Chapter 9 Functions of Several Variables
Exercise 1
(By analambanomenos) Let be in the span of and let be a scalar; we need to show that and are in the span of . We have
for some in and some scalars . Hence
shows that and are in the span of .
Exercise 2
(By analambanomenos) Let and . Let and let be a scalar. Then
which shows that is linear.
Let be invertible. Let and let be a scalar. Then there are such that
Then
which shows that is linear. Since is 1-1 and maps onto itself, it is also invertible.
Exercise 3
(By analambanomenos) Suppose . Then so that . Hence is 1-1.
Exercise 4
(By analambanomenos) Let and let be a scalar. Then
which shows that is a vector space.
Let , so that and , and let be a scalar. Then
which shows that is a vector space.
Exercise 5
(By analambanomenos) I’m going to show this by induction on . For the case , let , and let . If , then for some scalar , so .
Suppose the assertion is true for the case and let . Restricting to the subspace (spanned by ) yields an element of , so by the induction assumption there is a such that for all . If , let . If , then for some and some scalar . Then, since the scalar product of with every element of is 0, we have
If , then , so that . Let . Then and
Hence .
Exercise 6
(By analambanomenos) The usual rules of differentiaion show that the partial derivatives of at areAnd since is equal to 0 everywhere along the and axes, they also exist and are equal to 0 at .
For , . Since , is not continuous along the line at .
Exercise 7
(By analambanomenos) Suppose that in , for . Following the hint to mimic the proof of Theorem 8.21, fix and let . Since is open, there is an open ball , with center at and radius . Suppose , , put , and , for . Then
Since for and since is convex, the segments with endpoints and lie in . Since , the mean value theorem, Theorem 5.10, shows that the th summand in is equal to
for some . By , it follows that
which shows that is continuous at . Since was arbitrary, we have continuous on .
Exercise 8
(By analambanomenos) You can use Theorem 9.17 to express 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 such that . Let , and take the limit in the definition of derivative as approaches 0 through positive numbers. Then we get
Since the term in the limit is non-negative for small , this forces , or . Hence .
Exercise 9
(By analambanomenos) Fix . Since is open, there is a open ball containing . By the Corollary to Theorem 9.19, is constant on . Hence the set of all points such that is an open subset of . Similarly, the set is also open in (being the union of open sets on which has a constant value not equal to ). By Exercise 2.19(b), and are two separated sets whose union is . Since is connected, we must have , so is constant on .
Exercise 10
(By analambanomenos) Let satisfy the weaker condition: if and , then for all we have . For such points, define , then , so by the Mean Value Theorem must be constant on . Hence must equal this constant value for all such that , so that depends only on .
To see a counterexample, let be the square , , less the positive -axis, , . Define
Then is continuous, for all , but the value of does not depend only on .
Exercise 11
(By analambanomenos) We have
Exercise 12
(By analambanomenos) The range of is a torus with inner radius and outer radius , centered at the origin, whose plane is perpendicular to the -axis. To see this, the central circle of such a torus is the set of points . Then the circle of radius which intersects the half-plane that goes through the -axis and the point is the circle centered at that point and perpendicular to the --plane, with radius vector , which is described by Note that is a one-to-one mapping of the square given by , , onto the torus.
(a) Taking the partial derivatives of with respect to and to get , we want to solveRestricing ourselves to the square , the first component equals 0 when either , , , or . Since for all , the second component equals 0 only when or . Hence equals in only at the four pointswhich map onto the four pointsrespectively.
(b) Taking the partial derivatives of with respect to and to get , we want to solvewhich occurs in at the points , where or , and . The points , , are mapped by to the circle , and the points , , are mapped to the circle .
(c) At , corresponding to , where , attains its maximum value, , and at , corresponding to , where and , attains its minimum value, . At , corresponding to , where and , increases if you fix and vary , and decreases if you fix and vary . Similarly, at , corresponding to , where and , decreases if you fix and vary , and increases if you fix and vary .
The points , , correspond to the points where attains its maximum value of , and , , correspond to the points where attains its minimum value of .
(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 is a continuous mapping from to (both metric, or even just topological spaces). Then if is a dense subset of , then is a dense subset of . For if not, then there would be a nonempty open set in disjoint from . Then would be an open subset of disjoint from , contradicting the density of .
Hence, to show that the image of is dense in the torus , it suffices to show that its preimage under the mapping is dense in . Call this set , it is the set of points of the formFirst I want to show that the intersection of with the -axis is dense on the axis. If is a point in this set, then from the above, we that is a multiple of , so for some integers and . To see that this set is dense, I first show another general proposition.
If is an irrational number, then the set of points in which are equivalent to integral multiples of modulo 1 are dense in that interval. That is, let be an integer, then is in the interval for some integer , so that is between 0 and 1. These points are all distinct, since if , for some integers , then would be rational. Hence is an infinite set.
Another simple fact is that if are two points in then , so that is also in . Also, if for some positive integer , then is also in .
Let and let be an integer large enough so that . Divide into intervals (This isn’t strictly a partition, but I am ignoring the points on the boundaries, which are rational numbers which cannot be in .) Since is infinite, one of these intervals contains at least two points in , say . Then is a point such that . Hence each of the intervals , , contains a point of . If , then is in one of the , so there is a point such that . Hence is dense in .
Going back to the intersection of with the -axis, it is not hard to see that the intersection of with the interval is the image of the set above, with , under the mapping , hence is dense in the interval in the -axis. Since the other points of on the -axis are translates of this dense set by multiples of , we see that is dense in the entire axis.
For other lines in parallel to the -axis, if , that is, the intersection of with this line is the image of the dense interesection of with the -axis under the translation . Hence is also dense in this line, and so is dense in all of .
If , then which would only happen if and for some integers and . But that would imply which contradicts the irrationality of .
By Theorem 9.17 we have
If we let , so that , then we have by the chain rule
Baby Rudin 数学分析原理完整第九章习题解答