Spivak’s Calculus On Manifolds: Solutions

Manual

Thomas Hughes

August 2017

Chapter 1

Functions on EuclideanSpace

1.1 Prove that |x| ≤∑ni=1 |xi|

Proof. If {e1, e2, . . . , en} is the usual basis on Rn, then we can write

x = x1e1 + x2e2 + . . .+ xnen

and thus

|x| =

∣∣∣∣∣n∑i=1

xiei

∣∣∣∣∣ ≤n∑i=1

|xiei| =n∑i=1

|xi||ei| =n∑i=1

|xi|

1.2 When does equality hold in Theorem 1-1(3)?

Proof. Notice in the proof that we get

|x+ y|2 =n∑i=1

(xi)2 +

n∑i=1

(yi)2 + 2

n∑i=1

xiyi ≤ |x|2 + |y|2 + 2|x||y|

and so we have equality precisely when∑ni=1 xiyi = |

∑ni=1 xiyi| and x

and y are dependent. That is, when x and y are dependent and sgn(xi) =sgn(yi) for all i. That is, when one is a non-negative multiple of theother.

1.3 Prove that |x− y| ≤ |x|+ |y|. When does equality hold?

Proof.|x− y| = |x+ (−y)| ≤ |x|+ | − y| = |x|+ |y|

Conditions for equality are the same as in 1.2, for x and −y.

1

1.4 Prove that ||x| − |y|| ≤ |x− y|.

Proof. Notice|x| = |x− y + y| ≤ |x− y|+ |y|

Thus|x| − |y| ≤ |x− y|

Similarly,

|y| = |y − x+ x| ≤ |y − x|+ |x| = |x− y|+ |x|

Thus

|y| − |x| ≤ |x− y|−|x− y| ≤ |x| − |y|

So, combining these results yields

−|x− y| ≤ |x| − |y| ≤ |x− y|

which implies||x| − |y|| ≤ |x− y|

as desired.

1.5 The quantity |y − x| is called the distance between x and y. Prove andinterpret geometrically the “triangle inequality”:

|z − x| ≤ |z − y|+ |y − x|

Proof.|z − x| = |z − y + y − x| ≤ |z − y|+ |y − x|

Geometrically, we have

z

y

x

|z−y|

|y − x|

|z − x|

1.6 Let f and g be integrable on [a, b].

(a) Prove that∣∣∣∫ ba f · g∣∣∣ ≤ (∫ ba f2) 12 · (∫ ba g2) 12 .

2

(b) If equality holds, must f = λg for some λ ∈ R? What if f and g arecontinuous?

(c) Show that Theorem 1-1(2) is a special case of (a).

Proof. (a) One way to prove this would be to observe that 1-1(2) impliesthat∣∣∣∣∣

n∑k=1

f(tk)g(tk)∆xk

∣∣∣∣∣ ≤(

n∑k=1

(f(tk))2∆xk

) 12(

n∑k=1

(g(tk))2∆xk

) 12

Notice, by integrability, all the sums can be considered functions of

the tagged partition Ṗ, such that each will approach∫ baf · g,

∫ baf2

and∫ bag2, respectively, when

∣∣∣∣∣∣Ṗ∣∣∣∣∣∣ → 0. Thus, by continuity ofthe square root function and the absolute value, taking the limit as∣∣∣∣∣∣Ṗ∣∣∣∣∣∣→ 0 will give us the desired result.However, following Spivak’s hint, we observe that either there existsλ ∈ R such that

0 =

∫ ba

(f − λg)2

or, since (f − λg)2 is nonnegative, for all λ ∈ R

0 <

∫ ba

(f − λg)2

Notice from Lebesgue’s theorem of Riemann-integrability that in thefirst case we must have that f = λg almost everywhere on [a, b], andtherefore ∫ b

a

f · g = λ∫ ba

g2

and ∫ ba

f2 =

∫ ba

λ2g2

Which would give us that∣∣∣∣∣∫ ba

f · g

∣∣∣∣∣ =∣∣∣∣∣λ∫ ba

g2

∣∣∣∣∣ =√√√√λ2(∫ b

a

g2

)2=

√∫ ba

λ2g2∫ ba

g2 =

√∫ ba

f2∫ ba

g2

which can be rewritten as∣∣∣∣∣∫ ba

f · g

∣∣∣∣∣ =(∫ b

a

f2

) 12(∫ b

a

g2

) 12

3

In the second case, we have that

0 <

∫ ba

(f − λg)2 =∫ ba

f2 − 2λ∫ ba

f · g + λ2∫ ba

g2

On the right, we have a quadratic in λ which has no real roots sincethe inequality holds for all λ ∈ R. Thus

4

(∫ ba

f · g

)2− 4

∫ ba

f2∫ ba

g2 < 0

which implies (∫ ba

f · g

)2<

∫ ba

f2∫ ba

g2

which finally gives us

−

(∫ ba

f2

) 12(∫ b

a

g2

) 12

<

∫ ba

f · g <

(∫ ba

f2

) 12(∫ b

a

g2

) 12

Thus, together, the results from both cases imply that∣∣∣∣∣∫ ba

f · g

∣∣∣∣∣ ≤(∫ b

a

f2

) 12

·

(∫ ba

g2

) 12

as desired.

(b) No, in general, we may have f 6= λg and still have∣∣∣∫ ba f · g∣∣∣ =(∫ b

af2) 1

2(∫ b

ag2) 1

2

. For example, take

f =

{1, x = 1

0, elsewhere

and

g =

{1, x = 0

0, elsewhere

Then∣∣∣∫ 10 f · g∣∣∣ = (∫ 10 f2) 12 (∫ 10 g2) 12 = 0 while f 6= λg. However, if

f and g are continuous, then equality holds if and only if f = λg forsome λ ∈ R. This follows from the theorem which states that if f iscontinuous on [a, b] and f ≥ 0 with f(x0) > 0 for some x0 ∈ [a, b]then

∫ baf > 0. The proof of this can be found in Exercise 7.4.4 from

the solutions manual for Abbott’s Understanding Analysis.

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(c) Divide [a, b] into n sub-intervals so that on the ith sub-interval,[a+ b−an (i− 1), a+

b−an i), we define f to be

f(x) =xi√b−an

and g to be

g(x) =yi√b−an

Then ∫ ba

f · g =n∑i=1

xiyi∫ ba

f2 =

n∑i=1

x2i∫ ba

g2 =

n∑i=1

y2i

which, by (a), gives us the desired inequality.

5

Table 1.1: A poorly drawn graph of f

6

1.7 A linear transformation T : Rn → Rn is norm preserving if |T (x)| = |x|,and inner product preserving if 〈Tx, Ty〉 = 〈x, y〉.

(a) Prove that T is norm preserving if and only if T is inner productpreserving.

(b) Prove that such a linear transformation T is injective and T−1 is ofthe same sort.

Proof. (a) Suppose that T is norm preserving. Then, by the polarizationidentity, we have

〈x, y〉 = |x+ y|2 − |x− y|2

4

and since T is norm preserving, we have

=|T (x+ y)|2 − |T (x− y)|2

4

=〈T (x+ y), T (x+ y)〉 − 〈T (x− y), T (x− y)〉

4

and by linearity of T , we have

=〈T (x) + T (y), T (x) + T (y)〉 − 〈T (x)− T (y), T (x)− T (y)〉

4

and by bilinearity of the inner product, we have

= 〈Tx, Ty〉

Now, suppose that T is inner product preserving. Then

|T (x)| =√〈T (x), T (x)〉 =

√〈x, x〉 = |x|

(b) Suppose that T is norm preserving (and therefore, also inner productpreserving) and that

T (x) = T (y)

then

T (x)− T (y) = 0T (x− y) = 0|T (x− y)| = 0|x− y| = 0

which implies that

x− y = 0x = y

7

Therefore, T is injective. Since injectivity of a linear transformationis equivalent to surjectivity, it follows that T−1 exists. To show thatT−1 is of the same sort, we show that it is norm preserving. Observethat ∣∣T−1(y)∣∣ = |x| = |T (x)| = |y|

1.8 If x, y ∈ Rn are non-zero, the angle between x and y, denoted ∠(x, y),is defined as arccos

(〈x,y〉|x|·|y|

), which makes sense by Theorem 1-1(2). The

linear transformation T is angle preserving if T is injective, and forx, y 6= 0 we have ∠(Tx, Ty) = ∠(x, y).

(a) Prove that if T is norm preserving, then T is angle preserving.

(b) If there is a basis x1, . . . , xn of Rn and numbers λ1, . . . , λn such thatTxi = λixi, prove that T is angle preserving if and only if all |λi| areequal.

(c) What are all angle preserving T : Rn → Rn?

Proof. (a) If T is norm preserving, then it is also inner product preservingand injective. Thus, if x, y 6= 0,

∠(Tx, Ty) = arccos

(〈Tx, Ty〉|Tx| · |Ty|

)= arccos

(〈x, y〉|x| · |y|

)= ∠(x, y)

so that T is also angle preserving.

(b) This is not true. Take ((−1,−1), (1, 0)) = (x1, x2) as a basis for R2.If

T (x1) = 3x1

T (x2) = −3x2

Then T is injective and has that |λi| are all equal. However,

∠(x1, x2) =3π

4

while

∠(Tx1, Tx2) =π

4

1.9 If 0 ≤ θ < π, let T : R2 → R2 have the matrix[cos θ sin θ− sin θ cos θ

]Show that T is angle preserving and if x 6= 0, then ∠(x, Tx) = θ.

8

Proof. Observe that[cos θ sin θ− sin θ cos θ

] [x1x2

]=

[x1 cos θ + x2 sin θ−x1 sin θ + x2 cos θ

]and [

cos θ sin θ− sin θ cos θ

] [y1y2

]=

[y1 cos θ + y2 sin θ−y1 sin θ + y2 cos θ

]Now, some rather tedious calculations show that

〈Tx, Ty〉 = x1y1 + x2y2 = 〈x, y〉

and|Tx| = |x|

and|Ty| = |y|

which, together, imply that ∠(Tx, Ty) = ∠(x, y) so that T is angle pre-serving. Again, a rather tedious calculation shows that

〈x, Tx〉 = |x|2 cos θ

which implies that, for x 6= 0

∠(x, Tx) = arccos

(〈x, Tx〉|x| · |Tx|

)= arccos

(|x|2 cos θ|x|2

)= arccos(cos θ)

= θ

1.10∗ If T : Rm → Rn is a linear transformation, show that there is a numberM such that |T (h)| ≤M |h| for h ∈ Rm.

