
-
12
10
Ω
26
3. Let Γ1,Γ2 : R ⇒ R be two lhc (uhc) correspondences. Show that the correspondence Γdefined by
Γ(x) = Γ1(x) ∪ Γ2(x)
is also lhc (uhc).
Solution: We only prove the lhc case. Let x0 ∈ R and let U be an open set such thatΓ(x0) ∩ U 6= ∅. Then Γ1(x0) ∩ U 6= ∅ or Γ2(x0) ∩ U 6= ∅. Suppose, without loss ofgenerality, that Γ1(x0)∩U 6= ∅. Since Γ1 is lhc, there is an open set N1 with x0 ∈ N1 andsuch that for any x ∈ N1, Γ1(x)∩U 6= ∅, thus (Γ1(x)∪ Γ2(x))∩U 6= ∅, showing that Γ islhc.
4. Show that if Γ1,Γ2 : X ⇒ Rp are continuous and Γ1(x)∩Γ2(x) 6= ∅ for every x ∈ X, thenthe correspondence given by Γ(x) = Γ1(x) ∩ Γ2(x) is also continuous.
Solution: Let x0 ∈ X and let U be open with U ⊇ Γ(x0). Let Ui open such thatΓi(x0) ⊆ Ui, i = 1, 2 and U1 ∩ U2 ⊆ U . There are open sets N1 3 x0 and N2 3 x0 suchthat Γ1(X ∩N1) ⊆ U1 and Γ2(X ∩N2) ⊆ U2. Hence, for N = N1 ∩N2 which is open, wehave Γ(X ∩N) ⊆ U1 ∩ U2 ⊆ U , hence Γ is uhc.
Now, let U ∩ Γ(x0) 6= ∅. Let Ui be open such that Γi(x0) ∩ Ui 6= ∅, i = 1, 2 andU1 ∩ U2 ⊆ U . The there are open sets Ni 3 x0 such that Γi(X ∩ Ni) ∩ Ui 6= ∅ i = 1, 2.Hence, for N = N1 ∩N2 which is open, we have Γ(X ∩N) ∩ U 6= ∅, hence Γ is lhc.
5. (Sequential characterization of lower hemi–continuity). Let Γ : X ⊆ Rm ⇒ Rp be acorrespondence and let x0 ∈ X. Prove:
Γ is lhc at x0
if and only if
[∗] ∀y0 ∈ Γ(x0) ∀xn ∈ X with xn → x0, ∃n0 ∈ N and yn ∈ Γ(xn) for n ≥ n0 withyn → y0.
Solution:
⇒]
Let y0 ∈ Γ(x0) and let xn ∈ X with xn → x0. Let Bj = B(y0, 1/j) for j ≥ 1. Since Γ islhc at x0, ∃εNj open, x0 ∈ Nj, such that ∀x ∈ Nj ∩X, Bj ∩ Γ(x) 6= ∅. Since xn → x0,∀j ∃nj such that xn ∈ Nj for n ≥ j. One can take n1 < n2 < · · · < nj < · · · . By the lhcof Γ at x0 ∃yjn ∈ Bj ∩ Γ(xn) for n = nj, nj + 1, . . .. Let now yn = yjn if nj ≤ n < nj+1.Then yn ∈ Γ(xn) and yn → y0, since n→∞ implies j →∞.
⇐]
By contradiction, if Γ is not lhc at x0, then ∃y0 ∈ Γ(x0) and an open set U with U ∩Γ(x0) 6= such that ∀n ∃xn ∈ B(x0, 1/n) and U ∩ Γ(xn) = ∅. Let y0 ∈ Γ(x0). Sincexn → x0, there exists n0 and yn ∈ Γ(xn) for n ≥ n0 such that yn → y0, so there exists
27
n1 such that yn ∈ U for any n ≥ n1. Then, for n ≥ maxn0, n1 we get yn ∈ U ∩ Γ(xn)with xn ∈ B(x0, 1/n), that contradicts U ∩ Γ(xn) = ∅.
6. (Sequential characterization of upper hemi–continuity). Let Γ : X ⊆ Rm ⇒ Rp be acorrespondence with graph Ω and let x0 ∈ X. Prove:
Solution: We argue by contradiction supposing that Γ is not uhc at x0. Then ∃U openΓ(x0) ⊆ U , ∃xn ∈ B(x0, 1/n) and ∃yn ∈ Γ(xn) such that yn /∈ U . Since xn → x0, by [**]there exists a subsequence (yn(j)) of (yn) converging to y0 ∈ Γ(x0). But yn(j) ∈ Rm\Uwhich is closed, thus y0 /∈ U , contradicting Γ(x0) ⊆ U .
To show the converse, let x0 ∈ X and a sequence xn ∈ X with xn → x. The setK = (xn) ∪ x0 is compact, thus the image of K under Γ, Γ(K), is also compact sinceΓ is uhc and compact–valued (Proposition 2.18). Let yn ∈ Γ(xn), so that (yn) ⊆ Γ(K)and hence ∃(yn(j)) → y0 ∈ Γ(K) as j → ∞. In fact y0 ∈ Γ(x0): if not, it is possible tofind open sets U and W with U ∩W = ∅ and Γ(x0) ⊆ U , y0 ∈ W . Since Γ is uhc at x0,∃N open, x0 ∈ N Γ(x) ⊆ U ∀x ∈ N ∩ X. Since xn → x0, ∃n0 such that xn ∈ N ∩ X∀n ≥ n0, thus Γ(xn) ⊆ U ∀n ≥ n0 and hence yn(j) ∈ U , yn(j) /∈ W for j large enough.Since the set Rp\W is closed, y0 ∈ Rp\W and thus y0 /∈ W , a contradiction.
7. Let Γ : X ⊆ Rm ⇒ Rp be a correspondence with an open graph Ω. Show that Γ is lhc.
Solution: Let x0 ∈ X and let xn ∈ X with xn → x. Let y0 ∈ Γ(x0). Since Ω is open,there exists n′0 ∈ N such that for all n ≥ n′0 the ball B((x0,y0), 1/n) ⊆ Ω. Since xn → x,there exists n0 ∈ N and yn ∈ Γ(xn) such that (xn,yn) ∈ B((x0,y0), 1/n) for all n ≥ n0.Then yn → y0 and thus Γ is lhc at x0 in virtue of the sequential characterization of lowerhemi–continuity of correspondences.
8. (Budget correspondence) Consider m commodities x ∈ Rm++ with prices p ∈ Rm++. Aconsumer receives income I > 0. Let for (p, I) the set of affordable commodity vectors
B(p, I) =
x ∈ Rm++ : p · x ≡
m∑i=1
pixi ≤ I
.
The correspondence B : Rm+1++ ⇒ Rm++ is the budget correspondence. Show that B is
continuous.
Solution:
28
B is uhc:
Let a sequence (pn, In) ∈ Rm+1++ converging to (p0, I0)Rm+1
++ and let xn ∈ Rm++ a sequence ofcommodity vectors in the budget set, xn ∈ B(pn, In). If we prove that ∃(xn(j)) convergingto x0 ∈ B(p0, I0) we are done, since [**] of problem above is fulfilled, and this impliesthat B is uhc. Let us define
p∗i = infnpni , i = 1, . . . ,m,
I∗ = supnIn
Note that p∗i > 0, since pni > 0 and pni → p0i > 0 as n → ∞; analogously, 0 < I∗ < ∞.
We havep∗ · xn ≤ pn · xn ≤ In ≤ I∗,
thus xn ∈ B(p∗, I∗) for all n. Any budget set is compact, thus ∃xn(j) → x0 ∈ B(p∗, I∗)and by continuity of the scalar product of vectors
p0 · x0 = limj→∞
pn(j) · xn(j) ≤ limj→∞
In(j) = I0,
hence x0 ∈ B(p0, I0).
B is lhc:
Let (p0, I0) ∈ Rm+1++ , x0 ∈ B(p0, I0) an consider the ball U = B(x0, δ). Let x0 ∈ U ∩
B(p0, I0), so that p0x0 ≤ I0. Let ε > 0 such that εx0 ∈ U ; note that p0 · (εx0) ≤ εI0 < I0.By contradiction, if B is not lhc, then ∃pn → p0, ∃In → I0 and U ∩ B(pn, In) = ∅.Given that pn(εx0) − In → p0(εx0) − I0 < 0, for n large enough pn(εx0) − In < 0, thusx0 ∈ B(pn, In), hence U ∩ B(pn, In) 6= ∅, a contradiction.
9. Let f : R −→ R be defined by
f(x, y) = x− y − (x− y)2
2.
Define the correspondence Γ : R⇒ R by
Γ(x) = y ∈ R : y ≤ 2x.
Let v(x) = supy∈Γ(x) f(x, y) and Γ∗(x) the set of maximizers. Do the hypotheses of theMaximum Theorem hold for this problem? Verify, through direct calculation, whether theconclusions of the Maximum Theorem hold.
