Exercise 2
(By analambanomenos) The has support in the square , , and the term has support in the rectangle , , so has compact support in the square , . Each has a neighborhood small enough so that at most three of the terms in the sum are nonzero. Since these terms are continuous, is continous away from the origin.
Let be the maximum value of , attained at . Since , we have . Hence diverges to as , so is not continuous at and is unbounded in every neighborhood of .
We have
Exercise 3
(By analambanomenos)
(a) (Much of this solution just repeats the proof of Theorem 10.7.) Assume , and make the following induction hypothesis (which evidently holds for ):
is a neighborhood of , , , , and for ,
By , we have
where are real -functions in . Hence, for ,
Since are independent, we must have
Define, for ,
Then , is primitive, and by . The inverse function theorem shows therefore that there is an open set , with , such that is a 1-1 mapping of onto a neighborhood of , in which is continuously differentiable, and
Define , for , by
Then , , and by the chain rule. Also, for ,
so that, for , . Our induction hypothesis holds therefore with in place of .
Note that, for , we have
If we apply this with , we successively obtain
in some neighborhood of . By , is primitive, so we can let .
(b) Let be the mapping and suppose in some neighborhood of the origin, where
are primitive mappings. Then we would have
so that
which is impossible. Trying leads to a similar contradiction.
Exercise 4
(By analambanomenos) We have
The derivative matrices are
so that , hence . By the chain rule and the properties of determinants, we also have .
Let . Then, for near the origin,
Exercise 5
(By analambanomenos) We want to show: Suppose is a compact subset of a metric space , and is an open cover of . Then there exists such that
(a) for ;
(b) each has its support in some , and
(c) for every .
Repeating the proof of Theorem 10.8 in the text and following the hint, associate with each an index so that . Then there are open balls and centered at , with
Since is compact, there are points in such that
By Exercise 4.22, there are functions such that on , outside , and on , namely,
where is the distance from to the complement of , a closed set, and is the distance from to . Letting , and
for , 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 such that for and for all . Let . Then the function is also infinitely differentiable, equals 0 for and , and for all . Since it has compact support, we can define a function
which is infinitely differentiable, equals 1 for , equals 0 for , and for all .
Now let , and let and be open balls centered at with radii , respectively. Define the function for which is the distance between and , that is,
which is infinitely differentiable for . Then the function is infinitely differentiable on , equals 1 on , equals 0 on , and for all . We can use these functions in the proof of Theorem 10.8 to get infinitely differentiable functions .
Exercise 7
(By analambanomenos)
(a) First we need to show that is convex. Let , so that the components satisfy
Let , and let . Then
so that lies between and , and lies between and . Hence .
Let be a convex subset of containing ; we need to show that . We can consider for by letting the components with index greater than be 0. I am going to show that , by induction. Let . Then for , so that . Now suppose that and let . Then implies
so that
Hence since , which shows that .
(b) Let be vector spaces and let , for some , be an affine mapping from to . Let be a convex subset of , and let , be elements of for some and . Then for , we have so that
Hence is convex.
Exercise 8
(By analambanomenos) Since and , the linear part of the affine map is , so
Baby Rudin 数学分析原理不完整第十章习题解答