.
.
.
.
.
.
.
.
.
.
find
.
Since
, one has
.
.
.
.
.
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
.