[an, bn] in Rn be a closed n-cell and let f : K → R be integrable on K

33  Download (0)

Full text

(1)

The Integral as an Iterated Integral

Let K = [a1, b1] × · · · × [an, bn] in Rn be a closed n-cell and let f : K → R be integrable on K.

Then one can calculate the integral Z

K

f in terms of a n-fold “iterated integral”

Z bn

an



· · ·

Z b2

a2

Z b1

a1

f (x1, x2, . . . , xn) dx1

 dx2



· · ·

 dxn. For simplicity we shall justify this for n = 2.

Theorem Let K = [a, b] × [c, d]. If f : K → R is continuous on K, then Z

K

f = Z d

c

Z b a

f (x, y)dx

 dy =

Z b a

Z d c

f (x, y)dy

 dx.

Proof Let F be defined for y ∈ [c, d] by F (y) =

Z b a

f (x, y)dx.

Let

c = y0 < y1 < · · · < yr= d be a partition of [c, d]

a = x0 < x1 < · · · < xs= b be a partition of [a, b]

and let

P = {[xi−1, xi] × [yj−1, yj] | 1 ≤ i ≤ s, 1 ≤ j ≤ r} be a partition of K.

Let yj be any point in [yj−1, yj] and note that F (yj) =

Z b a

f (x, yj) dx =

s

X

i=1

Z xi

xi−1

f (x, yj) dx.

For each j and i, by the Fundamental Theorem of Calculus and the Mean Value Theorem, there exists a point xji ∈ [xi−1, xi] such that

F (yj) =

s

X

i=1

f (xji, yj)(xi− xi−1).

We multiply by (yj − yj−1) and add to obtain

r

X

j=1

F (yj)(yj − yj−1) =

r

X

j=1 s

X

i=1

f (xji, yj)(xi− xi−1)(yj− yj−1).

Now the expression on the left side of this formula is an arbitrary Riemann sum for the integral Z d

c

F (y) dy = Z d

c

Z b a

f (x, y) dx

 dy.

We have shown that an arbitrary Riemann sum for F on [c, d] is equal to a particular Riemann sum of f on K corresponding to the partition P. Since f is integrable on K, the existence of the iterated integral and its equality with the integral on K is established.

A minor modification of the proof given for the preceding theorem yields the following, slightly stronger, assertion.

(2)

Theorem Let K = [a, b] × [c, d]. Suppose that f : K → R is integrable on K and for each y ∈ [c, d], the function x 7→ f (x, y) of [a, b] into R is continuous except possibly for a finite number of points, at which it has one-sided limits. Then

Z

K

f = Z d

c

Z b a

f (x, y)dx

 dy.

Definition Let K = I1× · · · × In, where Ij = [aj, bj] for each j = 1, . . . , n.

(a) The (n-)content or volume c(K) of K is defined to be c(K) = (b1− a1) × · · · × (bn− an) =

n

Y

j=1

(bj− aj).

(b) A set Z ⊂ Rn has n-content or volume zero if ∀  > 0, ∃ afinite collection C = {Kj}mj=1 of n-cells such that

(1) Z ⊂

m

[

j=1

Kj, and (2)

m

X

j=1

c(Kj) < ε.

Remarks

(a) If K is a cell (not necessarily closed) in Rn, then the boundary ∂K of K is a set of n-content zero.

(b) (i) Let K be a cell in Rn and suppose that K =

m

[

j=1

Kj and Int Ki∩ Int Kj = ∅ for 1 ≤ i 6= j ≤ m.

Then

c(K) =

m

X

j=1

c(Kj).

(ii) Let K1, K2 be cells in Rn. Then

c(K1∪ K2) = c(K1\ K2) + c(K1∩ K2) + c(K2 \ K1).

(iii) Let x ∈ Rn, let K be a cell in Rn and let x + K = {x + z | z ∈ K}. Then x + K is a cell in Rn with c(x + K) = c(K),

i.e. the content of a cell is invariant under translations.

Examples

(a) Let Z = Q∩[0, 1], a (0-dim’l) subset of R. Then (1-dim’l) c(Z) 6= 0 since any finite collection C = {K1, . . . , Km} of 1-dimensional cells covering Z has

m

X

j=1

c(Kj) ≥ 1.

(b) Let Z = {(x, y) | |x| + |y| = 1}, a (1-dim’l) boundary of {(x, y) | |x| + |y| ≤ 1} ⊂ R2. Then the (2-dim’l) content c(Z) = 0.

(3)

Definition Let A ⊂ Rn be a bounded set and let f : A → R be a bounded function. Suppose that K is a closed cell containing A and let fK : K → R be an extension of f defined by

fK(x) =

(f (x) if x ∈ A, 0 if x ∈ K \ A.

We say that f is integrable on A if fK is integrable on K, and define Z

A

f = Z

K

fK.

Theorem Let A = {(x, y) : α(y) ≤ x ≤ β(y), c ≤ y ≤ d}, where α, β : [c, d] → [a, b] are continuous functions from [c, d] into [a, b]. If f : A → R is continuous on A, then f is integrable on A and

Z

A

f = Z d

c

(Z β(y) α(y)

f (x, y)dx )

dy.

Proof Let K be a closed cell containing A, and let fK be the extension of f to K defined by

fK(x) =

(f (x) if x ∈ A 0 if x ∈ K \ A

Since ∂A has content zero, fKis integrable on K. Now for each y ∈ [c, d] the function x 7→ fK(x, y) is continuous except possibly at the two points α(y) and β(y), at which it has one-sided limits.

It follows from the preceding theorem that Z

A

f = Z

K

fK = Z d

c

Z b a

fK(x, y)dx

 dy =

Z d c

(Z β(y) α(y)

f (x, y)dx )

dy.

Examples

(a) Given a prime number p, let A =

 (i

p,j

p) | 1 ≤ i, j ≤ p − 1, p is a prime number



⊆ K = [0, 1] × [0, 1].