Lemma. Let T ∈ L(Rm,Rn). Then there exists M > 0 such that for allx admitting |x| = 1, we have

|T (x)| ≤M

Proof of Lemma. Let |x| = 1. Given the standard basis, {e1, . . . , em}, onRm we may write

x = α1e1 + . . .+ αmem

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where |αi| ≤ 1 for all 1 ≤ i ≤ m. Thus

|T (x)| = |T (α1e1 + . . .+ αmem)|= |α1T (e1) + . . .+ αmT (em)|≤ |α1| |T (e1)|+ . . .+ |αm| |T (em)|≤ |T (e1)|+ . . .+ |T (em)|= M

Proof of 1.10. From the Lemma, we get that there exists M > 0 suchthat, if h 6= 0, then ∣∣∣∣T ( h|h|

)∣∣∣∣ ≤M∣∣∣∣ 1|h|T (h)∣∣∣∣ ≤M

|T (h)| ≤M |h|

1.11 If x, y ∈ Rn and z, w ∈ Rm, show that 〈(x, z), (y, w)〉 = 〈x, y〉+ 〈z, w〉 and|(x, z)| =

√|x|2 + |z|2. Note that (x, z) and (y, w) denote points in Rn+m.

Proof. Observe(x, z) = (x1, . . . , xn, z1, . . . , zm)

and(y, w) = (y1, . . . , yn, w1, . . . , wm)

and thus

〈(x, z), (y, w)〉 =n∑i=1

xiyi +

m∑i=1

ziwi = 〈x, y〉+ 〈z, w〉

and

|(x, z)| =

√√√√ n∑i=1

x2i +

m∑i=1

z2i =

√|x|2 + |z|2

1.12∗ Let (Rn)∗ denote the dual space of the vector space Rn. If x ∈ Rn, defineφx ∈ (Rn)∗ by φx(y) = 〈x, y〉. Define T : Rn → (Rn)∗ by T (x) = φx.Show that T is an injective linear transformation and conclude that everyφ ∈ (Rn)∗ is φx for a unique x ∈ Rn.

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Proof. The linearity of the inner product implies the linearity of T . Toshow that T is injective, by linearity of T , it is sufficient to show thatker(T ) = {0}. To that end, suppose that

T (x) = 0

then

φx = 0

so that

φx(x) = 0

〈x, x〉 = 0

if and only if

x = 0

Thus ker(T ) = {0} and T is injective. Moreover, since dim(Rn)∗ =dimRN , injectivity of T implies surjectivity. Thus T is a bijection be-tween Rn and (Rn)∗. Thus each φ ∈ (Rn)∗ corresponds to a φx for someunique x ∈ Rn.

1.13∗ If x, y ∈ Rn, then x and y are called perpendicular (or orthogonal) if〈x, y〉 = 0. If x and y are perpendicular, prove that

|x+ y|2 = |x|2 + |y|2

Proof.

|x+ y|2 = 〈x+y, x+y〉 = 〈x, x〉+2〈x, y〉+〈y, y〉 = 〈x, x〉+〈y, y〉 = |x|2+|y|2

1.14∗ Prove that the union of any (even infinite) number of open sets is open.Prove that the intersection of two (and hence infinitely many) open setsis open. Give a counterexample for infinitely many open sets.

Proof. Let x ∈ ∪λOλ with each Oλ open. Then there exists λ such thatx ∈ Oλ. Since Oλ is open, there exists an open rectangle, A, such thatx ∈ A ⊂ Oλ ⊂ ∪λOλ. Thus the arbitrary union of open sets is open. Asimple but tedious argument could be made to show that the intersectionof two open rectangles is, again, an open rectangle. Then the intersectionof two open sets being, again open is a simple corollary. Notice that

∞⋂n=1

(− 1n,

1

n

)= {0}

Clearly, {0} does not contain any open rectangles, and is therefore notopen.

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1.15 Prove that {x ∈ Rn : |x− a| < r} is open.

Lemma. There exists an open rectangle, U , such that U ⊂ B0(r) = {x ∈Rn : |x| < r}

Proof. Let U = Πni=1(− �√n ,r√n

). Then

|x|2 ≤n∑i=1

|xi|2 <n∑i=1

(r√n

)2= r2

implying |x| < r. Therefore U ⊂ B0(r).

Corollary. There exists an open rectangle, U , such that U ⊂ Ba(r) ={x ∈ Rn : |x− a| < r} is open.

Proof. By the transformation T (x) = x−a and the open rectangle U fromthe Lemma.

Corollary. Ba(r) is open.

Proof. Let x ∈ Ba(r). Then, from the Corollary, we get that there existsan open rectangle, U , such that

U ⊂ Bx (r − |x− a|) ⊂ Ba(r)

Therefore, Ba(r) is open.

It is not to hard to apply the same idea backwards: every open rectanglecontains an open ball. Thus a set, A, is open if and only if for each x ∈ Athere exists Bx(r) such that Bx(r) ⊂ A. This characterization of opensets is often easier to work with than the characterization provided in thebook. We will consider this as given for future exercises.

1.16 Find the interior, exterior, and boundary of the sets.

{x ∈ Rn : |x| ≤ 1}{x ∈ Rn : |x| = 1}

{x ∈ Rn : each xi is rational.}

Proof. For the first, we have that the interior is {x ∈ Rn : |x| < 1}, theexterior is {x ∈ Rn : |x| > 1}, and the boundary is {x ∈ R : |x| = 1}.For the second, we have that the interior is empty (every open rectanglein Rn will contain some points outside this set), the exterior is {x ∈ Rn :|x| 6= 1}, and the boundary is the set itself.For the third, we have that both the interior and exterior are empty, andthe boundary is again the set itself.

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1.17 Construct a set A ⊂ [0, 1]× [0, 1] such that A contains at most one pointon each horizontal and each vertical line but boundary A = [0, 1]× [0, 1].

Proof. Let A be the set of all points with rational coordinates in (0, 1)×(0, 1). Then clearly boundary(A) ⊂ [0, 1]× [0, 1]. If x ∈ [0, 1]× [0, 1] thenx = (x1, x2) where 0 ≤ x1, x2 ≤ 1. Thus, if B is an open set containingx, it follows that there exist 0 ≤ �1, �2 such that [x1 − �1, x1 + �1]× [x2 −�2, x2 + �2] ⊂ B ∩ [0, 1]× [0, 1]. Density of Q in R and the existence of anirrational between any two reals implies that [0, 1]× [0, 1] ⊂ boundary(A).Thus boundary(A) = [0, 1]× [0, 1].

1.18 If A ⊂ [0, 1] is the union of open intervals (ai, bi) such that each rationalnumber in (0, 1) is contained in some (ai, bi), show that boundary(A) =[0, 1] \A.

Proof. Notice, by openness of A, x ∈ boundary(A) implies x 6∈ A. Fur-thermore, by openness of [0, 1]c, x ∈ boundary(A) implies x 6∈ [0, 1]c. Ifx ∈ [0, 1] \A and if B is open with x ∈ B, then B ∩Ac 6= ∅. Furthermore,by density of Q in R we also get that there exists a rational r ∈ B ∩A, sothat B ∩A 6= ∅. Thus boundary(A) = [0, 1] \A.

1.19∗ If A is a closed set that contains every rational number r ∈ [0, 1], showthat [0, 1] ⊂ A

Proof. Suppose, to the contrary, that there exists x ∈ [0, 1] ∩ Ac. SinceA is closed, it follows that Ac is open and that therefore, there exists anopen B such that x ∈ B ⊂ Ac. But, by density of Q in R, we knowthat there exists a rational r ∈ A ∩ B contradicting with B ⊂ Ac. Thus[0, 1] ⊂ A.

1.20 Show that a compact set in Rn is closed and bounded.

Proof. If K ⊂ Rn is compact then clearly it is bounded, for if it wereunbounded, then {Πn1 (−k, k) : k ∈ N} would be an open cover of K whichadmits no finite subcover. To show that K is closed, let x ∈ Kc andconsider the open cover of K given by{

Bk

(|x− k|

2

): k ∈ K

}By compactness, there exists a finite subcover{

Bki

(|x− ki|

2

): 1 ≤ i ≤ n

}Thus Bx

(mini|x−ki|

2

)⊂ Kc. Therefore, Kc is open.

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1.21∗ (a) If A is closed and x 6∈ A, prove that there is a number d > 0 suchthat |y − x| ≥ d for all y ∈ A.

(b) If A is closed, B is compact, and A∩B = ∅, prove that there is d > 0such that |y − x| ≥ d for all y ∈ A and x ∈ B.

(c) Give a counterexample in R2 if A and B are closed but neither iscompact.

Proof. (a) Let A be closed and x ∈ Ac. Since A is closed, there exists anopen B ⊂ Ac with x ∈ B. So there exists an open ball Bx(r) ⊂ Ac.Thus, if y ∈ A, then y ∈ (Bx(r))c. Thus

|y − x| ≥ r

(b) Let A be closed and B be compact with A∩B = ∅. Since A is closedand A and B are disjoint, by (a), we know that for every x ∈ B thereexists dx > 0 such that for all y ∈ A we have |y − x| ≥ dx. So{

Bx

(dx2

): x ∈ B

}forms an open cover of B. By compactness of B, there exists a finitesubcover {

Bx1

(dx12

), . . . , Bxm

(dxm

2

)}So for all y ∈ A and 1 ≤ i ≤ m

|y − xi| ≤ |y − x|+ |x− xi|

Thus, if x ∈ B then x ∈ Bxi(dxi2

)for some i, and the above inequal-

ity implies

min1≤i≤m

dxi2≤ dxi

2= dxi −

dxi2≤ |y − xi| − |x− xi| ≤ |y − x|

(c) If

A = {(0, n) : n ∈ N}

and

B =

{(0, n− 1

n) : n ∈ N

}then it is clear that both A and B are closed, but not bounded andtherefore not compact. Furthermore, A∩B = ∅ and for all n ∈ N wecan find y ∈ A and an x ∈ B such that

|y − x| < 1n

14

1.22∗ If U is open and C ⊂ U is compact, show that there is a compact set Dsuch that C ⊂ interior(D) and D ⊂ U .

