Real Analysis Math 131AH Rudin, Chapter #1 1.1. 6= 0) and...Real Analysis Math 131AH Rudin, Chapter...

Real AnalysisMath 131AH

Rudin, Chapter #1

Dominique Abdi

1.1. If r is rational (r 6= 0) and x is irrational, prove that r + x and rx areirrational.

Solution. Assume the contrary, that r+x and rx are rational. Since the rationalnumbers form a field, axiom (A5) guarantees the existence of a rational number−r so that, by axioms (A4) and (A3), we have

x = 0 + x = (−r + r) + x = −r + (r + x).

Both −r and r + x are rational by assumption, so x is rational by axiom (A1),contradicting that x is irrational.

Similarly, axiom (M5) guarantees the existence of a rational number 1/r sothat, by axioms (M4) and (M3), we have

x = 1x =(

1r· r

)x =

1r

(rx).

Both 1/r and rx are rational by assumption, so x is rational by axiom (M1),contradicting that x is irrational.

1.2. Prove that there is no rational number whose square is 12.

Solution. Assume the contrary, that there is a rational number p such thatp2 = 12. Then there are integers m and n with p = m/n and for which 3 isnot a common factor. By assumption, m2 = 12n2, so m2 is divisible by 3, andtherefore m is divisible by 3. But then m2 is divisible by 9. This means that12n2 is divisible by 9, so 4n2 is divisible by 3. Thus n2 is divisible by 3, andtherefore n is divisible by 3. We have shown that 3 is a common factor of mand n, a contradiction.

1.3. Prove Proposition 1.15.

Solution. Suppose x, y, and z are elements of a field and x 6= 0. The axioms(M) imply

y = 1y =(

1x· x

)y =

1x

(xy) =1x

(xz) =(

1x· x

)z = 1z = z,

proving part (a). Part (b) follows from part (a) by setting z = 1, and part (c)follows from part (a) by setting z = 1/x. Since

x · 1x

= 1,

replacing x with 1/x in part (c) gives part (d).

1.4. Let E be a nonempty subset of an ordered set; suppose α is a lower boundof E and β is an upper bound of E. Prove that α ≤ β.

Solution. Since E is nonempty, there exists a point x ∈ E. Note that α is alower bound of E and β is an upper bound of E by the definition of supremumand infimum. Thus α ≤ x by the definition of lower bounds and x ≤ β by thedefinition of upper bounds, and therefore α ≤ β by Definition 1.5(ii).

1.5. Let A be a nonempty set of real numbers which is bounded below. Let−A be the set of all numbers −x, where x ∈ A. Prove that

inf A = − sup (−A) .

Solution. By the least-upper-bound property, α = inf A exists. By the definitionof infimum, α ≤ x for all x ∈ A, and if β > α, then there exists a y ∈ A suchthat β > y. Therefore Proposition 1.18(a) implies −α ≥ −x for all x ∈ A,and if γ < −α, then there exists a y ∈ A such that γ < −y. Thus −α is anupper bound of −A, and if γ < −α, then γ is not an upper bound of −A.By the definition of supremum, it follows that −α = sup (−A), and thereforeinf A = − sup (−A).

1.6. Fix b > 1.

(a) If m,n, p, q are integers, n > 0, q > 0, and r = m/n = p/q, prove that

(bm)1/n = (bp)1/q .

Hence it makes sense to define br = (bm)1/n.

(b) Prove that br+s = brbs if r and s are rational.

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(c) If x is real, define B(x) to be the set of all numbers bt, where t is rationaland t ≤ x. Prove that

br = supB(r)

when r is rational. Hence it makes sense to define

bx = supB(x)

for every real x.

(d) Prove that bx+y = bxby for all real x and y.

Solution. Fix b > 1.

(a) If m/n = p/q, then mq/nq = np/nq, and therefore mq = np. For simplic-ity, let α = (bm)1/n and β = (bp)1/q. Then

α = (αnq)1/nq = (bmq)1/nq = (bnp)1/nq = (βnq)1/nq = β,

so (bm)1/n = (bp)1/q. Thus br = (bm)1/n is well-defined, since any tworepresentations of r yield the same value.

(b) Suppose r = m/n and s = p/q for some integers m,n, p, q with n > 0 andq > 0. Then r + s = (mq + np)/nq, and by part (a) and the corollary toTheorem 1.21, we have

brbs =(bm/n

) (bp/q

)=

(bmq/nq

) (bnp/nq

)= (bmq)1/nq (bnp)1/nq

=(bmq+np

)1/nq = b(mq+np)/nq = br+s.(c) Suppose r = p/q for some integers p and q > 0, and fix bs ∈ B(r). Then

s is rational and s ≤ r, so br−s ≥ 1 since b > 1. Multiplying through bybs yields br ≥ bs, so br is an upper bound of B(r).Since r ≤ r, we have br ∈ B(r). Thus if γ < br, then γ is not an upperbound of B(r). Hence br = supB(r).

The definition bx = supB(x) for every real x makes sense because theequation br = supB(r) holds for every rational r by the argument above.

(d) Suppose x and y are real, r and s are rational, r ≤ x, and s ≤ y. Thenr + s ≤ x+ y, so

brbs = br+s ≤ bx+y.Thus taking the supremum over all rationals r ≤ x and s ≤ y givesbxby ≤ bx+y.Now suppose t is rational and t ≤ x+y. Then there exist rational numbersr and s such that t = r + s, r ≤ x, and s ≤ y. Thus

bt = br+s = brbs ≤ bxby,

and taking the supremum over all rationals t ≤ x+ y gives bx+y ≤ bxby.Hence bx+y = bxby.

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1.8. Prove that no order can be defined in the complex field that turns it intoan ordered field.

Solution. By Definition 1.27, we have i 6= 0, and by Theorem 1.28 and Propo-sition 1.18(a,d), we have i2 = −1 < 0. This contradicts Proposition 1.18(d),and therefore no order can be defined in the complex field that turns it into anordered field.

1.9. Suppose z = a + bi, w = c + di. Define z < w if a < c, and also if a = cbut b < d. Prove that this turns the set of all complex numbers into an orderedset. Does this ordered set have the least-upper-bound property?

Solution. The real numbers form an ordered set, so one and only one of thefollowing holds: a < c, a = c, a > c. Similarly, either b < d, b = d, or b > d.There are 5 possible cases:

(1) If a < c, then z < w.

(2) If a > c, then z > w.

(3) If a = c and b < d, then z < w.

(4) If a = c and b > d, then z > w.

(5) If a = c and b = d, then z = w.

Thus one and only one of z < w, z = w, and z > w holds, so Definition 1.5(i)holds.

Let z1, z2, z3 be complex numbers satisfying z1 < z2 and z2 < z3. Then oneand only one of the following holds,

(1) Re(z1) < Re(z2),

(2) Re(z1) = Re(z2) and Im(z1) < Im(z2),

and similarly, one and only one of the following holds,

(1) Re(z2) < Re(z3),

(2) Re(z2) = Re(z3) and Im(z2) < Im(z3).

This yields 2 possible cases:

(1) If Re(z1) < Re(z2) or if Re(z2) < Re(z3), then Re(z1) < Re(z3).

(2) If Re(z1) = Re(z2) = Re(z3), then Im(z1) < Im(z2) < Im(z3).

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Thus z1 < z3 in all cases, so Definition 1.5(ii) holds.Hence the set of all complex numbers is an ordered set.

Now let E = {in : n ∈ Z}. Then E is a nonempty subset of C which isbounded above by 1. Suppose α+ iβ = supE. Then α ≥ 0 since α is an upperbound. But if α > 0, then α/2 is a smaller upper bound of E, contradictingthat α + iβ = supE. Thus α = 0, and so iβ = supE. But for any β, thereexists an integer n > β, contradicting that iβ is an upper bound of E. HencesupE does not exist, so the least-upper-bound property does not hold.

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Rudin, Chapter #2

Dominique Abdi

2.1. Prove that the empty set is a subset of every set.

Solution. Assume the contrary, that there is a set E such that the empty set isnot a subset of E. Then there is an element x ∈ ∅ such that x /∈ E, but thiscontradicts that the empty set is empty. Hence ∅ ⊂ E.

2.2. A complex number z is said to be algebraic if there are integers a0, . . . , an,not all zero, such that

a0zn + a1zn−1 + · · ·+ an−1z + an = 0.

Prove that the set of all algebraic numbers is countable.

Solution. Let p be a polynomial of degree n with integer coefficients and let

Zp := {z ∈ C : p(z) = 0} .

By the fundamental theorem of algebra, Zp has at most n elements.Now define the set

Pn(Z) :=

{n∑k=0

akzk : ak ∈ Z, an 6= 0

},

consisting of all polynomials of degree n with integer coefficients, and let

En := {(a0, a1, . . . , an) : ak ∈ Z, an 6= 0} .

Example 2.5 can be easily modified to show that the set of non-zero integers iscountable. Thus by Example 2.5 and Theorem 2.13, En is countable. Define

f : En → Pn(Z), f(a0, a1, . . . , an) :=n∑k=0

akzk.

Every polynomial of degree n is uniquely determined by its n + 1 coefficients,so f is one-to-one and onto. Hence Pn(Z) is countable, and by Theorem 2.12,the set

P (Z) :=∞⋃n=0

Pn(Z),

consisting of all polynomials with integer coefficients, is countable.Since the algebraic numbers are defined to be the roots of polynomials with

integer coefficients, the set A of all algebraic numbers can be written as

A =⋃

p∈P (Z)

Zp,

a union of a countable collection of finite (hence at most countable) sets. Bythe corollary to Theorem 2.12, A is at most countable. But for any n ∈ Z,

p(z) = z − n

is a polynomial with integer coefficients for which z = n is a root, so Z ⊂ A.We have shown that A is an at most countable set which contains an infinite

subset. Hence A is infinite and at most countable, and thus A is countable.

2.3. Prove that there exist real numbers which are not algebraic.

Solution. Assume that every real number is algebraic. Then R is countable byExercise 2. This contradicts the corollary to Theorem 2.43, which states that Ris uncountable.

2.4. Is the set of all irrational real numbers countable?

Solution. Assume that the set Qc of all irrational real numbers is countable.Since R = Q ∪ Qc and Q is countable by the corollary to Theorem 2.13, itfollows that R is countable by the corollary to Theorem 2.12. This contradictsthe corollary to Theorem 2.43, which states that R is uncountable.

2.5. Construct a bounded set of real numbers with exactly three limit points.

Solution. Define the sets

E0 :={

1n

: n ∈ N}, E1 :=

{n− 1n

: n ∈ Z}, E2 :=

{2n− 1n

: n ∈ N},

and E := E0 ∪ E1 ∪ E2. Then E is bounded since E ⊂ (0, 2).We now show that E′ = {0, 1, 2}. Fix � > 0 and choose N such that N > 1/�.

Then 0 < 1/n < � for all n ≥ N , and in particular,

1n∈ (−�, �), n− 1

n∈ (1− �, 1 + �), 2n− 1

n∈ (2− �, 2 + �)

for all n ≥ N . This shows that an open interval of arbitrarily small radius � > 0centered at 0, 1, or 2 intersects E at infinitely many points, so {0, 1, 2} ⊂ E′.

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Now suppose x /∈ {0, 1, 2}. If x < 0, then (2x, 0) ∩ E = ∅, so x /∈ E′. Ifx > 0, then there are three possible cases:

Case 1 : If 0 < x < 1, then choose k ∈ N such that

1k + 1

< x ≤ 1k,

and define δ to be the smallest non-zero element of the set{|x| ,

∣∣∣∣x− 1k + 1∣∣∣∣ , ∣∣∣∣x− 1k

∣∣∣∣ , |x− 1|} .Case 2 : If 1 < x < 2, then choose k ∈ N such that

k − 1k

< x ≤ kk + 1

,

and define δ to be the smallest non-zero element of the set{|x− 1| ,

∣∣∣∣x− k − 1k∣∣∣∣ , ∣∣∣∣x− kk + 1

∣∣∣∣ , |x− 2|} .Case 3 : If x > 2, then define δ := x− 2.In all three cases, we have (x−δ, x+δ)∩E = ∅ by the choice of δ, so x /∈ E′.

Hence E′ = {0, 1, 2}.

2.6. Let E′ be the set of all limit points of a set E. Prove that E′ is closed.Prove that E and E have the same limit points. (Recall that E = E ∪E′.) DoE and E′ always have the same limit points?