Solution: The only condition of the Maximum Theorem that is not fulfilled is thatΓ is not compact valued. However, the value function is continuous and the optimalcorrespondence is uhc, indeed it is a continuous function. To find v, consider first thepossibility that the maximizing argument is interior. Then
∂f
∂y(x, y) = 0⇔ −1 + (x− y) = 0⇔ y = x− 1.
29
Now, y = x− 1 is admissible if and only if x− 1 ≤ 2x, if and only if x ≥ −1. Thus, forx ≥ −1, Γ∗(x) = x− 1, since
∂2f
∂y2(x, y) = −1 < 0 ∀(x, y)
implies that f is strictly concave with respect to y, so the critical point of f is automat-ically the (unique) global maximizer of f . For x < −1 (−1 + x > 0), y = x − 1 is notadmissible and
∂f
∂y(x, y) = −1 + x− y > −y > 0,
since y ≤ 2x ≤ −2. Thus, f is increasing in y, and takes the greatest value at y = 2x.Hence we have found that
Γ∗(x) =
2x, if x < −1;x− 1, if x ≥ −1.
which is a continuous functions (hence in particular a uhc correspondence). The valuefunction is
v(x) =
−x−x2
2, if x < −1;
1
2, if x ≥ −1;
which is continuous. Thus, although not all the hypotheses of the theorem hold, theconclusions still hold. By the way, notice that v is concave, as expected, since f isconcave and Ω is convex.
10. Show that for the problem v(x) = supy∈Γ(x) f(x, y) with X = Y = [0, 1], f : X × Y → Rgiven by f(x, y) = xy and Γ(x) = [0, 1], the optimal correspondence Γ∗ is not lhc. Thisshows that the lhc of Γ∗ does not follows from the Maximum Theorem.
Solution: The hypotheses of the Maximum Theorem are fulfilled, thus Γ∗ is uhc andcompact valued. In fact
Γ∗(x) =
1, if x > 0;[0,1], if x = 0,
which is not lhc at 0.
11. Study the problem maxy∈Γi(x) y, where the correspondences Γi for i = 1, 2, 3 are defined inProblem 2–1.
Solution: This problem shows that the hypotheses of the Maximum Theorem cannotbe weakened to attain the conclusions.
The function f(x, y) = y is continuous.
For Γ1 we have that Γ∗1(0) = ∅ (need of compact valuedness of the correspondence). ForΓ2, which is not lhc at 0, we have
Γ∗2(x) =
1, if x = 0;0, otherwise,
30
which is not uhc at 0. For Γ3, which is not lhc at 0, we have
Γ∗3(x) =
0, if x = 0;1, otherwise,
which is neither lhc nor uhc at 0. Note also that the value function in this case is notcontinuous at 0, but it is lsc, since Γ3 is lhc.
12. Check that for the following problem the optimal correspondence is not uhc and the valuefunction is not continuous. What hypotheses of the Maximum Theorem fail in this case?
X = Y = [0, 1], Γ(x) = [0, x] and f defined as
f(x, y) =
1− 2y, if 0 ≤ y < 1
2;
3− 2y, if 12≤ y ≤ 1,
for all x ∈ X.
Solution: Observe the graph of function f against y.
6
-
AAAAAA
AAAAAA
12
0
1
2
1
For x < 1/2 the maximum of f on [0, x] is attained at 0, since y < 1/2 and for x ≥ 1/2the maximum of f on [0, x] is attained at 1/2. Thus
Γ∗(x) =
0, if 0 ≤ x < 1
2;
12, if 1
2≤ x ≤ 1,
which is not uhc at 1/2. This is because function f is not lsc at points (x, 1/2). Thevalue function is
v(x) = f(x,Γ∗(x)) =
1, if 0 ≤ x < 1
2;
2, if 12≤ x ≤ 1
which is not continuous at 1/2 (but it is usc, as we know).
13. Let X ⊆ Rm a nonempty convex set and let Γ : X ⇒ Rp be a correspondence withnonempty values and a convex graph Ω ⊆ Rm × Rp. Let f : Ω → R be a quasiconcavefunction, that is, for any λ ∈ R, the upper level set Sλ(f) = (x,y) ∈ Ω : f(x,y) ≥ λis convex. Assuming that the value function v(x) = supy∈Γ(x) f(x,y) is finite valued forall x ∈ X, show that v is also quasiconcave and that for all x ∈ X, Γ∗(x) is convex.
31
Solution: Let λ ∈ R be arbitrary and let xi ∈ Sλ(v), i = 1, 2. Let θ ∈ [0, 1]. We wantto show that xθ = θx1 + (1− θ)x2 ∈ Sλ(v). By definition of v as a supremum, and giventhat it is finite valued for any x ∈ X, for any ε > 0 there exists yi ∈ Γ(xi) such that
λ ≤ v(xi)− ε < f(xi,yi), i = 1, 2. (3)
Let yθ = θy1 + (1− θ)y2; since Ω is convex, (xθ,yθ) ∈ Ω, thus yθ ∈ Γ(xθ), so that
v(xθ) = supy∈Γ(xθ)
f(xθ,y) ≥ f(xθ,yθ) > λ+ ε.
where the last inequality is due to (3) and given that f is quasiconcave. Since v(xθ) > λ+εis true for any ε > 0, we have v(xθ) ≥ λ, thus xθ ∈ Sλ(v). Since λ is arbitrary, we haveshown that v is quasiconcave. The assertion about Γ∗(x) is a standard result in convexoptimization (the maximizers of a quasiconcave function, if they exist, form a convex set).
14. Let X ⊆ Rm a nonempty convex set and let Γ : X ⇒ Rp be a constant correspondence,that is, Γ(x) = Y ⊆ Rp for all x ∈ X. Suppose that for any fixed y ∈ Y , the mappingx 7→ f(x,y) is convex on X. Assuming that the value function is finite valued, show thatit is convex. Is Γ∗ convex–valued?
Solution: Let xi ∈ X, i = 1, 2 and for θ ∈ [0, 1] consider xθ ∈ X as in problem above.Given ε > 0, there exists yε ∈ Y such that
v(xθ) < f(xθ,yε) + ε
≤ θf(x1,yε) + (1− θ)f(x2,yε) + ε
≤ θv(x1) + (1− θ)v(x2) + ε.
The second line is due to the (partial) convexity of f and the third line is by definitionof v. Since this is true for any ε, we have that v is convex. Γ∗ is not convex–valued ingeneral, as f is not concave with respect y.
32
3 Fixed Point Theorems in Rm
3.1 Theorems
Definition 3.1. Given a mapping f : X −→ X, x ∈ X is a fixed point of f if f(x) = x.
Definition 3.2. Given a correspondence Γ : X ⇒ X, x ∈ X is a fixed point of Γ if x ∈ Γ(x).
Note that both definitions coincide when Γ is single valued.
Theorem 3.3 (Brower’s Fixed Point Theorem). Let X ⊆ Rm be nonempty, convex and compactand let f : X → X be continuous. Then f has a fixed point.
Theorem 3.4 (Kakutani’s Fixed Point Theorem). Let X ⊆ Rm be nonempty, convex andcompact and let Γ : X ⇒ X be an uhc non–empty–valued, closed–valued and convex–valuedcorrespondence. Then Γ has a fixed point.
Note that Brower’s Theorem is a particular case of Kakutani’s Theorem. In the next chapterwe will see another fixed point theorem, due to Banach.
3.2 Application to Game Theory
A finite game in normal form is a 3–tuple G = (N, (Si)i, (ri)i) where N is the finite set/numberof players, i denotes a generic player, Si is the finite set of pure strategies of player i andri : S −→ S is the reward function of player i and S ≡
∏j Sj.
A Nash Equilibrium (NE) is a profile of strategies s∗ = (s∗1, . . . , s∗N) such that
∀i ∈ N, ∀si ∈ Si, ri(si, s∗−i) ≤ ri(s
∗),
where −i denotes “everyone but i”.The mixed extension of game G is a 3–tuple M(G) = (N, (Σi)i, (ri)i)) where
Σi =
σi : Si → [0, 1] :
∑i∈Si
σi(si) = 1
is the set of probability distributions over Si or mixed strategies (it can be identified with thesimplex of Rni−1, where ni is the cardinality of Si) and ri is now
ri(σ) =∑s∈S
(∏j
σj(sj)
)ri(s).
Note that ri(σ) can be rewritten as a convex combination of the payoffs of player i attained athis/her pure strategies
ri(σ) =∑si∈Si
σi(si)ri(si, σ−i). (4)
This follows from the identity
ri(si, σ−i) =∑
s−j∈S−j
(∏j 6=i
σj(sj)
)ri(s),
33
where S−j has the obvious meaning. From (4) for any σ ∈ Σ ≡∏
j Σj, for any i ∈ N thereexists some si ∈ Si such that
σi(si) > 0 and ri(si, σ−i) ≤ ri(σ). (5)
Otherwise, if for all si ∈ Si with σ(si) > 0 we had ri(σ) < ri(si, σ−i), then
ri(σ) <∑si∈Si
σi(si)ri(si, σ−i) = ri(σ),
a contradiction.