Proof. Let U be open and C ⊂ U be compact. It follows that U c is closedand that U c ∩ C = ∅. By the previous exercise, we know then that thereexists a d > 0 such that

|x− y| ≥ d

for all x ∈ C and y ∈ U c. We observe that since

{Bx(d/2) : x ∈ C}

forms an open cover of C, there is a finite subcover {Bx1 , . . . , Bxm}. So ifwe denote

Bx(d/2) =

{z : |x− z| ≤ d

2

}then

m⋃i=1

Bxi(d/2) = D

is compact andC ⊂ D ⊂ U

If x ∈ C then it follows x ∈ Bx(d/2) ⊂ D which implies C ⊂ Int(D)

1.23 If f : A → Rm and a ∈ A, show that limx→a f(x) = b if and only iflimx→a fi(x) = bi for 1 ≤ i ≤ m.

Proof. We first show sufficiency by contrapositive. Suppose that there is1 ≤ k ≤ m such that

limx→a

fk(x) 6= bk

Then there is a sequence (xr) ⊂ A \ {a} and � > 0 such that xr → a andfor all r ∈ N we have

|fk(xr)− bk| ≥ �

which, in turn, implies

(fk(xr)− bk)2 ≥ �2

Thus, for the same � we have that for every δ > 0 there is r ∈ N such that0 < |xr − a| < δ while

|f(xr)− b|2 =m∑i=1

(fk(xr)− bk)2 ≥m∑i=1i 6=k

(fk(xr)− bk)2 + �2 ≥ �2

which implies that for all δ > 0 there exists r ∈ N such that 0 < |xr − a| <δ while

|f(xr)− b| ≥ �

15

so that limx→a f(x) 6= b. To demonstrate necessity, let

limx→a

fi(x) = bi

for every 1 ≤ i ≤ m. Given � > 0, we may choose δ > 0 so that 0 <|x− a| < δ implies

|fi(x)− bi| <�

m

for every 1 ≤ i ≤ m. Thus for 0 < |x− a| < δ we have

|f(x)− b| ≤m∑i=1

|fi(x)− bi| ≤m∑i=1

�

m= �

Therefore,limx→a

f(x) = b

1.24 Prove that f : A→ Rm is continuous at a if and only if each fi is.

Proof. Notice f is continuous at a if and only if limx→a f(x) = f(a) if andonly if limx→a fi(x) = fi(a) for every 1 ≤ i ≤ m.

1.25 Prove that a linear transformation T : Rn → Rm is continuous.

Proof. From 1.10 we know that there exists an M > 0 such that

|T (h)| ≤M |h|

So, if � > 0 is given, and h ∈ Rn then whenever 0 < |x− h| < �M we getthat

|T (x)− T (h)| = |T (x− h)| ≤M |x− h| < M �M

= �

Therefore T is continuous.

1.26 Let A = {(x, y) ∈ R2 : x > 0 and 0 < y < x2}

(a) Show that every straight line through (0, 0) contains an intervalaround (0, 0) which is in Ac.

(b) Define f : R2 → R by f(x) = 0 if x 6∈ A and f(x) = 1 if x ∈ A. Forh ∈ R2 define gh : R → R by gh(t) = f(th). Show that each gh iscontinuous at 0, but f is not continuous at (0, 0).

Proof. (a) Let s be a straight line through (0, 0). If s is just the verticalline through x = 0 then s ⊂ Ac. Similarly, if s(x) = mx with m ≤ 0then s ⊂ Ac. If s(x) = mx with m > 0 , then {(x,mx) : x ≤ m} ⊂s ∩Ac.

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(b) First, we demonstrate that f is not continuous at (0, 0). Observethat f((0, 0)) = 0. Notice that the sequence

(1n ,

1n

)→ (0, 0) and

yet f((

1n ,

1n

))= 1 for all n implying that f is not continuous at

(0, 0). Now, if h is of the form (0, h2) then gh(t) = 0 and is thereforecontinuous at 0. If h is not of the form (0, h2) then gh(0) = f(0) = 0.Notice that th = t(h1, h2) is simply the line s(x) =

h2h1x. By (a), we

know that for δ small enough f(th) = 0 whenever |t| < δ. Thereforegh is continuous at 0.

1.27 Prove that Ba(r) is open by considering the function f : Rn → R withf(x) = |x− a|.Lemma. |x| : Rn → R is continuous.

Proof of Lemma. Let � > 0 be given and c ∈ Rn. Then |x− c| < �, alongwith 1.4, imply that

||x| − |c|| ≤ |x− c| < �

Corollary. f(x) = |x− a| is continuous.

Proof of 1.27. NoticeBa(r) = f−1 ({f(x) ∈ R : f(x) < r}). Clearly {f(x) ∈

R : 0 < f(x) < r} is open, so by continuity of f and definition of the norm,we have that Ba(r) \ {a} is open. Furthermore, since a ∈ Ba(r) ⊂ Ba(r)it follows Ba(r) is open.

1.28 If A ⊂ Rn is not closed, show that there is a continuous function f : A→ Rwhich is unbounded.

Proof. If A is not closed then Ac is not open and so there exists a sequence(am) ⊂ A such that am → x ∈ Ac. Thus f : A→ R defined by

f(a) =1

|a− x|

is continuous and unbounded.

1.29 If A is compact, prove that every continuous function f : A→ R takes ona maximum and minimum value.

Proof. From compactness of A and continuity of f , it follows f(A) iscompact. Thus f(A) is closed and bounded. Thus inff(A), supf(A) ∈f(A) implying f takes on a maximum and minimum value.

1.30 Let f : [a, b] → R be an increasing function. If x1, x2, . . . , xn ∈ [a, b] aredistinct, show that

∑ni=1 o(f, xi) < f(b)− f(a).

17

Proof. Let x1 < x2 < . . . < xn ∈ [a, b] and letδ = min{|xi−xj |:i 6=j,1≤i,j≤n}4 .If 1 ≤ i < n then, by our choice of δ, we have

xi + δ < xi + 2δ < xi+1 − δ

Since f is increasing, it follows

M(xi, f, δ) ≤ f(xi + δ) < f(xi + 2δ) < f(xi+1 − δ) ≤ m(xi+1, f, δ)

which establishes that

M(xi, f, δ) < m(xi+1, f, δ)

Furthermore, we observe the following

M(xn, f, δ) ≤ f(b)f(a) ≤ m(x1, f, δ)

and, for all 1 ≤ i ≤ n

o(f, xi) ≤M(xi, f, δ)−m(xi, f, δ)

Thus it follows that

n∑1

o(f, xi) ≤n∑1

M(xi, f, δ)−m(xi, f, δ) < f(b)−m(xn, f, δ)−n−1∑1

m(xi+1, f, δ)−m(xi, f, δ)

which reduces to

n∑1

o(f, xi) < f(b)−m(x1, f, δ) ≤ f(b)− f(a)

which was to be shown.

18

Chapter 2

Differentiation

2.1∗ Prove that if f : Rn → Rm is differentiable at a, then it is continuous ata.

Proof. We shall present two proofs. Let ∆fa(h) = f(a + h) − f(a). Forthe first proof, let � > 0 be given. We observe that differentiability at aimplies that

∆fa(h) = Dfa(h) + r(h)

where limh→0|r(h)||h| = 0. Note that this then implies limh→0 |r(h)| = 0.

So then|∆fa(h)| = |Dfa(h) + r(h)| ≤ |Dfa(h)|+ |r(h)|

From linearity of Dfa and exercise 1.10, we know that there exists λ suchthat, for all h, |Dfa(h)| ≤ λ |h|. Thus by choosing δ small enough, |h| < δimplies

|∆fa(h)| ≤ |Dfa(h)|+ |r(h)| < λ |h|+ |r(h)| < λ�

2λ+�

2= �

which implies thatlimh→0

f(a+ h) = f(a)

so that f is continuous at a.

For the second proof, we observe that given � > 0 we can choose δ > 0 sothat 0 < |h| < δ implies

||∆fa(h)| − |Dfa(h)|| < |∆fa(h)−Dfa(h)| < � |h|

Thus−� |h| < |∆fa(h)| − |Dfa(h)| < � |h|

so that|Dfa(h)| − � |h| < |∆fa(h)| < |Dfa(h)|+ � |h|

19

Linearity of Dfa along with exercise 1.10 implies that, by taking the limitas h → 0 and applying the squeeze theorem, we get that f is continuousat a.

2.2 A function f : R2 → R is independent of the second variable if for eachx ∈ R we have f(x, y1) = f(x, y2) for all y1, y2 ∈ R. Show that f isindependent of the second variable if and only if there is a function g :R→ R such that f(x, y) = g(x). What is f ′(a, b) in terms of g′?

Proof. If f is independent of the second variable, let x ∈ R and set

g(x) = f(x, x) = f(x, y)

Thus f(x, y) = g(x). On the other hand, if g : R → R is a function suchthat f(x, y) = g(x), then

f(x, y1) = g(x) = f(x, y2)

implying that f is independent of the second variable. Now, assumingthat f is differentiable at (a, b), we claim

f ′(a, b) =[g′(a) 0

]To demonstrate, we see that there is a unique linear λ : R2 → R such that

0 = lim(h,k)→0

|f(a+ h, b+ k)− f(a, b)− λ(h, k)||(h, k)|

= limh→0

|g(a+ h)− g(a)− λ(h, k)||h|

which implies then that g′(a) exists and

λ(h, k) = g′(a) · h =[g′(a) 0

] [hk

]Thus

f ′(a, b) =[g′(a) 0

]

2.3 Define when a function f : R2 → R is independent of the first variableand find f ′(a, b) for such f . Which functions are independent of the firstvariable and also of the second variable?

Proof. A function f : R2 → R is independent of the first variable if andonly if there is a g : R→ R such that g(y) = f(x, y). As before, it can beshown that, if f is differentiable at a point (a, b), then

f ′(a, b) =[0 g′(b)

]Functions which are independent of the first and the second variable areconstant functions.

20

2.4 Let g be a continuous real-valued function on the unit circle {x ∈ R2 :|x| = 1} such that g(0, 1) = g(1, 0) = 0 and g(−x) = −g(x). Definef : R2 → R by

f(x) =

{|x| g

(x|x|

), x 6= 0

0 , x = 0

(a) If x ∈ R2 and h : R → R is defined by h(t) = f(tx), show that h isdifferentiable.

(b) Show that f is not differentiable at (0, 0) unless g = 0.

Proof. (a) If x = 0 ∈ R2 then there is nothing to show. So, let x 6= 0,then

lims→0

h(t+ s)− h(t)s

= lims→0

|(t+ s)x| g(

(t+s)x|(t+s)x|

)− |tx| g

(tx|tx|

)s

since t > 0 and s can always be chosen small enough, we get that

=

lims→0

|t+ s| g(

(t+s)x|(t+s)x|

)− |t| g

(tx|tx|

)s

|x|= |x| g

(x

|x|

)To justify the last equality, we observe that

t > 0⇒ s can be chosen so that t+ s > 0

which will produce the last equality.

t < 0⇒ s can be chosen so that t+ s < 0

which will produce the last equality

t = 0⇒ regardless of sgn(s) we get lims→0

sg(x|x|

)s

which will produce the last equality.

(b) Suppose that Df(0, 0) exists. Then there is λ ∈ L(R2,R) such that

0 = lim(h,k)→(0,0)

|f(h, k)− f(0, 0)− λ(h, k)||(h, k)|

= lim(h,k)→(0,0)

|f(h, k)− λ(h, k)||(h, k)|

21

Approaching along the x-axis and y-axis separately, along with thelinearity of λ, implies that λ(x, y) = 0. That is, that Df(0, 0) = 0.Thus

0 = lim(h,k)→(0,0)

|f(h, k)||(h, k)|

= lim(h,k)→(0,0)

∣∣∣∣g( (h, k)|(h, k)|)∣∣∣∣

Now, let � > 0 be given and let (x, y) ∈ S1. For |α| > 0 small enough,we will have that

|g(x, y)| =∣∣∣∣g( α(x, y)|α| |(x, y)|

)∣∣∣∣ < �So |g(x, y)| < � for every � > 0. Thus g(x, y) = 0. Therefore g = 0.