(i) Show that each horizontal and each vertical line in R2 intersects A in a finite number (often zero) of points and that A does not have content zero. Let f : K → R be defined by

f (x, y) =

(1 if (x, y) ∈ A 0 if (x, y) ∈ K \ A.

(ii) Show that f is not integrable on K. However, the iterated integrals exist and satisfy Z 1

0

Z 1 0

f (x, y)dx

 dy =

Z 1 0

Z 1 0

f (x, y)dy

 dx.

(4)

Proof Let {Kj}mj=1 be a collection of 2-cells covering A. Since every r ∈ Q ∩ [0, 1] is a limit point of A,

K = [0, 1] × [0, 1] = ¯A ⊆ ∪mj=1j =⇒

m

X

j=1

c(Ki) ≥ 1.

This implies that A does not have content zero and f is not integrable on K.

However, since each horizontal and each vertical line in R2 intersects A in a finite number (often zero) of points, f is continuous on each horizontal and each vertical line in R2 except possibly at a finite number of points there and hence the iterated integrals exist and satisfy

Z 1 0

Z 1 0

f (x, y)dx



dy = 0 = Z 1

0

Z 1 0

f (x, y)dy

 dx.

(b) Let K = [0, 1] × [0, 1] and let f : K → R be defined by

f (x, y) =

0 if either x or y is irrational, 1

n if y is rational and x = m

n where m and n > 0 are relatively prime integers.

Show that

Z

K

f = Z 1

0

Z 1 0

f (x, y)dx



dy = 0, but that R1

0 f (x, y)dy does not exist for rational x.

Proof Divide [0, 1] into n equal length subintervals. Note that if x = p

q is a rational in a subinterval, since j − 1

n ≤ p q ≤ j

n for some 1 ≤ j ≤ n, we have q ≥ n. This implies that the Riemann sum of f with respect to partition P = i − 1

n , i n



× j − 1 n , j

n



| 1 ≤ i, j ≤ n

 satisfies that

|SP(f, K)| = |

n

X

i=1 n

X

j=1

f (xi, yj) · 1 n2| ≤

n

X

i=1 n

X

j=1

1 n · 1

n2 = 1

n → 0 as n → ∞.

Hence f is integrable on K. Similarly, f (x, y) is integrable on [0, 1] for each y ∈ [0, 1] and hence

Z

K

f = Z 1

0

Z 1 0

f (x, y)dx



dy = 0.

For each x = m

n, since the upper and lower sums of f (x, y) with respect to any partition of [0, 1] are 1

n and 0, repectively, f (x, y) is not integrable with respect to y on [0, 1].

(c) Let R denote the triangular region in the first quadrant bounded by the lines y = x, y = 0, and x = 1. Show that

Z 1 0

Z 1 y

sin x

x dxdy = Z 1

0

Z x 0

sin x x dydx.

Proof For each (x, y) ∈ R, consider the function

f (x, y) =

 sin x

x if x 6= 0 1 if x = 0.

(5)

Since lim

x→0

sin x

x = 1, f is a continuous function on R and hence f is integable on R and Z

R

f = Z 1

0

Z 1 y

sin x

x dxdy = Z 1

0

Z x 0

sin x

x dydx = Z 1

0

sin xdx = − cos x|10.

(d) Show that Z 2

0

Z 1 y/2

ye−x3 dx dy = 2

3(1 − e−1).

Proof Let R denote the triangular region in the first quadrant bounded by the lines y = 2x, y = 0, and x = 1. Since f (x, y) = ye−x3 is continuous on R, f is integrable on R and

Z

R

f = Z 2

0

Z 1 y/2

ye−x3 dx dy = Z 1

0

Z 2x 0

ye−x3 dy dx.

Now

Z 1 0

Z 2x 0

ye−x3 dy dx = Z 1

0

y2

2e−x3|2x0 dx

= Z 1

0

2x2e−x3dx

= −2e−x3 3 |10 = 2

3(1 − e−1).

(e) For β > α > 0, let R = [0, ∞) × [α, β] and f (x, t) = e−tx. Show that Z

0

e−αx− e−βx

x dx = logβ α. Proof Since f (x, t) is continuous on R,

0 < f (x, t) ≤ e−αx ∀ (x, t) ∈ R and 0 <

Z

R

e−αx = Z β

α

Z 0

e−αxdx dt = β − α α < ∞, f (x, t) is integrable on R by the M -Test and

Z

R

f = Z β

α

Z 0

e−txdx dt = Z

0

Z β α

e−txdt dx.

Hence logβ

α = Z β

α

1 t dt =

Z β α

Z 0

e−txdx dt = Z

0

Z β α

e−txdt dx = Z

0

e−αx− e−βx

x dx.

The Jacobian Theorem Let Ω ⊆ Rp be open. Suppose that

φ : Ω → Rp belongs to class C1(Ω) is injective on Ω, Jφ(x) 6= 0 for x ∈ Ω,

A has content and ¯A ⊂ Ω.

If ε > 0 is given, then there exists r > 0 such that if K is a closed cell with center x ∈ A and side length less than 2r, then

|Jφ(x)|(1 − ε)p ≤ c(φ(K))

c(K) ≤ |Jφ(x)|(1 + ε)p.

(6)

Proof For each x ∈ Ω, let Lx = (dφ(x))−1, since

1 = det(Lx◦ dφ(x)) = (det Lx) (det dφ(x)), it follows that

det Lx = 1 Jφ(x).

Let Ω1 be a bounded open subset of Ω such that A ⊂ Ω¯ 1 ⊂ ¯Ω1 ⊆ Ω and dist(A, ∂Ω1) = 2δ > 0.

Since φ ∈ C1(Ω), Lx is continuous on the compact subset ¯Ω and there exists a constant M > 0 such that

kLxk ≤ M for all x ∈ Ω1.

Given 0 < ε < 1, since dφ is uniformly continuous on Ω1, there exists β with 0 < β < δ such that

if x1, x2 ∈ Ω1 and kx1− x2k ≤ β, then kdφ(x1) − dφ(x2)k ≤ ε M√

p. This implies that if x ∈ A and if kzk ≤ β, then x, x + z ∈ Ω1 and

kφ(x + z) − φ(x) − dφ(x)(z)k ≤ kzk sup

0≤t≤1

kdφ(x + tz) − dφ(x)k ≤ ε M√

pkzk.