Solution. By Definition 2.26 and Theorem 2.27(b), to prove that E′ is closed,it suffices to show that E′ contains its limit points. Let x be a limit point of E′

and let Nr(x) be a neighborhood of x. Then there exists a point y ∈ Nr(x)∩E′with x 6= y. Let r′ := r − d(x, y) and note that Nr′(y) ∩ E contains a pointz ∈ E since y is a limit point of E. Moreover, x 6= z since d(x, z) > 0. Thetriangle inequality gives

d(x, z) ≤ d(x, y) + d(y, z) < d(x, y) + r′ = r,

so z ∈ Nr(x). Thus x is a limit point of E, so therefore (E′)′ ⊂ E′ and henceE′ is closed.

We now show that E and E have the same limit points. If x ∈ E′ and r > 0,then there exists a point y ∈ Nr(x) ∩ E with x 6= y. Since E ⊂ E, we havey ∈ Nr(x) ∩ E, and therefore x ∈ (E)′. Conversely, suppose that x ∈ (E)′ andr > 0. Then there exists a point y ∈ Nr(x) ∩ E with x 6= y. If y /∈ E, theny ∈ E′ since E = E ∪ E′. Setting r′ := r − d(x, y), then there exists a z ∈ Ewith z 6= y and z 6= x (since d(x, z) > 0). In either case, Nr(x) contains a pointof E distinct from x, so x ∈ E′. Thus E′ = (E)′.

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The sets E and E′ need not have the same limit points. Consider the setE defined in the solution to Exercise 2.5. We showed that E′ = {0, 1, 2}. Ifx ∈ E′, then (x− 1, x+ 1) ∩ E′ = {x}, and if x /∈ E′, then for

δ := min {|x− y| : y ∈ E′} ,

we have (x− δ, x+ δ) ∩ E′ = ∅. Thus (E′)′ = ∅, and therefore E′ 6= (E′)′.

2.7. Let A1, A2, A3, . . . be subsets of a metric space.

(a) If Bn =⋃ni=1Ai, prove that Bn =

⋃ni=1Ai, for n = 1, 2, 3, . . ..

(b) If B =⋃∞i=1Ai, prove that B ⊃

⋃∞i=1Ai.

Show, by an example, that the inclusion can be proper.

Solution.

(a) Fix x ∈⋃ni=1Ai. Then x ∈ Ai = Ai ∪ A′i for some i. If x ∈ Ai, then

x ∈ Bn since Ai ⊂ Bn ⊂ Bn. If x ∈ A′i, then for every r > 0, there is apoint y ∈ Ai ⊂ Bn with x 6= y such that d(x, y) < r. Thus x ∈ B′n ⊂ Bn,so⋃ni=1Ai ⊂ Bn.

Fix x /∈⋃ni=1Ai. Then

x ∈

(n⋃i=1

Ai

)c=

n⋂i=1

(Ai)c.

Thus there exist r1, . . . , rn > 0 such that Nri(x) ⊂ (Ai)c for each i. Letr := min {r1, . . . , rn}. Then Nr(x) ⊂

⋂ni=1Nri(x), so Nr(x) ⊂

⋂ni=1(Ai)

c.In particular, Nr(x) contains no points of any Ai, and therefore no pointsof Bn. If Nr(x) contains a point y ∈ B′n, then letting r′ := r− d(x, y), wesee that Nr′(y) contains a point z ∈ Bn by the definition of limit points,and z ∈ Nr′(y) ⊂ Nr(x) by the triangle inequality, contradicting thatNr(x) ∩Bn = ∅. Thus

x ∈ Bcn ∩ (B′n)c = (Bn ∪B′n)c = (Bn)c,

so therefore Bn ⊂⋃ni=1Ai, and by the result proved above, Bn =

⋃ni=1Ai.

(b) Fix x ∈⋃∞i=1Ai. Then x ∈ Ai = Ai ∪ A′i for some i. If x ∈ Ai, then

x ∈ B since Ai ⊂ B ⊂ B. If x ∈ A′i, then for every r > 0, there is apoint y ∈ Ai ⊂ B with x 6= y such that d(x, y) < r. Thus x ∈ B′ ⊂ B, soB ⊃

⋃∞i=1Ai.

For all i ∈ N, let Ai := {1/i}. Then Ai = Ai since finite sets are closed bythe corollary to Theorem 2.20. However, B = {1/i : i ∈ N}, and therefore

B ={

1i

: i ∈ N}∪ {0} ⊃ {1/i : i ∈ N} =

∞⋃i=1

Ai,

and the inclusion is proper since 0 /∈⋃∞i=1Ai.

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2.8. Is every point of every open set E ⊂ R2 a limit point of E? Answer thesame question for closed sets in R2.

Solution. Yes, every point of every open set E ⊂ R2 a limit point of E. LetE ⊂ R2 be open and fix x ∈ E. Then x is an interior point of E, so thereexists an r > 0 such that Nr(x) ⊂ E. Let Ns(x) be an arbitrary neighborhoodof x. If s ≥ r, then Nr(x) ⊂ Ns(x), so Nr(x) ⊂ Ns(x) ∩ E. If s < r, thenNs(x) ⊂ Nr(x) ⊂ E, so Ns(x) = Ns(x)∩E. In either case, Ns(x)∩E is itself aneighborhood of x in R2 and therefore contains infinitely many points, and inparticular, some point y 6= x. Thus x is a limit point of E.

The same result does not hold for closed sets in R2. A set consisting of asingle point x ∈ R2 is closed since its complement is open, but finite sets haveno limit points by the corollary to Theorem 2.20.

2.9. Let E◦ denote the set of all interior points of a set E.

(a) Prove that E◦ is always open.

(b) Prove that E is open if and only if E = E◦.

(c) If G ⊂ E and G is open, prove that G ⊂ E◦.

(d) Prove that the complement of E◦ is the closure of the complement of E.

(e) Do E and E always have the same interiors?

(f) Do E and E◦ always have the same closures?

Solution.

(a) Fix x ∈ E◦. Then there exists an r > 0 such thatNr(x) ⊂ E. If d(x, y) < rand r′ = r − d(x, y), then d(y, z) < r′ implies

d(x, z) ≤ d(x, y) + d(y, z) < r,

so y ∈ E◦. This shows that Nr(x) ⊂ E◦, so x is an interior point of E◦.Thus E◦ is open.

(b) If E = E◦, then E is open by part (a). If E is open, then every point ofE is an interior point, so E ⊂ E◦. Since every interior point of a set mustbe contained in the set, the reverse inclusion holds as well, so E = E◦.

(c) Suppose G ⊂ E and G is open, and fix x ∈ G. Then x is an interior pointof G, so there exists an r > 0 such that Nr(x) ⊂ G ⊂ E. Thus x is aninterior point of E, so G ⊂ E◦.

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(d) Fix x ∈ (E◦)c. Then x is not an interior point of E, so every neighborhoodNr(x) intersects Ec. Thus x is in Ec or x is a limit point of Ec, so

x ∈ Ec ∪ (E′)c = E.

Now suppose x /∈ (E◦)c. Then x ∈ E◦, so there exists an r > 0 such thatNr(x) ⊂ E. Thus x ∈ E and x is not a limit point of Ec, so

x ∈ E ∩ (E′)c = (Ec)c ∩ (E′)c = (E ∪ E′)c = (E)c.

(e) No, E and E do not always have the same interiors. For example, notethat Q◦ = ∅ because every neighborhood of a rational number containsirrational numbers. However, Q = R, so (Q)◦ = R since R is open.

(f) No, E and E◦ do not always have the same closures. For example, notethat Q◦ = ∅ because every neighborhood of a rational number containsirrational numbers. Therefore Q◦ = ∅ = ∅ since the empty set is closed,but Q = R.

2.10. Let X be an infinite set. For p ∈ X and q ∈ X, define

d(p, q) =

{1 (if p 6= q)0 (if p = q).

Prove that this is a metric. Which subsets of the resulting metric space areopen? Which are closed? Which are compact?

Solution. Parts (a) and (b) of Definition 2.15 follow immediately from the def-inition of d. If p = q, then part (c) is trivial. If p 6= q, then for every r ∈ X,either p 6= r or q 6= r. Thus d(p, r) + d(q, r) ≥ 1, and since d(p, q) ≤ 1, part (c)holds in this case as well. Thus d is a metric.

Fix E ⊂ X and x ∈ E. Then

N1/2(x) = {y ∈ X : d(x, y) < 1/2} = {x} ,

so every point is a neighborhood of itself. By Theorem 2.19, every point is open,and since

E =⋃x∈E{x} ,

every set is open by Theorem 2.24(a).If E ⊂ X, then Ec is open by what we just proved, so therefore every set is

closed by the corollary to Theorem 2.23.Suppose K ⊂ X is compact. Then the open cover {{x} : x ∈ K} has a finite

subcover, so K must be finite. Conversely, if K is finite, say K = {x1, . . . , xn},then given any open cover {Gα}α∈A of K, there are indices αi such that xi ∈ Gαifor all i = 1, . . . , n. Thus {Gα1 , . . . , Gαn} is a finite subcover of K, so K iscompact. Hence the compact subsets of X are the finite subsets.

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2.11. For x ∈ R1 and y ∈ R1, define

d1(x, y) = (x− y)2,

d2(x, y) =√|x− y|,

d3(x, y) =∣∣x2 − y2∣∣ ,

d4(x, y) = |x− 2y| ,

d5(x, y) =|x− y|

1 + |x− y|.

Determine, for each of these, whether it is a metric or not.

Solution.

(a) Since

d1(2, 0) = (2− 0)2 = 4,d1(2, 1) = (2− 1)2 = 1,d1(1, 0) = (1− 0)2 = 1,

it follows that

d1(2, 0) = 4 > 2 = d1(2, 1) + d1(1, 0),

so d1(x, y) is not a metric by Definition 2.15(c).

(b) Since |x− y| > 0 if x 6= y, we have d2(x, y) =√|x− y| > 0 if x 6= y. But

if x = y, then d2(x, y) = 0. Thus Definition 2.15(a) holds.

Since |x− y| = |y − x|, we have

d2(x, y) =√|x− y| =

√|y − x| = d2(y, x),

so Definition 2.15(b) holds.

By the triangle inequality, we have

|x− z| ≤ |x− y|+ |y − z|

≤ |x− y|+ 2√|x− y|

√|y − z|+ |y − z|

=(√|x− y|+

√|y − z|

)2,

and taking square roots gives d2(x, z) ≤ d2(x, y) + d2(y, z), so Defini-tion 2.15(c) holds.

Hence d2(x, y) is a metric.

(c) Sinced3(1,−1) =

∣∣12 − (−1)2∣∣ = |1− 1| = 0,Definition 2.15(a) fails, so d3(x, y) is not a metric.

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(d) Sinced4(2, 1) = |2− 2(1)| = |2− 2| = 0,

Definition 2.15(a) fails, so d4(x, y) is not a metric.

(e) Since |x− y| defines a metric, we have

d5(x, y) =|x− y|

1 + |x− y|≥ 0,

with equality holding if and only if x = y. Thus Definition 2.15(a) holds.

Since |x− y| = |y − x|, we have

d5(x, y) =|x− y|

1 + |x− y|=

|y − x|1 + |y − x|

= d5(y, x),

so Definition 2.15(b) holds.

Definition 2.15(c) is satisfied if and only if

d5(x, y) +d5(y, z)−d5(x, z) =|x− y|

1 + |x− y|+|y − z|

1 + |y − z|− |x− z|

1 + |x− z|≥ 0.

Let a := |x− y|, b := |y − z|, and z := |x− z|. Then taking

(1 + a)(1 + b)(1 + c)

as the common denominator in the fraction above, the numerator is

a+ ab+ ac+ abc+ b+ ba+ bc+ abc− c− ac− cb− abc= a+ b− c+ 2ab+ abc ≥ (2 + c) ab ≥ 0.

Hence d5(x, y) is a metric.

2.12. Let K ⊂ R1 consist of 0 and the numbers 1/n, for n = 1, 2, 3, . . .. Provethat K is compact directly from the definition (without using the Heine-Boreltheorem).