Note: Continuity of g was not needed in 2.4

2.5 Let f : R2 → R be defined by

f(x, y) =

{ x|y|√x2+y2

, (x, y) 6= (0, 0)

0 , (x, y) = (0, 0)

Show that f is a function of the kind considered in 2.4, so that f is notdifferentiable at (0, 0).

Proof. Set g(x, y) = x |y|. Note then that g is defined on S1. It is simpleto show that

g(0, 1) = g(1, 0) = 0 and g(−x) = −g(x) and g 6= 0

We also see that for (x, y) 6= (0, 0) we have that

√x2 + y2g

((x, y)√x2 + y2

)=√x2 + y2

x |y|x2 + y2

=x |y|√x2 + y2

= f(x, y)

Thus, by 2.4, f is not differentiable at (0, 0).

2.6 Let f : R2 → R be defined by f(x, y) =√|xy|. Show that f is not

differentiable at (0, 0).

Proof. Notice, for (x, y) 6= (0, 0)

|(x, y)| f(

(x, y)

|(x, y)|

)=√|xy| = f(x, y)

22

Thus, we have

f(x, y) =

{|(x, y)| f

((x,y)|(x,y)|

), (x, y) 6= (0, 0)

0 , (x, y) = (0, 0)

Furthermore, the conclusion in 2.4 (b) still holds when g(−x) = g(x).Thus, observing that

f(0, 1) = f(1, 0) = 0 and f(−(x, y)) = f(x, y) and f 6= 0

it follows from 2.4 (b) that f is not differentiable at (0, 0).

2.7 Let f : Rn → R be a function such that |f(x)| ≤ |x|2. Show that f isdifferentiable at 0.

Proof. We claim that Df(0, 0) = 0. First, observe that

0 ≤ |f(0)| ≤ |0|2 = 0

implying that f(0) = 0. Now, let � > 0 be given. Then 0 < |h| < � implies

|f(h)− f(0)− λ(h)||h|

=|f(h)||h|

≤ |h| < �

2.8 Let f : R → R2. Prove that f is differentiable at a ∈ R if and only if f1and f2 are, and that in this case

f ′(a) =

[f ′1(a)f ′2(a)

]Proof.

0 = limh→0

|f(a+ h)− f(a)− λ(h)||h|

= limh→0

|(f1(a+ h)− f1(a)− λ1(h), f2(a+ h)− f2(a)− λ2(h))||h|

= limh→0

∣∣∣∣(f1(a+ h)− f1(a)− λ1(h)h , f2(a+ h)− f2(a)− λ2(h)h)∣∣∣∣

if and only if

(0, 0) = limh→0

(f1(a+ h)− f1(a)− λ1(h)

h,f2(a+ h)− f2(a)− λ2(h)

h

)

23

by Exercise 1.23, if and only if

0 = limh→0

f1(a+ h)− f1(a)− λ1(h)h

and

0 = limh→0

f2(a+ h)− f2(a)− λ2(h)h

This establishes that Df(a) exists if and only if

f ′(a) =

[(f1)

′(a)(f2)

′(a)

]

2.9 Two functions f, g : R→ R are equal up to nth order at a if

limh→0

f(a+ h)− g(a+ h)hn

= 0

(a) Show that f is differentiable at a if and only if there is a function gof the form g(x) = a0 + a1(x− a) such that f and g are equal up tothe first order at a

(b) If f ′(a), . . . , f (n)(a) exist, show that f and the function g defined by

g(x) =

n∑i=0

f (i)(a)

i!(x− a)i

are equal up to nth order at a.

Proof. (a) ⇒: If f is differentiable at a then simply set g(x) = f(a) +c(x− a) where f ′(a) = c.

⇐: We disagree, and offer a counter-example:

f(x) =

{x , x 6= 01 , x = 0

and g(x) = x. Then

limh→0

f(0 + h)− g(0 + h)h

= limh→0

h− hh

= 0

Yet f is not continuous at 0, and so, a fortiori, is not differentiableat 0. However, if we insist that f be continuous at a, then the prooffollows from the fact that

limh→0

f(a+ h)− g(a+ h)h

= 0

24

implies thatlimh→0

f(a+ h) = a0

which, together with continuity, implies

f(a) = a0

which gives the desired result.

(b)

2.10 Use theorems of this section to find f ′ for the following:

(a) f(x, y, z) = xy

(b) f(x, y, z) = (xy, z)

(c) f(x, y) = sin(x sin(y))

(d) f(x, y, z) = sin(x sin(y sin(z)))

(e) f(x, y, z) = xyz

(f) f(x, y, z) = xy+z

(g) f(x, y, z) = (x+ y)z

(h) f(x, y) = sin(xy)

(i) f(x, y) = sin(xy)cos(3)

(j) f(x, y) = (sin(xy), sin(x sin(y)), xy)

Proof. (a) Notice xy = eln(xy). Thus f(x, y, z) = exp{π2 ln(π1)}(x, y, z).

So, if

g(x) = ex

and

h(x, y, z) = π2 ln(π1)(x, y, z)

then f(x, y, z) = (g ◦ h)(x, y, z). It easily follows that

g′(h(a, b, c)) = g′(b · ln(a)) =[ab]

On the other hand

Dh(a, b, c) = [ln(π1)Dπ2 + π2D ln(π1)] (a, b, c)

= ln(a)Dπ2(a, b, c) + bD ln(π1)(a, b, c)

= ln(a)π2 + bD ln(π1)(a, b, c)

25

and

D ln(π1)(a, b, c) = D ln(π1(a, b, c)) ◦Dπ1(a, b, c)= D ln(a) ◦ π1

=1

a◦ π1

=1

aπ1

Thus

h′(a, b, c) =[

0 ln(a) 0]

+[

ba 0 0

]=[

ba ln(a) 0

]Therefore,

f ′(a, b, c) = g′(a, b, c)·h′(a, b, c) =[ab]·[

ba ln(a) 0

]=[bab−1 ab ln(a) 0

](b) From part (a) we know that

[xy]′(a, b, c) =

[bab−1 ab ln(a) 0

]By linearity, we also have

[z]′(a, b, c) =

[0 0 1

]and so

f ′(a, b, c) =

[bab−1 ab ln(a) 0

0 0 1

](c) Let f(x, y) = sin(x sin(y)) then, if

g(x) = sin(x)

andh(x, y) = x sin(y) = π1 sin(π2)

then f(x, y) = (g ◦ h)(x, y). So then

g′(h(a, b)) = [cos(a sin(b))]

and

Dh(a, b) = sin(b)Dπ1(a, b) + aD sin(π2)(a, b) = sin(b)π1 + a cos(b)π2

so that

f ′(a, b) =[

sin(b) cos(a sin(b)) a cos(b) cos(a sin(b))]

26

(d) Let f(x, y, z) = sin(x sin(y sin(z))). If

g(x) = sin(x)

and

h(x, y, z) = x sin(y sin(z)) = π1 sin ◦(π2(sin ◦π3))(x, y, z)

Thusg′1(g2(a, b, c)) = [cos(a sin(b sin(c)))]

and

Dh(a, b, c) = [sin ◦(π2(sin ◦π3))] (a, b, c)Dπ1(a, b, c)+π1(a, b, c)D sin ◦(π2(sin ◦π3))(a, b, c)

and so by (c) and by an argument similar to those in exercises (2.2)and (2.3) we get

g′2(a, b, c) = sin(b sin(c))[

1 0 0]+a[

0 sin(c) cos(b sin(c)) b cos(c) cos(b sin(c))]

so that

g′2(a, b, c) =[

sin(b sin(c)) a sin(c) cos(b sin(c)) ab cos(c) cos(b sin(c))]

Therefore,

f ′(a, b, c) = g′1(g2(a, b, c)) · g′2(a, b, c)= [cos(a sin(b sin(c)))]

[sin(b sin(c)) a sin(c) cos(b sin(c)) ab cos(c) cos(b sin(c))

](e) Let f(x, y, z) = xy

z

. Let

g(x, y, z) = xy

and

h(x, y, z) = (x, yz, 1)

= (π1(x, y, z), ππ32 (x, y, z), 1)

Thus

(g◦h)(x, y, z) = g(h(x, y, z)) = g((x, yz, 0)) = x(yz)1 = xy

z

= f(x, y, z)

ThusDf(a, b, c) = Dg(h(a, b, c)) ◦Dh(a, b, c)

Recalling from part (a)

g′(h(a, b, c)) = g′((x, yz, 1)) =[bcab

c−1 abc

ln(a) 0]

27

and

Dh(a, b, c) = (Dπ1(a, b, c), Dππ32 (a, b, c), D1(a, b, c))

= (π1, Dππ32 (a, b, c), 0)

Again, from part (a), we get that

(ππ32 )′(a, b, c) =

[0 cbc−1 bc ln(b)

]which gives us that

h′(a, b, c) =

1 0 00 cbc−1 bc ln(b)0 0 0

Therefore,

f ′(a, b, c) = g′(h(a, b, c)) · h′(a, b, c)

=[bcab

c−1 abc

ln(a) 0] 1 0 00 cbc−1 bc ln(b)

0 0 0

=[bcab

c−1 cbc−1abc

ln(a) abc

ln(a)bc ln(b)]

2.11 Find f ′ for the following (where g : R→ R is continuous):

(a) f(x, y) =∫ x+ya

g

(b) f(x, y) =∫ x·ya

g

(c) f(x, y, z) =∫ sin(x sin(y sin(z)))a

g

Remark : It’s not too hard to show that if

f(x) =

∫ h(x)a

g

with g continuous, then

Df(c) = g(h(c)) ◦Dh(c)

Proof. (a) Let h(x, y) = x+ y = π1 + π2. Then by our remark, we get

Df(c, d) = g(c+ d) ◦D(π1 + π2)(c, d)

which implies

f ′(c, d) = [g(c+ d)][

1 1]

=[g(c+ d) g(c+ d)

]28

(b) Let h(x, y) = xy = [π1 · π2](x, y). Then by our remark, we get

Df(c, d) = g(cd) ◦D(π1π2)(c, d)

which impliesf ′(c, d) =

[dg(cd) cg(cd)

](c) Let h(x, y, z) = sin(x sin(y sin(z))). Then by our remark, we get

Df(a, b, c) = g(sin(a sin(b sin(c)))) ◦Dh(a, b, c)

whose Jacobian is not worth writing out (see 2.10 (d)).

2.12 A function f : Rn × Rm → Rp is bilinear if for x, x1, x2 ∈ Rn, y, y1, y2 ∈Rm, and a ∈ R we have

f(ax, y) = af(x, y) = f(x, ay)f(x1 + x2, y) = f(x1, y) + f(x2, y)f(x, y1 + y2) = f(x, y1) + f(x, y2)

(a) Prove that if f is bilinear, then

lim(h,k)→0

|f(h, k)||(h, k)|

= 0

(b) Prove that Df(a, b)(x, y) = f(a, y) + f(x, b)

(c) Show that the formula for Dp(a, b) in Theorem 2-3 is a special caseof (b).