For each x ∈ A, let ψ(z) be defined by

(∗) ψ(z) = Lx[φ(x + z) − φ(x)] for kzk ≤ β.

Since

kψ(z) − zk = kLx[φ(x + z) − φ(x) − dφ(x)(z)]k

≤ M kzk sup

0≤t≤1

kdφ(x + tz) − dφ(x)k

≤ ε

√pkzk for kzk ≤ β, this implies that

ψ(z) ∈ ¯Bεr(z) for kzk ≤ β = √ pr.

If K1 is a closed cell with center 0 and contained in the open ball Bβ(0) for β =√

pr, since ψ is an injective, open mapping for kzk ≤ β, there exist closed cells Ki, Ko both centered 0 and with side lengths 2(1 − ε)r and 2(1 + ε)r, respectively, such that

Ki ⊂ ψ(K1) ⊂ Ko, and

(∗∗) (1 − ε)p ≤ c(ψ(K1))

c(K1) ≤ (1 + ε)p. In general, if K = x + K1 is a closed cell with center x ∈ A, since

c(K1) = c(K),

c(ψ(K1)) = | det Lx|c(φ(x + K1) − φ(x)) by (∗)

= 1

Jφ(x)c(φ(K) − φ(x))

= 1

Jφ(x)c(φ(K))

(7)

and by substituting these into (∗∗), we have (1 − ε)p ≤ c(φ(K))

Jφ(x) c(K) ≤ (1 + ε)p =⇒ |Jφ(x)|(1 − ε)p ≤ c(φ(K))

c(K) ≤ |Jφ(x)|(1 + ε)p.

Change of Variables Theorem Let Ω ⊆ Rp be open. Suppose that φ : Ω → Rp belongs to class C1(Ω) is injective on Ω,

Jφ(x) = det dφ(x) 6= 0 for x ∈ Ω, A has content and ¯A ⊂ Ω and

f : φ(A) → R is bounded and continuous.

Then Z

φ(A)

f = Z

A

(f ◦ φ)|Jφ|.

Examples

(a) Let C ⊂ R3 be the circular helix parametrized by

C : φ(t) = (x(t), y(t), z(t)) = (cos t, sin t, t) for t ∈ [0, 2π].

Evaluate Z

C

y sin z ds, where s = s(t) = Z t

0

0(u)| du.

Solution Since φ : [0, 2π] → C is C1 injective with |Dφ(t)| = |φ0(t)| > 0 for all t ∈ [0, 2π], we have

Z

C

y sin z ds = Z

φ([0,2π])

y sin z ds

= Z

0

y sin z|(x,y,z)=φ(t)0(t)| dt

= Z

0

sin2tp

sin2t + cos2t + 1 dt

= √

2π.

(b) Let S1(r) = {(x, y) ∈ R2 | x2 + y2 = r2} be the counteclockwise oriented circle of radius r with centere at (0, 0). For each k ∈ Z, evaluate

Z

S1(r)

zkdz, where z = x + iy and dz = dx + idy.

Solution By identifying (x, y) ∈ R2 with z = x + iy ∈ C, S1(r) is parametrized by S1(r) : φ(t) = reit = r(cos t + i sin t) for t ∈ [0, 2π].

(8)

Since φ : [0, 2π] → S1(r) is C1 injective with Dφ(t) = φ0(t) = i r eit 6= 0 for all t ∈ [0, 2π], we have

Z

S1(r)

zkdz = Z

φ([0,2π])

zkdz

= Z

0

zk|z=φ(t) φ0(t) dt

= Z

0

rk+1ei(k+1)ti dt

=

(0 if k 6= −1, 2πi if k = −1.

(c) Let S = {(x, y) | x > 0, y > 0, 1 < xy < 3, 1 < x2− y2 < 4}. Evaluate Z Z

S

x2+ y2 dA Solution Let K = {(u, v) | 1 < u < 4, 1 < v < 3} and let φ : S → K be defined by

φ(x, y) = (x2− y2, xy) for (x, y) ∈ S.

Then φ : S → K is C1 bijective with Jφ(x, y) = 2(x2 + y2) > 0 for all (x, y) ∈ S. Hence we have

Z Z

S

x2+ y2 dA = Z Z

φ−1(K)

1

2Jφ(x, y) dA

= Z Z

K

1

2Jφ(x, y)|(x,y)=φ−1(u,v) |Jφ−1(u, v)| du dv

= Z 3

1

Z 4 1

1

2 du dv since Jφ(x, y)Jφ−1(u, v) = 1 ∀ (u, v) ∈ K

= 3.

(d) Show that Z

0

e−x2dx =

√π 2 .

Solution Let Q1 = {(x, y) | x > 0, y > 0}, K = {(r, θ) | r > 0, 0 < θ < π/2} and let φ : K → Q1 be defined by

φ(r, θ) = (x, y) = (r cos θ, r sin θ) for (r, θ) ∈ K.

Then φ : K → Q1 is C1 bijective with Jφ(r, θ) = r ≥ 0 for all (r, θ) ∈ K. Hence we have

Z 0

e−x2dx

2

= Z

0

Z 0

e−x2−y2dx dy

= Z Z

Q1

e−x2−y2dx dy

= Z Z

K

e−x2−y2|(x,y)=φ(r,θ) |Jφ(r, θ)| dr dθ

=

Z π/2 0

Z 0

re−r2dr dθ

= π

4.

(9)

Partition of Unity

Let f : R → R be defined by

f (x) =

(e−1/x if x > 0,

0 otherwise.

x y

Then f is smooth, i.e. infinitely differentiable, on R. For a < b, let g : R → R be defined by g(x) = f (x − a)f (b − x). Then g is smooth and

g(x) =

(e−1/(x−a)e−1/(b−x) if a < x < b,

0 otherwise.

x y

Let h : R → R be defined by

h(x) = R

x g(t) dt R

−∞g(t) dt for x ∈ R.

x y

Then 





h(x) = 1 if x ∈ (−∞, a], 0 < h(x) < 1 if x ∈ (a, b), h(x) = 0 if x ∈ [b, ∞).