Solution. Let {Gα}α∈A be an open cover of K. Then there is an index α0 ∈ Asuch that 0 ∈ Gα0 . Since Gα0 is an open set containing 0, it contains anopen interval (−�, �) for some � > 0. Let N be the largest integer such thatN ≤ 1/�. Note that for all n > N , we have 0 < 1/n < �, so 1/n ∈ Gα0 , andfor all positive n ≤ N , there exist indices αn ∈ A such that 1/n ∈ Gαn . Then{Gα0 , Gα1 , . . . , GαN } is a finite subcover of K, so K is compact.

2.13. Construct a compact set of real numbers whose limit points form a count-able set.

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Solution. Define a set

E := {1/m+ 1/n : m,n ∈ N} ∪ {1/n : n ∈ N} ∪ {0} .

Then E′ = {1/n : n ∈ N} ∪ {0}, so E′ is countable and E′ ⊂ E, so E is closed.Moreover, E ⊂ [0, 2], so E is bounded. By the Heine-Borel theorem, it followsthat E is compact.

2.14. Give an example of an open cover of the segment (0, 1) which has nofinite subcover.

Solution. For every n ∈ N, define Gn := (1/n, 1). Then each Gn is an openinterval in R, and

⋃∞n=1Gn = (0, 1), so {Gn}

∞1 is an open cover of (0, 1).

Suppose {Gn}∞1 has a finite subcover C of (0, 1). Then there is a largest integerN such that GN ∈ C. But then

⋃Gn∈C Gn = (1/N, 1), so every � satisfying

0 < � < 1/N is an element of (0, 1) but is not covered by C. This contradictsthat C is a cover, so therefore {Gn}∞1 has no finite subcover of (0, 1).

2.15. Show that Theorem 2.36 and its Corollary become false (in R1, for ex-ample) if the word “compact” is replaced by “closed” or by “bounded.”

Solution. For every n ∈ N, define En := [n,∞) ⊂ R. Then En is closed, andfor any finite subcollection C, let N be the largest integer such that EN ∈ C.Then

⋂En∈C En = [N,∞) 6= ∅, but

⋂∞n=1En = ∅. Thus Theorem 2.36 and its

corollary are false if “compact” is replaced by “closed.”For every n ∈ N, define En := (0, 1/n) ⊂ R. Then En is bounded, and

for any finite subcollection C, let N be the largest integer such that EN ∈ C.Then

⋂En∈C En = (0, 1/N) 6= ∅, but

⋂∞n=1En = ∅. Thus Theorem 2.36 and its

corollary are false if “compact” is replaced by “bounded.”

2.16. Regard Q, the set of all rational numbers, as a metric space, withd(p, q) = |p− q|. Let E be the set of all p ∈ Q such that 2 < p2 < 3. Show thatE is closed and bounded in Q, but that E is not compact. Is E open in Q?

Solution. Fix x ∈ Ec. Since√

2 and√

3 are irrational, either x2 > 3 or x2 < 2.Assume without loss of generality that x2 > 3 (the proof of the other case issimilar). Let r := x−

√3 and note that

Nr(x) ={p ∈ Q : 3 < p2 < (x+ r)2

}⊂ Ec,

proving that x is an interior point of Ec. Thus Ec is open, and by the corollaryto Theorem 2.23, E is closed.

9

Since E ⊂ [0, 2], it follows that E is bounded.For every integer n ≥ 10, define Gn := (

√2 + 1/n,

√3) ∩ Q. If p ∈ E, then

2 < p2 < 3, so there is an integer N such that (√

2 + 1/N)2 < p. Thus p ∈ GN ,and therefore {Gn}∞10 is an open cover of E. If C is a finite subcollection of{Gn}∞10, then there is a largest integer k such that Gk ∈ C. Choose p ∈ Q suchthat

√2 < p <

√2 + 1/k. Then p /∈

⋃Gn∈C Gn = (

√2 + 1/k,

√3) ∩ Q, so C is

not a cover of E. Thus {Gn}∞10 contains no finite subcover of E, and thereforeE is not compact.

Yes, E is open in Q. Since E = (√

2,√

3)∩Q, E is open by Theorem 2.30.

2.19.

(a) If A and B are disjoint closed sets in some metric space X, prove thatthey are separated.

(b) Prove the same for disjoint open sets.

(c) Fix p ∈ X, δ > 0, define A to be the set of all q ∈ X for which d(p, q) < δ,define B similarly, with > in place of 0 and Nr(x) ⊂ B by the triangle inequality, so B is open.Moreover, A ∩B = ∅ since d(p, q) cannot be both greater than δ and lessthan δ. By part (b), A and B are separated.

(d) Suppose X is a metric space which is at most countable. Fix x ∈ X. Thenthere exists an r > 0 such that {y ∈ X : d(x, y) = r} = ∅, for otherwise,we would have an injective function from (0,∞) into X. The sets

A := {y ∈ X : d(x, y) < r} and B := {y ∈ X : d(x, y) > r}

separate X by part (c).

10

2.20. Are closures and interiors of connected sets always connected? (Look atsubsets of R2.)

Solution. The closure of a connected set is always connected. Suppose E isconnected and E ⊂ A∪B, where A∩B = ∅ = A∩B. Then E ∩B = ∅ withoutloss of generality, for otherwise, A and B separate E. Thus E ⊂ A, and byTheorem 2.27(c), E ⊂ A. Since A ∩ B = ∅, it follows that E ∩ B = ∅. Thus Eis not a union of two nonempty separated sets.

The interior of a connected set need not be connected. Define

E :={x ∈ R2 : d(x,±1) ≤ 1

}.

Then E is connected, but

E◦ ={x ∈ R2 : d(x,−1) < 1

}∪{x ∈ R2 : d(x, 1) < 1

},

which is a union of two nonempty separated sets by Exercise 2.19(b).

2.21. Let A and B be separated subsets of some Rk, suppose a ∈ A, b ∈ B,and define

p(t) = (1− t)a + tb

for t ∈ R1. Put A0 = p−1(A), B0 = p−1(B). [Thus t ∈ A0 if and only ifp(t) ∈ A.]

(a) Prove that A0 and B0 are separated subsets of R1.

(b) Prove that there exists t0 ∈ (0, 1) such that p(t0) /∈ A ∪B.

(c) Prove that every convex subset of Rk is connected.

Solution.

(a) Fix t ∈ A0. Then p(t) ∈ A, and since A ∩ B = ∅, p(t) /∈ B. Thus thereis an r > 0 such that Nr(p(t)) ⊂ B

c. In particular, if s ∈ Nr(t), then

s /∈ B0. Assume that there exists an s ∈ Nr(t) such that s ∈ B0. Ifr′ := r − |s− t|, then there exists a point s′ ∈ Nr′(s) ∩B0. But

|t− s′| ≤ |t− s|+ r′ < r

by the triangle inequality, so therefore s′ ∈ Nr(t), contradicting thatNr(t) ⊂ Bc0. We have thus shown that Nr(t) ⊂ (B0)c. The same ar-gument applied to a point t′ ∈ B0 shows that there is a neighborhood oft′ which is contained in (A0)c. Thus A ∩B = ∅ = A ∩B.

11

(b) Assume the contrary, that p(t) ∈ A ∪ B for all t ∈ (0, 1). Then we have[0, 1] ⊂ A0 ∪B0, and by part (a),(

A0 ∩ [0, 1])∩(B0 ∩ [0, 1]

)= (A0 ∩B0) ∩ [0, 1] = ∅ ∩ [0, 1] = ∅,(

A0 ∩ [0, 1])∩(B0 ∩ [0, 1]

)= (A0 ∩B0) ∩ [0, 1] = ∅ ∩ [0, 1] = ∅.

The sets A0 ∩ [0, 1] and B0 ∩ [0, 1] are nonempty since 0 ∈ A0 ∩ [0, 1]and 1 ∈ B0 ∩ [0, 1], and they are disjoint since A0 and B0 are disjointby part (a). Thus A0 ∩ [0, 1] and B0 ∩ [0, 1] separate [0, 1], contradictingTheorem 2.47.

(c) Assume that there exists a convex set E ⊂ Rk such that E = A ∪ B fortwo nonempty separated sets A and B. Fix a ∈ A and b ∈ B. By thedefinition of convexity,

p(t) := (1− t)a + tb ∈ E = A ∪B

for every t ∈ (0, 1). This contradicts part (b). Hence every convex set inRk is connected.

12


Rudin, Chapter 3

Dominique Abdi

3.1. Prove that the convergence of {sn} implies convergence of {|sn|}. Is theconverse true?

Solution. We first prove the reverse triangle inequality from Exercise 1.13. Ifa, b ∈ C (or Rk), then

|a| = |a− b+ b| ≤ |a− b|+ |b| ,

so |a| − |b| ≤ |a− b|. Interchanging a and b, we see that |b| − |a| ≤ |a− b|, andtherefore ∣∣ |a| − |b| ∣∣ ≤ |a− b| .

Now suppose sn → s. Fix � > 0 and N such that |sn − s| < � for all n ≥ N .The reverse triangle inequality then yields∣∣ |sn| − |s| ∣∣ ≤ |sn − s| < �for all n ≥ N , so |sn| → |s|.

The converse is not true. Consider the sequence sn := (−1)n. This sequencediverges, but since |sn| =

∣∣(−1)n∣∣ = 1 for all n ∈ N, we have |sn| → 1.

3.2. Calculate limn→∞

(√n2 + n− n

).

Solution. By elementary algebra, we have(√n2 + n− n

)=

n√n2 + n+ n

=1√

1 + 1/n+ 1.

Since 1/n → 0, Theorem 3.3 and the continuity of the square root function onthe positive real numbers yields

limn→∞

(√n2 + n− n

)= lim

n→∞

1√1 + 1/n+ 1

=12.

3.3. If s1 =√

2, and

sn+1 =√

2 +√sn (n = 1, 2, 3, . . .),

prove that {sn} converges, and that sn < 2 for n = 1, 2, 3, . . ..

Solution. First, note that

s1 =√

2 <√

2 +√

2 = s2 and s2 =√

2 +√

2 < 2.

Suppose s1 < s2 < · · · < sn < 2. Then

sn =√

2 +√sn−1 <

√2 +√sn = sn+1.

Moreover, sn < 2 implies that√sn < 2, and therefore 2 +

√sn < 4. Taking

square roots, we have

sn+1 =√

2 +√sn < 2,

and therefore s1 < · · · < sn < sn+1 < 2. By induction, {sn} is a monotoni-cally increasing sequence which is bounded above by 2, so {sn} converges byTheorem 3.14.

3.4. Find the upper and lower limits of the sequence {sn} defined by

s1 = 0; s2m =s2m−1

2; s2m+1 =

12

+ s2m.

Solution. We can prove by induction that

s2m =12− 1

2mand s2m+1 = 1−

12m−1

for all m ∈ N. Since the subsequence of even terms converges to 1/2 and thesubsequence of odd terms converges to 1, both 1/2 and 1 are subsequentiallimits of {sn}. If x /∈ {1/2, 1}, then let r := 12 min {|x− 1| , |x− 1/2|} and notethat sn ∈ Nr(x) for at most finitely many n ∈ N. Thus 1/2 and 1 are the onlysubsequential limits of {sn}, so therefore

lim supn→∞

sn = 1 and lim infn→∞

sn =12.

3.5. For any two real sequences {an}, {bn}, prove that

lim supn→∞

(an + bn) ≤ lim supn→∞

an + lim supn→∞

bn,

provided the sum on the right is not of the form ∞−∞.

2

Solution. Assume that the sum on the right is not of the form ∞−∞.Suppose one of the terms on the right is −∞, say lim sup an = −∞. Then

an → −∞, and for any M > 0, there are only finitely many n ∈ N such thatbn > M . Thus an + bn → −∞, so lim sup(an + bn) = −∞.

Now suppose one of the terms on the right is ∞, say lim sup an =∞. Thenfor any M > 0, there are infinitely many n ∈ N such that an > M , but onlyfinitely many n ∈ N such that bn < −M . Thus there are infinitely many n ∈ Nsuch that an + bn > M , so lim sup(an + bn) =∞.

Finally, suppose α := lim sup an and β := lim sup bn are both finite and fix� > 0. By Theorem 3.17(b), there exist N1 and N2 such that n ≥ N1 impliesan < α+ �/2, and n ≥ N2 implies bn < β + �/2. Let N := max {N1, N2}. Thenn ≥ N implies

an + bn < α+ β + �.