Lemma. If f : Rn ×Rm → Rp is bilinear, then there is M > 0 such thatfor all (a, b) ∈ Rn × Rm with |(a, b)| ≤ 1

|f(a, b)| ≤M

Proof of Lemma. Let {e1, . . . , en} and {s1, . . . , sm} be the standard basesof Rn and Rm respectively and let (a, b) be such that |(a, b)| ≤ 1. Bilin-earity of f yields

|f(a, b)| =

∣∣∣∣∣∣f n∑i=1

αiei,

m∑j=1

βjsj

∣∣∣∣∣∣ ≤n∑i=1

m∑j=1

|αi| |βj | |f(ei, sj)|

Setting M ′ = maxi,j{|f(ei, sj)|} gives us that

|f(a, b)| ≤n∑i=1

m∑j=1

|αi| |βj |M ′

29

We observe that |a| , |b| ≤ |(a, b)| ≤ 1, which implies that |αi| , |βj | ≤ 1 forall i and j. So we have

|f(a, b)| ≤n∑i=1

m∑j=1

M ′ = n ·m ·M ′ = M

Proof of Exercise. (a) Let � > 0 be given and 0 < |(h, k)| < �M where Mis as defined in the Lemma. Let γ = |(h, k)|. Then

|f(h, k)|γ

=γ2∣∣∣f ( 1γ (h, k))∣∣∣

γ

= γ

∣∣∣∣f ( 1γ (h, k))∣∣∣∣

≤ γM< �

(b) Notice that

f(a+ h, b+ k)− f(a, b)− (f(a, k) + f(h, b)) = f(a+ h, b+ k)− f(a, b)− f(a, k)− f(h, b)

which, by bilinearity, yields

= f(a, b) + f(a, k) + f(h, b) + f(h, k)− f(a, b)− f(a, k)− f(h, b)

= f(h, k)

which, by (a), will produce the result.

(c) It is a simple matter to show that p : R×R→ R defined by p(x, y) =x · y is bilinear. Part (b) in the above implies then that

Dp(a, b)(x, y) = p(a, y) + p(x, b) = ay + bx = bx+ ay

as desired.

2.13 Define IP : Rn × Rn → R by IP (x, y) = 〈x, y〉

(a) Find D(IP )(a, b) and (IP )′(a, b)

(b) If f, g : R → Rn are differentiable and h : R → R is defined byh(t) = 〈f(t), g(t)〉, show that

h′(a) = 〈f ′(a)T , g(a)〉+ 〈f(a), g′(a)T 〉

(c) If f : R → Rn is differentiable and |f(t)| = 1 for all t, show that〈f ′(t)T , f(t)〉 = 0

30

(d) Exhibit a differentiable function f : R → R such that the function|f | defined by |f | (t) = |f(t)| is not differentiable.

Proof. (a) It is a simple matter to show that IP is bilinear. So it followsfrom the previous exercise that

DIP (a, b)(x, y) = IP (a, y) + IP (x, b) = 〈a, y〉+ 〈x, b〉

which implies(IP )′(a, b) =

[b a

]where b · x = IP (b, x) and a · y = IP (a, y)

(b) Note that f(t) = (f1(t), f2(t), . . . , fn(t)) and (g1(t), g2(t), . . . , gn(t)).So

h(t) = 〈f(t), g(t)〉 =n∑i=1

fi(t)gi(t)

which implies

h′(a) =

n∑i=1

(fi · gi)′(a) =n∑i=1

[gi(a)] (fi)′(a) + [fi(a)] (gi)

′(a)

and so, considering [f ′(a)]T and [g′(a)]T as elements of Rn, then weget

h′(a) = 〈f ′(a)T , g(a)〉+ 〈f(a), g′(a)T 〉

(c) Notice that |f(t)| = 1 for all t implies that 〈f(t), f(t)〉 = 1 for allt. Then if g(t) = 〈f(t), f(t)〉, by Theorem 2-3 and (b) from thisexercise, we get that for all a

0 = g′(a) = 〈f ′(a)T , f(a)〉+ 〈f(a), f ′(a)T 〉 = 2〈f ′(a), f(a)〉

which implies that 〈f ′(t)T , f(t)〉 = 0.(d) f(x) = x3 is such a function.

1. Let Ei, i = 1, 2, . . . , k be Euclidean spaces of various dimensions. A func-tion f : E1×. . .×Ek → Rp is called multilinear if for each choice of xj ∈ Ej ,j 6= i the function g : Ei → Rp defined by g(x) = f(x1, . . . , xi−1, x, xi+1, . . . , xk)is a linear transformation.

(a) If f is multilinear and j 6= i, show that for h = (h1, h2, . . . , hk), withhl ∈ El, we have

limh→0

|f(a1, . . . , hi, . . . , hj , . . . , ak)||h|

= 0

31

(b) Prove that

Df(a1, . . . , ak)(x1, . . . , xk) =

k∑i=1

f(a1, . . . , ai−1, xi, ai+1, . . . , ak)

Proof. (a) It is a simple matter to show that if g(x, y) is defined by

g(x, y) = f(a1, . . . , x, . . . , y, . . . , ak)

then g is bilinear. Thus, by Exercise 2-12, we have

limh→0

|f(a1, . . . , hi, . . . , hj , . . . , ak)||h|

= limh→0

|g(hi, hj)||h|

= 0

(b) First, a simple inductive argument establishes that for n ≥ 2 we willhave

limh→0

|f(h1, . . . , hn)||(h1, . . . , hn)|

= 0

To show this, we first observe that

f(h1, . . . , hi, . . . , hn) = f

(h1, . . . ,

d∑k=1

αkek, . . . , hn

)

=

d∑k=1

αkf(h1, . . . , ei, . . . , hn)

where αk(hi)→ 0 as hi → 0. So, if the claim holds for n then, by theinductive hypothesis along with our previous observation, we have

limh→0

|f(h1, . . . , hi, . . . , hn+1)||(h1, . . . , hi, . . . , hn+1)|

≤ limh→0

d∑k=1

|αk||f(h1, . . . , ek, . . . , hn+1)||(h1, . . . , ek, . . . , hn+1)|

= 0

Now, it’s not too hard to show that the difference

= f(a1 + h1, . . . , ak + hk)− f(a1, . . . , ak)−k∑i=1

f(a1, . . . , hi, . . . , ak)

= f(a1, . . . , ak) +

k∑i=1

f(a1, . . . , hi, . . . , ak) + σ − f(a1, . . . , ak)

−k∑i=1

f(a1, . . . , hi, . . . , ak)

= σ(h1, . . . , hk)

32

where

σ =∑

f(a1, . . . , hi1 , . . . , hi2 , . . . , ak)+∑f(a1, . . . , hi1 , . . . , hi2 , . . . , hi3 . . . , ak) + . . .+ f(h1, . . . , hk)

To be clear, σ is simply the sum of all images by f where there aretwo or more h’s in the argument. Thus,

= limh→0

∣∣∣f(a1 + h1, . . . , ak + hk)− f(a1, . . . , ak)−∑ki=1 f(a1, . . . , hi, . . . , ak)∣∣∣|h|

= limh→0

|σ||h|

and by our previous considerations, we get

= 0

2.15 Regard an n×n matrix as a point in the n-fold product Rn× . . .×Rn byconsidering each row as a member of Rn.

(a) Prove that det : Rn × . . .× Rn → R is differentiable and

D(det)(a1, . . . , an)(x1, . . . , xn) =

n∑i=1

det

a1...xi...an

(b) If aij : R→ R are differentiable and f(t) = det(aij(t)), show that

f ′(t) =

n∑j=1

det

a11(t) . . . a1n(t)...

...a′j1(t) . . . a

′jn(t)

......

an1(t) . . . ann(t)

(c) If det(aij(t)) 6= 0 for all t and b1, . . . , bn : R → R are differentiable,

let s1, . . . , sn : R→ R be the functions such that s1(t), . . . , sn(t) arethe solutions of the equations

n∑j=1

aji(t)sj(t) = bi(t) i = 1, 2, . . . , n

33

Show that si is differentiable and find s′i(t).

Proof. (a) It is well-known that the determinant, det, is multilinear inboth the rows and the columns. Thus, the result follows immediatelyfrom the previous exercise part (b).

(b) Let g : R→ Rn × . . .× Rn be defined by

g(t) =

a11(t) . . . a1n(t)... ...an1(t) . . . ann(t)

=g1(t)...gn(t)

Then

f = det ◦ g

and thusDf(k) = Ddet(g(k)) ◦Dg(k)

Now, we know

Ddet(g(k))(x1, . . . , xn) =

n∑i=1

det

g1(k)...xi...

gn(k)

which, by Theorem 2-3, implies that

Ddet(g(k)) ◦D(g(k)) =n∑i=1

det

g1(k)...

Dgi(k)...

gn(k)

which, by Ddet(c) = det′(c) and, again, Theorem 2-3, implies that

f ′(t) =

n∑i=1

det

g1(k)...

g′i(k)...

gn(k)

=n∑i=1

det

a11k . . . a1n(k)...

...a′i1(k) . . . a

′in(k)

......

an1(k) . . . ann(k)

(c) The first part follows from Cramer’s rule. To find s′(t) it is enough

to apply (b) and the quotient rule.

34

2.16 Suppose f : Rn → Rn is differentiable and has a differentiable inversef−1 : Rn → Rn. Show that (f−1)′(a) =

[f ′(f−1(a))

]−1.

Proof.f ◦ f−1(x) = I(x) = x

impliesIn = (f ◦ f−1)′(a) = f ′(f−1(a)) · (f−1)′(a)

which implies

(f−1)′(a) =[f ′(f−1(a))

]−1

2.17 Find the partial derivatives of the following functions:

2.18 Find the partial derivatives of the following functions (where g : R → Ris continuous):

(a) f(x, y) =∫ x+ya

g

(b) f(x, y) =∫ xyg

(c) f(x, y) =∫ ∫ y

bg

ag

Proof. (a) For 1 ≤ i ≤ 2

Dif(c, d) = limh→0

∫ c+d+ha

g −∫ c+da

g

h

= limh→0

G(c+ d+ h)−G(c+ d)h

= g(c+ d)

(b)

D1f(c, d) = D1[G(x)−G(y)](c, d)= g(c)

D2f(c, d) = D2[G(x)−G(y)](c, d)= −g(d)

(c)

D1f(c, d) = 0

D2f(c, d) = g(G(d)−G(b)) · g(d)

35

2.19 If f(x, y) = xxxx

y

+log(x)(arctan(arctan(arctan(sin(cosxy)−log(x+y)))))find D2f(1, y).