For each p ∈ Rn and for any 0 < r < t, let Br(p) and Bt(p) be concentric balls of radius r and t, respectively. Let ζ : R → R be the linear function such that ζ(r2) = a and ζ(t2) = b, and let

ψ(x) = h(ζ(kx − pk2)) for x ∈ Rn.

(10)

Then ψ : Rn→ R is smooth, 0 ≤ ψ(x) ≤ 1 for all x ∈ Rn and

ψ(x) =

(1 if x ∈ Br(p), 0 if x /∈ Bt(p).

Theorem Suppose K is a compact subset of Rn, and {Vα} is an open cover of K. Then there exist functions ψ1, . . . , ψs∈ C(Rn), the space of smooth functions on Rn, such that

(a) 0 ≤ ψi ≤ 1 for 1 ≤ i ≤ s;

(b) each ψi has its support in some Vα, i.e. {x ∈ Rn| ψi(x) 6= 0}⊂ Vα, and (c)

s

X

i=1

ψi(x) = 1 for every x ∈ K.

Because of (c), {ψ} is called a partition of unity, and (b) is sometimes expressed by saying that {ψi} is subordinate to cover {Vα}.

Corollary If f ∈C (Rn) is a continuous function in Rn and the support of f lies in K, then

f =

s

X

i=1

ψif.

Each ψif has its support in some Vα.

Proof For each x ∈ K, since {Vα} is an open cover of K, there exist Vα(x) ∈ {Vα}, open balls B(x) and W (x), centered at x, such that

(∗) B(x) ⊂ W (x) ⊂ W (x) ⊂ Vα(x).

Since K is compact and {B(x) | x ∈ K} is an open cover of K, there are points x1, . . . , xs ∈ K such that

K ⊂ B(x1) ∪ · · · ∪ B(xs).

By (∗), there are functions φ1, . . . , φs ∈C (Rn), such that

φi(x) =

(1 if x ∈ B(xi) 0 if x ∈ Rn\ W (xi) and 0 ≤ φi(x) ≤ 1 for all x ∈ Rn for each 1 ≤ i ≤ s.

Vα

W (xi) W (xj) B(xi) B(xj)

xi xj

(11)

Define

ψ1 = φ1

(†) ψi+1 = (1 − φ1) · · · (1 − φii+1 for i = 1, . . . , s − 1.

Properties (a) and (b) are clear. The relation

(††) ψ1+ · · · + ψi = 1 − (1 − φ1) · · · (1 − φi)

is trivial for i = 1. If (†) holds for some i < s, addition of (†) and (††) yields (††) with i + 1 in place of i. It follows that

s

X

i=1

ψi(x) = 1 −

s

Y

i=1

1 − φi(x)

for x ∈ Rn.

If x ∈ K, then x ∈ B(xj) for some 1 ≤ j ≤ s, hence φj(x) = 1,

s

Y

i=1

1 − φi(x)

= 0 and

s

X

i=1

ψi(x) = 1. This proves (c).

Theorem Let A ⊂ Rn and let O be a collection of open subsets of Rn covering A. Then there is a collection Φ of continuous functions ϕ defined in an open set containing A, with the following properties:

(1) For each x ∈ A we have 0 ≤ ϕ(x) ≤ 1.

(2) For each x ∈ A there is an open set V containing x such that all but finitely many of ϕ ∈ Φ are 0 on V.

(3) For each x ∈ A we have X

ϕ∈Φ

ϕ(x) = 1 (by (2) for each x this sum is finite in some open set containing x).

(4) For each ϕ ∈ Φ there is an open set U ∈ O such that ϕ = 0 outside of some closed set contained in U.

A collection Φ satisfying (1) to (3) is called a continuouspartition of unity for A.If Φ also satisfies (4), it is said to besubordinate to the cover O.In this chapter we will only use continuity of the functions ϕ.

Proof

Case 1. A is compact.

We use an alternative method to construct a continuous partition of unity as follows.

Since A ⊂ Rn is compact, there exists an open ball BR(0) of radius R centered at 0 such that A ⊂ BR(0). By taking U ∩BR(0) for each U ∈O, we may assume that O is a collection of bounded open subsets covering A. Again since A is compact, there exist open sets U1, . . . , Um ∈ O such that

(∗) A ⊂

m

[

j=1

Uj and A \ U1 ∪ · · · ∪ bUi∪ · · · ∪ Um 6= ∅ ∀ 1 ≤ i ≤ m,

where bUi means the term Ui is omitted.

(12)

Since A is compact and by (∗), the set C1 = A \

m

[

j=2

Uj is a compact subset of U1 with

r1 = d(∂U1, C1) = inf

x∈∂U1, y∈C1

d(x, y) > 0.

Let

D1 = {x ∈ U1 | d(x, ∂U1) = inf

y∈∂U1

d(x, y) ≥ r1/2}, W1 = {x ∈ U1 | d(x, ∂U1) = inf

y∈∂U1

d(x, y) ≥ r1/4},

ψ1(x) =





1 if x ∈ D1,

0 if x /∈ W1,

0 ≤ ψ1(x) ≤ 1 ∀ x ∈ Rn.

Note that D1 is a compact subset of U1, C1 ⊂ Int D1 and A ⊂ Int D1 ∪

m

[

j=2

Uj.

Suppose that D1, . . . , Dk have been chosen so that A ⊂

k

[

j=1

Int Dj ∪

m

[

j=k+1

Uj. Let

Ck+1 = A \ (Int D1∪ · · · ∪ Int Dk∪ Uk+2∪ · · · ∪ Um).

Then Ck+1 ⊂ Uk+1 is a compact with

rk+1 = d(∂Uk+1, Ck+1) = inf

x∈∂Uk+1, y∈Ck+1

d(x, y) > 0.