Definition 3.5. A NE in mixed strategies is a profile of mixed strategies σ∗ = (σ∗1, . . . , σ∗N)
such that∀i ∈ N, ∀σi ∈ Σi, ri(σi, σ
∗−i) ≤ ri(σ
∗).
Theorem 3.6 (Nash Theorem). Every finite game in normal form has a NE.
Proof. The proof uses Brower’s Theorem. The set Σ is compact and convex in some Euclideanspace RM for suitable M . For si ∈ Si and σ ∈ Σ let
ϕi,si(σ) = max0, ri(si, σ−i)− ri(σ)
and define a mapping f over Σ where the component i acting over the pure strategy si ∈ Si is
fi(σ)(si) =σi(si) + ϕi,si(σ)∑
di∈Si σi(di) + ϕi,di(σ)=
σi(si) + ϕi,si(σ)
1 +∑
di∈Si ϕi,di(σ).
intuitively, f maps a strategy profile σ to a new profile in which each player’s pure strategiesthat are better responses to σ receive increased probability mass. Observe that we can considereach fi(σ) as a probability distribution over Si, since 0 ≤ fi(σ) ≤ 1 and
∑si∈Si fi(σ)(si) = 1.
Thus, f : Σ → Σ. Moreover, a fixed point of f is a NE. To show this, note that if σ∗ = f(σ∗)for some σ∗, then by the definition of f we have
σ∗i (si)∑di∈Si
ϕi,di(σ∗) = ϕi,si(σ
∗).
By (5) and the definition of ϕi,si , there exists si ∈ Si with σ∗i (si) > 0 such that ϕi,si(σ∗) = 0.
From the previous identity with si = si it must be∑
di∈Si ϕi,di(σ∗) = 0, so that ϕi,di(σ
∗) = 0for all di ∈ Si, what means
ri(di, σ∗−i) ≤ ri(σ
∗).
for all di ∈ Si. Take any σi ∈ Σi and multiply the above inequality by σi(di); summing up indi ∈ Si we get ri(σi, σ
∗−i) ≤ ri(σ
∗), hence σ∗ is a NE. It remains to check that the hypotheses ofBrower’s Theorem are fulfilled. That Σ is compact has been commented above. Clearly, it isalso a convex set, as the Cartesian product of convex sets. Finally, f is continuous since eachri is continuous and hence ϕi,si is continuous. Thus Brower’s Theorem assures the existence ofa fixed point.
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There exists a proof based in Kakutani’s Theorem that uses the concept of best response.Given σ−i ∈ Σ−i, the best response of player i is the set of strategies that maximize his/herreward once the remainder players choose to play σ−i
We make the dependence of BRi explicit on σi to simplify notation. Note that σ∗ is a NE ifand only if
∀i ∈ N σ∗i ∈ BRi(σ∗).
Thus if we define the correspondence BR : Σ ⇒ Σ by BR(σ) = (BR1(σ), . . . ,BRN(σ)), a NEis simply a fixed point of BR and conversely. Let us check that all conditions of Kakutani’sTheorem are fulfilled.
• Σ is compact, convex: already known.
• BR is an uhc, non–empty–valued and compact–valued correspondence: The best responsecorrespondence of player i is the optimal correspondence of the parametric optimizationproblem
maxσi∈Σi
ri(σi, σ−i).
The parameter is σ−i ∈ Σ−i and the decision variable belongs to the compact and convexset Σi, which defines a constant correspondence. The function ri is continuous, thus bythe Maximum Theorem BRi has non–empty–values, it is compact–valued and an uhccorrespondence.
• BR is convex–valued. Suppose that σ′i, σ′′i ∈ BRi. Let θ ∈ [0, 1] and let σθi = θσ′i+(1−θ)σ′′i .
Then by (4)
ri(σθi ) =
∑si∈Si
σθi (si)ri(si, σ−i) = θ∑si∈Si
σ′i(si)ri(si, σ−i)+(1−θ)∑si∈Si
σ′′i (si)ri(si, σ−i) = max ri,
thus σθi ∈ BRi.
3.3 Application to strictly positive matrices
Given a square matrix A ∈ Mm(R), A = (aij), we say that A is strictly positive if aij > 0 forall i, j = 1, . . . ,m. Given a row vector x ∈ Rm, (Ax)i denotes the ith–row of the product Atimes x. We say that a vector x is positive is all its components are positive.
Theorem 3.7 (Perron’s Theorem). A strictly positive matrix A has a positive eigenvalue anda positive eigenvector.
Proof. Let X = x ∈ Rm+ :∑m
i=1 = 1, which is compact and convex. Note that necessarily∑mi=1(Ax)i > 0. Let the function f defined over X as
f(x) =Ax∑m
i=1(Ax)i.
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Note that f(X) ⊆ X, since the ith component of the vector f(x) is (f(x))i = (Ax)i∑mi=1(Ax)i
≥ 0
and the sum of all its components is∑m
i=1(f(x))i = 1. Since f is clearly continuous, accordingto Brower’s Theorem there exists x∗ ∈ X such that f(x∗) = x∗, hence
Ax∗ =
(m∑i=1
(Ax∗)i
)x∗.
Then λ =∑m
i=1(Ax∗)i > 0 is a positive eigenvalue and x∗ is a positive eigenvector of matrixA.
3.4 Problems
1. It is 3:00 p.m. Find the elapsed time until the minute hand and the hour hand coincide,considering the problem as a fixed point problem.
Solution: A standard clock is divided into 60 parts. The ratio of angular velocities ofthe hour hand to the minute hand is ωh/ωm = 1/12, since in one hour the hour handruns 5 parts of the clock and the minute hand does a full circumference. Let h (m) bethe position of the hour (minute) hand over the clock after t minutes. Then
h = 15 + tωh,
m = 0 + tωm.
Eliminating t from the two equations we find
h = 15 +ωhωm
m = 15 +1
12m.
Obviously, the problem is solved when h = m, thus we have the fixed point problem
m = 15 +1
12m.
The solution is m = 15×1211
.
2. Let f : [0, 1] −→ [0, 1]. Show that f has at least a fixed point in the following cases.
(a) f is continuous.
(b) f is monotonous non–decreasing.
Solution: Notice that0 ≤ f(0), f(1) ≤ 1. (6)
(a) Let g(x) = f(x)−x. Function g is continuous and g(0) = f(0) ≥ 0, g(1) = f(1)− 1.If f(0) = 0 or f(1) = 1 we have finished. Otherwise, f(0) > 0 and f(1) < 1 by (6),hence g(0) > 0 and g(1) < 0. By Bolzano’s Theorem, there exists c ∈ (0, 1) suchthat g(c) = 0, thus f(c) = c.
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(b) This is a particular case of a much more general result known as Tarski’s Theorem.The proof is as follows. From (6) we know that 0 ≤ f(0), thus the set
C = x ∈ [0, 1] : x ≤ f(x)
is non–empty. Note that x ∈ C implies f(x) ≤ f(f(x)) since f is nondecreasing,thus f(x) ∈ C also. We have shown
x ∈ C ⇔ f(x) ∈ C. (7)
Let c = supC, which exists since C ⊆ [0, 1] and is therefore bounded. For everyx ∈ C we have x ≤ c, thus f(x) ≤ f(c) since f is nondecreasing and hence x ≤f(x) ≤ f(c), thus f(c) is un upper bound of the set C; since c is the least upperbound, c ≤ f(c) follows, hence c ∈ C. But then f(c) ∈ C by (7), hence f(c) ≤ cbecause c is the maximum of set C, therefore f(c) = c and c is a fixed point of f .
Note that the result is not true for decreasing mappings. The function f(x) = 1 ifx ∈ [0, 1
2] and f(x) = 0 if x ∈ (1
2, 1] is decreasing, maps [0, 1] into [0, 1] but has no
fixed points.
3. Let Γ : [0, 1] ⇒ [0, 1] be nonempty-valued, compact-valued and continuous. Show that Γhas a fixed point.
Solution: We cannot apply Kakutani’s Theorem since Γ is not convex-valued. Considerinstead the function
v(x) = maxy∈Γ(x)
y.
By the Maximum Theorem, v is continuous and obviously, v(x) ∈ Γ(x) for all x ∈ [0, 1],hence v : [0, 1]→ [0, 1], thus by Brower’s Theorem v(x) = x for some x ∈ [0, 1], so Γ hasa fixed point.
4. Show by means of counterexamples that the hypotheses of Brower Theorem are needed.
Solution:
• Continuity of f : let f(x) = 1/2 if x ∈ [0, 1/2), f(x) = 0 if x ∈ [1/2, 1]; then,X = [0, 1] is compact and convex, f([0, 1]) ⊆ [0, 1] but f has no fixed points in [0, 1].