Proof. f(1, y) = 1 implying D2f(1, y) = 0.

2.20 Find the partial derivatives of f in terms of the derivatives of g and h if

(a) f(x, y) = g(x)h(y)

(b) f(x, y) = g(x)h(y)

(c) f(x, y) = g(x)

(d) f(x, y) = g(y)

(e) f(x, y) = g(x+ y)

Proof. (a) D1f(a, b) = h(b)g′(a) and D2f(a, b) = g(a)h

′(b).

(b) D1f(a, b) = h(b)g(a)h(b)g′(a) and D2f(a, b) = g(a)

h(b) ln(g(a))h′(b)

(c) D1f(a, b) = g′(a) and D2f(a, b) = 0

(d) D1f(a, b) = 0 and D2f(a, b) = h′(b)

(e) D1f(a, b) = g′(a+ b) = D2f(a, b)

2.21 Let g1, g2 : R2 → R be continuous. Define f : R2 → R by

f(x, y) =

∫ x0

g1(t, 0)dt+

∫ y0

g2(x, t)dt

(a) Show that D2f(x, y) = g2(x, y).

(b) How should f be defined so that D1f(x, y) = g1(x, y)?

(c) Find a function f : R2 → R such that D1f(x, y) = x and D2f(x, y) =y. Find one such that D1f(x, y) = y and D2f(x, y) = x.

Proof. (a) Let us define

f1(x) =

∫ x0

g1(t, 0)dt

f2(y) =

∫ y0

g2(x, t)dt

Then f(x, y) = f1(x) + f2(y) which implies that

D2f(x, y) = f′2(y) = g2(x, y)

(b) If f(x, y) =∫ x0g1(t, y)dt+

∫ y0g2(0, t)dt then D2f(x, y) = g1(x, y).

36

(c) Let f(x, y) = x2

2 +y2

2 . Then

D1f(x, y) = x and D2f(x, y) = y

If f(x, y) = xy then

D1f(x, y) = y and D2f(x, y) = x

2.22∗ If f : R2 → R and D2f = 0, show that f is independent of the secondvariable. If D1f = D2f = 0, show that f is constant.

Proof. Let f : R2 → R such that D2f = 0. Given x, if we set g(y) =f(x, y) then g′(y) = D2f(x, y) = 0 for all y, implying, by MVT, that g(y)is constant. But then

f(x, y1) = g(y1) = g(y2) = f(x, y2)

so that f is independent of the second variable. Now, if D1f = D2f = 0then D1f = 0 implies that f is independent of the first variable andD2f = 0 implies that f is independent of the second variable. So

f(x1, y1) = f(x2, y2)

implying that f is constant.

2.23∗ Let A = {(x, y) ∈ R2 : x < 0, or x ≥ 0 and y 6= 0}

(a) If f : A→ R and D1f = D2f = 0, show that f is constant.(b) Find a function f : A→ R such that D2f = 0 but f is not indepen-

dent of the second variable.

Proof. (a) Let (x1, y1), (x2, y2) ∈ A. Notice there exists a sequence oflines, each parallel to one of the axes, going from (x1, y1) to (x2, y2)so that no line passes through the origin. So, if (x(i), y(i)) is theappropriate endpoint of the ith line, then from Exercise 2.22, wehave

f(x1, y2) = f(x′, y′) = f(x′′, y′′) = . . . = f(x2, y2)

so that f is constant.

(b) Notice that

f(x) =

{1 , (0, y) where y > 0

0 , elsewhere

is such a function.

37

2.24 Define f : R2 → R by

f(x, y) =

{xy x

2−y2x2+y2 , (x, y) 6= 0

0 , (x, y) = 0

(a) Show that D2f(x, 0) = x for all x and D1f(0, y) = −y for all y(b) Show that D1,2f(0, 0) 6= D2,1f(0, 0).

Proof. (a) It is a simple matter to show that

D2f(x, 0) = limh→0

f(x, h)− f(x, 0)h

= x

and

D1f(0, y) = limh→0

f(h, y)− f(0, y)h

= −y

(b) It is a simple matter to show that

D1,2f(0, 0) = −1 6= 1 = D2,1f(0, 0)

2.25∗ Define f : R→ R by

f(x) =

{e−x

−2, x 6= 0

0 , x = 0

Show that f is a C∞ function, and f (i)(0) = 0 for all i.

Proof.

2.26∗ Too bored. Will return later.

Proof.

2.27 Too bored. Will return later.

Proof.

2.28 Find expressions for the partial derivatives of the following functions:

(a) F (x, y) = f(g(x)k(y), g(x) + h(y))

(b) F (x, y, z) = f(g(x+ y), h(y + z))

(c) F (x, y, z) = f(xy, yz, zx)

(d) F (x, y) = f(x, g(x), h(x, y))

38

Proof. (a) Let us define

q1(x, y) = g(x)k(y)

q2(x, y) = g(x) + h(y)

Then

q′1(x, y) =[k(y)g′(x) g(x)k′(y)

]q′2(x, y) =

[g′(x) h′(y)

]and

F (x, y) = f(q1(x, y), q2(x, y))

By Theorem 2.9, we get that

D1F (a, b) = D1f(q1(a, b), q2(a, b))k(b)g′(a) +D2f(q1(a, b), q2(a, b))g

′(a)

D2F (a, b) = D1f(q1(a, b), q2(a, b))g(a)k′(b) +D2f(q1(a, b), q2(a, b))h

′(b)

(b) We see that

(g ◦ (π1 + π2))′(x, y, z) =[g′(x+ y) g′(x+ y) 0

]and

(h ◦ (π2 + π3))′(x, y, z) =[0 h′(y + z) h′(y + z)

]and so if r = (g(a+ b), h(b+ c)), then

D1F (a, b, c) = D1f(r)g′(a+ b)

D2F (a, b, c) = D1f(r)g′(a+ b) +D2f(r)h

′(b+ c)

D3F (a, b, c) = D2f(r)h′(b+ c)

(c) Skipping the details, if r = (ab, bc, ca) then

D1F (a, b, c) = D1f(r)bab−1 +D3f(r) ln(c)c

a

D2F (a, b, c) = D1f(r) ln(a)ab +D2f(r)cb

c−1

D3F (a, b, c) = D2f(r) ln(b)bc +D3f(r)ac

a−1

(d) Skipping the details, if r = (a, g(a), h(a, b)) then

D1F (a, b) = D1f(r) +D2f(r)g′(a) +D3f(r)D1h(a, b)

D2F (a, b) = D3f(r)D2h(a, b)

39

2.29 Let f : Rn → R. For x ∈ Rn, the limit

limt→0

f(a+ tx)− f(a)t

if it exists, is denoted Dxf(a), and called the directional derivative off at a, in the direction of x.

(a) Show that Deif(a) = Dif(a).

(b) Show that Dtxf(a) = tDxf(a).

(c) If f is differentiable at a, show that Dxf(a) = Df(a)(x) thereforeDx+yf(a) = Dxf(a) +Dyf(a).

Proof. (a) Notice that a+ hei = (a1, . . . , ai + h, . . . , an). Specifically,

Deif(a) = limh→0

f(a+ hei)− f(a)h

= limh→0

f(a1, . . . , ai + h, . . . , an)− f(a1, . . . , an)h

= Dif(a)

(b)

Dαxf(a) = limt→0

f(a+ tαx)− f(a)t

= limt→0

f(a+ tααx)− f(a)tα

= α limt→0

f(a+ tx)− f(a)t

= αDxf(a)

(c) Let f be differentiable at a. If g(t) = a+ tx then

Dxf(a) = limt→0

f(a+ tx)− f(a)t

= limt→0

(f ◦ g)(t)− (f ◦ g)(0)t

= D(f ◦ g)(0)= Df(g(0)) ◦Dg(0)= Df(a)(x)

By linearity of Df(a) we get the desired result.

2.30 Let f be defined as in Problem 2.4. Show that Dxf(0, 0) exists for all x,but if g 6= 0, then Dx+yf(0, 0) = Dxf(0, 0) +Dyf(0, 0) is not true for allx and y.

40

Proof. We see from 2.4(a) that Dxf(0, 0) = h′(0) = |x| g

(x|x|

)for x 6=

0 and D(0,0)f(0, 0) = 0. If g 6= 0, then there exists z 6= 0 such thatg(z|z|

)6= 0. If x = (π1(z), 0) and y = (0, π2(z)), then

Dx+yf(0, 0) = Dzf(0, 0) = |z| g(z

|z|

)6= 0

while, by our assumptions on g, we have

Dxf(0, 0) +Dyf(0, 0) = 0 + 0 = 0

which shows that

Dx+yf(0, 0) 6= Dxf(0, 0) +Dyf(0, 0)

2.31 Let f : R2 → R be defined as in Problem 1.26. Show that Dxf(0, 0) existsfor all x, although f is not even continuous at (0, 0).

Proof. Recall that from 1.26(a), we have that every straight line through(0, 0) contains an interval about (0, 0) which is contained in R2 \A, whereA is as defined in 1.26. This means Dxf(0, 0) exists for all x and

Dxf(0, 0) = 0

2.32 (a) Let f : R→ R be defined by

f(x) =

{x2 sin

(1x

), x 6= 0

0 , x = 0

Show that f is differentiable at 0 but that f ′ is not continuous at 0.

(b) Let f : R2 → R be defined by

f(x, y) =

(x2 + y2) sin(

1√x2+y2

), (x, y) 6= 0

0 , (x, y) = 0

Show that f is differentiable at (0, 0) but that Dif is not continuousat (0, 0).

Proof. (a) We see thatf ′(0) = 0

and that, for x 6= 0

f ′(x) = 2x sin(1/x)− cos(1/x)

which is clearly not continuous at 0.

41

(b) We see that

limh→0

f(0 + h)− f(0)|h|

= limh→0

|h|2 sin (1/ |h|)|h|

= limh→0|h| sin(1/ |h|) = 0

so that Df(0, 0) = 0. With a little calculation, we can see that for(x, y) 6= (0, 0) we get

Dif(x, y) = 2 · πi(x, y) · sin

(1√

x2 + y2

)− πi(x, y) ·

cos

(1√x2+y2

)√x2 + y2

It is not too hard to show that when (x, y) → (0, 0) the first termtends to zero but that the term on the end is pathological (Checklimx→0+ D1f(x, 0) for example). Therefore, Dif is not continuous at(0, 0).

2.33 Show that continuity of D1fi at a may be eliminated from the hypothesisof Theorem 2-8.

Proof. This is because existence of D1fj(a) will imply that

limh→0

|fj(a+ h)− fj(a)−∑i=1Difj(a)(hi)|

|h|=

limh→0

|fj(a1 + h1, a2 . . . , an)− fj(a1, . . . , an)−D1fj(a)(h1)||h|

+|∑ni=2[Difj(ci)−Difj(a)]hi|

|h|

so that continuity of only the other partials is need to obtain the desiredlimit.