Let

Dk+1 = {x ∈ Uk+1 | d(x, ∂Uk+1) = inf

y∈∂Uk+1

d(x, y) ≥ rk+1/2}, Wk+1 = {x ∈ U1 | d(x, ∂U1) = inf

y∈∂U1

d(x, y) ≥ rk+1/4},

ψk+1(x) =





1 if x ∈ Dk+1,

0 if x /∈ Wk+1,

0 ≤ ψk+1 ≤ 1 ∀ x ∈ Rn.

Note that Dk+1 is a compact subset of Uk+1, Ck+1 ⊂ Int Dk+1 and A ⊂

k+1

[

j=1

Int Dj

m

[

j=k+2

Uj. We obtain a collection of compact subsets {Di}mi=1, {Wi}mi=1 and a collection of nonnegative continuous functions {ψi}mi=1 such that

A ⊂

m

[

i=1

Int Di

m

[

i=1

Di

m

[

i=1

Int Wi

m

[

i=1

Wi

m

[

i=1

Ui, (ψi(x) > 0 if x ∈ Di,

ψi(x) = 0 if x /∈ Wi. =⇒

m

X

j=1

ψj(x) > 0 for all x ∈

m

[

i=1

Di.

(13)

∂Ui

Wi

Ci

Di

For each 1 ≤ i ≤ m, and for x ∈

m

[

i=1

Di, if we let

ϕi(x) = ψi(x)

ψ1(x) + · · · + ψm(x),

then Φ = {ϕ1, . . . , ϕm} is the desired partition of unity since {D1, . . . , Dm} covers A.

Case 2. A = A1∪ A2∪ A3∪ · · · , where each Ai is compact and Ai ⊂ Int Ai+1. For each i ∈ N, let

Bi =

(A1 if i = 1,

Ai\ Int Ai−1 if i ≥ 2, and

Oi =

({U ∩ Int A3 | U ∈O} if 1 ≤ i ≤ 2, {U ∩ (Int Ai+1\ Ai−2) | U ∈O} if i ≥ 3.

· · · A1

Ai−2

Ai−1

Ai

Ai+1

· · · Bi

Then

(i) each Bi is compact and A =

[

i=1

Bi,

(ii) each Oi is a collection of bounded open sets covering Bi, i.e. Bi ⊂ [

U ∈Oi

U.

(14)

(iii) U ∩ V = ∅ for all U ∈Oi, V ∈Oj with j > i + 2 ⇐⇒ j − 1 > i + 1.

Thus, by case 1, there is a partition of unity Φi for Bi, subordinate to Oi.

Note that for each x ∈ A, since x ∈ Bi for some Bi and since Φiis a partition of unity subordinate toOi, so there exists ϕ ∈ Φi such that ϕ(x) > 0, and by (iii),

ϕ(x) = 0 ∀ ϕ ∈ Φj with j > i + 2 ⇐⇒ j − 1 > i + 1, and the sum

σ(x) = X

ϕ∈Φi, all i

ϕ(x)

is a finite sum in some open set containing x, and σ(x) > 0 for all x ∈ A.

Let

Φ =

[

i=1

 ϕ(x)

σ(x) | ϕ ∈ Φi

 . Then Φ is a partition of unity subordinate to the open coverO.

Case 3. A is open.

Let

Ai = {x ∈ A | kxk ≤ i and d(x, ∂A) ≥ 1 i},

where d(x, ∂A) = the distance from x to the boundary ∂A. Note that Aiis compact, Ai ⊂ Int Ai+1 for all i ≥ 1, and

[

i=1

Ai = lim

i→∞Ai = A.

By applying the case 2, we obtain a partition of unity subordinate to the open cover O.

Case 4. A is arbitrary.

Let B be the union of all U in O. By case 3 there is a partition of unity for B; this is also a partition of unity for A.

Remarks

(a) An important consequence of condition (2) of the theorem should be noted. Let C ⊂ A be compact. For each x ∈ C there is an open set Vx containing x such that only finitely many ϕ ∈ Φ are not 0 on Vx. Since C is compact, finitely many such Vx cover C. Thus only finitely many ϕ ∈ Φ are not 0 on C.

(b) One important application of partitions of unity will illustrate their main role-piecing to- gether results obtained locally.

Definition Let A ⊆ Rn be an open subset and let O be an open cover of A. Then O is an admissible open cover if every U ∈O is contained in A.

LetO be an admissible open cover of the open set A ⊆ Rn, Φ be a partition of unity for A subordinate toO, and let f : A → R be bounded in some open set around each point of A.

Then we say f is integrable (in the extended sense) on A if X

ϕ∈Φ

Z

A

ϕ · |f | converges. This

(15)

implies convergence of X

ϕ∈Φ

| Z

A

ϕ · f |, and hence absolute convergence ofX

ϕ∈Φ

Z

A

ϕ · f, which we define to be

Z

A

f, i.e. if X

ϕ∈Φ

Z

A

ϕ · |f | converges, then Z

A

f :=X

ϕ∈Φ

Z

A

ϕ · f.

Furthermore,

• if {x | f is discontinuous at x} has measure 0, since {x | ϕ · |f | is discontinuous at x}

is a compact subset of supp ϕ which has content 0, each Z

A

ϕ · |f | exists,

• if {x | f is discontinuous at x} has measure 0, then f is integrable if and only if |f | is integrable on any closed cell K ⊂ A. So, f is integrable (in the extended sense) if X

ϕ∈Φ

Z

A

ϕ · |f | converges.

• these definitions do not depend on O or Φ.

Remarks

(a) Recall that a subset A of Rn has (n-dimensional)content 0if for every ε > 0 there is afinite cover {K1, . . . , Km} of A by closed (n-dimensional) cells such that

m

X

j=1

c(Ki) < ε. A subset A of Rn has (n-dimensional)measure 0 if for every ε > 0 there is a cover {K1, K2, K3, . . .}

of A by closed (n-dimensional) cells such that

X

j=1

c(Ki) < ε.

A bounded set C whose boundary has measure 0 is calledJordan-measurable.