But � > 0 is arbitrary, so therefore

an + bn ≤ α+ β = lim supn→∞

an + lim supn→∞

bn

for all but finitely many n. Thus every subsequential limit of {an + bn} isbounded above by lim sup an + lim sup bn, and in particular,

lim supn→∞

(an + bn) ≤ lim supn→∞

an + lim supn→∞

bn

by Theorem 3.17(a).

3.8. If∑an converges, and if {bn} is monotonic and bounded, prove that∑

anbn converges.

Solution. By Theorem 3.14, {bn} converges to some number b. Define

cn :=

{b− bn if {bn} is monotonically increasing,bn − b if {bn} is monotonically decreasing.

Then {cn} is monotonically decreasing and lim cn = 0. Moreover, for all N ∈ N,

N∑n=1

anbn =N∑

n=1

ancn + bN∑

n=1

an

if {bn} is decreasing, for example, and similarly if {bn} is increasing. By Theo-rem 3.42,

∑ancn converges, and therefore

∑anbn converges.

3.20. Suppose {pn} is a Cauchy sequence in a metric space X, and some sub-sequence {pni} converges to a point p ∈ X. Prove that the full sequence {pn}converges to p.

3

Solution. Fix � > 0. Choose N1 such that ni ≥ N1 implies d(pni , p) < �/2,and choose N2 such that m,n ≥ N2 implies d(pm, pn) < �/2. Now defineN := max {N1, N2}. Then n ≥ N implies

d(pn, p) ≤ d(pn, pni) + d(pni , p) < �

for some ni ≥ N . Thus lim pn = p.

3.21. Prove the following analogue of Theorem 3.10(b): If {En} is a sequence ofclosed nonempty and bounded sets in a complete metric space X, if En ⊃ En+1,and if

limn→∞

diamEn = 0,

then⋂∞

1 En consists of exactly one point.

Solution. For every n ∈ N, choose a point xn ∈ En. Fix � > 0 and choose Nsuch that diamEN < �. Then m,n ≥ N implies xm, xn ∈ EN since En ⊃ En+1for all n, so therefore

d(xm, xn) ≤ diamEN < �.

Thus {xn} is a Cauchy sequence, and since X is complete, {xn} converges tosome point x ∈ X. For every N ∈ N, {xn}∞N ⊂ EN and EN is closed, sotherefore x ∈ EN . Hence x ∈

⋂∞1 En.

If x, y ∈⋂∞

1 En and x 6= y, then d(x, y) > 0. Then diamEn ≥ d(x, y) > 0for all n, contradicting that lim diamEn = 0.

4


Rudin, Chapter 4

Dominique Abdi

4.1. Suppose f is a real function defined on R1 which satisfies

limh→0

[f(x+ h)− f(x− h)] = 0

for every x ∈ R1. Does this imply that f is continuous?

Solution. No. Define f : R→ R by

f(x) :=

{0 if x 6= 0,1 if x = 0.

.

Thenlimh→0

[f(h)− f(−h)] = 0,

but f is not continuous at x = 0.

4.2. If f is a continuous mapping of a metric space X into a metric space Y ,prove that

f(E) ⊂ f(E)

for every set E ⊂ X. Show, by an example, that f(E) can be a proper subsetof f(E).

Solution. Fix x ∈ E and a sequence {xn}∞1 ⊂ E such that limxn = x. Then{f(xn)}∞1 is a sequence in f(E), and since f is continuous, lim f(xn) = f(x).If f(x) /∈ f(E) ∪ f(E)′ = f(E), then there is a neighborhood Nr(f(x)) whichdoes not intersect f(E). By Theorem 4.8, f−1(Nr(f(x))) is open, so thereis a neighborhood Ns(x) which does not intersect E. This contradicts thatlimxn = x, so f(x) ∈ f(E) and therefore f(E) ⊂ f(E).

Define f : N→ R by f(n) = 1/n. Since N = N, we have

f(N) ={

1n

: n ∈ N}⊂{

1n

: n ∈ N}∪ {0} = f(N).

4.3. Let f be a continuous real function on a metric space X. Let Z(f) (thezero set of f) be the set of all p ∈ X at which f(p) = 0. Prove that Z(f) isclosed.

Solution. Let x be a limit point of Z(f) and let {xn}∞1 ⊂ Z(f) be a sequencesuch that limxn = x. Then

f(x) = limn→∞

f(xn) = limn→∞

0 = 0,

so x ∈ Z(f). Thus Z(f) contains its limit points, so Z(f) is closed.

4.4. Let f and g be continuous mappings of a metric space X into a metricspace Y , and let E be a dense subset of X. Prove that f(E) is dense in f(X).If g(p) = f(p) for all p ∈ E, prove that g(p) = f(p) for all p ∈ X. (In otherwords, a continuous mapping is determined by its values on a dense subset ofits domain.)

Solution. Fix y ∈ f(X) and choose x ∈ X such that f(x) = y. Since E isdense in X, there is a sequence {xn}∞1 ⊂ X such that limxn = x. Consider thesequence {f(xn)}∞1 ⊂ f(E). Since f is continuous, lim f(xn) = f(x) = y, soy ∈ f(E) ∪ f(E)′ = f(E). Thus f(X) ⊂ f(E), so f(E) is dense in f(X).

Fix p ∈ X and choose a sequence {pn} ⊂ E such that lim pn = p. Since fand g are continuous and f(pn) = g(pn), we have

f(p) = limn→∞

f(pn) = limn→∞

g(pn) = g(p),

proving that f(p) = g(p) for all p ∈ X.

4.5. If f is a real continuous function defined on a closed set E ⊂ R1, provethat there exist continuous real functions g on R1 such that g(x) = f(x) forall x ∈ E. (Such functions g are called continuous extensions of f from E toR1.) Show that the result becomes false if the word “closed” is omitted. Extendthe result to vector-valued functions. Hint : Let the graph of g be a straightline on each of the segments which constitute the complement of E (compareExercise 29, Chap. 2). The result remains true if R1 is replaced by any metricspace, but the proof is not so simple.

Solution. Since E is closed, Ec is the union of an at most countable collectionof disjoint open intervals C = {In} by Exercise 2.29. If (−∞, c) is one of theseintervals for some c ∈ R, then define g(x) := f(c) for all x ∈ (−∞, c), andsimilarly if (c,∞) is one of the intervals. On the remaining intervals In =(an, bn), define

g(x) := f(an) +(f(bn)− f(an)

bn − an

)(x− an) .

2

Finally, define g(x) := f(x) for all x ∈ E. Then clearly g extends f to R and gis continuous on E◦ and on Ec. It remains to show that g is continuous at theendpoints of the intervals In ∈ C.

Let x be the endpoint of some interval In ∈ C. If {xj} is a sequence inE converging to x, then lim g(xj) = g(x) = f(x). If {xj} is a sequence in Ecconverging to x = an, then

limj→∞

g(xj) = limj→∞

[f(an) +

(f(bn)− f(an)

bn − an

)(xj − an)

]= f(an),

and similarly if x = bn. Thus g is continuous on R.If the word “closed” is omitted, let E := (0, 1), and define f(x) := 1/x for all

x ∈ E. If f has a continuous extension g to R, then g is continuous on [−1, 1],and therefore bounded by the intermediate value theorem. However, f is notbounded in any neighborhood of x = 0, so therefore g is not bounded either, acontradiction.

If f : E → Rk is continuous and E ⊂ R is closed, then f has a continuousextension g : R→ Rk. Each component function fj : R→ R of f = (f1, . . . , fk)extends to a continuous function gj : R→ R by the proof above, and it followsthat g = (g1, . . . , gk) is a continuous extension of f to R.

4.6. If f is defined on E, the graph of f is the set of points (x, f(x)), for x ∈ E.In particular, if E is a set of real numbers, and f is real-valued, the graph of fis a subset of the plane.

Suppose E is compact, and prove that f is continuous on E if and only ifits graph is compact.

Solution. Suppose f is continuous and let {Uα × Vβ} be an open cover of thegraph of f . Then {Uα} is an open cover of E, so there exist α1, . . . , αm suchthat {Uαi}m1 is a finite subcover of E. Similarly, {Vβ} is an open cover of f(E),and since f(E) is compact by Theorem 4.14, there exist β1, . . . , βn such that{Vβj}n1 is a finite subcover of f(E). Thus{

Uαi × Vβj : 1 ≤ i ≤ m, 1 ≤ j ≤ n}

is a finite subcover of {Uα × Vβ}, proving that the graph of f is compact.Now suppose f is not continuous. Then there is a sequence {xn}∞1 ⊂ E such

that limxn = x ∈ E, but f(xn) does not converge to f(x). Thus there is an� > 0 such that d(f(xn), f(x)) ≥ � for infinitely many n, and by restricting to asubsequence, we can assume without loss of generality that d(f(xn), f(x)) ≥ �for all n. If the graph of f is compact, then the sequence {(xn, f(xn))}∞1 containsa convergent subsequence by Theorem 3.6(a). That is, (xnk , f(xnk))→ (x, f(x))for some x ∈ E, contradicting that d(f(xn), f(x)) ≥ � for all n. The graph of fis therefore not compact.

3

4.7. If E ⊂ X and if f is a defined on X, the restriction of f to E is thefunction g whose domain of definition is E, such that g(p) = f(p) for p ∈ E.Define f and g on R2 by: f(0, 0) = g(0, 0) = 0, f(x, y) = xy2/(x2 + y4),g(x, y) = xy2/(x2 + y6) if (x, y) 6= (0, 0). Prove that f is bounded on R2, thatg is unbounded in every neighborhood of (0, 0), and that f is not continuous at(0, 0); nevertheless, the restrictions of both f and g to every straight line in R2are continuous!

Solution. First, note that

0 ≤ (x− y2)2 = x2 − 2xy2 + y4

implies x2 + y4 ≥ 2xy2. Thus

|f(x, y)| =∣∣∣∣ xy2x2 + y4

∣∣∣∣ ≤ ∣∣∣∣ xy22xy2∣∣∣∣ = 12 ,

so f is bounded on R2.Along the curve y = x1/3, we have

limx→0

∣∣∣g(x, x1/3)∣∣∣ = limx→0

∣∣∣∣x5/32x2∣∣∣∣ = limx→0

∣∣∣∣ 12x1/3∣∣∣∣ =∞,

proving that g is unbounded in every neighborhood of (0, 0).Along the curve y = x1/2, we have

limx→0

f(x, x1/2) = limx→0

x2

2x2=

126= 0 = f(0, 0),

so f is not continuous at (0, 0).It is clear that f and g are continuous away from (0, 0) and along the axes,

so it suffices to consider lines of the form y = cx for some fixed constant c ∈ R.Since

limx→0

f(x, cx) = limx→0

cx3

x2 + c4x4= limx→0

cx

1 + c4x2= 0,

limx→0

g(x, cx) = limx→0

cx3

x2 + c6x6= limx→0

cx

1 + c6x4= 0,

the restrictions of both f and g to every straight line in R2 are continuous.

4.8. Let f be a real uniformly continuous function on the bounded set E inR1. Prove that f is bounded on E.

Show that the conclusion is false if boundedness of E is omitted from thehypothesis.

4

Solution. Fix � > 0 and choose δ > 0 such that |f(x), f(y)| < � whenever|x− y| < δ. The collection {Nδ(x) : x ∈ E} is an open cover of E. In fact, thisis also an open cover of E, for otherwise, there is a limit point p of E withthe property that p /∈ Nδ(x) for all x ∈ E, a contradiction. Since E is closedand bounded, it is compact by the Heine-Borel theorem, so there exist pointsx1, . . . , xn ∈ E such that {Nδ(xj)}n1 is a finite subcover of E. But then for everyx ∈ E, there is a j such that |x− xj | < δ, and therefore |f(x)− f(xj)| < �.That is, {N�(f(xj))}n1 is a finite cover of f(E), so diam f(E) ≤ n�. Thus f isbounded on E.

If E = R and f(x) := x on E, then f is uniformly continuous since forevery � > 0, |x− y| < � implies that |f(x)− f(y)| < �. However, f is notbounded.

4.9. Show that the requirement in the definition of uniform continuity can berephrased as follows, in terms of diameters of sets: To every � > 0 there existsa δ > 0 such that diam f(E) < � for all E ⊂ X with diamE < δ.

