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Solution to Principles of Mathematical Analysis Chapter 9 Part A


Chapter 9 Functions of Several Variables

Exercise 1

(By analambanomenos) Let x,y be in the span of S and let c be a scalar; we need to show that x+y and cx are in the span of S. We have
x=c1x1++cmxmy=d1y1++dnyn for some x1,,xm,y1,,yn in S and some scalars c1,,cm,d1,,dn. Hence
x+y=c1x1++cmxm+d1y1++dnyncx=cc1x1++ccmxmshows that x+y and cx are in the span of S.


Exercise 2

(By analambanomenos) Let AL(X,Y) and BL(Y,Z). Let x,yX and let c be a scalar. Then
(BA)(x+y)=B(A(x+y))=B(A(x)+A(y))=B(A(x))+B(A(y))=(BA)(x)+(BA)(y)(BA)(cx)=B(A(cx))=B(cA(x))=cB(A(x))=c(BA)(x)which shows that BA is linear.

Let AL(X) be invertible. Let x,yX and let c be a scalar. Then there are x,yX such that
A(x)=x,A(y)=y,A1(x)=x,A1(y)=yThen
A1(x+y)=A1(A(x)+A(y))=A1(A(x+y))=x+y=A1(x)+A1(y)A1(cx)=A1(cA(x))=A1(A(cx))=cx=cA1(x)which shows that A1 is linear. Since A1 is 1-1 and maps X onto itself, it is also invertible.


Exercise 3

(By analambanomenos) Suppose A(x1)=A(x2). Then A(x1x2)=A(x1)A(x2)=0 implies that x1x2=0 so that x1=x2. Hence A is 1-1.


Exercise 4

(By analambanomenos) Let x,yN(A) and let c be a scalar. Then
A(x+y)=A(x)+A(y)=0+0=0 so that x+yN(A)A(cx)=cA(x)=c0=0 so that cxN(A) which shows that N(A) is a vector space.

Let x,yR(A), so that x=A(x) and y=A(y), and let c be a scalar. Then
x+y=A(x)+A(y)=A(x+y) so that x+yR(A)cx=cA(x)=A(cx) so that cxR(A)which shows that R(A) is a vector space.


Exercise 5

(By analambanomenos) I’m going to show this by induction on n. For the case n=1, let Ae1=k, and let y=ke1. If xR1, then x=xe1 for some scalar x, so Ax=xAe1=ck=xy.

Suppose the assertion is true for the case n and let AL(Rn+1,R1). Restricting A to the subspace Rn (spanned by e1,,en) yields an element of L(Rn,R1), so by the induction assumption there is a yRn such that Ax=xy for all xRn. If Aen1=k, let y=y+ken+1. If xRn+1, then x=x+xen+1 for some xRn and some scalar x. Then, since the scalar product of en+1 with every element of Rn is 0, we have
Ax=Ax+xAen+1=xy+xk=(x+xen+1)y+(x+xen+1)(ken+1)=xyIf |x|1, then |Ax|=|xy||x||y||y|, so that A|y|. Let y~=y/|y|. Then |y~|=1 and
|Ay~|=|yy||y|=|y|2|y|=|y|.Hence A=|y|.


Exercise 6

(By analambanomenos) The usual rules of differentiaion show that the partial derivatives of R2 at (x,y)(0,0) areD1f(x,y)=y(y2x2)(x2+y2)2D2f(x,y)=x(x2y2)(x2+y2)2And since f is equal to 0 everywhere along the x and y axes, they also exist and are equal to 0 at (0,0).

For x0, f(x,x)=x2/(x2+x2)=1/2. Since f(0,0)=0, f is not continuous along the line y=x at (0,0).


Exercise 7

(By analambanomenos) Suppose that |Djf|<M in E, for j=1,,n. Following the hint to mimic the proof of Theorem 8.21, fix xE and let ε>0. Since E is open, there is an open ball SE, with center at x and radius r<(Mn)1. Suppose h=hjej, |h|<r, put v0=0, and vk=h1e1++hkek, for 1kn. Then
(1)f(x+h)f(x)=j=1n(f(x+vj)f(x+vj1)).Since |vk|<r for 1kn and since S is convex, the segments with endpoints x+vj1 and x+vj lie in S. Since vj=vj1+hjej, the mean value theorem, Theorem 5.10, shows that the jth summand in (1) is equal to
hj(Djf)(x+vj1+θjhjej) for some θj(0,1). By (1), it follows that
|f(x+h)f(x)|j=1n|f(x+vj)f(x+vj1)|=j=1n|hj(Djf)(x+vj1+θjhjej)|<j=1n|hj|M<(Mn)r<εwhich shows that f is continuous at x. Since xE was arbitrary, we have f continuous on E.