(b) Theorem If A = A1∪ A2∪ A3∪ · · · and each Ai has measure 0, then A has measure 0.

Proof Let ε > 0. Since Ai has measure 0, there is a cover {Ki,1, Ki,2, Ki,3, . . .} of Ai by closed (n-dimensional) cells such that

X

j=1

c(Ki,j) < ε 2i.

Then the collection of all Ki,j is a cover of A. By considering the array K1,1, K2,1, K1,2, K3,1K2,2, K1,3, K4,1, . . . we see that this collection can be arranged in a sequence

U1, U2, U3, . . . . Clearly

X

i=1

c(Ui) ≤

X

i=1

ε 2i = ε.

(c) Theorem If A is compact and has measure 0, then A has content 0.

Proof Let ε > 0. Since A has measure 0, there is a cover {K1, K2, . . .} of A by open rectangles such that

X

i=1

v(Ki) < ε. Since A is compact, a finite number K1, . . . , Kn of the

Ki also cover A and surely

n

X

i=1

v(Ki) < ε.

Remark Consider the example A = [0, 1] ∩ Q. Note that A is not compact, A has measure 0, and A does not have content 0.

(16)

Theorem

(1) If Ψ is another partition of unity, subordinate to an admissible coverF of A, thenX

ψ∈Ψ

Z

A

ψ ·

|f | also converges, and

X

ϕ∈Φ

Z

A

ϕ · f =X

ψ∈Ψ

Z

A

ψ · f.

(2) If A and f are bounded, then f is integrable in the extended sense.

(3) If A is Jordan-measurable and f is bounded, then this definition of Z

A

f agrees with the old one.

Proof

(1) Since ϕ · f = 0 except on some compact set C, then there are only finitely many ψ ∈ Ψ, which are non-zero on C, and we can write

X

ϕ∈Φ

Z

A

ϕ · f =X

ϕ∈Φ

Z

A

X

ψ∈Ψ

ψ · ϕ · f =X

ϕ∈Φ

X

ψ∈Ψ

Z

A

ψ · ϕ · f.

This result, applied to |f |, shows the convergence of X

ϕ∈Φ

X

ψ∈Ψ

Z

A

ψ · ϕ · |f |,

and hence of

X

ϕ∈Φ

X

ψ∈Ψ

| Z

A

ψ · ϕ · f |.

This absolute convergence justifies interchanging the order of summation in the above equation;

the resulting double sum clearly equals X

ψ∈Ψ

Z

A

ψ · f so that

X

ϕ∈Φ

Z

A

ϕ · f =X

ϕ∈Φ

X

ψ∈Ψ

Z

A

ψ · ϕ · f = X

ψ∈Ψ

X

ϕ∈Φ

Z

A

ϕ · ψ · f =X

ψ∈Ψ

Z

A

ψ · f.

Finally, this result applied to |f | proves convergence of X

ψ∈Ψ

Z

A

ψ · |f |.

(2) If A is contained in the closed cell B and |f (x)| ≤ M for x ∈ A, and F ⊂ Φ is finite, then X

ϕ∈F

Z

A

ϕ · |f | ≤X

ϕ∈F

M Z

A

ϕ ≤ M Z

A

X

ϕ∈F

ϕ ≤ M v(B),

since X

ϕ∈F

ϕ ≤ 1 on A.

(3) If ε > 0, since A is Jordan-measurable, there is a compact Jordan-measurable C ⊂ A such

that Z

A\C

1 < ε,

(17)

and there are only finitely many ϕ ∈ Φ which are non-zero on C. If F ⊂ Φ is any finite collection which includes these, and

Z

A

f has its old meaning, then

Z

A

f −X

ϕ∈F

Z

A

ϕ · f

≤ Z

A

f −X

ϕ∈F

ϕ · f

≤ M Z

A

1 −X

ϕ∈F

ϕ

!

= M Z

A

X

ϕ∈Φ\F

ϕ

≤ M Z

A\C

1

≤ M ε.

Sard’s Theorem Let U ⊂ Rn be open, g : U → Rn∈ C1(U ), and let B = {x ∈ U | det Dg(x) = 0}.

Then g(B) has measure 0.

Proof Let K ⊂ U be a closed (n-dimensional) cell such that all sides of K have length r, say.

Let ε > 0. If N ∈ N is sufficiently large and K is divided into Nn cells, with sides of length r N, then for each of these cells S, if x ∈ S, we have

(∗) kDg(x)(y − x) − g(y) − g(x)k < εky − xk ∀ y ∈ S.

If S ∩ B 6= ∅, we can choose x ∈ S ∩ B, since det Dg(x) = 0, the kernel space of Dg(x) is not trivial and the range set lies in an (n − 1)-dimensional subpace V of Rn, i.e.

{Dg(x)(y − x) | y ∈ S} ⊂ V ⊆ Rn−1, and, by (∗), the set

{g(y) − g(x) | y ∈ S} lies within ε√ nr N of V, so that

{g(y) | y ∈ S} lies within ε√ nr

N of V + g(x).

On the other hand, since g ∈ C1(U ) and by the Mean Value Theorem, there is an M > 0 such that

kg(x) − g(y)k < M kx − yk ≤ M√ nr

N ∀ x, y ∈ S ⊂ K.

Thus if S ∩ B 6= ∅, the set {g(y) | y ∈ S} is contained in a cylinder

• whose height is < 2ε√ nr N , and

• whose base is an (n − 1)-dimesional sphere of radius < M√ nr

N .

(18)

Since this cylinder has volume < C

r N

n

ε for some constant C, and there are at most Nnsuch cells S, so

g(K ∩ B) lies in a set of volume < Cr N

n

εNn= Crnε.

Since this is true for all ε > 0, the set g(K ∩ B) has measure 0. Since we cover U with a sequence of such cells K, g(B) = g(U ∩ B) has measure 0.