• Compactness of X: let X = [0, 1) and let f(x) = x.
• Convexity of X: let X = [0, 1] ∪ 2 and let f(x) = 2 if x ∈ [0, 1], f(2) = 0. Then,f is continuous on X (why?), f(X) ⊆ X but f has no fixed points.
5. Show by means of counterexamples that the hypotheses of Kakutani Theorem are needed.
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6. Let K ⊆ Rm be compact and let f : K −→ K be continuous such that ‖f(x) − f(y)‖ <‖x− y‖ for all x,y ∈ K. Prove that f has a fixed point.
Solution: By contradiction, suppose that f has no fixed points in K. Then the functiondefined as g(x) = ‖x− f(x)‖ is strictly positive over K. Moreover, g is continuous, thusattains a global minimum over K, z. Let u = f(z) ∈ K. We have
Here we generalize the concept of norm and metric from Rm to more general sets/vector spaces.Let E be any set. Remind that E × E = (x, y) : x, y ∈ E. Notice that from now on, wewrite elements in a metric space without boldface.
Definition 4.1. A metric or distance function on E is a function d : E ×E −→ R+ such thatfor every x, y, and z in E we have
Definition 4.2. A metric space is a pair (E, d), where E is a set and d is a metric defined onit.
Some sets E have an algebraic structure that allows us to operate with their elements. Inparticular we will consider vector spaces. We are interested not only in measuring the distancebetween points, but also in giving a meaning to the length of a vector.
Definition 4.3. Let E be a vector space over the reals (hence, in particular, 0 ∈ E). A normon E is a function ‖ ‖ : E −→ R+ such that for every x, y in E and for every scalar λ ∈ R wehave
(i) ‖x‖ ≥ 0,
(ii) ‖x‖ = 0⇔ x = 0,
(iii) ‖x+ y‖ ≤ ‖x‖+ ‖y‖ (triangle inequality),
(iv) ‖λx‖ = |λ|‖x‖ (homogeneity).
Definition 4.4. A normed space is a pair (E, ‖ ‖) where E is a vector space and ‖ ‖ is a norm.
Remark 4.5. Note that a normed space becomes a metric space if we define d(x, y) = ‖x− y‖.
Example 4.6 (Examples of metric spaces). 1. Every set E is metrizable; just define d(x, x) =0, d(x, y) = 1 (x 6= y). This is the discrete metric (not a very interesting one).
2. We already know that the Euclidean space Rm has metric
In the case m = 1, the real line, we have d(x, y) = |x− y|.
3. In Rm with the Euclidean norm consider d(x,y) = ‖x‖+‖y‖ when x 6= y and d(x,x) = 0.Then (Rm, d) is a metric space.
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4. Given a set X, the set of real bounded functions f on X (that is, |f(x)| ≤ M for everyx ∈ X) is denoted B(X),
B(X) = f : X −→ R : f bounded .
The space B(X) is a vector space. The sup norm is defined as
‖f‖∞ = supx∈X|f(x)|
and the sup metric is defined as
d∞(f, g) = supx∈X|f(x)− g(x)| = ‖f − g‖∞.
(B(X), ‖ · ‖∞) is a norm space, and (B(X), d∞) is a metric space.
5. Given a metric space X, the set space of real bounded and continuous functions f on Xis denoted Cb(X),
Cb(X) = f : X −→ R : f bounded and continuous .
It is a subspace of B(X), thus it can be given the sup norm (and metric).
Note that if X is compact, then every continuous function is bounded on it (Weierstrass’Theorem) and the space is then denoted C(X).
6. Let X = [a, b], a < b. Consider the set of continuous functions C([a, b]) with the norm
‖f‖1 =
∫ b
a
|f(t)| dt, f ∈ C([a, b]).
The associated metric is
d1(f, g) =
∫ b
a
|f(t)− g(t)| dt, f, g ∈ C([a, b]).
7. Let X = [a, b], a < b. Consider the set of continuous functions C([a, b]) with the norm
‖f‖2 =
√∫ b
a
|f(t)|2 dt, f ∈ C([a, b]).
The associated metric is
d2(f, g) =
√∫ b
a
|f(t)− g(t)|2 dt, f, g ∈ C([a, b]).
8. Given the interval [a, b], the metric space of real functions of class C1 is denoted C1([a, b]),
C1([a, b]) = f : [a, b] −→ R : f of class C1 on [a, b].
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Of course, it is a metric space with the usual sup norm, but in some applications it canbe interesting to know how different values take two functions and their derivatives. Tothis end we can define either of the following distance functions
d(f, g) = max
supx∈X|f(x)− g(x)|, sup
x∈X|f ′(x)− g′(x)|
,
d(f, g) = supx∈X
(|f(x)− g(x)|+ |f ′(x)− g′(x)|),
d(f, g) = |f(a)− g(a)|+ supx∈X|f ′(x)− g′(x)|.
Definition 4.7. Let (E, d) be a metric space. The open ball of center x0 and radius r > 0 is
B(x, r) = x ∈ E : d(x, x0) < r.
The closed ball of center x0 and radius r > 0 is
B(x, r) = x ∈ E : d(x, x0) ≤ r.
Definition 4.8. Let (E, d) be a metric space. A sequence xn of the metric space E is a mapfrom N to E assigning to every natural number an element of E. The sequence xn convergesto x0 ∈ E if
d(xn, x0)→ 0, as n→∞,that is, for every ε > 0 there exists n0(ε) such that
d(xn, x0) < ε ∀n ≥ n0(ε) as n→∞.
We say that x0 is the limit of xn.
Lemma 4.9. Let (E, d) be a metric space.
1. The limit is unique;
2. If xn → x and n(1) < n(2) < · · · < n(j) < · · · , then xn(j) → x as j →∞.
It is possible to express the notion of convergence in terms of balls: xn converges to x0 iffor every ε > 0 there exists n0(ε) such that xn ∈ B(x0, ε) for every n ≥ n0, so except for a finitenumber of elements, the sequence is contained in an open ball of radius as small as required.
Notice that the limit is unique, because if y0 ∈ E, y0 6= x0, is also a limit, then 0 ≤d(x0, y0) ≤ d(x0, xn) + d(xn, y0)→ 0 as n→∞ and so d(x0, y0) = 0, a contradiction.
Example 4.10 (Convergence in the sup norm). Consider the space of bounded functions onthe set X with the supremum norm, (B(X), ‖ · ‖∞). Convergence of a sequence fn to f impliespunctual convergence, that is
limn→∞
fn(x) = f(x) for every x ∈ X.
However, the reciprocal is not true, that is, punctual convergence does not imply convergence inthe sup norm. Consider the following example. Let fn(x) = xn and X = [0, 1]. The punctuallimit function of fn is f given by f(x) = 0, x 6= 1, f(1) = 1. However, the convergence is notin norm, because for x = (1/2)(1/n), |fn(x) − f(x)| = 1
2for every n and then ‖fn − f‖ ≥ 1/2.
Convergence in the sup norm in the space B(X) or Cb(X) is also called uniform convergenceof functions.
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Definition 4.11. Let (E, d) be a metric space. A set U ⊆ E is open if whenever x ∈ U , thereexists an ε > 0 such that B(x, ε) ⊆ U .
Definition 4.12. Let (E, d) be a metric space. A set F ∈ E is closed if whenever xn ∈ F foreach n and xn → x as n→∞, then x ∈ F .
Lemma 4.13. Let (E, d) a metric space. A subset U of E is open if and only if its complementE\U is closed.
Lemma 4.14. Let (E, d) be a metric space. Consider the collection τ of open sets in E.
1. ∅ ∈ τ , E ∈ τ ;
2. If Uα ∈ τ for all α ∈ A, then⋃α∈A Uα ∈ τ ;
3. If U1, U2, . . . , Un ∈ τ , then⋂nj=1 Uj ∈ τ .
Lemma 4.15. Let (E, d) be a metric space. Consider the collection F of closed sets in E.
1. ∅ ∈ F , E ∈ F ;
2. If Fα ∈ F for all α ∈ A, then⋂α∈A Fα ∈ F ;
3. If F1, F2, . . . , Fn ∈ F , then⋃nj=1 Fj ∈ F .
Definition 4.16. Let (E, d) and (G, ρ) be metric spaces. We say that a function f : E → G iscontinuous at x ∈ E if, given ε > 0, we can find δ(x, ε) such that, if y ∈ E and d(x, y) < δ(x, ε),we have
ρ(f(x), f(y)) < ε.
If f is continuous at every point x ∈ E, then we say that f is a continuous on E.
Lemma 4.17. Let (E, d) and (G, ρ) be metric spaces. The function f : E → G is continuousat x ∈ E if and only if for any sequence xn ∈ E that converges to x as n→∞, f(xn)→ f(x)as n→∞.