Exercise 8

(By analambanomenos) You can use Theorem 9.17 to express f as a sum of the partial derivatives and easily reduce the problem to the the single-variable case, Theorem 5.8. However, I thought I’d use the new definition of derivative (commonly called a Fr\’echet derivative, by the way) instead.

By Exercise 5, there is a yRn such that f(x)=yx. Let h=hy/|y|, and take the limit in the definition of derivative as h approaches 0 through positive numbers. Then we get
0=limh0|f(x+h)f(x)f(x)h||h|=limh0+|f(x+h)f(x)y(hy)/|y||h|y|/|y|=limh0+|f(x+h)f(x)h|y||h(since yy=|y|2)=limh0+f(x)f(x+h)+h|y|h(since f(x+h)f(x) and h|y| are 0)=|y|+limh0+f(x)f(x+h)hSince the term in the limit is non-negative for small h, this forces |y|0, or y=0. Hence f(x)=0.


Exercise 9

(By analambanomenos) Fix xE. Since E is open, there is a open ball SE containing x. By the Corollary to Theorem 9.19, f is constant on S. Hence the set E of all points zE such that f(z)=f(x) is an open subset of E. Similarly, the set EE is also open in E (being the union of open sets on which f has a constant value not equal to f(x)). By Exercise 2.19(b), E and EE are two separated sets whose union is E. Since E is connected, we must have E=E, so f is constant on E.


Exercise 10

(By analambanomenos) Let E satisfy the weaker condition: if (x,x2,,xn)E and (y,x2,,xn)E, then for all x<z<y we have (z,x2,,xn)E. For such points, define fx2,,xn(z)=f(z,x2,,xn), then fx2,,xn(z)=(D1f)(z,x2,,xn)=0, so by the Mean Value Theorem fx2,,xn must be constant on [x,y]. Hence f(z,x2,,xn) must equal this constant value for all z such that (z,x2,,xn)E, so that f(x) depends only on x2,,xn.

To see a counterexample, let ER2 be the square (x,y), 1<x<1, 1<y<1 less the positive y-axis, (0,y), 0<y<1. Define
f(x,y)={01<x<1,1<y<0y1<x<0,0y<1y0<x<1,0y<1Then f is continuous, (D1f)(x,y)=0 for all (x,y)E, but the value of f does not depend only on y.


Exercise 11

(By analambanomenos) We have (fg)=i=1n(Di(fg))ei=i=1n(f(Dig)+g(Dif))ei=fi=1n(Dig)ei+gi=1n(Dif)ei=fg+gf0=(1)=(ff1)=f(f1)+f1ff(f1)=f1f(f1)=f2f


Exercise 12

(By analambanomenos) The range of f is a torus with inner radius ba and outer radius b+a, centered at the origin, whose plane is perpendicular to the z-axis. To see this, the central circle of such a torus is the set of points (bcost,bsint,0). Then the circle of radius a which intersects the half-plane that goes through the z-axis and the point (bcost,bsint,0) is the circle centered at that point and perpendicular to the x-y-plane, with radius vector a(cost,sint,0), which is described by (bcost,bsint,0)+a(cost,sint,0)coss+a(0,0,1)sins=(f1(s,t),f2(s,t),f3(s,t)).Note that f is a one-to-one mapping of the square S given by (x,y), 0x<2π, 0y<2π onto the torus.

(a) Taking the partial derivatives of f1 with respect to s and t to get f1, we want to solve(f1)(s,t)=(asinscost,(b+acoss)sint)=(0,0).Restricing ourselves to the square S, the first component equals 0 when either s=0, s=π, t=π/2, or t=3π/2. Since b+acoss>0 for all s, the second component equals 0 only when t=0 or t=π. Hence f1 equals 0 in S only at the four points(0,0),(0,π),(π,0),(π,π),which map onto the four points(b+a,0,0),(ba,0,0),(ba,0,0),(ab,0,0),respectively.

(b) Taking the partial derivatives of f3 with respect to s and t to get f3, we want to solve(f3)(s,t)=(acoss,0)=(0,0).which occurs in S at the points (s,t), where s=π/2 or s=3π/2, and 0t<2π. The points (π/2,t), 0t<2π, are mapped by f to the circle (bcost,bsint,a), and the points (3π/2,t), 0t<2π, are mapped to the circle (bcost,bsint,a).