Solution. Suppose f is uniformly continuous, fix � > 0, and choose δ > 0 suchthat d(x, y) < δ implies d(f(x), f(y)) < �. Given a set E with diamE < δ, wehave d(x, y) < δ for all x, y ∈ E. Thus d(f(x), f(y)) < � for all x, y ∈ E, sotherefore diam f(E) < �.

Conversely, fix � > 0 and choose δ > 0 such that diam f(E) < � for allE ⊂ X with diamE < δ. If d(x, y) < δ, let E := {x, y}. Then diamE < δ andf(E) = {f(x), f(y)}. By assumption, diam f(E) < �, that is, d(f(x), f(y)) < �,proving that f is uniformly continuous.

4.10. Complete the details of the following alternate proof of Theorem 4.19: Iff is not uniformly continuous, then for some � > 0 there are sequences {pn}, {qn}in X such that dX(pn, qn)→ 0 but dY (f(pn), f(qn)) > �. Use Theorem 2.37 toobtain a contradiction.

Solution. Suppose f is not uniformly continuous. Then there exists an � > 0such that for any n ∈ N, there are points pn, qn ∈ X such that dX(pn, qn) < 1/n,but dY (f(pn), f(qn)) > �. Thus {pn} and {qn} are sequences in X such thatdX(pn, qn)→ 0, but dY (f(pn), f(qn)) > �.

By Theorem 4.14, f(X) is compact. If the sequences {pn} and {qn} are bothfinite, then pn = qn for all n sufficiently large, contradicting that

dY (f(pn), f(qn)) > �

for all n. Therefore the set E consisting of the sequences {pn} and {qn} isinfinite, so it has a limit point x by Theorem 2.37. It follows that there are

5

increasing sequences of positive integers {nk} and {n′k} such that lim pnk = xand lim qn′k = x. Then

limk→∞

f(pnk) = f(x) = limk→∞

f(qn′k)

since f is continuous, and dX(pnk , qn′k)→ 0. Choose k sufficiently large so that

dY (f(pnk), f(x)) <�

2and dY (f(qn′k), f(x)) <

�

2.

Then

dY (f(pnk), f(qn′k)) ≤ dY (f(pnk), f(x)) + dY (f(qn′k), f(x)) < �,

contradicting thatdY (f(pnk), f(qn′k)) > �

for all n.

4.11. Suppose f is a uniformly continuous mapping of a metric space X intoa metric space Y and prove that {f(xn)} is a Cauchy sequence in Y for everyCauchy sequence {xn} is in X. Use this result to give an alternative proof ofthe theorem stated in Exercise 13.

Solution. Fix � > 0 and choose δ > 0 such that d(f(x), f(y)) < � wheneverd(x, y) < δ. Now choose N such that d(xm, xn) < δ for all m,n ≥ N . Thend(f(xm), f(xn)) < � for all m,n ≥ N , so {f(xn)} is a Cauchy sequence.

Let E be a dense subset of a metric space X, and let f be a uniformlycontinuous real function defined on E. For every x ∈ X, there is a sequence{xn}∞1 ⊂ E such that limxn = x. Since every convergent sequence is Cauchy,the result above shows that {f(xn)}∞1 is a Cauchy sequence in R, and thusconverges. If limxn = x and lim yn = x, then

limn→∞

f(xn) = f(x) = limn→∞

f(yn)

by the continuity of f , so we can extend f to X by defining f(x) = lim f(xn)for any sequence {xn}∞1 ⊂ E converging to x. Moreover, by Theorem 4.2, theextension of f is continuous on X.

4.12. A uniformly continuous function of a uniformly continuous function isuniformly continuous.

State this more precisely and prove it.

6

Solution. Let X,Y, Z be metric spaces and suppose f : X → Y and g : Y → Zare uniformly continuous. Then g ◦ f is uniformly continuous.

Proof. Fix � > 0 and choose η > 0 such that dY (y1, y2) < η implies

dZ(g(y1), g(y2)) < �.

Now choose δ > 0 such that dX(x1, x2) < δ implies dY (f(x1), f(x2)) < η. ThendX(x1, x2) < δ implies dY (f(x1), f(x2)) < η, which in turn implies

dZ(g(f(x1), g(f(x2))) < �.

Thus g ◦ f is uniformly continuous.

4.14. Let I = [0, 1] be the closed unit interval. Suppose f is a continuousmapping of I into I. Prove that f(x) = x for at least one x ∈ I.

Solution. The function g : I → I defined by g(x) := f(x) − x is continuous byTheorem 4.4. It suffices to show that g(x) = 0 for some x ∈ I. If g(0) = 0 org(1) = 1, then there is nothing to prove. Otherwise, g(0) > 0 since 0 < f(0) ≤ 1,and g(1) < 0 since 0 ≤ f(1) < 1. By the intermediate value theorem, there is apoint x ∈ I such that g(x) = 0.

7


Rudin, Chapter 5

Dominique Abdi

5.1. Let f be defined for all real x, and suppose that

|f(x)− f(y)| ≤ (x− y)2

for all real x and y. Prove that f is constant.

Solution. If x 6= y, dividing by |x− y| and taking the limit as y → x gives

limy→x

∣∣∣∣f(x)− f(y)x− y∣∣∣∣ ≤ limy→x |x− y| = 0.

Thus f ′(x) exists and f ′(x) = 0 for every x ∈ R. By Theorem 5.11(b), f isconstant.

5.2. Suppose f ′(x) > 0 in (a, b). Prove that f is strictly increasing in (a, b),and let g be its inverse function. Prove that g is differentiable, and that

g′(f(x)) =1

f ′(x)(a < x < b).

Solution. Fix x, y ∈ (a, b) with x < y. By the mean value theorem, there is anx0 ∈ (a, b) with

f(y)− f(x) = f ′(x0)(y − x) > 0.Thus f is strictly increasing.

Fix p = f(x) in the range of f , let {pn} be a sequence in the range of fconverging to p, and choose xn ∈ (a, b) such that f(xn) = pn. Then

limn→∞

g(pn)− g(p)pn − p

= limn→∞

g(f(xn))− g(f(x))f(xn)− f(x)

= limn→∞

xn − xf(xn)− f(x)

=1

f ′(x).

5.3. Suppose g is a real function on R1, with bounded derivative (say |g′| ≤M).Fix � > 0, and define f(x) = x+ �g(x). Prove that f is one-to-one if � is smallenough. (A set of admissible values of � can be determined which depends onlyon M .)

Solution. First, note that f is differentiable by Theorem 5.3 because g is differ-entiable. Moreover,

f ′(x) = 1 + �g′(x).

If � < 1/M , then

|�g′(x)| < |g′(x)|M

≤ 1,

so f ′(x) > 0. By the mean value theorem, if x < y, then there is an x0 ∈ (x, y)such that

f(y)− f(x) = f ′(x0)(y − x) > 0,so f is strictly increasing, and therefore one-to-one.

5.4. If

C0 +C12

+ · · ·+ Cn−1n

+Cnn+ 1

= 0,

where C0, . . . , Cn are real constants, prove that the equation

C0 + C1x+ · · ·+ Cn−1xn−1 + Cnxn = 0

has at least one real root between 0 and 1.

Solution. Define

f(x) := C0x+C12x2 + · · ·+ Cn

n+ 1xn+1.

Then f is a polynomial, and is therefore differentiable, and

f ′(x) = C0 + C1x+ · · ·+ Cn−1xn−1 + Cnxn.

Since f(0) = 0 and

f(1) = C0 +C12

+ · · ·+ Cn−1n

+Cnn+ 1

= 0

by assumption, by the mean value theorem, there is an x ∈ (0, 1) such that

0 = f(1)− f(0) = f ′(x).

5.5. Suppose f is defined and differentiable for every x > 0, and f ′(x)→ 0 asx→ +∞. Put g(x) = f(x+ 1)− f(x). Prove that g(x)→ 0 as x→ +∞.

Solution. Fix � > 0 and choose N such that x > N implies |f ′(x)| < �. Thenfor any x ≥ N , by the mean value theorem, there is an x0 ∈ (x, x+ 1) such that

g(x) = f(x+ 1)− f(x) = f ′(x0)(x+ 1− x) = f ′(x0) < �.

Thus g(x)→ 0 as x→ +∞.

2

5.6. Suppose

(a) f is continuous for x ≥ 0,

(b) f ′(x) exists for x > 0,

(c) f(0) = 0,

(d) f ′ is monotonically increasing.

Put

g(x) =f(x)

x(x > 0)

and prove that g is monotonically increasing.

Solution. Since f is differentiable and x > 0, g is differentiable by Theorem 5.3.It suffices to prove that g′ is non-negative by Theorem 5.11(a). Moreover, since

g′(x) =xf ′(x)− f(x)

x2,

it suffices to prove that xf ′(x) ≥ f(x).Fix x > 0. By the mean value theorem, there is an x0 ∈ (0, x) such that

f(x) = f(x)− f(0) = f ′(x0)(x− 0) = xf ′(x0) ≤ xf ′(x),

where the first equality follows by (c), and the last inequality follows by (d).

5.7. Suppose f ′(x), g′(x) exist, g′(x) 6= 0, and f(x) = g(x) = 0. Prove that

limt→x

f(t)

g(t)=f ′(x)

g′(x).

(This also holds for complex functions.)

Solution.

f ′(x)

g′(x)= lim

t→x

f(t)− f(x)t− x

· t− xg(t)− g(x)

= limt→x

f(t)− f(x)g(t)− g(x)

= limt→x

f(t)

g(t)=f(x)

g(x).

5.8. Suppose f ′ is continuous on [a, b] and � > 0. Prove that there exists δ > 0such that ∣∣∣∣f(t)− f(x)t− x − f ′(x)

∣∣∣∣ < �whenever 0 < |t− x| < δ, a ≤ x ≤ b, a ≤ t ≤ b. (This could be expressed bysaying that f is uniformly differentiable on [a, b] if f ′ is continuous on [a, b].)Does this hold for vector-valued functions too?

3

Solution. Fix � > 0. By Theorem 4.19, f ′ is uniformly continuous on [a, b],so there exists a δ > 0 such that |x− y| < δ implies |f ′(x)− f ′(y)| < �. Fixx ∈ [a, b], and without loss of generality, fix t ∈ [a, b] with 0 < t−x < δ. By themean value theorem, there is an x0 ∈ (x, t) such that

f(t)− f(x)t− x

= f ′(x0).

Then 0 < |x− x0| < δ, so∣∣∣∣f(t)− f(x)t− x − f ′(x)∣∣∣∣ = |f ′(x0)− f ′(x)| < �.

Now suppose f′ is continuous on [a, b] for some function f : Rn → R, f =(f1, . . . , fn). Then each component fj is differentiable, f

′ = (f ′1, . . . , f′n), and

each f ′j is continuous on [a, b]. Fix � > 0, and by the result above, choose δj > 0such that ∣∣∣∣fj(t)− fj(x)t− x − f ′j(x)

∣∣∣∣ < �√nwhenever 0 < |t− x| < δj , a ≤ x ≤ b, a ≤ t ≤ b, for j = 1, . . . , n. Then∣∣∣∣ f(t)− f(x)t− x − f′(x)

∣∣∣∣ = ∣∣∣∣(f1(t)− f1(x)t− x − f ′1(x), . . . , fn(t)− fn(x)t− x − f ′n(x))∣∣∣∣

=

n∑j=1

(fj(t)− fj(x)

t− x− f ′j(x)

)21/2

<

n∑j=1

(�√n

)21/2 = �whenever 0 < |t− x| < δj , a ≤ x ≤ b, a ≤ t ≤ b.

5.9. Let f be a continuous real function on R1, of which it is known that f ′(x)exists for all x 6= 0 and that f ′(x) → 3 as x → 0. Does it follow that f ′(0)exists?

Solution. Fix � > 0 and choose δ > 0 such that |f ′(x)− 3| < � whenever0 < |x| < δ. Now fix t ∈ R such that 0 < |t| < δ. By the mean value theorem,there is an x0 ∈ (0, t) (or x0 ∈ (t, 0) if t < 0) such that

f(t)− f(0)t

= f ′(x0).

Then 0 < |x0| < δ, so therefore∣∣∣∣f(t)− f(0)t − 3∣∣∣∣ = |f ′(x0)− 3| < �.4

This proves that

limt→0

f(t)− f(0)t

= 3,

so therefore f ′(0) exists.

5.10. Suppose f and g are complex differentiable functions on (0, 1), f(x)→ 0,g(x)→ 0, f ′(x)→ A, g′(x)→ B as x→ 0, where A andB are complex numbers,B 6= 0. Prove that

limx→0

f(x)

g(x)=A

B.