Lemma 4.18. Let (E, d) and (G, ρ) be metric spaces. The function f : E → G is continuousif and only if for any open set O in G, the inverse image f−1(O) is open in E.
4.1 Complete metric spaces
Definition 4.19. Let (E, d) be a metric space. A sequence xn ∈ E is Cauchy if given any ε > 0there exists n0(ε) such that d(xp, xq) < ε whenever p, q ≥ n0(ε).
Every convergent sequence is a Cauchy sequence: since xn converges to x0, for every ε > 0one can find n0(ε) with d(xp, x0) < ε/2 for p ≥ n0)ε). By the triangle inequality
d(xp, xq) ≤ d(xp, x0) + d(x0, xq) <ε
2+ε
2= ε, for all p, q ≥ n0(ε).
The reciprocal is not true: take e.g. E = (0, 1), d(x, y) = |x − y| and xn = 1/n, which is aCauchy sequence (take p > q; then |1/p − 1/q| < 1/q < ε for suitable large q); however, 1/ndoes not converges to any element in E (note that 0 /∈ E).
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Definition 4.20. A metric space (E, d) is complete if every Cauchy sequence converges. Anorm space (E, ‖ · ‖) is a Banach space if it is complete.
The condition of being complete depends on the metric. It could be that a set is completewith some metric but not with another.
The following result is interesting because it provides examples of metric spaces which arecomplete and of metric spaces which are not complete.
Proposition 4.21. Let (E, d) be a complete metric space. If A is a subset of X, then (A, d) iscomplete if and only if A is closed in (E, d).
Proof. Any sequence xn ∈ A that is a Cauchy sequence in E converges to some x ∈ E. SinceA is closed, x ∈ A, thus A is complete.
Reciprocally, suppose that (A, d) is complete and let xn ∈ A a convergent sequence to somex ∈ E. Then the sequence xn is Cauchy in A and must converge to some y ∈ A. By uniquenessof the limit, x = y and x ∈ A, showing that A is closed in (E, d).
We already know that the Euclidean space is complete. Thus, the closed interval [a, b] iscomplete, but [a, b), (a, b] or (a, b) are not complete.
Here is another interesting example of a complete space.
Lemma 4.22. The space Cb(X) is complete.
Proof. Let d be the metric defined on the set X. Let fn be a Cauchy sequence in Cb(X), thatis, for every ε > 0 there exists n0(ε) such that
supx∈X|fp(x)− fq(x)| < ε, whenever p, q ≥ n0(ε).
this implies that for all x ∈ X fixed
|fp(x)− fq(x)| < ε, whenever p, q ≥ n0(ε), (8)
that is, the sequence of real numbers fn(x) is Cauchy and so by the Cauchy’s Principle ofConvergence it converges to a real number, denoted by f(x).
Since any Cauchy sequence is bounded, there exists a constant K such that ‖fn‖∞ ≤ K forall n. Hence
as m→∞. Thus |f(x)− fn(x)| ≤ supp,q≥n0‖fp − fq‖∞ for all x ∈ E, hence
‖f − fn‖∞ ≤ supp,q≥n0
‖fp − fq‖∞
43
for all n ≥ n0. Since the sequence fn is Cauchy, we see that ‖f − fn‖∞ → 0 as n→ 0.Finally, we must show that the limit function f is continuous. Consider the inequality
and fix ε > 0. Since that the convergence of fp to f is independent of x, there exists n0(ε) suchthat both |f(x)− fp(x)| < ε and |fp(x0)− f(x0)| < ε for all p ≥ n0(ε). Since fp is continuous,we can find δ(ε) such that |fp(x) − fp(x0)| < ε whenever x is such that d(x, x0) < δ(ε). Inconsequence, |f(x)− f(x0)| < ε+ ε+ ε and f is continuous.
Example 4.23. The normed space (C([a, b]), ‖·‖1) defined in 6 of Example 4.6 is not complete.For, consider w.l.g. [a, b] = [−1, 1]. Let
fn(x) = −1 for − 1 ≤ x ≤ −1/n,fn(x) = nx for − 1/n ≤ x ≤ 1/n,fn(x) = 1 for 1/n ≤ x ≤ 1.
If m ≥ n, then
‖fn− fm‖1 =
∫ 1
−1
|fn(x)− fm(x)| dx ≤∫ 1/n
−1/n
dx =2
n→ 0
as n→∞. Hence fn is Cauchy. Suppose that there exists f ∈ C([−1, 1]) such that ‖fn−f‖1 →0 as n→∞. For n ≥ N we have∫ 1
1/N
|f(x)− 1| dx =
∫ 1
1/N
|f(x)− fn(x)| dx =≤∫ 1
−1
|fn(x)− f(x)| dx→ 0
as n→∞. Thus ∫ 1
1/N
|f(x)− 1| dx = 0.
Since function f is continuous, this means that f(x) = 1 for x ∈ [1/N, 1]. Since N is arbitrary,f(x) = 1 for all x ∈ (0, 1]. A similar reasoning shows that f(x) = 0 in [−1, 0), hence f cannotbe continuous at 0, a contradiction.
4.2 Fixed points. Contractions and Banach’s Theorem
A very useful concept is that of fixed point of an operator.
Definition 4.24. Let T : E −→ E be a mapping. We say that x is a fixed point of T ifT (x) = x.
It turns out that many problems of the applied mathematics can be formulated as fixed pointproblems for suitable operators. As an example, consider the problem of finding a solution ofthe scalar equation f(x) = 0, where f is a given function (e.g. maybe we are interested infinding the roots of the polynomial f(x) = x3 + x − 1). Construct the auxiliary functionT (x) = x+ f(x). It is obvious that a fixed point of T is a root of f and reciprocally. Note thatthere exists many ways of getting T from f . Another selection is T (x) = x+ g(x)f(x) where gis an auxiliary function satisfying suitable conditions.
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Definition 4.25. Let (E, d) be a metric space. The operator T : (E, d) −→ (E, d) is a con-traction of parameter k if
d(Tx, Ty) ≤ kd(x, y), for every x, y ∈ E,
where 0 ≤ k < 1.
Theorem 4.26 (Banach’s Theorem). Let (E, d) be a complete metric space and T a contrac-tion operator. Then T admits a unique fixed point, which can be approached from successiveiterations of the operator T from any element x1 ∈ E
Proof.Uniqueness. Assume that x0 and y0 are fixed points. Then
d(x0, y0) = d(Tx0, T y0) ≤ kd(x0, y0).
If d(x0, y0) > 0, then k ≥ 1, a contradiction. Thus d(x0, y0) = 0 and x0 = y0.Existence. Let x1 be an arbitrary element of E. Define the sequence xn by the formulaxn+1 = T (xn). Then
d(xn+1, xn) = d(T (xn), T (xn−1)) ≤ kd(xn−1, xn).
Iterating this inequality, we obtain
d(xn+1, xn) ≤ kn−1d(x2, x1), ∀n ∈ N.
It follows from the triangle inequality that for n > m
as m → 0, hence xn is Cauchy and because E is complete, it converges to some elementx0 ∈ E. We now show that x0 is a fixed point of T . We have
d(T (x0), x0) ≤ d(Tx0, xn) + d(xn, x0)
= d(Tx0, T (xn−1)) + d(xn, x0)
< kd(x0, xn−1) + d(xn, x0) < ε
for n ≥ n0. Thus d(T (x0), x0) = 0 and T (x0) = x0.
It is really interesting to note that, as shown in the proof, the fixed point can be approachedfrom successive iterations of the operator T from any element x1 ∈ E, that is, the sequenceT n(x1) converges to the fixed point x0 as n→∞. Here, T n = T T n−1 for n ∈ N, n > 1; note
45
that in the proof of the theorem xn = T n(x1). We even have an upper estimate of the distanceof the approaching sequence T n(x1) to the fixed point x0 that can be obtained as follows:
d(x0, xn) = d(T (x0), T (xn−1)) ≤ kd(x0, xn−1) = kd(T (x0), T (xn−2))
≤ k2d(x0, xn−2) ≤ · · · ≤ knd(x0, x1).
More useful is the following bound, where knowledge of the fixed point x0 is not required. Fromthe inequality
Example 4.27. Consider the mapping f(x) = 12(x + 2
x) and E = [1,∞). Show that f is a
contraction in E with contraction parameter 12
and which fixed point is√
2. Form the sequence
xn with xn+1 = f(xn) and x1 = 1 and show that |xn−√
2| ≤ 2−n. How much iterations mustbe done to get three correct decimal places of
√2?
Solution. We will use that for a function f that is continuous on [a, b] and differentiable on(a, b), there exists a < θ < b such that f(b) − f(a) = f ′(θ)(b − a) and will try to bound thederivative f ′.