2.34 A function f : Rn → R is homogeneous of degree m if f(tx) = tmf(x)for all x. If f is also differentiable, show that

n∑i=1

xiDif(x) = mf(x)

Proof. If we let g(t) = f(tx) and h(t) = tx, then

g(t) = (f ◦ h)(t)

From this it is not too difficult to show that

g′(1) =

n∑i=1

xiDif(x)

42

Now, if f is homogeneous of degree m, then

g(t) = f(tx) = tmf(x)

so thatg′(t) = mf(x)tm−1

andg′(1) = mf(x)

2.35 If f : Rn → R is differentiable and f(0) = 0, prove that there existgi : Rn → R such that

f(x) =

n∑i=1

xigi(x)

Proof. The author’s hint is very revealing. Let hx(t) = f(tx). So then∫ 10

h′x(t)dt = hx(1)− hx(0) = f(x)− f(0) = f(x)

However, we can see that

h′x(t) =

n∑i=1

xiDif(tx)

which implies that

f(x) =

∫ 10

h′x(t)dt

=

n∑i=1

xi

∫ 10

Dif(tx)dt

which implies that if gi(x) =∫ t0Dif(tx)dt, we obtain the desired result.

2.36 Let A ⊂ Rn be an open set and f : A→ Rn a continuously differentiable1-1 function such that detf ′(x) 6= 0 for all x. Show that f(a) is an openset and f−1 : f(A)→ A is differentiable. Show also that f(B) is open forany open set B ⊂ A.

Proof. Note that the assumptions on f and A imply that the InverseFunction Theorem holds on f for all x ∈ A. So, given y ∈ f(A) thereare open Wy ⊂ f(A) and Vx ⊂ A and a function f̃−1 : Wy → Vx whichis continuous and differentiable. But by injectivity of f , we know thatf−1 : f(A) → A exists and f−1(z) = f̃−1(z) for z ∈ Wy. So f−1 iscontinuous and differentiable at y. Therefore, f−1 is differentiable andcontinuous on f(A). Continuity of f−1 implies both that f(A) is open,and that f(B) is open whenever B ⊂ A is (Willard).

43

2.37 (a) Let f : R2 → R be a continuously differentiable function. Show thatf is not 1-1.

(b) Generalize this result to the case of a continuously differentiable func-tion f : Rn → Rm with m < n.

Proof. (a) Assuming f is continuously differentiable (though we coulduse a weaker assumption than this) and injective, it follows thatf(x, y) is differentiable on R2 and therefore continuous on R2. Sof(x, 0) and f(0, y) are continuous and injective on the interval [−1, 1] ⊂R. By connectedness of [−1, 1] and continuity of f(x, 0) and f(0, y),the images of [−1, 1] by f(x, 0) and f(0, y) are both connected. f(0, 0)being a maximum or minimum of either function on [−1, 1] contra-dicts with f being injective. If f(0, 0) is not a maximum or a mini-mum, then it follows by connectedness that both images contain someopen interval about f(0, 0), which, again, contradicts with injectivityof f .

(b) Let f : Rn → Rm, m < n, be continuously differentiable. Supposefor contradiction that f is injective. If we define g : Rn → Rn by

g(x1, . . . , xn) = (x1, x2, . . . , xn−m, f(x1, . . . , xn))

then we know that

g′(x) =

[Id 0

∗ ∂f∂~y (x)

]where

∂f

∂~y(x) =

Dn−m+1f1(x) . . . Dnf1(x)... ...Dn−m+1fm(x) . . . Dnfm(x)

and ∗ is unimportant since detg′(x) 6= 0 iff det∂f∂~y (x) 6= 0. Now, ifdet∂f∂~y (x) = 0 then there exists v ∈ R

m such that v 6= 0 and[∂f

∂~y(x)

]v = 0

which implies that if v∗ ∈ Rn admitting v∗ = (0, . . . , 0, v) then weget that

Dv∗f(x) = Df(x)(v∗) =

[∂f

∂~y(x)

]v = ~0 ∈ Rm

2.38 (a) If f : R → R satisfies f ′(a) 6= 0 for all a ∈ R, show that f is notinjective (on all R).

44

(b) Define f : R2 → R2 by f(x, y) = (ex cos(y), ex sin(y)). Show thatdetf ′(x, y) 6= 0 for all (x, y) ∈ R2 but f is not injective.

Proof. (a) This is a simple consequence of the Mean Value Theorem.

(b) We see that

f ′(x, y) =

[cos(y)ex −ex sin(y)sin(y)ex ex cos(y)

]so that

detf ′(x, y) = e2x(cos2(y) + sin2(y)) = e2x 6= 0 ∀(x, y)

and yet f(0, 0) = (1, 0) = f(0, 2π).

2.39 Use the function f : R→ R defined by

f(x) =

{x2 + x

2 sin(1x

)x 6= 0

0 x = 0

To show that continuity of the derivative cannot be eliminated from thehypotheses in theorem 2-11.

Proof. Observe that

f ′(0) = limh→0

f(h)

h= limh→0

1

2+ h sin

(1

h

)=

1

2

and, for x 6= 0

f ′(x) =1

2+ 2x sin

(1

x

)− cos

(1

x

)so that f ′(0) 6= 0 but f ′(x) is continuously differentiable only when x 6= 0.One could argue by relatively elementary means that

f

(1

(2n+ 1)π

)< f

(1

(n+ 1)2π

)< f

(1

n2π

)which implies that f is not injective on any interval about 0.

2.40 Use the implicit function theorem to re-do Problem 2-15(c).

Proof. The assumptions in 2-15(c) are not strong enough to enable usto use the Implicit Function Theorem. If we assume that each aij(t)and bi(t) are continuously differentiable, then letting [A(t)]ij = aij(t),s = (s1, . . . , sn) ∈ Rn and b : R → Rn with b(t) = (b1(t), . . . , bn(t)), wecan define f : R× Rn → Rn by

f(t, s) = A(t)T s− b(t)

45

By non-singularity of A(t), we know that for every t there is s such thatf(t, s) = 0. Furthermore, by aij(t) and bi(t) being continuously differen-tiable, we know that D1f(t, s) exists and is continuous. It is easy to showthat for all i ≥ 2, Dif(t, s) exists and is continuous since f(t, s) is affine ins. Indeed, if M is as defined in the theorem, then M = A and so is non-singular for all t. Thus, by the Implicit Function Theorem, we will havethat for all t, we can find s(t) so that f(t, s(t)) = 0 with s differentiableat t.

2.41 Let f : R × R → R be differentiable. For each x ∈ R define gx : R → Rby gx(y) = f(x, y). Suppose that for each x there is a unique y withg′x(y) = 0; let c(x) be this y.

(a) If D2,2f(x, y) 6= 0 for all (x, y), show that c is differentiable and

c′(x) = −D2,1f(x, c(x))D2,2f(x, c(x))

(b) Show that if c′(x) = 0, then for some y we have

D2,1f(x, y) = 0

D2f(x, y) = 0

(c) Let f(x, y) = x(y log y − y)− y log(x). Find

max12≤x≤2

(min

13≤y≤1

f(x, y)

)

Proof. (a) If we just assume that D2f(x, y) is continuously differen-tiable, then we will get that, for M as defined in Theorem 2-12,M = D2,2f(x, c(x)) so that det M 6= 0 for all x ∈ R. The ImplicitFunction Theorem then implies that c′(x) exists for all x ∈ R. Tocompute c′(x) we simply see that if we let h(x) = (x, c(x)), then[D2f ] ◦ h = 0. This implies

0 = (D2f ◦ h)′ (x)

= [D2,1f(x, c(x)) D2,2f(x, c(x))]

[1

c′(x)

]= D2,1f(x, c(x)) +D2,2f(x, c(x))c

′(x)

which implies the desired result.

(b) Note that g′x(y) = D2f(x, y). So, if c′(x) = 0, then, letting y = c(x)

yields

g′x(y) = D2f(x, y) = 0 = −D2,1f(x, y)

D2,2f(x, y)

(c) Maybe some other time.

46

Chapter 3

Integration

3.1 Let f : [0, 1]× [0, 1]→ R be defined by

f(x, y) =

{0 if 0 ≤ x < 121 if 12 ≤ x < 1

Show that f is integrable and that∫[0,1]×[0,1] f =

12

Proof. We see that for any partition P = (P1, P2) that admits{0,

1

2, 1

}⊂ P1

We have

L(f, P ) = U(f, P ) =1

2

Thus, if we have an arbitrary partition, P , then, with the proper refine-ment and by Corollary 3-2, we obtain∫

[0,1]×[0,1]f =

1

2

3.2 Let f : A→ R be integrable and let g = f except at finitely many points.Show g is integrable and

∫Af =

∫Ag.

Proof. Let us denote those values of x where g and f differ by x1, . . . , xm.We will show that ∀U(f, P ), ∃P ′U(g, P ′) ≤ U(f, P ), and similar for thelower sums. This will imply that ∀P , ∃P ′ such that

L(f, P ) ≤ L(g, P ′) ≤ U(g, P ′) ≤ U(f, P )

47

which, in turn, implies that g is integrable and that∫Ag =

∫Af . So, given

a partition, P we can first refine it by adding enough subrectangles so thateach subrectangle contains at most one xi. Let us denote this partitionP ∗. We can refine P ∗ by including subrectangles around each xi withsufficiently small volume. This refinement we denote by, P ′. Now, if S isa subrectangle in P ∗ that does not contain any xi then

MS(f)v(S) = MS(g)v(S)

If R is a subrectangle in P ∗ that does include an xi, then, in P′, we have

R = ∪jRj where Rj is a subrectangle in P ′ and xi ∈ Rk for some k. NoteRj does not contain any xi for j 6= k. But then MRj (g) ≤ MR(f) for allj 6= k, and

∑j 6=k v(Rj) < v(R) so that∑

j 6=k

MRj (g)v(Rj) < MR(f)v(R)

Since v(Rk) was sufficiently small, in the sense that

MRk(g)v(Rk) < MR(f)v(R)−∑j 6=k

MRi(g)v(Ri)

we get

0 < MR(f)v(R)−∑j

MRi(g)v(Ri)

But then U(f, P ∗) ≤ U(f, P ) and

U(f, P ∗)− U(g, P ′) =∑S∪R

MS∪R(f)v(S ∪R)−∑S∪R′

MS∪R′(g)v(S ∪R′)

=∑R

MR(f)v(R)−∑R′

MR′(g)v(R′)

> 0

which impliesU(g, P ′) ≤ U(f, P )

Similar reasoning produces an analogous result for the lower sums, whichimplies the conclusion.

3.3 Proof. (a) Notice that for all x ∈ S we have

mS(f) +mS(g) ≤ f(x) + g(x) and f(x) + g(x) ≤MS(f) +MS(g)

48

which implies that

L(f, P ) + L(g, P ) =∑S

mS(f)v(S) +∑S

mS(g)v(S)

=∑S

(mS(f) +mS(g))v(S)

≤∑S

mS(f + g)v(S)

= L(f + g, P )

and similar for the upper sums.