Tensors on Vector Space

Definition Let V be a vector space over R, we will denote a k-fold product V × · · · × V by Vk. A function T : Vk → R is calledmultilinear if for each i with 1 ≤ i ≤ k, for all v1, . . . , vk, wi ∈ V and for all a ∈ R, we have

• T (v1, . . . , vi+ wi, . . . , vk) = T (v1, . . . , vi, . . . , vk) + T (v1, . . . , wi, . . . , vk),

• T (v1, . . . , avi, . . . , vk) = aT (v1, . . . , vi, . . . , vk).

A multilinear function T : Vk→ R is called ak-tensor on V and the set of all k-tensors, denoted byT k(V ), becomes a vector space (over R) if for S, T ∈ Tk(V ), and a ∈ R, we define

• (S + T )(v1, . . . , vk) = S(v1, . . . , vk) + T (v1, . . . , vk),

• (aS)(v1, . . . , vk) = a · S(v1, . . . , vk).

There is also an operation connecting the various spacesT k(V ). If S ∈T k(V ) and T ∈ T `(V ), we define the tensor product S ⊗ T ∈Tk+`(V ) by

S ⊗ T (v1, . . . , vk, vk+1, . . . , vk+`) = S(v1, . . . , vk) · T (vk+1, . . . , vk+`)

for all v1, . . . , vk+` ∈ V. Note that the order of the factors S and T is crucial here since S ⊗ T and T ⊗ S are far from equal. It is easy to check that ⊗ satisfies the following properties.

• (S1+ S2) ⊗ T = S1⊗ T + S2⊗ T,

• S ⊗ (T1+ T2) = S ⊗ T1+ S ⊗ T2,

• (aS) ⊗ T = S ⊗ (aT ) = a(S ⊗ T ),

• (S ⊗ T ) ⊗ U = S ⊗ (T ⊗ U ).

Due to the above associative law of ⊗, both (S ⊗ T ) ⊗ U and S ⊗ (T ⊗ U ) are usually denoted S ⊗ T ⊗ U ; higher-order products T1 ⊗ · · · ⊗ Tr are defined similarly.

Theorem Let V be a finite dimensional vector space over R and let V denote the dual space of V. Then

• T1(V ) = V.

• Let v1, . . . , vn be a basis for V, and let ϕ1, . . . , ϕn be the dual basis such that ϕi(vj) = δij =

(1 if i = j, 0 if i 6= j.

Then the set of all k-fold tensor products

i1 ⊗ · · · ⊗ ϕik | 1 ≤ i1, . . . , ik ≤ n}

is a basis ofT k(V ), which therefore has dimension nk.

(19)

Proof If w1, . . . , wk are k vectors with wi =

n

X

j=1

aijvj and T ∈Tk(V ), then

T (w1, . . . , wk) =

n

X

j1,...,jk=1

a1,j1· · · ak,jkT (vj1, . . . , vjk)

=

n

X

i1,...,ik=1

T (vi1, . . . , vik) · ϕi1 ⊗ · · · ⊗ ϕik(w1, . . . , wk).

Thus for each T ∈T k(V ), T =

n

X

i1,...,ik=1

T (vi1, . . . , vik) · ϕi1 ⊗ · · · ⊗ ϕik. Consequently, {ϕi1 ⊗ · · · ⊗ ϕik | 1 ≤ i1, . . . , ik ≤ n} span T k(V ).

Suppose now that there are numbers ai1,...,ik such that

n

X

i1,...,ik=1

ai1,...,ik· ϕi1 ⊗ · · · ⊗ ϕik = 0.

Applying both sides of this equation to (vj1, . . . , vjk) yields aj1,...,jk = 0.

Thus {ϕi1 ⊗ · · · ⊗ ϕik | 1 ≤ i1, . . . , ik ≤ n} are linearly independent.

Let V and W be vector spaces over R and let f : V → W be a linear transformation. Then f

induces a linear transformation

• f : W → V is defined by

fT (v) = T (f(v)) for T ∈ W =T1(W ) and v ∈ V.

• f :T k(W ) →T k(V ) is defined by

fT (v1, . . . , vk) = T (f(v1), . . . , f(vk))) for T ∈Tk(W ) and v1, . . . , vk ∈ V.

It is easy to verify that f(S ⊗ T ) = fS ⊗ fT.

Note that theinner product h , i on Rnis a symmetric, positive definite 2-tensor in T2(Rn). We define an inner product on a vector space V over R, to be a 2-tensor T ∈ T 2(V ) such that T is symmetric, that is T (v, w) = T (w, v) for v, w ∈ V and such that T is positive definite, that is, T (v, v) > 0 if v 6= 0.

Theorem If T is an inner product on V, there is a basis v1, . . . , vn for V such that T (vi, vj) = δij. (Such a basis is called orthonormal with respect to T.) Consequently there is an isomorphism

f : Rn→ V such that

T (f(x), f(y)) = hx , yi for x, y ∈ Rn. In other words fT = h , i.

(20)

Proof Let w1, . . . , wn be any basis for V. Define w10 = w1

w20 = w2− T (w2, w01) T (w10, w01) · w10 w30 = w3− T (w3, w01)

T (w10, w01) · w10 − T (w3, w02) T (w02, w02)· w02

· · · · w0n = wn

n−1

X

i=1

T (wn, wi0) T (wi0, w0i) · w0i It is easy to check that

(T (wi0, w0j) = 0 if i 6= j, T (wi0, w0j) > 0 if i = j.

Now define vi = wi0

pT (w0i, wi0). The isomormorphism f : Rn→ V may be defined by f (ei) = vi. Despite its importance, the inner product plays a far lesser role than another familiar, seemingly ubiquitous function , the tensor det ∈ Tn(Rn). In attempting to generalize this function, we recall that interchanging two rows of a matrix changes the sign of its determinant. This suggests the following definition. A k-tensor w ∈Tk(V ) is called alternating k-tensor on V if

w(v1, . . . ,vi, . . . ,vj, . . . , vk) = −w(v1, . . . ,vj, . . . ,vi, . . . , vk) ∀ v1, . . . , , vk ∈ V.