(c) At (b+a,0,0), corresponding to (s,t)=(0,0), where coss=cost=1, f1 attains its maximum value, b+a, and at (ba,0,0), corresponding to (s,t)=(0,π), where coss=1 and cost=1, f1 attains its minimum value, ba. At (ba,0,0), corresponding to (s,t)=(π,0), where coss=1 and cost=1, f1 increases if you fix t=0 and vary s, and decreases if you fix s=π and vary t. Similarly, at (ab,0,0, corresponding to (s,t)=(0,π), where coss=1 and cost=1, f1 decreases if you fix t=π and vary s, and increases if you fix s=π and vary t.

The points (bcost,bsint,a), 0t<2π, correspond to the points (s,t)=(π/2,t) where f3 attains its maximum value of a, and (bcost,bsint,a), 0t<2π, correspond to the points (s,t)=(3π/2,t) where f3 attains its minimum value of a.

(d) To show that the “irrational winding of the torus” (it even has its own Wikipedia page) is dense, I first show a general proposition about continuous functions. Suppose F is a continuous mapping from X to Y (both metric, or even just topological spaces). Then if D is a dense subset of X, then F(D) is a dense subset of F(Y). For if not, then there would be a nonempty open set G in F(Y) disjoint from F(D). Then F1(D) would be an open subset of X disjoint from D, contradicting the density of D.

Hence, to show that the image of g is dense in the torus K, it suffices to show that its preimage under the mapping f is dense in R2. Call this set K, it is the set of points of the form(s+m2π,λs+n2π),s real, m and n integers.First I want to show that the intersection of K with the s-axis is dense on the axis. If (s,0) is a point in this set, then from the above, we that λs is a multiple of 2π, so s=(nλ1+m)2π for some integers m and n. To see that this set is dense, I first show another general proposition.

If α is an irrational number, then the set of points D in [0,1] which are equivalent to integral multiples of α modulo 1 are dense in that interval. That is, let n be an integer, then nα is in the interval (m,m+1) for some integer m, so that xn=nαm is between 0 and 1. These points are all distinct, since if xn1=n1αm1=xn2=n2αm2, for some integers n1n2, then α=(m1m2)/(n2n1) would be rational. Hence D is an infinite set.

Another simple fact is that if xn1<xn2 are two points in D then 0<xn2xn1<1, so that xn2xn1=(n2n1)α(m2m1) is also in D. Also, if 0<jxn<1 for some positive integer j, then jxn=jnαjm is also in D.

Let ε>0 and let k be an integer large enough so that k1<ε. Divide [0,1] into k intervals Ij=[jk1,(j+1)k1]j=0,,(k1)(This isn’t strictly a partition, but I am ignoring the points on the boundaries, which are rational numbers which cannot be in D.) Since D is infinite, one of these intervals contains at least two points in D, say xn1<xn2. Then xn2xn1 is a point xnD such that 0<xn<k1. Hence each of the intervals Ij, j=0,,(k1), contains a point jxn of D. If x[0,1], then x is in one of the Ij, so there is a point xnD such that |xxn|<k1<ε. Hence D is dense in [0,1].

Going back to the intersection of K with the s-axis, it is not hard to see that the intersection of K with the interval [0,2π] is the image of the set D above, with α=λ1, under the mapping s2πs, hence K is dense in the interval [0,2π] in the s-axis. Since the other points of K on the s-axis are translates of this dense set by multiples of 2π, we see that K is dense in the entire axis.

For other lines in R2 parallel to the s-axis, (s,t)K if (sλ1t,0)K, that is, the intersection of K with this line is the image of the dense interesection of K with the s-axis under the translation (s,0)(s+λ1t,t). Hence K is also dense in this line, and so is dense in all of R2.

If g(t1)=g(t2), then f(t1,λt1)=f(t2,λt2) which would only happen if (t1t2)=n2π and λ(t1t2)=m2π for some integers m and n. But that would imply λ=m/n which contradicts the irrationality of λ.

By Theorem 9.17 we have
f(s,t)e1=((D1f1)(s,t),(D1f2)(s,t),(D1f3)(s,t))=(asinscost,asinssint,acoss)f(s,t)e2=((D2f1)(s,t),(D2f2)(s,t),(D2f3)(s,t))=((b+acoss)sint,(b+acoss)cost,0). If we let γ(t)=(t,λt), so that γ(t)=e1+λe2, then we have by the chain rule
g(t)=f(γ(t))γ(t)=f(t,λt)(e1+λe2)=(asintcosλλ(b+acost)sinλ,asintsinλt+λ(b+acost)cosλt,acost)|g(t)|2=a2sin2tcos2λt+2λa(b+acost)sintcosλtsinλt+λ2(b+acost)2sin2λt+=a2sin2tsin2λt2λa(b+acost)sintsinλtcosλt+λ2(b+acost)2cos2λt+=a2cos2t=a2+λ2(b+acost)2.

Baby Rudin 数学分析原理完整第九章习题解答

Linearity

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