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find .
Since , one has .
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True since the first term depends only on .
One could let .
One could let .
Find one such that and .
One could let .
By the mean value theorem, one knows that if a function of one variable has zero derivative on a closed interval , then it is constant on that interval. Both assertions are immediate consequences of this result.
Suppose and are arbitrary points of . Then the line segment from to , from to , and from to are all contained in . By the proof of Problem 2-22, it follows that , , and . So .
One could let .
One has when , and since . Further, one has and so for all and (because ). The assertions follow immediately by substituting into one formula and in the other.
Using part (a), one has and .
Consider the function defined by
where is a polynomial. Then, one has for . By induction on , define and . Then it follows that for all . In particular, function is for all .
Now, suppose by induction on , that . We have . To show that this limit is zero, it suffices to show that for each integer . But this is an easy induction using L'Hôpital's rule: .
For points other than 1 and -1, the result is obvious. At each of the exceptional points, consider the derivative from the left and from the right, using Problem 2-25 on the side closest to the origin.
Following the hint, let be as in Problem 2-25 and Now use Problem 2-25 to prove that works.
Show that is a function which is positive on
and zero elsewhere.
This follows from part (a).
Let be the distance between and the complement of , and choose For each , let be the open rectangle centered at with sides of length . Let be the function defined for this rectangle as in part (c). Since the set of these rectangles is an open cover of the compact set , a finite number of them also cover ; say the rectangles corresponding to form a subcover. Finally, let .
Since we have a subcover of , we have positive on . The choice of guarantees that the union of the closures of the rectangles in the subcover is contained in and is clearly zero outside of this union. This proves the assertion.
Let be as in part (d). We know that for all . Since is compact, one knows that is attains its minimum (Problem 1-29). As suggested in the hint, replace with where is the function of part (b). It is easy to verify that this new satisfies the required conditions.
Show that the maximum of on is either the maximum of or the maximum of on .
This is obvious because .