Compare with Example 5.18. Hint :

f(x)

g(x)=

{f(x)

x−A

}· xg(x)

+A · xg(x)

.

Apply Theorem 5.13 to the real and imaginary parts of f(x)/x and g(x)/x.

Solution. Define f1 := Re(f), f2 := Im(f), g1 := Re(g), and g2 := Im(g). Then

f ′1 = Re(f′), f ′2 = Im(f

′), g′1 = Re(g′), g′2 = Im(g

′),

so therefore

f ′1(x)→ Re(A), f ′2 → Im(A), g′1 → Re(B), g′2 → Im(B)

as x→ 0. Moreover, fi → 0 and gi → 0 for i = 1, 2 as x→ 0. By Theorem 5.13,

limx→0

f(x)

x= lim

x→0

(f1(x)

x+ i

f2(x)

x

)= lim

x→0

f1(x)

x+ i lim

x→0

f2(x)

x

= limx→0

f ′1(x) + i limx→0

f ′2(x) = Re(A) + i Im(A) = A,

and similarly, g(x)/x → B as x → 0. Since B 6= 0, it follows from Theo-rem 4.4(c) that

limx→0

f(x)

g(x)= lim

x→0

f(x)

x· xg(x)

=A

B.

5.11. Suppose f is defined in a neighborhood of x, and suppose f ′′(x) exists.Show that

limh→0

f(x+ h) + f(x− h)− 2f(x)h2

= f ′′(x).

Show by an example that the limit may exist even if f ′′(x) does not.Hint : Use Theorem 5.13.

5

Solution. Since f ′′(x) exists, so does f ′(x), and by Theorem 5.2, f is continuousat x. Therefore f(x+h)→ f(x) and f(x−h)→ f(x) as h→ 0, so the numeratorand denominator of the fraction

f(x+ h) + f(x− h)− 2f(x)h2

both converge to 0 as h→ 0. By Theorem 5.13,

limh→0

f(x+ h) + f(x− h)− 2f(x)h2

= limh→0

f ′(x+ h)− f ′(x− h)2h

= limh→0

f ′(x+ h)− f ′(x) + f ′(x)− f ′(x− h)2h

=1

2limh→0

f ′(x+ h)− f ′(x)h

+1

2limh→0

f ′(x)− f ′(x− h)h

= f ′′(x).

Define f : R→ R by

f(x) :=

{1/x if x 6= 0,0 if x = 0.

Then f is not continuous at x = 0, so f ′(0) and f ′′(0) do not exist by Theo-rem 5.2. However, for any h 6= 0, we have

f(h) + f(−h)− 2f(0)h2

= 0,

so the fraction above converges to 0 as h→ 0.

5.12. If f(x) = |x|3, compute f ′(x), f ′′(x) for all real x, and show that f (3)(0)does not exist.

Solution. If x > 0, then f(x) = x3, so f ′(x) = 3x2 and f ′′(x) = 6x. If x < 0,then f(x) = −x3, so f ′(x) = −3x2 and f ′′(x) = −6x. Since

limt→0

f(t)

t= lim

t→0

|t|3

sgn(t) |t|= lim

t→0sgn(t) |t|2 = 0,

we have f ′(0) = 0, and since

limt→0

f ′(t)

t= lim

t→0

3 |t|2

sgn(t) |t|= lim

t→0sgn(t)3 |t| = 0,

we have f ′′(0) = 0. However, f (3)(0) does not exist because

limt→0+

f ′′(t)

t= lim

t→0

6t

t= 6, lim

t→0−

f ′′(t)

t= lim

t→0

−6tt

= −6,

shows that the limit of f ′′(t)/t as t→ 0 does not exist.

6

Math 131C: Homework 1 Solutions

(From Rudin, Chapter 7)

Problem 14:The continuity of Φ follows from the continuity of x and y, which follows from the Weierstrass M -test

(Theorem 7.10, Rudin) and the continuity and boundedness of f .To show that Φ maps the Cantor set onto I2, we follow the hint. Note that to as given belongs to

the Cantor set (its ternary expansion contains no 1’s). We have

3kt0 =∞∑

i=1

3k−i−1(2ai)

= 2k−1∑

i=1

3k−i−1ai +∞∑

j=0

3−j−1(2aj+k)

≡∞∑

j=0

3−j−1(2aj+k) (mod 2),

so that

f(3kt0) = f( ∞∑

j=0

3−j−1(2aj+k))

by the 2-periodicity of f . Now suppose ak = 0; then

0 ≤∞∑

j=0

3−j−1(2aj+k) ≤ 13 =⇒ f( ∞∑

j=0

3−j−1(2aj+k))

= f(3kt0) = 0 = ak.

(The lower bound follows from setting ai = 0 for i > k, and the upper bound follows from setting ai = 1for i > k.) Similarly, if ak = 1, we have

23≤

∞∑

j=0

3−j−1(2aj+k) ≤ 1 =⇒ f( ∞∑

j=0

3−j−1(2aj+k))

= f(3kt0) = 1 = ak.

The hint and surjectivity of Φ follow immediately. �

Problem 15:Such an f must be constant on [0,∞). Fix arbitrary nonnegative numbers x and y. Let ² > 0 be

arbitrary, and choose δ > 0 s.t.

|x− y| < δ ⇒ |fn(x)− fn(y)| < ² ∀x, y ∈ [0, 1], ∀n ∈ N.

Now choose n ∈ N s.t. |x−y|n < δ and xn , yn ≤ 1 . Then | xn − yn | < δ, so

|f(x)− f(y)| =∣∣∣∣ fn

(x

n

)− fn

(y

n

)∣∣∣∣ < ².

1

Since ² was arbitrary, we have f(x) = f(y). But x and y were arbitrary, so f is constant. �

Problem 16:Take ² > 0. Choose δ > 0 as in the definition of equicontinuity for the family {fn}. The collection

{Bδ(x) |x ∈ K} (where Bδ(x) is the open ball of radius δ centered at x) is an open cover of K. Bycompactness of K, there exists a finite subcover, so there exist x1, . . . , xm ∈ K s.t. K =

⋃n1 Bδ(xi). Say

fn → f pointwise, and choose N ∈ N s.t. n > N ⇒ |fn(xi)− f(xi)| < ², 1 ≤ i ≤ m.Now take any x ∈ K. x ∈ Bδ(xi) for some i, so |fn(x) − fn(xi)| < ² for all n by equicontinuity.

This in turn implies |f(x)− f(xi)| ≤ ² (by continuity of the absolute value and pointwise convergence of{fn}). Finally, we have

n > N =⇒ |fn(x)− f(x)| ≤ |fn(x)− fn(xi)|+ |fn(xi)− f(xi)|+ |f(xi)− f(x)|< ² + ² + ².

Since x was arbitrary, fn → f uniformly on K. �

Problem 18:By the Fundamental Theorem of Calculus and the uniform boundedness of {fn}, the sequence {Fn}

has uniformly bounded derivatives. By the Mean Value Theorem, this implies that {Fn} is uniformlyLipschitz; i.e., there exists M s.t.

|Fn(x)− Fn(y)||x− y| ≤ M

for all x, y ∈ [a, b] and all n. But then |Fn(x) − Fn(y)| ≤ M |x − y| for all n, x, and y, so the family{Fn} is equicontinuous. Furthermore, {Fn} is pointwise (in fact, uniformly) bounded by the uniformboundedness of {fn}. By the Arzelà-Ascoli Theorem (Theorem 7.25, Rudin), {Fn} contains a uniformlyconvergent subsequence. �

Problem 20:Since f is continuous on a compact set it is bounded; say |f(x)| ≤ M ∀x. The complex conjugate

f(x) is continuous, so by the Weierstrass Approximation Theorem, there is a sequence of complex poly-nomials Pn s.t. Pn → f̄ uniformly on [0, 1]. Since f is bounded, fPn → ff̄ = |f |2 uniformly as well(since ‖fPn − ff̄‖∞ ≤ M‖Pn − f̄‖∞ → 0). By Theorem 7.16 in Rudin,

0 = limn→∞

∫ 10

f(x)Pn(x) dx =∫ 1

0

limn→∞

f(x)Pn(x) dx =∫ 1

0

|f(x)|2 dx.

By continuity of f , this implies f ≡ 0 on [0,1]. �

Problem 21:To see that A vanishes at no point of K and separates points, simply notice that f(eiθ) = eiθ is in

A . Proving the hint is a pretty straightforward exercise in integration and application of Theorem 7.16.Finally, consider g(eiθ) = e−iθ, which is a continuous function on K. We have

∫ 2π0

g(eiθ)eiθ dθ = 2π 6= 0,

which shows that g is not in the uniform closure of A by the hint. �

2



Problem 3:Suppose Ax = Ay. Then A(x− y) = 0 ⇒ x = y. �

Problem 4:These results follow immediately from the definition of a vector space and linearity. You don’t have

to verify all the vector space axioms; since the range and kernel are subsets of vector spaces, it sufficesto show they are subspaces (i.e. that they are closed under addition and scalar multiplication).

Problem 5:Given A ∈ L(Rn,R), let yA = (Ae1, Ae2, . . . , Aen) (where {e1, . . . , en} is the standard basis for Rn).

By linearity, Ax = x · yA for all x ∈ Rn. The uniqueness of yA is immediate.

Now by Cauchy-Schwarz, we have

‖A‖ = sup|x|≤1

x · yA ≤ sup|x|≤1

|x||yA| = |yA|.

But if x = yA|yA| , we have |x| = 1 and Ax = |yA|, so ‖A‖ ≥ |yA|. So ‖A‖ = |yA|. �

Problem 6:For (x, y) 6= (0, 0), we can compute (Djf)(x, y) via the usual differentiation formulas just as in Math

32A or the like. At (0, 0), we consider the definition of partial derivatives:

(D1f)(0, 0) = limh→0

f(0 + h, 0)− f(0, 0)h

= limh→0

0− 0h

= 0.

Similarly, (D2f)(0, 0) = 0. So the partial derivatives of f exist everywhere in R2.

However, f is not continuous at (0, 0), since f(a, a) = 12 for all a ∈ R. (So there is no neighbor-hood of the origin on which |f(x, y)− f(0, 0)| = |f(x, y)| < 12 ). �

Problem 7:Fix x ∈ E and ² > 0. Since the partial derivatives of f are bounded in E (say by M), the Mean

Value Theorem implies that |f(a+hej)− f(a)| < hM for all a ∈ E and all h ∈ R such that a+hej ∈ E(1 ≤ j ≤ n). Now choose δ < ²nM s.t. Bδ(x) ⊂ E (we can choose such a δ since E is open).

Now pick an arbitrary y in Bδ(x). Write y − x =∑n

1 hjej , v0 = 0, and vk =∑k

1 hjej . (Notevn = y − x). It’s pretty easy to see that hj < δ for all j. Now we have

1

|f(y)− f(x)| = |f(x + (y − x))− f(x)|

= |n∑

j=1

[f(x + vj)− f(x + vj−1)]|

≤n∑

j=1

|f(x + vj)− f(x + vj−1)|

≤n∑

j=1

hjM

<

n∑

j=1

δM = ²,

where we’ve used the Mean Value Theorem result from the first paragraph. So |y − x| < δ⇒ |f(y)− f(x)| < ², and f is continuous at x. �

Problem 8:Let x = (x1, . . . , xn), and let Ej be the j-cross-section of E through x; i.e.,

Ej = {x ∈ R | (x1, . . . , xj−1, x, xj+1, . . . , xn) ∈ E}.Now define fj : Ej → R by fj(x) = f(x1, . . . , xj−1, x, xj+1, . . . , xn). Since f has a local maximum at x,fj has a local maximum at xj ∈ Ej for all j. By single-variable calculus, this implies f ′j(xj) = 0 for allj. But since f ′j(xj) = Djf(x) (VERIFY!), this implies f

′(x) = 0 by Theorem 9.17 in Rudin. �

Problem 9:To show f is constant, fix some x0 ∈ E and consider the set A = {x ∈ E | f(x) = f(x0)}. We want

to show that A = E, but since E is connected and A is nonempty, it suffices to show that A is both openand closed in the relative topology on E. Since E is open, the relative topology coincides with the usualtopology on Rn.