(In this equation vi and vj are interchanged and all other v’s are left fixed.) The set of all alternating k-tensors is clearly a subspace Λk(V ) of T k(V ). Since it requires considerable work to produce the determinant, it is not surprising that alternating k-tensors are difficult to write down . There is, however, a uniform way of expressing all of them. Recall that the sign of a permutation σ, denoted sgn σ, is defined by

sgn σ =

(+1 if σ is even,

−1 if σ is odd.

If T ∈T k(V ), we define Alt (T ) by

Alt (T )(v1, . . . , vk) = 1 k!

X

σ∈Sk

sgn σ · T (vσ(1), . . . , vσ(k)),

where Sk is the set of all permutations of numbers 1 to k.

Theorem

(1) If T ∈ Tk(V ), then Alt (T ) ∈ Λk(V ).

(2) If w ∈ Λk(V ), then Alt (w) ∈ Λk(V ).

(3) If T ∈ Tk(V ), then Alt (Alt (T )) = Alt (T ).

(21)

Proof

(1) Let (i, j) be the permutation that interchanges i and j and leaves all other numbers fixed. If σ ∈ Sk, let σ0 = σ · (i, j). Since Sk = {σ0 = σ · (i, j) | σ ∈ Sk}, we have

Alt (T )(v1, . . . , vj, . . . , vi, . . . , vk)

= 1

k!

X

σ∈Sk

sgn σ · T (vσ(1), . . . , vσ(j), . . . , vσ(i), . . . , vσ(k))

= 1

k!

X

σ∈Sk

sgn σ · T (vσ0(1), . . . , vσ0(i), . . . , vσ0(j), . . . , vσ0(k))

= 1

k!

X

σ0∈Sk

−sgn σ0· T (vσ0(1), . . . , vσ0(i), . . . , vσ0(j), . . . , vσ0(k))

= −Alt (T )(v1, . . . , vk) (2) Note that if w ∈ Λk(V ), and σ = (i, j), then

w(vσ(1), . . . , vσ(k)) = sgn σ · w(v1, . . . , vk) ∀ v1, . . . , , vk ∈ V.

Since every σ ∈ Sk is a product of permutations of the form (i, j), this equation holds of all σ ∈ Sk. Therefore

Alt (w)(v1, . . . , vk) = 1 k!

X

σ∈Sk

sgn σ · w(vσ(1), . . . , vσ(k))

= 1

k!

X

σ∈Sk

sgn σ · sgn σ · w(v1, . . . , vk)

= w(v1, . . . , vk) (3) follows immediately from (1) and (2).

Definition If w ∈ Λk(V ) and η ∈ Λ`(V ), we define the wedge product w ∧ η ∈ Λk+`(V ) by w ∧ η = (k + `)!

k! `! Alt (w ⊗ η).

It is easy to check that ∧ satisfies the following properties.

• (w1+ w2) ∧ η = w1∧ η + w2∧ η,

• w ∧ (η1+ η2) = w ∧ η1+ w ∧ η2,

• aw ∧ η = w ∧ aη = a(w ∧ η),

• w ∧ η = (−1)k`η ∧ w,

Proof Consider τ ∈ Sk+` defined by

τ (j) =

 1 · · · k k + 1 · · · k + `

` + 1 · · · ` + k 1 · · · k



=

(` + j if 1 ≤ j ≤ k,

j − k if k + 1 ≤ j ≤ k + `,

(22)

and note that sgn τ = (−1)k` and Sk+`= {σ · τ | σ ∈ Sk+`}. Then Alt (η ⊗ w)(v1, . . . , vk+`) = 1

(k + `)!

X

σ·τ ∈Sk+`

sgn (σ · τ ) · η ⊗ w(vσ·τ (1), . . . , vσ·τ (k+`))

= 1

(k + `)!

X

σ∈Sk+`

sgn τ sgn σ · w ⊗ η(vσ(1), . . . , vσ(k+`))

= (−1)k`Alt (w ⊗ η)(v1, . . . , vk+`)

• f(w ∧ η) = f(w) ∧ f(η).

Proof For any v1, . . . , vk+` ∈ V,

f(w ∧ η)(v1, . . . , vk+`) = w ∧ η (f(v1), . . . , f(vk+`))

= 1

(k + `)!

X

σ∈Sk+`

sgn σ · w ⊗ η(f(vσ(1)), . . . , f(vσ(k+`)))

= 1

(k + `)!

X

σ∈Sk+`

sgn σ · fw ⊗ fη(vσ(1), . . . , vσ(k+`))

= f(w) ∧ f(η)(v1, . . . , vk+`)

Theorem

(1) If S ∈T k(V ) and T ∈ T`(V ) and Alt (S) = 0, then

Alt (S ⊗ T ) = Alt (T ⊗ S) = 0.

(2) Alt (Alt (w ⊗ η) ⊗ θ) = Alt (w ⊗ η ⊗ θ) = Alt (w ⊗ Alt (η ⊗ θ)).

(3) If w ∈ Λk(V ), η ∈ Λ`(V ), and θ ∈ Λm(V ), then

(w ∧ η) ∧ θ = w ∧ (η ∧ θ) = (k + ` + m)!

k! `! m! Alt (w ⊗ η ⊗ θ).

Theorem If {vi}ni=1 is a basis for V and {ϕi}ni=1 is the dual basis such that ϕi(vj) = δij =

(1 if i = j, 0 if i 6= j.

Then the set of all

i1 ∧ · · · ∧ ϕik | 1 ≤ i1 < i2 < . . . < ik ≤ n}

is a basis of Λk(V ), which therefore has dimension

n k



= n!

k! (n − k)!.

Proof Since {ϕi1 ⊗ · · · ⊗ ϕik | 1 ≤ i1, . . . , ik≤ n} is a basis of T k(V ), and since ϕi1 ∧ · · · ∧ ϕik = k! Alt (ϕi1 ⊗ · · · ⊗ ϕik),

and ϕi∧ ϕi = 0 for all 1 ≤ i ≤ n, so the set of all

i1 ∧ · · · ∧ ϕik | 1 ≤ i1 < i2 < . . . < ik ≤ n}

is a basis of Λk(V ).

Figure

Updating...

References

Related subjects :