The derivative of f is f ′(x) = 12
(1− 2
x2
). Observe that 1 − 2
x2< 1 and that x > 1 implies
− 1x2> −1, hence 1− 2
x2> −1. In consequence∣∣∣∣1− 2
x2
∣∣∣∣ < 1, ∀x ≥ 1
and thus supx≥1 |f ′(x)| ≤ 12. On the other hand, f([1,∞)) ⊂ [1,∞), thus f is a contraction of
parameter 12. The fixed point is the solution of the equation f(x) = x, which gives
√2 as the
only positive solution. The sequence
xn+1 =1
2
(xn +
2
xn
)converges to
√2 and the estimate
|xn −√
2| ≤ 2−(n−1)|x2 − 1| = 2−n,
since x2 = 12. To get three correct decimal places we need to take n such that |xn−
√2| < 0.001,
hence it suffices 2−n < 0.001. Since 210 = 1024, then n = 10 does the job.
Since that a closed subset of a complete metric space is itself complete, we have the followinguseful result.
Corollary 4.28. Let S be a closed subset of the complete metric space (E, d) and let T be acontraction on S. Then T has a unique fixed point that belongs to S that can be approachedstarting form any point of S.
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4.3 Application to the existence and uniqueness of solutions of or-dinary differential equations
Consider the initial value problem or Cauchy problem
x(t) = f(t,x(t)), t ∈ [t0, t1], (10)
x(t0) = z0 ∈ Rm, (11)
where f : [t0, t1]× Rm −→ Rm and the solution x : [t0, t1] −→ Rm is of class C1 on [t0, t1]. Theproblem is obviously equivalent to the following formulation
Given f and x0 as above, find a function x is of class C1 on [t0, t1] such that forevery t ∈ [t0, t1]
x(t) = z0 +
∫ t
t0
f(s,x(s)) ds.
Define for a function x of class C1 the operator T by
T (x)(t) = z0 +
∫ t
t0
f(s,x(s)) ds. (12)
It is obvious T (x) is also a function of class C1. Even this is true if the function x is onlycontinuous, hence T (C([t0, t1])) ⊆ C1([t0, t1]) ⊆ C([t0, t1]) and we can consider that T is definedover the complete metric space C([t0, t1]). The essential observation now is that a fixed pointof T is a solution of the Cauchy problem (10)–(11) and reciprocally.
Remark 4.29. In the following, the euclidean norm in Rm will be denoted with | · |. Hence theeuclidean norm of the vector x(t) will be denoted |x(t)| instead of ‖x(t)‖ in order to distinguishit from the norm of the function x, ‖x‖.
Theorem 4.30 (Picard–Lindelof). Assume f : [t0, t1]×Rm −→ Rm is continuous and satisfiesthe Lipschitz condition
|f(t,x)− f(t,y)| ≤ `|x− y|, (13)
for all t ∈ [t0, t1], for all x,y ∈ Rm, where ` is a constant. Then the Cauchy problem (10)–(11)admits a unique solution.
Proof. For convenience, define on E = C([t0, t1]) the weighted norm
‖x‖L = maxt∈[t0,t1]
e−L(t−t0)|x(t)|,
where L > ` is a fixed constant. It easily follows that (E, ‖ · ‖L) is a Banach space. Take x1,
47
x2 in E. Then
e−L(t−t0)|T (x1)(t)− T (x2)(t)| ≤∫ t
t0
e−L(t−t0)|f(s,x1(s))− f(s,x2(s))| ds
≤ `
∫ t
t0
e−L(t−t0)|x1(s)− x2(s)| ds
= `
∫ t
t0
e−L(t−s)e−L(s−t0)|x1(s)− x2(s)| ds
≤ `‖x1 − x2‖L∫ t
t0
e−L(t−s) ds
≤ `
L‖x1 − x2‖L.
Thus ‖T (x1)− T (x2)‖L ≤ (`/L)‖x1− x2‖L and T is a contraction, because `/L < 1. Then theTheorem of Banach assures the existence of a unique fixed point, which is the solution we arelooking for.
Notice that the Lipschitz condition means that the quotient
|f(t,x)− f(t,y)||x− y|
is bounded in [t0, t1]×Rm. This requirement is too much stringent, but consider the followingexample.
Example 4.31. Consider
f(t, x) =sin t2
1 + x2.
f is globally Lipschitz, because
|f(t, x)− f(t, y)| =∣∣∣∣ sin t2
1 + x2− sin t2
1 + y2
∣∣∣∣≤ |x2 − y2|
(1 + x2)(1 + y2)
=|x+ y|
(1 + x2)(1 + y2)|x− y|.
On the other hand, the function|x+ y|
(1 + x2)(1 + y2)
is continuous and has limit 0 as ‖(x, y)‖ → ∞, hence it is bounded by `, say and f is Lipschitz.We can conclude that the Cauchy problem x′ = sin t2/(1 + x2), x(t0) = x0, admits a uniquesolution on every interval [t0, t1].
48
A similar computation shows that the function
f(t, x) =t
1 + x2
is not globally Lipschitz in R × Rm. However, it is Lipschitz in every subset of the form[t0, t1]× Rm. As for the application of the Picard–Lindelof Theorem we need a closed interval,[t0, t1], from the point of view of the existence and uniqueness of solutions to first order ODEsthere is no difference with the above function.
There exists a useful criterium to deduce whether the function f satisfies a Lipschitz con-dition. Suppose that for every fixed t, f is of class C1 on Rm and that |∂fi/∂xj| ≤ C for alli, j = 1, . . . , n. Then it is possible to prove that for every x,y ∈ Rm
|f(t,x)− f(t,y)| ≤ n32C|x− y|,
hence f is globally Lipschitz. This argument also works for regions of the form [t0, t1]×D withD ⊆ Rm a compact set, leading to local Lipschitzianity of f .
Remark 4.32. The theorem implies also a computational method to approximate the solution,starting from any initial admissible function. The method is the following.
1. Choose a continuous function, x1 (it may be a constant, usually the null function);
2. Compute
x2(t) = T (x1)(t) = z0 +
∫ t
t0
f(s,x1(s)) ds;
3. Repeat the process for calculating x3 from x2 and so on for n steps:
xn+1(t) = T (xn)(t) = z0 +
∫ t
t0
f(s,xn(s)) ds;
4. The norm of the error in the approximation after n steps is given by (9).
Example 4.33. Find by the method of Picard–Lindelof an approximate solution to the Cauchyproblem
x′(t) = t− x(t), x(0) = 0,
in the interval [0, 1].
Note that we are in condition to apply the theorem and a unique solution exists on everyinterval [−h, h]. We choose x1 ≡ 0.
x1(t) = 0,
x2(t) =
∫ t
0
s− x1(s) ds =t2
2,
x3(t) =
∫ t
0
s− x2(s) ds =
∫ t
0
s− s2
2ds =
t2
2− t3
6,
x4(t) =
∫ t
0
s− x3(s) ds =
∫ t
0
s− s2
2+s3
6ds =
t2
2− t3
6+t4
24,
...
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4.4 Problems
1. Let (E, d) be a metric space.
(a) Show that for every x, y, z ∈ E, |d(y, z)− d(x, z)| ≤ d(x, y).
(b) Prove that ρ = d/(1 + d) is also a metric. (Hint: the function a 7→ a/(1 + a) isstrictly increasing).
Solution:
(a) By the triangle inequality, d(y, z) ≤ d(x, z) + d(x, y), thus
d(y, z)− d(x, z) ≤ d(x, y).
In the same way, d(x, z) ≤ d(y, z) + d(x, y), hence
−(d(y, z)− d(x, z)) ≤ d(x, y).
From both inequalities, |d(y, z)− d(x, z)| ≤ d(x, y).
(b) Triangle inequality: ρ(x, y) = d(x, y)/(1+d(x, y)) ≤ (d(x, z)+d(z, y))/(1+d(x, z)+d(z, y)) because d satisfies the triangle inequality and a/(1 + a) is increasing. Now,
d(x, z) + d(z, y)
1 + d(x, z) + d(z, y)≤ d(x, z)
1 + d(x, z)+
d(z, y)
1 + d(z, y)= ρ(x, z) + ρ(z, y).
2. Show that if x and y are two points of a metric space and if xn and yn are sequenceswhich converge to x and y respectively, d(xn, yn) converges to d(x, y).
Solution: From the inequality
|d(xn, yn)− d(x, y)| ≤≤d(yn,y)
|d(xn, yn)− d(xn, y)| +≤d(xn,x)
|d(xn, y)− d(x, y)|≤ d(yn, y) + d(xn, x)
we have that d(xn, yn) converges to d(x, y).
3. Let N = 1, 2, . . . the set of positive integers and define the function d as
d(n,m) =
0, if n = m,
1
n+
1
m, if n 6= m.
(a) Prove that d is a distance and describe the open balls of radius r > 0.
(b) Show that xnn∈N with xn = n is a Cauchy sequence, but it is not convergent.
