If you find any mistakes, please make a comment! Thank you.

Solution to Principles of Mathematical Analysis Chapter 10


Chapter 10 Integration of Differential Forms

Exercise 2

(By analambanomenos) The φi(x)φi(y) has support in the square 2i<x<21+1, 2i<y<21+1, and the φi+1(x)φi(y) term has support in the rectangle 2i1<x<21, 2i<y<21+1, so f has compact support in the square 0<x<1, 0<y<1. Each (x,y)(0,0) has a neighborhood small enough so that at most three of the terms in the sum are nonzero. Since these terms are continuous, f is continous away from the origin.

Let Mi be the maximum value of φi, attained at xi(2i,2i+1). Since 1=φi<Mi2i, we have Mi>2i. Hence f(xi,xi)=Mi2>2i+1 diverges to as i, so f is not continuous at (0,0) and is unbounded in every neighborhood of (0,0).

We have
dyf(x,y)dx=i=1(φi(y)dy)(φi(x)dxφi+1(x)dx)=i=110=0dxf(x,y)dy=(φ1(x)dx)(φ1(y)dy)+i=2(φi(x)dx)(φi(y)dyφi1(y)dy)=11+i=210=1


Exercise 3

(By analambanomenos)

(a) (Much of this solution just repeats the proof of Theorem 10.7.) Assume 1mn1, and make the following induction hypothesis (which evidently holds for m=1):

Vm is a neighborhood of 0, FmC(Vm), Fm(0)=0, Fm(0)=I, and for xVm,
(1)Pm1Fm(x)=Pm1x.By (1), we have
Fm(x)=Pm1x+i=mnαi(x)ei,where αm,,αn are real C-functions in Vm. Hence, for j=m,,n,
ej=Fm(0)ej=i=mn(Djαi)(0)ei.Since em,en are independent, we must have
(2)(Dmαm)(0)=1(Dm+1αm)(0)==(Dnαm)(0)=0.Define, for xVm,
Gm(x)=x+(αm(x)xm)emThen GmC(Vm), Gm is primitive, and Gm(0)=I by (2). The inverse function theorem shows therefore that there is an open set Um, with 0UmVm, such that Gm is a 1-1 mapping of Um onto a neighborhood Vm+1 of 0, in which Gm1 is continuously differentiable, and
Gm1(0)=Gm(0)1=I.Define Fm+1(y), for yVm+1, by
Fm+1(y)=FmGm1(y).Then Fm+1C(Vm+1), Fm+1(0)=0, and Fm+1(0)=I by the chain rule. Also, for xUm,
PmFm+1(Gm(x))=PmFm(x)=Pm(Pm1x+αm(x)em+)=Pm1x+αm(x)em=PmGm(x)so that, for yVm+1, PmFm+1(y)=Pmy. Our induction hypothesis holds therefore with m+1 in place of m.

Note that, for y=Gm(x), we have
Fm+1(Gm(x))=Fm(x).If we apply this with m=1,,n1, we successively obtain
F1=F2G1=F3G2G1==FnGn1G1in some neighborhood of 0. By (1), Fn is primitive, so we can let Gn=Fn.

(b) Let F be the mapping (x,y)(y,x) and suppose F=G2G1 in some neighborhood of the origin, where
G1(x,y)=(f(x,y),y)G2(u,v)=(u,g(u,v))are primitive mappings. Then we would have
(y,x)=G2G1(x,y)=G2(f(x,y),y)=(f(x,y),g(f(x,y),y))so that
y=f(x,y)x=g(f(x,y),y)=g(y,y)which is impossible. Trying F=G1G2 leads to a similar contradiction.


Exercise 4

(By analambanomenos) We have
G2G1(x,y)=G2(excosy1,y)=(excosy1,excosytany)=(excosy1,exsiny)=F(x,y)The derivative matrices are
G1(x,y)=(excosyexsiny01)G2(u,v)=(10tanv(1+u)cos2v)so that G1(0,0)=G2(0,0)=I, hence JG1(0,0)=JG2(0,0)=1. By the chain rule and the properties of determinants, we also have JF(0,0)=1.