First we show A is open. Take any x ∈ A. Since E is open, there is an open ball Br(x) ⊂ E.Now Br(x) is convex, so Theorem 9.19 (or its corollary) in Rudin implies that f is constant on Br(x).So, since x ∈ A, this implies Br(x) ⊂ A. So A is open.

To show A is closed, use an exactly analogous argument to show that Ac is open. So A is bothclosed and open, which implies that A = E. �

Problem 10:(This whole problem is easier to think about if you draw a picture in two dimensions.) Fix any

x0 = (x(0)1 , . . . , x

(0)n ) ∈ E. We want to show that for all x = (x1, x(0)2 , . . . , x(0)n ) ∈ E, f(x) = f(x0). As

in Problem 8, define f1(x) = f(x, x(0)2 , . . . , x

(0)n ). Note that f ′1(x) = (D1f)(x, x

(0)2 , . . . , x

(0)n ), so f ′1(x) = 0

for all x at which f1 is defined. Now take any x1 s.t. (x1, x(0)2 , . . . , x

(0)n ) ∈ E. By convexity, we have

(x, x(0)2 , . . . , x(0)n ) ∈ E for all x(0)1 ≤ x ≤ x1, so f1 is defined for all such x. Since f ′1(x) = 0 for all these

2

x, the Mean Value Theorem shows that f1(x(0)1 ) = f1(x1). So f(x0) = f(x1, x

(0)2 , . . . , x

(0)n ), as desired. �

Note that the proof just given only required that E be “convex in the first variable”; i.e., we canrelax the convexity condition to

(a, x2, . . . , xn) ∈ E and (b, x2, . . . , xn) ∈ E =⇒ (x, x2, . . . , xn) ∈ E ∀ a ≤ x ≤ b.

But, as noted, the proof doesn’t work for arbitrary connected regions. Draw the union of the followingthree regions in R2:

A = {(x, y) | − 2 < x < 2, 0 < y < 1}B = {(x, y) | − 2 < x < −1, −2 < y ≤ 0}

C = {(x, y) | 1 < x < 2, −2 < y ≤ 0}.Take

f(x, y) =

0, (x, y) ∈ A−y, (x, y) ∈ B

y, (x, y) ∈ CThen clearly ∂f∂x ≡ 0 on the union of these regions, but f does not depend only on y. The problem isthat we can’t apply the Mean Value Theorem because of the “gap” between B and C.

3



Problem 18:(a) If we define f(x, y) = (u(x, y), v(x, y)), then the range of f is R2. The slickest way to see this

is to note that if z = x + iy, then u =

0 or 3. To show that we can solve for x, y, u, in terms of z, apply the Implicit Function Theorem:Let f(x, y, z, u) = (3x + y − z + u2, x− y + 2z + u, 2x + 2y − 3z + 2u) = (f1, f2, f3). Then define

Ax,y,u =

∂f1∂x

∂f1∂y

∂f1∂u

∂f2∂x

∂f2∂y

∂f2∂u

∂f3∂x

∂f3∂y

∂f3∂u

=

3 1 2u1 −1 12 2 2

Then detAx,y,u = 8u− 12 6= 0 since u = 0 or u = 3, so the Implicit Function Theorem gives the desiredresult. The next two claims follow by analogous reasoning.

Finally, to show that we cannot solve for x, y, z in terms of u, note that the system can be expressedas

A

xyz

=

−u2−u−2u

,

where

A =

3 1 −11 −1 22 2 −3

.

But detA = 0, so the system cannot be solved.

Problem 21:(a) Compute ∇f = 6(x2 − x, y2 + y). Then clearly the gradient is zero at precisely the points (0, 0),

(1, 0), (0,−1), (1,−1). (1, 0) is a local minimum and (0,−1) is a local maximum, while the other two aresaddle points. This can easily be seen either by examining the Hessian of f or by examining f directlynear the points in question.

(b) We’ll first try to describe the set S. With some cleverness, you can notice that f(x,−x) = 0 forall x ∈ R. This should suggest that we can factor f as

f(x, y) = (x + y)P (x, y),

where P is a polynomial. P turns out to be

P (x, y) = 2x2 − 3x− 2xy + 3y + 2y2.

The set of zeroes of P is an ellipse; so that S is the union of this ellipse and the line y=−x. (Try plottingthis in Mathematica or some other such program.)

Now by the Implicit Function Theorem, the only candidates for points that have no neighborhoodsin which f(x, y) = 0 can be solved for y in terms of x (or vice versa) will be points at which D1f = 0and D2f = 0, i.e. ∇f = 0. (This is a NOT(p OR q) ⇐⇒ (NOT p) AND (NOT q) thing). Using part(a), it’s easy to see that the only points of S at which ∇f = 0 are (0, 0) and (1,−1). These both lie onthe line y =−x and the ellipse P (x, y) = 0, so the curves “cross” at these points. If you examine a plotof the points in S, it’s pretty clear that it is neither a graph of y as a function of x nor a graph of x asa function of y. So these are the points we’re looking for. As an example of how to be more rigorousabout this, we’ll consider the point (0, 0). Pick a positive ε ¿ 1. Then for any x0 ∈ (−ε, ε), the point(x0,−x0) satisfies f(x, y) = 0. But there is also a y0 s.t. P (x0, y0) = 0 ⇒ f(x0, y0) = 0. It’s pretty easy

2

to see that this y0 must have the same sign as x0, so y0 6= −x0. So, in short, we can’t solve for y interms of x in some small neighborhood of the (0, 0). An analogous argument shows we can’t solve forx in terms of y, and the point (1,−1) is treated similarly. (I realize this is probably confusing; it reallyhelps to draw a picture).

Problem 24:Compute the Jacobian matrix of f :

[f ′(x, y)] =1

(x2 + y2)2

[4xy2 −4x2y

y3 − x2y x3 − y2x]

.

Noticing that the first column is −xy times the second, we see that f′(x, y) has rank 1. The range of f is

an ellipse, as you can see from the relation

f21 + 4f22 = 1.

Problem 25:(a) By the hint, SA is a projection in Rn. The zj used to define S are linearly independent (VER-

IFY!), so S is injective. Then it’s immediate that N (SA) = N (A). Finally, SAzj = zj so thatR(SA) = R(S), since clearly R(SA) ⊆ R(S).

(b) This is immediate from the Rank-Nullity Theorem, but we can do the problem as Rudin suggestswithout using this result. We’ll use the following result from the proof of 9.31 (b) in Rudin:

P a projection in Rn =⇒ dimN (P ) + dimR(P ) = n.Since we noted in part (a) that the zj are linearly independent, we know that rank(S) = r = rank(A).The desired result now follows immediately from part (a).

Problem 28:To show ϕ is continuous, it suffices to consider points (x, t) with t ≥ 0 since ϕ is odd in t. The

following is just a sketch of the continuity argument; I expect a bit more detail on the actual homeworks.First consider a point (x0, t0) with 0 < x0 <

√t0. Argue that you can find an ε > 0 such that 0 < x <

√t

for all (x, y) ∈ Bε((x0, y0)). Then ϕ(x, t) = x on this ball, and the continuity follows immediately. Treatthe cases

√t0 < x < 2

√t0 and 2

√t0 < x similarly. Now consider a point of the form (

√t0, t0) (i.e.

x =√

t0). Since the functions ϕ1(x, t) = x and ϕ2(x, t) = −x + 2√

t are both continuous, choose δ > 0so that both |ϕ1(x, t) − ϕ1(

√t0, t0)| < ε and |ϕ2(x, t) − ϕ2(

√t0, t0)| < ε for a given ε > 0. This δ will

“work” for ϕ in the usual sense. Treat the other “boundary” cases similarly.Showing that (D2ϕ)(x, 0) = 0 for all x is routine; just use the definition of partial derivatives.Now |t| < 14 ⇒ 2

√|t| < 1, so we have

f(t) =∫ 1−1

ϕ(x, t) dx =∫ 1

0

ϕ(x, t) dx = sgn(t)[ ∫ √|t|

0

x dx +∫ 2√|t|√|t|

(−x + 2√|t|) dx

]= t,

where

sgn(t) =

−1, t < 0

0, t = 01, t > 0.

3

So this implies f ′(t) = 1 for |t| < 14 , and hence

1 = f ′(0) 6=∫ 1−1

(D2ϕ)(x, 0) dx = 0.

This suggests you should take care when differentiating under the integral sign.

4

PRINCIPLES OF MATHEMATICAL ANALYSIS.

WALTER RUDIN

Disclaimer: these solutions were typed at warp speed, often with little orno preparation. And very little proofreading. Consequently, there are bound to beloads of mistakes. Consider it part of the challenge of the course to find these errors.

(And when you do find some, please let me know so I can fix them!)

7. Sequences and Series of Functions

1. Prove that every uniformly convergent sequence of bounded functions is uni-formly bounded.{fn} is uniformly Cauchy, so |fn(x)−fm(x)| < 1, for n,m bigger than some

large N . Then∣∣∣|fn(x)| − |fN(x)|

∣∣∣ ≤ |fn(x)− fN(x)| < 1, so|fn(x)| < |fN(x)|+ 1.

Let M be a bound for fN . Then

|fn(x)| < M + 1,so {fn}∞n=N is uniformly bounded by M + 1. Define

K := max{sup |f1|, sup |f2|, . . . , sup |fN−1|,M + 1}.Then {fn}∞n=1 is uniformly bounded by K.

2. Show that {fn}, {gn} converge uniformly on E implies {fn + gn} convergesuniformly on E. If, in addition, {fn}, {gn} are sequences of bounded functions,prove that {fngn} converges uniformly.

supx∈E

{|(fn + gn)(x)− (f + g)(x)|} = supx∈E

{|fn(x)− f(x) + gn(x)− g(x)|}≤ sup

x∈E{|fn(x)− f(x)|+ |gn(x)− g(x)|}

≤ supx∈E

|fn(x)− f(x)|+ supx∈E

|gn(x)− g(x)|n→∞−−−−−→ 0 + 0.

When {fn}, {gn} are bounded sequences,supx∈E

{|(fngn)(x)− (fg)(x)|}

March 23, 2006. Solutions by Erin P. J. Pearse.

1

Principles of Mathematical Analysis — Walter Rudin

= supx∈E

{|fngn(x)− fgn(x) + fgn(x)− fg(x)|}≤ sup

x∈E{|fn(x)− f(x)||gn(x)|+ |f(x)||gn(x)− g(x)|}

≤ supx∈E

|fn(x)− f(x)||gn(x)|+ supx∈E

|f(x)||gn(x)− g(x)|≤ sup

x∈E|fn(x)− f(x)| ·Mg + sup

x∈EMf · |gn(x)− g(x)|.

We can introduce the uniform bounds Mf and Mg by problem 1 and theadditional hypothesis. Then it is clear that the last line goes to 0 as n →∞.

3. Construct sequences {fn}, {gn} which converge uniformly on a set E, but suchthat {fngn} does not converge uniformly.

Work on I = (0, 1). Let fn(x) =1x

+ xn

and gn(x) = − 11−x − 1−xn so thatgn is the horizontal reflection of fn, translated 1 to the right. (Graph them tosee it.)

fn(x) =1x

+ xn

n→∞−−−−−→ 1x

= f(x) sup0≤x≤1

|fn − f | = supx∈I

xn

= 1n→ 0

gn(x) = − 11−x − 1−xnn→∞−−−−−→ − 1

1−x = g(x) sup0≤x≤1

|gn − g| = supx∈I

1−xn

= 1n→ 0

fngn(x) = −(n + (x− 1)2)(n + x2)

n2x(x− 1)n→∞−−−−−→ 1

x− x2 = fg(x)

sup0≤x≤1

|fngn − fg| = supx∈I

∣∣∣∣x2(x− 1)2 + n + 2nx(x− 1)

n2x(x− 1)

∣∣∣∣ = ∞, for any n.To see the sup is infinite, check x = 0, 1.

4. Consider the sum

f(x) =∞∑

n=1

1

1 + n2x.

For what x does the series converge absolutely? On what intervals does itconverge uniformly? On what intervals does it fail to converge uniformly? Isf bounded?

For xk = − 1k2 , we get

f(xk) =∞∑

n=1

1

1− (nk

)2 .

The kth term of the sum is undefined, so f is undefined at xk = − 1k2 , k =1, 2, . . .

To be completed.