Solution:
50
(a) To show d is a metric is easy. To describe open balls, notice that d(n∗,m) < r implies1/n∗+ 1/m < r for m 6= n∗; if r ≤ 1/n∗, then B(n∗, r) = n∗, and if r > 1/n∗, then
B(n∗, r) = m ∈ N : m > (r − 1/n∗)−1 = [n∗r] + 1, . . .,
where n∗r = (r − 1/n∗)−1, and [x] denotes the integer part of x ∈ R+ ([2] = 2,[2.21] = 2). Examples: B(1, 1) = 1, B(2, 1) = N− 1, 2, B(2, 2) = N.
(b) d(xn, xm) = 1/n + 1/m → 0 as n,m → ∞. Thus, (xn) = (n) is Cauchy. Suppose,by contradiction, it converges to some n∗ ∈ N. Then, for ε > 0 there is n0 such thatfor any n ≥ n0, xn = n ∈ B(n∗, ε). Take ε < 1/n∗. By (a), B(n∗, ε) = n∗ thus,n∗ = n for every n ≥ n0, a contradiction. Hence, the sequence is not convergent,and the space is not complete.
4. Consider the extended real number system, R = R ∪ ±∞, that consists of all realnumbers together with two ideal elements +∞ and −∞, with the property that −∞ <x < +∞ for every real number x. Define
f(x) =x
1 + |x|, f(+∞) = 1, f(−∞) = −1,
and d(x, y) = |f(x) − f(y)|. Show that d is a metric on R and describe the open ballB(−∞, r) and the closed ball B(−∞, r), r > 0.
Solution: Notice that f is strictly increasing and it is bounded, −1 ≤ f(x) ≤ 1 for everyx ∈ R. Obviously, d is non negative, satisfies symmetry and d(x, y) = 0 implies x = y.To prove the triangle inequality, consider x, y, z ∈ R. Then
B(−∞, r) = x ∈ R : |f(−∞)− f(x)| < r = x ∈ R : | − 1− f(x)| < r.
it is convenient to draw function f . The inequality defining B(−∞, r) decouples into
−1− r < f(x) < −1 + r.
The lower inequality is always fulfilled for r > 0, and the upper one holds for every x ∈ Rwhenever r > 2. Let 0 < r ≤ 2. Since f is increasing,
B(−∞, r) = [−∞, xr),
where f(xr) = −1 + r. We distinguish two cases: i) r ≤ 1, ii) 1 < r ≤ 2. In case i)xr = r−1
rand in case ii) xr = r−1
2−r . Resuming:
B(−∞, r) =
[−∞, r − 1
r
), 0 < r ≤ 1
=
[−∞, r − 1
2− r
), 1 < r ≤ 2
= [−∞,+∞] , r > 2.
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5. Consider the space of continuous functions on [a, b] with metric
d(f, g) =
√∫ b
a
|f(t)− g(t)|2 dt.
(a) Prove that the sequence of continuous functions xn defined on [−1, 1] by
xn(t) = 0, −1 ≤ t ≤ 0
= nt, 0 ≤ t ≤ 1
n
= 1,1
n≤ t ≤ 1,
is a Cauchy sequence.
(b) Show that the above sequence does not converge.
Solution: The sequence is Cauchy, because for m > n,
d2(fm, fn) = (m− n)2
∫ 1m
0
t2 dt+
∫ 1n
1m
(1− nt)2 dt =(m− n)2
3m2n<
1
3n< ε
if n > 13ε−1.
If this function converged to a continuous function f , then∫ 1
−1|fn(t)−f(t)|2 dt→ 0. This
would imply that any partial integral in [−1, 1] also tends to zero. In particular,∫ 0
−1
|fn(t)− f(t)|2 dt→ 0.
Since fn(t) = 0 for t ≤ 0 the continuous function f satisfies∫ 0
−1|f(t)|2 dt = 0, hence
x(t) = 0 when t ≤ 0.
Now, if a > 0, ∫ 1
a
|fn(t)− f(t)|2 dt→ 0, as n→∞.
Choosing n > 1/a, ∫ 1
a
|1− f(t)|2 dt→ 0, as n→∞.
Thus by continuity of x, x(t) = 1 for t ≥ a. Since a can be chosen as near to zero as weplease, we have that x must be discontinuous at zero, so the Cauchy sequence does notconverge to point in the space and the space is not complete.
6. If (E, d) is a complete metric space and T : E −→ E is a mapping such that T n is acontraction for some positive integer n, show that T has a unique fixed point that can befound by successive approximations.
Solution: By Banach’s Theorem, there exists a unique fixed point x0 ∈ E of T n. Now,T (x0) = T (T n(x0)) = T n+1(x0) = T n(T (x0)), thus T (x0) is a fixed point of T n and by
52
uniqueness, T (x0) = x0, it is also a fixed point of T . It is unique, because in the contrarycase, suppose that y0 ∈ E is another fixed point of T with x0 6= y0. Then
a contradiction since k < 1 (the parameter of contraction of T n).
Finally, consider T = T n. Then, for any x1 ∈ E, the sequence (T k(x1))k converges to x0,by Banach’s Theorem. Hence (T nk(x1))k converges to x0.
7. Let φ be continuous on the interval [a, b]. Let K be continuous on the square [a, b]× [a, b].Show that for all sufficiently small values of |λ|, the integral equation
f(x) = φ(x) + λ
∫ b
a
K(x, y)f(y) dy
has a unique solution, continuous on [a, b].
Solution: We consider the supremum distance (or supremum norm) d∞ in C([a, b]),which is then a complete metric space. Let the mapping defined as
Tf(x) = φ(x) + λ
∫ b
a
K(x, y)f(y) dy.
Obviously, Tf is continuous. Since φ and K are continuous, they are bounded; let|φ(x)| ≤ m, |K(x, y)| ≤M . Then
|Tf(x)− Tg(x)| = |λ|∣∣∣∣∫ b
a
K(x, y)(f(y)− g(y) dy
∣∣∣∣≤ |λ|M(b− a)d∞(f, g).
Hence d∞(Tf, Tg) ≤ |λ|M(b− a)d∞(f, g). Thus, T is a contraction with 0 < k < 1 whenk = |λ|M(b− a) < 1. Then the functional equation has a unique continuous solution.
8. Consider a function of two variables, W : [0,∞) × [0,∞) −→ [0,∞), which satisfies thefollowing assumptions:
(A1) For any function f bounded over [0,∞), the function of x, W (x, f(x)), is bounded;
(A2) There is some constant 0 < β < 1 such that, for every x ∈ [0,∞), and for anyy, y′ ∈ [0,∞)
|W (x, y)−W (x, y′)| ≤ β|y − y′|.
For an arbitrary constant a > 0 consider the operator T defined by
(Tf)(x) = W (x, f(ax)).
(a) Show that T is a contraction mapping in the space (B[0,∞), d∞) of bounded functionson [0,∞), and thus the functional equation W (x, f(ax)) = f(x) admits a uniquesolution in this space.
53
(b) Check that W (x, y) = u(x) + βy, where u ∈ B([0,∞)), and 0 < β < 1, satisfies theassumptions (A1), (A2) above.
(c) Give the solution of the functional equation for W defined in (b), in terms of aninfinite series.
Solution:
(a) For a real bounded function f , Tf is also bounded, by assumption A1, hence T :B[0,∞) −→ B[0,∞). On the other hand, for f, g ∈ B[0,∞), and for any x ∈ [0,∞)
The first inequality is due to A2. Thus, taking supremum in x ∈ [0,∞) we have
d∞(Tf, Tg) ≤ βd∞(f, g).
In consequence, since (B[0,∞), d∞) is complete, T has a unique fixed point in theclass of bounded functions, which is the unique solution of the equationW (x, f(ax)) =f(x).
(b) A1: Since u is bounded, W (x, f(x)) = u(x) + βf(x) is also bounded whenever f isbounded; A2: trivial.
(c) Given any bounded f , T nf converges to the fixed point, f ∗. Choosing f = 0 weget f1 = T0 = u, and f2 = Tf1 = T 20, that is, f2(x) = (Tu)(x) = u(x) + βu(ax).In general, fn(x) = (T n0)(x) =
∑n−1i=0 β
iu(aix). Taking limits as n → ∞, the fixedpoint is f ∗(x) =
∑∞i=0 β
iu(aix).
9. Find an approximated solution of the equation (1+x)3−x = 0 by the method of successiveiterations.
Solution: This is equivalent to a fixed point problem x = f(x), where f(x) = (x +1)3. The derivative f ′(x) = 3(x + 1)2 < 1 only in the interval I = (−1/
√3, 1/√
3) ≈(−0.5774, 0.5744). But
f(0) = 1 /∈ I,
hence f is not a contraction in I. Let us locate the roots of the equation with Bolzanos’Theorem. Let g(x) = x− (1 + x)3; g is continuous and g(−2) = −1 < 0, g(−3) = 5 > 0,thus g has a root in [−3,−2] (equivalently, f has a fixed point in this interval; however,the derivative of f is not smaller by 1 and f([−3,−2]) is not contained in [−3,−2], thusf has no the contraction property in this interval either).