Let h(u,v)=v2e2u1. Then, for (x,y) near the origin,
H1H2(x,y)=H1(x,exsiny)=(e2xsin2ye2x1,exsiny)=(excosy1,exsiny)=F(x,y)


Exercise 5

(By analambanomenos) We want to show: Suppose K is a compact subset of a metric space X, and {Vα} is an open cover of K. Then there exists ψ1,,ψsC(X) such that

(a) 0ψi1 for 1is;
(b) each ψi has its support in some Vα, and
(c) ψi(x)++ψs(x)=1 for every xK.

Repeating the proof of Theorem 10.8 in the text and following the hint, associate with each xK an index α(x) so that xVα(x). Then there are open balls B(x) and W(x) centered at x, with
B(x)W(x)W(x)Vα(x).Since K is compact, there are points x1,,xs in K such that
KB(x1)B(xs).By Exercise 4.22, there are functions φ1,φsC(X) such that φi(x)=1 on B(xi), φi(x)=0 outside W(xi), and 0φi(x)1 on X, namely,
φi(x)=ρi1(x)ρi1(x)+ρi2(x)where ρi1(x) is the distance from x to the complement of W(xi), a closed set, and ρi2(x) is the distance from x to B(xi). Letting ψ1=φ1, and
ψi+1=(1φ1)(1φi)φi+1for i=1,,s1, the remainder of the proof follows exactly as in the proof of Theorem 10.8.


Exercise 6

(By analambanomenos) Following the hint, recall that Exercise 8.1 defined an infinitely differentiable function on R1 such that f(x)=0 for x0 and 0f(x)<1 for all x. Let a<b. Then the function ga,b(x)=f(xa)f(bx) is also infinitely differentiable, equals 0 for xa and xb, and 0ga,b(x)<1 for all x. Since it has compact support, we can define a function
ha,b(x)=1Axga,b(t)dtwhereA=ga,b(t)dtwhich is infinitely differentiable, equals 1 for xa, equals 0 for xb, and 0ha,b(x)1 for all x.

Now let xRn, and let B(x) and W(x) be open balls centered at x with radii a<b, respectively. Define the function r(y) for yRn which is the distance between x and y, that is,
r(y)=(xiyi)2which is infinitely differentiable for yx. Then the function φ=ha,br is infinitely differentiable on Rn, equals 1 on B(x), equals 0 on W(x), and 0φ(y)1 for all y. We can use these functions in the proof of Theorem 10.8 to get infinitely differentiable functions ψi.


Exercise 7

(By analambanomenos)

(a) First we need to show that Qk is convex. Let x,yQk, so that the components satisfy
xi0,yi0,xi1,yi1.Let 0λ1, and let z=λx+(1λ)y. Then
zi=λxi+(1λ)yizi=λxi+(1λ)yiso that zi lies between xi and yi, and zi lies between xi and yi. Hence zQk.

Let C be a convex subset of Rk containing 0,e1,,ek; we need to show that QkC. We can consider QiQj for i<j by letting the components with index greater than i be 0. I am going to show that QiC, i=1,,k by induction. Let xQ1. Then x=x1e1+(1x1)0 for 0x11, so that xC. Now suppose that Qi1C and let xQi. Then x1++xi1 implies
(x1++xi1)1xi1.so that
x=(1xi)1(x1,,xi1,0,,0)Qi1C.Hence x=(1xi)x+xieiC since 0xi1, which shows that QiC.

(b) Let X,Y be vector spaces and let f=f(0)+A, for some AL(X,Y), be an affine mapping from X to Y. Let C be a convex subset of X, and let y1=f(x1), y2=f(x2) be elements of f(C) for some x1C and x2C. Then for 0λ1, we have λx1+(1λ)x2C, so that
λy1+(1λ)y2=f(0)+λA(x1)+(1λ)A(x2)=f(0)+A(λx1+(1λ)x2)f(C).Hence f(C) is convex.


Exercise 8

(By analambanomenos) Since (3,2)=(1,1)+(2,1) and (2,4)=(1,1)+(1,3), the linear part of the affine map is A(u,v)=(2u+v,u+3v), so
T(u,v)=(1,1)+A(u,v)=(2u+v+1,u+3v+1)JT=|2113|=5Hexydxdy=0101e(2u+v+1)(u+3v+1)JTdudv=5(01eudu)(01e2vdv)=52(ee1+e21)

Baby Rudin 数学分析原理不完整第十章习题解答

Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.
Close Menu
Close Menu