2

Solutions by Erin P. J. Pearse

5. Define a sequence of functions by

fn(x) =

0(x < 1

n+1

),

sin2 πx

(1

n+1< x ≤ 1

n

),

0(

1n

< x).

Show that the series {fn} converges to a continuous function, but not uni-formly. Use the series

∑fn to show that absolute convergence, even for all x,

does not imply uniform convergence.For x ≤ 0, fn(x) = 0 for every n, so lim fn(x) = 0. For x > 0,

n > N :=[

1x

]=⇒ 1

n< x =⇒ fn(x) = 0.

Thus fn(x)pw−−−→ f(x) := 0 for x ∈ R.

To see that the convergence is not uniform, consider

f ′n(x) = 2(sin π

x

) (cos π

x

) (− πx2

).

f ′n(x) = 0 when• sin π

x= 0, in which case x = 1

kfor some k ∈ Z, or

• cos πx

= 0, in which case x = 22k+1

for some k ∈ Z.For each fn, only a few of these values occur where fn is not defined to be 0,so checking these values of x,

fn(

1n

)= sin(nπ) = 0

fn(

1n+1

)= sin((n + 1)π) = 0

fn(

22n+1

)= sin

(2n+1

2π)

= 1.

So Mn = supx{|fn(x)− f(x)|} = 1 for each n, and clearly Mn → 1 6= 0.The series

∑∞n=1 fn(x) converges absolutely for all x ∈ R: for any fixed x,

there is only one nonzero term in the sum.The series

∑∞n=1 fn(x) does not converge uniformly: check partial sums.

supx

∣∣∣∣∣N∑

n=1

fn(x)−N+1∑n=1

fn(x)

∣∣∣∣∣ = supx |fN+1(x)| = 1,

so the sequence of partial sums is not Cauchy (in the topology of uniformconvergence), hence cannot converge.

6. Prove that the series ∞∑n=1

(−1)n x2 + n

n2

converges uniformly in every bounded interval, but does not converge abso-lutely for any value of x.

To see uniform convergence on a bounded interval,

supa


={∣∣∣

∑∞n=N+1

(−1)n c2+nn2

∣∣∣}

(7.1)

where c := max{|a|, |b|}. We will use the alternating series test to show∑∞n=1(−1)n c

2+nn2

converges. From this, it will follow that (7.1) goes to 0 asN →∞.(i) For all x ∈ R, x2+n

n2> 0. So the sum alternates.

(ii) For fixed x, lim x2+nn2

= lim 12n

= 0. (L’Hôp)(iii) To check monotonicity, prove

x2 + n + 1

(n + 1)2≤ x

2 + n

n2

by cross-multiplying.

7. For n = 1, 2, 3, . . . , and x ∈ R, put fn(x) = x1+nx2 . Show that {fn} convergesuniformly to a function f , and that the equation f ′(x) = limn→∞ f ′n(x) iscorrect if x 6= 0 but false if x = 0.

8. If

I(x) =

{0 (x ≤ 0),1 (x > 0),

if {xn} is a sequence of distinct points in (a, b), and if∑ |cn| converges, then

prove that the series

f(x) =∞∑

n=1

cnI(x− xn) (a ≤ x ≤ b)

converges uniformly, and that f is continuous for every x 6= xn.

4


9. {fn} are continuous and fn unif−−−→ f on E. Show lim fn(xn) = f(x) for everysequence {xn} ⊆ E with xn → x.

Pick N1 such that

n ≥ N1 =⇒ supx∈E

|fn(x)− f(x)| < ε/2.

Then surely |fn(xn) − f(xn)| < ε/2 holds for each n ≥ N1. By Thm. 7.12, fis continuous, so pick N2 such that

n ≥ N2 =⇒ |f(xn)− f(x)| < ε/2.Then we are done because

|fn(xn)− f(x)| ≤ |fn(xn)− f(xn)|+ |f(xn)− f(x)|.

11. {fn}, {gn} are defined on E with:(a) {∑Nn=1 fn} uniformly bounded, (b) gn

unif−−−→ g on E, and (c) gn(x) ≤gn−1(x) ∀x ∈ E, ∀n. Prove

∑fngn converges uniformly on E.

Define a := supx∈E |fn(x)| and bn := supx∈E |gn(x)|. Then

supx∈E

∣∣∣∣∣∞∑

N+1

fn(x)gn(x)

∣∣∣∣∣ ≤∞∑

N+1

supx∈E

|fn(x)gn(x)| ≤∞∑

N+1

anbn → 0,

by Thm. 3.42.

13. {fn} is monotonically increasing on R, and 0 ≤ fn(x) ≤ 1(a) Show ∃f, {nk} such that f(x) = limk→∞ fnk(x),∀x ∈ R.

By Thm. 7.23, we can find a subsequence {fni} such that {fni(r)} con-verges for every rational r. Thus we may define f(x) := supr≤x f(r),where the supremum is taken over r ∈ Q. It is clear that f is monotone,because

x < y =⇒ {r ≤ x} ⊆ {r ≤ y}and the supremum can only increase on a larger set. Thus, f has at mosta countable set of discontinuities, by Thm. 4.30, pick x such that f iscontinuous at x.We want to show fni(x) → f(x). Fix ε > 0. Since f is continuous at x,choose δ such that |x − y| < δ =⇒ |f(x) − f(y)| < ε/3. Now pick arational number r ∈ [x− δ

3, x]. Then

|fni(x)− f(x)| ≤ |fni(x)− fni(r)|+ |fni(r)− f(r)|+ |f(r)− f(x)|. (7.2)On the RHS of (7.2): the last term is less than ε/3 by the choice of δ;and the middle term is less than ε/3 whenever i ≥ N1 for some large N1,because the subsequence converges on the rationals. It remains to showthe first term gets small.

5


Pick some rational s ∈ [x, x + δ/3]. Then r ≤ x ≤ s and the continuityof f at x shows

|r − s| < δ =⇒ |f(r)− f(s)| < ε/3.Also, since the fni are monotone,

fni(r) ≤ fni(x) ≤ fni(s). (7.3)With

|fni(r)− fni(s)| ≤ |fni(r)− f(r)|+ |f(r)− f(s)|+ |f(s)− fni(s)|,some large N2, i ≥ N2 gives |fni(r) − fni(s)| < ε. By (7.3), this shows|fni(r)− fni(x)| < ε.

(b) If f is continuous, show fnk → f uniformly on compact sets.Let K be compact. Fix ε > 0. Since f is uniformly continuous onK, pick δ such that |x − y| < δ =⇒ |f(x) − f(y)| < ε/3. Since Kis compact, we can find {x1, . . . xJ} such that K ⊆

⋃Jj=1 Bδ(xj), where

Bδ(xj) := (xj − δ, xj + δ).

16. {fn} is equicontinuous on a compact set K, fn pw−−−→ f on K. Prove {fn}converges uniformly on K.

Define f by f(x) := lim fn(x). Fix ε. From equicontinuity, find δ such that

|x− y| < δ =⇒ |fn(x)− fn(y)| < ε/3, ∀n, x, y.Letting n →∞, this gives

|x− y| < δ =⇒ |f(x)− f(y)| < ε/3, ∀x, y.Since K is compact, we can choose a finite set {x1, . . . , xJ} such that

K ⊆⋃J

j=1Bδ(xj)

where Bδ(xj) := (xj − δ, xj + δ). For each xj, we know fn(xj) → f(xj), sopick N big enough that

n ≥ N =⇒ |fn(xj)− f(xj)| < ε/3, ∀j = 1, . . . , J.For any x ∈ K, x ∈ Bδ(xj) for some j. Thus for all x ∈ K,|fn(x)− f(x)| ≤ |fn(x)− fn(xj)|+ |fn(xj)− f(xj)|+ |f(xj)− f(x)|.

6


18. Let {fn} be uniformly bounded and Fn(x) :=∫ x

af(t) dt for x ∈ [a, b]. Prove

∃{Fnk} which converges uniformly on [a, b]We need to show {Fn} is equicontinuous. Then by Thm. 6.20, each Fn is

continuous; and by Thm. 7.25(b), we’re done. So fix ε > 0, let x < y, and letM be the uniform bound on the {fn}.

|Fn(x)− Fn(y)| =∣∣∣∣∫ y

x

fn(t) dt

∣∣∣∣ ≤∫ y

x

|fn(t)| dt ≤ M(y − x)

Then pick any δ < ε/M and

|x− y| < δ =⇒ |Fn(x)− Fn(y)| ∀n,so {Fn} is equicontinuous.

7


20. f is continuous on [0, 1] and∫ 10

f(x)xn dx = 0, n = 0, 1, 2, . . . . Prove thatf(x) ≡ 0.

Let g be any polynomial. Then g(x) = a0 + a1x + a2x2 + · · · + aKxK . By

linearity of the integral,

∫ 10

f(x)g(x) dx =K∑

k=0

ak

∫ 10

f(x)xk dx =K∑

k=0

0 = 0.

By the Weierstrass theorem, let {fn}∞n=1 be a sequence of polynomials whichconverge uniformly to f on [0, 1]. Then

∫ 10

f 2(x) dx = limn→∞

∫ 10

f(x)fn(x) dx = limn→∞

0 = 0.

Then by Chap. 6, Exercise 2, f 2(x) ≡ 0. Thus f(x) ≡ 0.

21. Let K be the unit circle in C and define

A :={

f(eiθ) =N∑

n=0

cneinθ ... cn ∈ C, θ ∈ R

}.

To see that A separates points and vanishes at no point, note that A con-tains the identity function f(eiθ) = eiθ.

To see that there are continuous functions on K that are not in the uniformclosure of A, note that

∫ 2π0

f(eiθ)eiθ dθ = 0 ∀f ∈ A, (7.4)

and hence for g = lim gn (uniform limit) with gn ∈ A,∫ 2π

0

g(eiθ)eiθ dθ = limn→∞

∫ 2π0

gn(eiθ)eiθ dθ = 0,

by Thm. 7.16. Thus, all functions in the closure of A satisfy (7.4). However,if we choose an h which is not in A, like

h(eiθ) = e−iθ,

then h is clearly continuous on K, and

∫ 2π0

h(eiθ)eiθ dθ =

∫ 2π0

1 dθ = 2π.

Thus h is not in the uniform closure of A.

8


22. Assume f ∈ R(α) on [a, b] and prove that there are polynomials Pn such that

limn→∞

∫ ba

|f − Pn|2dα = 0.

We need to find {Pn} such that ‖f−Pn‖2 n→∞−−−−−→ 0. Fix ε > 0. By Chap. 6,Exercise 12, we can find g ∈ C[a, b] such that ‖f − g‖2 < ε/2. Note that

∫ ba

|g − P |2 ≤∫ b

a

sup |g − P |2 = sup |g − P |2(b− a).

Then by the Weierstrass theorem, we can find a polynomial P such that

‖g − P‖2 ≤ sup |g − P |(b− a) < ε/2.By Chap. 6, Exercise 11, this gives ‖f − P‖2 < ε.

23. Put P0 = 0 and define Pn+1(x) := Pn(x) + (x2 − P 2n(x)) /2 for n = 0, 1, 2, . . . .

Prove that limn→∞ Pn(x) = |x| uniformly on [−1, 1].Note that if Pn is even, the definition will force Pn+1 to be even, also. Now

P1 =x2

2, so assume 0 ≤ Pn−1 ≤ 1 for |x| ≤ 1. Then

Pn = Pn−1 +x2

2− P

2n−12

=x2

2+ Pn−1

(1− Pn−1

2

).

By elementary calculus,

0 ≤ y ≤ 1 =⇒ f(y) = y(1− y2) takes values in [0, 1

2]. (7.5)

Since |x| ≤ 1 also implies x22∈ [0, 1], this gives 0 ≤ Pn ≤ 1. Then

0 ≤ |x|+ Pn(x)2

≤ 1

0 ≤ 1− |x|+ Pn(x)2

≤ 1. (7.6)To see Pn(x) ≤ |x|, consider that for x ≥ 0, the inequality

x− P1(x) = x− x2/2 ≥ 0holds in virtue of the positivity of f in (7.5). Then by the symmetry of evenfunctions, this is true for |x| ≤ 1. Now suppose Pn−1 ≤ |x|, i.e., |x|−Pn−1(x) ≥0. Then the given identity, and (7.6), give

|x| − Pn = (|x| − Pn−1)(1− 1

2(|x|+ Pn−1)

) ≥ 0.We have established

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