54
Consider the equation rewritten as follows:
x = 3√x− 1 ≡ g(x).
Then, g is increasing and g([−3,−2]) ⊂ [−3,−2]. Moreover, g′(x) = (1/3)x−2/3, and inthe interval [−3,−2], |g′(x)| ≤ (1/3)(1/4)1/3 < 1. Thus, according to the MVT, g is acontraction in [−3,−2]. Starting with x1 = −2, x6 gives approximately −2.325 as thesolution of the equation.
10. Consider the mapping f : R2 −→ R2 given by f(x, y) = (1 + 2y, 3 + x/8). Show that f isnot a contraction with the Euclidean metric d2, but it is with the metric
(It is not needed to check that ρ is indeed a distance).
Solution: For the points A = (8, 0), B(8, 4), f(A) = (1, 4) and f(8, 4) = (9, 4). Thend2(A,B) = 2 and d2(f(A), f(B)) = 8, hence f is not a contraction with d2. However
Show that it has a unique solution and find an approximation with three correct decimalplaces.
Solution: It is convenient to have in mind that for a differentiable vector function f :Rn −→ Rm, f = (f1, . . . , fm), the Mean Value Inequality is
‖f(x)− f(y)‖ ≤
(m∑k=1
‖∇fk(z‖
)‖x− y‖,
where ‖ · ‖ is the usual norm, and z is in the segment joining x and y. If the partialderivatives are bounded, then
‖f(x)− f(y)‖ ≤M‖x− y‖, M = supz∈Rn
m∑k=1
‖∇fk(z‖.
55
Solving the system it is equivalent to finding a fixed point of the mapping
f(x, y) = (f1(x, y), f2(x, y)) =(1
3sinx+
1
3cos y + 2,
1
6cosx− 1
2sin y + 1
).
Hence,
∇f1(x, y) =(1
3cosx,−1
3sin y
),
∇f2(x, y) =(− 1
6sinx,
1
2cos y
),
and
‖∇f1(x, y)‖ =1
3
√cos2 x+ sin2 y ≤
√2
3,
‖∇f2(x, y)‖ =1
2
√sin2 x
9+ cos2 y ≤ 1
6
√10.
Hence M =√
2/3 +√
10/6 =√
2(2 +√
5)/6 ≈ 0.9985 < 1 (!). Thus, f is a contraction inthe Euclidean norm, and it has a unique fixed point, which is the unique solution of oursystem.
The approximated solution x0 = 2.48138437, y0 = 0.59012392 has been obtained frominitial condition (x1 = y1 = 0) after 30 iterations (a commercial spreadsheet has beenused).
4.5 Homework
1. Consider the following functions d1, d∞ : R2 × R2 −→ R
Show that both are metrics and draw the open ball B(0, 1) for each one.
Solution: Let us check that d1 is a metric. The case d∞ is similar. Obviously, d1(x,y) ≥ 0and d1(x,y) = d1(y,x). Moreover, d1(x,y) = 0 iff |x1 − y1| + |x2 − y2| = 0 iff x1 = y1
The ball B(0, 1) in the metric d∞is given by points (x, y) satisfying
max|x|, |y| < 1⇔ (x < 1,−x < 1, y < 1,−y < 1).
2. Let (E, d) be a metric space.
(a) Show that
ρ(x, y) =d(x, y)
1 + d(x, y)
defines a new metric on E (Hint: prove first that the mapping d 7−→ d/(1 + d) isincreasing).
(b) Show that d and ρ have the same open sets.
Solution:
It is important to note that the mapping d 7→ d/(1 + d) is increasing.
(a) Done in class.
(b) This is because for any δ > 0, Bd(x0, δ) ⊆ Bρ(x0,δ
1+δ) and reciprocally, for any
0 < r < 1, Bρ(x0, r) ⊆ Bd(x0,r
1−r ).
3. Over E = R consider the mappings d(x, y) =√|x− y| and δ(x, y) = |x− y|2. Show that
d is a metric but δ is not.
Solution: The only problematic axiom is the triangle inequality. But for a, b, c ∈ R+ it isobvious that b+ c ≤ b+ c+ 2
√bc = (
√b+√c)2. Hence, if a ≤ b+ c, then
√a ≤√b+√c.
The triangle inequality for d follows now taking a = |x − y|, b = |x − z| and c = |z − y|for x, y, z ∈ R. Hence d is a metric. Mapping δ is not a metric: take x = 0, y = 1 andz = 1/2. Then δ(x, y) = 1, δ(x, z) = 1/4 and δ(z, y) = 1/4 and 1 1/4 + 1/4 = 1/2,thus the triangle inequality is not fulfilled.
4. Let (E, d) a metric space. Given non–empty subsets A, B of E, we define the distancebetween A and B as
d(A,B) = infx∈A,y∈B
d(x, y).
(a) Show that A ∩B 6= ∅ implies d(A,B) = 0.
(b) Let E = R with the Euclidean metric and let A = N, B =n− 1
n: n ∈ N
. Show
that A ∩B = ∅ but d(A,B) = 0.
57
(c) (Return to the general framework). Given A ⊆ E non–empty, let fA : E −→ R begiven by
fA(x) = infy∈A
d(x, y).
Using the triangle inequality show that fA is continuous.
Solution:
(a) Let x ∈ A ∩B. Then d(A,B) ≤ d(x, x) = 0, thus d(A,B) = 0.
(b) d(A,B) = infn d(n, n− 1/n) = infn 1/n = 0.
(c) Let z ∈ A and let x, y ∈ E. Then d(x, z) ≤ d(x, y) + d(y, z). Taking the infimum inz ∈ A we get
d(x,A) = infz∈A
d(x, z) ≤ infz∈A
(d(x, y) + d(y, z)) = d(x, y) + infz∈A
d(y, z)
= d(x, y) + d(y, A).
Exchanging the roles of x and y we get the similar inequality
d(y, A) ≤ d(x, y) + d(x,A),
thus|fA(x)− fA(y)| = |d(x,A)− d(y, A)| ≤ d(x, y)
which clearly implies the fA is continuous (in fact uniformly continuous).
5. Let (E, d) a metric space and let f : E −→ E be continuous with respect to the metric d.Show that the composition g : E −→ R defined by g(x) = d(x, f(x)) is continuous.
Solution: The distance function d : E2 −→ R is continuous with respect to the metricd! To show this, let (x0, y0) and arbitrary point in E2 and let ε > 0. Then taking δ = εone gets that for any (x, y) ∈ B((x0, y0), δ)
|d(x, y)− d(x0, y0)| < δ = ε.
Now, g is the composition of two continuous functions (w.r.t. the metric d), so it iscontinuous.
6. Let E = C[0, 1] be the Banach space of continuous functions x : [0, 1] −→ R with thesupremum norm, ‖x‖ = maxt∈[0,1] |x(t)|. Let T : E −→ E be the operator defined by
Tx(t) =
∫ t
0
e−δsf(x(s)) ds, 0 ≤ t ≤ 1,
where δ > 0 and f : R −→ R is continuous satisfying
|f(x1)− f(x2)| ≤ `|x1 − x2|
for every x1, x2 ∈ R.
58
(a) Show that T maps E into E.
(b) Find sufficient conditions on δ and ` such that the functional equation
x(t) =
∫ t
0
e−δsf(x(s)) ds, 0 ≤ t ≤ 1,
admits a unique solution x∗ in the space E.
(c) Check whether f(x) = 2x and δ = 1 fulfil your conditions. In this case, T has thefixed point x∗(t) = 0 all t. Is there some contradiction with (b) above?
Solution:
(a) Since f is continuous, Tx is continuous for any continuous mapping x (by the Fun-damental Theorem of Integral Calculus it is even differentiable).
(b) Let us force T to be a contraction mapping (if possible). For any x1, x2 in E wehave
|Tx1(t)− Tx2(t)| =∣∣∣∣∫ t
0
e−δsf(x1(s)) ds−∫ t
0
e−δsf(x2(s)) ds
∣∣∣∣≤∫ t
0
e−δs |f(x1(s))− f(x2(s))| ds
≤ `
∫ t
0
e−δs |x1(s)− x2(s)| ds
≤ `
∫ 1
0
e−δs‖x1 − x2‖ ds
=`
δ
(1− e−δ
)‖x1 − x2‖.
Then, T is a contraction iff
k =`
δ
(1− e−δ
)< 1.
this is the relationship we are looking for. Now, Banach’s Theorem assures that Tadmits a unique fixed point, i.e., that the functional equation
x(t) =
∫ t
0
e−δsf(x(s)) ds
admits a unique solution in the class of continuous functions x.
(c) With f(x) = 2x we have ` = 2, thus k = 2(1 − e−1) ≈ 1.2642 > 1, hence T is nota contraction in the supremum metric. There is no contradiction, since Banach’sTheorem establishes sufficient conditions for existence of a unique fixed point, notnecessary.