(b) Note that, for � > 0 given, for well-chosen P , we have∫A

f +

∫A

g − � < L(f, P ) + L(g, P )

≤ L(f + g, P )≤ U(f + g, P )≤ U(f, P ) + U(g, P )

<

∫A

f +

∫A

g + �

which implies that∫Af + g exists. Now, if for some P

U(f + g, P ) <

∫A

f +

∫A

g

then there is P ′ such that

U(f + g, P ) < L(f, P ′) + L(g, P ′) <

∫A

f +

∫A

g

which would imply U(f+g, P ) < L(f+g, P ′) which is absurd. Com-bined with the above inequality, we get that for all � > 0, there existsP such that∫

A

f +

∫A

g < U(f + g, P ) <

∫A

f +

∫A

g + �

Therefore,∫Af + g =

∫Af +

∫Ag.

(c) Two cases, too tedious.

3.4 Let f : A → R and let P be a partition of A. Show that f is integrableif and only if for each subrectangle S the function f |S is integrable, andthat in this case ∫

A

f =∑S

∫S

f |S

49

Proof. If PS is the portion of the partition P which produces subrectangleS, then ∑

S

U(f |S, PS) = U(f, P )∑S

L(f |S, PS) = L(f, P )

So, there exists a refinement, P ′, of P such that

U(f, P ′)− L(f, P ′) < �

which implies ∑S

U(f |S, P ′S)− L(f |S, P ′S) < �

which implies∑S

∫Sf |S exists. Furthermore

0 < U(f, P ′)−∫A

f < �

which implies

0 <∑S

U(f |S, P ′S)−∫A

f < �

which implies that ∑S

∫S

f |S =∫A

f

3.5 Let f, g : A→ R be integrable and suppose f ≤ g. Show that∫Af ≤

∫Ag.

Proof. Notice that for any P

U(f, P ) ≤ U(g, P )

By integrability of both f and g, we get∫A

f = infPU(f, P ) ≤ inf

PU(g, P ) =

∫A

g

3.6 If f : A→ R is integrable, show that |f | is integrable and∣∣∫Af∣∣ ≤ ∫

A|f |.

Proof. We wish to show that for any partition P ,

MS(|f |)−mS(|f |) ≤MS(f)−mS(f)

Notice that this is trivially true if ∀x∈Sf(x) ≥ 0 or f(x) ≤ 0. Otherwise,we observe that

50

(a) mS(f) < 0 ≤ ms(|f |) ≤MS(f)(b) Either MS(|f |) = MS(f) or MS(|f |) = −mS(f).

If MS(|f |) = MS(f) then

MS(|f |)−mS(|f |) = MS(f)−mS(|f |) ≤MS(f)−mS(f)

where the inequality comes from (a). If MS(|f |) = −mS(f) then

MS(|f |)−mS(|f |) ≤MS(|f |) = −mS(f) ≤MS(f)−mS(f)

ThusMS(|f |)−mS(|f |) ≤MS(f)−mS(f)

which implies that |f | is integrable and∣∣∣∣∫A

f

∣∣∣∣ ≤ ∫A

|f |

follows from the fact that |MS(f)| ≤MS(|f |), the integrability of |f |, andthe triangle inequality.

3.7 Let f : [0, 1]× [0, 1]→ R be defined by

f(x, y) =

0 x is irrational

0 x is rational, y is irrational1q x is rational, y =

pq in lowest terms

Show that f is integrable and∫[0,1]×[0,1] f = 0.

Proof.

3.8 Prove that [a1, b1]× . . .× [an, bn] does not have content zero if ai < bi foreach i.

Proof. Similar to 3-5.

3.9 (a) Show that an unbounded set cannot have content 0.

(b) Give an example of a closed set with measure 0 which does not havecontent 0.

Proof. (a) Notice that given {U1, . . . , Un} we will get that ∪ni=1Ui isbounded and therefore cannot cover an unbounded set.

51

(b) N is closed but unbounded and therefore cannot have content 0. How-ever, given 0 < � < 1, if we take Ui to be the rectangle, centered ati, such that v(Ui) = (

�2 )i then

∞∑i=1

v(Ui) =

∞∑i=1

( �2

)i=

�

2− �< �

3.10 (a) If C is a set of content 0, show that the boundary of C has content0.

(b) Give an example a bounded set C of measure 0 such that the bound-ary of C does not have measure 0.

Proof. (a) If {U1, . . . , Un} is a cover of C then ∪ni=1Ui is closed and con-tains C. Therefore, it contains the boundary of C. So {U1, . . . , Un}is also a cover for the boundary of C. Therefore, the boundary alsohas content 0.

(b) Notice that Q ∩ [0, 1] is countable and therefore has measure 0. Onthe other hand the boundary is [0, 1] which does not have measure 0.

3.11 Let A be the set in problem 1.18. If∑∞i=1(bi − ai) < 1, show that the

boundary of A does not have measure 0.

Proof. Suppose to the contrary. From 1.18, we know that boundaryA =[0, 1] \ A. So [0, 1] = A ∪ boundaryA. Since we can use open rectangles,we have a cover of A, {(ai, bi) : i ∈ N}, and a cover of boundaryA,{U1, U2, . . .}, such that

∞∑i=1

v(Ui) < 1−∞∑i=1

(bi − ai)

But then {(ai, bi) : i ∈ N} ∪ {U1, U2, . . .} forms a cover of [0, 1]. But

∞∑i=1

v(Ui) +

∞∑i=1

(bi − ai) < 1

which would contradict with it being a cover of [0, 1].

3.12 Let f : [a, b]→ R be an increasing function. Show that {x : f(x) is discontinuous at x}has measure 0.

52

Proof. First we will show that {x : o(f, x) > 1n} is finite for any given n.To this end, we suppose to the contrary, that there exists an n ∈ N sothat we have {x : o(f, x) > 1n} is infinite. Then we can select a countablesequence of distinct elements from it:

x0 < x1 < x2 < . . .

So, by f increasing, we get that for every k

f(xk)− f(x0) =k∑i=1

f(xi)− f(xi−1) ≥k∑i=1

1

n=k

n

which would imply that f is not bounded from above on [a, b] contradictingwith f(x) ≤ f(b) for all x ∈ [a, b]. Thus{

x : ∃n such that o(f, x) > 1n

}is at most countable, and therefore has measure 0. But, by Theorem 1.10,f is continuous at x iff o(f, x) = 0. Therefore, {x : f(x) is discontinuous at x}has measure 0.

3.13

3.14 Show if f, g : A→ R are integrable, then so is f · g.

Proof. Notice that (f ·g)(x) is continuous wherever both f and g are. Sinceboth f and g are continuous, it follows that both {x : f is not continuous at x}and {x : g is not continuous at x} have measure zero. This implies {x :f · g is not continuous at x} has measure zero. Therefore f · g is inte-grable.

3.15 Show that if C has content 0, then C ⊂ A for some closed rectangle Aand C is Jordan-measurable and

∫AχC = 0.

Proof. Since C has content zero, C ⊂ ∪ni=1Ui ⊂ A for some rectangle A,where the latter inclusion follows from boundedness of the union. Noticethat ∪ni=1Ui contains the closure of C and therefore its boundary. There-fore, the boundary of C has content 0 and so C is Jordan-measurable.Finally, notice that if P is a partition containing the cover {Ui}, thenU(f, P ) =

∑v(Ui) < �, implying

∫AχC = 0.

3.16 Give an example of a bounded set C of measure 0 such that∫AχC does

not exist.

53

Proof. If C = Q∩ [0, 1] then clearly C is bounded. Moreover, since C ⊂ Qand Q has measure 0, it follows C has measure 0. Furthermore, boundaryC = [0, 1] which clearly does not have measure 0. Therefore, χC is notintegrable, and

∫AχC does not exist.

3.17 If C is a bounded set of measure 0 and∫AχC exists, show that

∫AχC = 0.

Proof. We claim that if U is a (non-degenerate) rectangle, then U 6⊂ C,for otherwise, by problem 3.8, C would not have measure 0. Therefore, itfollows that if U is a rectangle then U contains points not in C. ThereforeL(χC , P ) = 0 for every partition P which, by integrability of χC , impliesthat

∫AχC = 0.

3.18 If f : A → R is nonnegative and∫Af = 0, show that {x : f(x) 6= 0} has

measure 0.

Proof. Observe that our hypotheses on f imply that for any subrectangle,S, we have that mS(f) = 0. In order to obtain a contradiction, let ussuppose that there is c ∈ A such that f(c) 6= 0 and f is continuous at c.But then there exists some subrectangle S such that x ∈ S implies

|f(x)− f(c)| < f(c)2

which further implies that 0 < f(c)2 ≤ f(x) for all x ∈ S, contradicting withmS(f) = 0. Therefore {x : f(x) 6= 0} ⊂ {x : f is not continuous at x}.Integrability of f and Theorem 3-8 imply the result.

3.19 Let U be the open set of Problem 3.11. Show that if f = χU except on aset of measure 0, then f is not integrable on [0, 1].

3.20 Show that an increasing function f : [a, b]→ R is integrable on [a, b].

Proof. Let B be defined as in Theorem 3.8. Notice, by f increasing, wehave

f(a) +

∞∑i=1

o(f, βi) ≤ f(b)

for any countable subcollection {βi : β ∈ B, i ∈ N}. But then

Bn =

{β : o(f, β) ≥ 1

n

}is finite for every n ≥ 2. But then

B =

∞⋃n=2

Bn

is at most countable. Therefore, by Theorem 3.8, it follows that f isintegrable.

54

3.21 If A is a closed rectangle, show that C ⊂ A is Jordan-measurable ifand only if for every � > 0 there exists a partition P of A such that∑S∈S1 v(S)−

∑S∈S2 < � where S1 consists of all subrectangles intersect-

ing C and S2 all subrectangles contained in C.

Proof. Notice thatS1 = S2 ∪ S3

where S3 is the collection of subrectangles that intersect with C but arenot contained in C. Thus the expression∑

S∈S1

v(S)−∑S∈S2

v(S) < �

is equivalent to∑S∈S2∪S3

v(S)−∑S∈S2

v(S) =∑S∈S2

v(S)+∑S∈S3

v(S)−∑S∈S2

v(S) =∑S∈S3

v(S) < �

which is equivalent to C having boundary with measure zero, and there-fore, Jordan-measurable.

3.22 If A is a Jordan-measurable set and � > 0, show that there is a compactJordan-measurable set C ⊂ A such that

∫A\C 1 < �.

Proof. Set C = S2, where S2 is as defined in problem 3.21.

3.23 Let C ⊂ A× B be a set of content 0. Let A′ ⊂ A be the set of all x ∈ Asuch that {y ∈ B : (x, y) ∈ C} is not of content 0. Show that A′ is a setof measure 0.

55