Solution to GTM249 Chapter 2 Part I

2.1 Maximal Functions

2.1.1

(a) Similar to the proof of Theorem 2.1.6, except for the only difference

(b)

2.1.2

(a) Select a subset of with the minimal cardinality such that . Assume that is connected, otherwise consider every connected component.
Write . Let . We claim that is a family of mutually disjoint intervals. Otherwise there exists such that . If , then , so can be removed; if , then , so can be removed. In both cases the minimality of fails.
Similarly we can prove that is a family of mutually disjoint intervals.
Similar to the proof of Theorem 2.1.6, except for the only difference (b) The first inequality is equivalent to This inequality holds because the integrand is nonpositive whether or .
Set , the following notation follows from the proof of part (a).
Let , where is a compact subset of . We haveTake in the first inequality and getAdd up the two inequalities, we obtainTake the supremum in , and then the required conclusion follows.
When and , , where is the positive root ofSo we have(c) Multiplying the last inequality of part (b) by , and integrating with respect to from to yieldsLet , thenSo
(d) According to part (b), is the positive root ofwhich means .
Consider , where .
It is easy to see that

2.1.3

(a) Obviously,
For containing ,Take the supremum to get
For a ball containing , denotes the minimal cube containing , thenTake the supremum to get .
For a cube containing , denotes the minimal ball containing , thenTake the supremum to get .
Similar process can be applied to obtain the third inequality.
For ,So .

For ,So .
Similarly, has the same property.

2.1.4

(a) Write .
Notice that , so .
From Theorem 2.1.6,So and hence(b) Write .
Taking in part (a) yields , soThen is also integrable over , as .

(c) From Theorem 2.1.20, there exists a collection of disjoint open cubes such that and , soFrom Exercise 2.1.3,(d) Let for some . For , we haveConsequently, for , , and thuswhich implies that Using polar coordiantes, the determinant of the map is . Thus we haveFrom Exercise 2.1.3, we conclude that (e) For ,Therefore we havewhere the second inequality is from part (c).
From part (d),and then the required conclusion follows.

2.1.5

We haveThe second result follows from taking , then .

2.1.6

(a) Define It is obvious that .
Thus, (b) Consider where .
From Exercise 2.1.2 (c), , so .
(c) Consider , then .
For , write , we haveThusYet let , by the Lebesgue monotone convergence theorem and Tonelli’s theorem, So is not of weak type .

2.1.7

Write , and .
Notice that

2.1.8

Let be the characteristic function of the unit ball in . Note thatwhere for .
For ,This means half of the unit ball is contained in , implying thatSo Let . Then and , so on some interval . Therefore,

2.1.9

Consider the Besicovitch Set (see Elias M. Stein, Harmonic Analysis, X.1, p.435), which satisfies
(1) is a union of rectangles , each having side lengths and , and , and
(2) the ’s are mutually disjoint, . Here the is defined to be the rectangle obtained by translating by two units along the longer side of .
So , and for , , which implies that Let , and then we can see the unboundedness.
Notice that is weak type , if is weak type , from the Marcinkiewicz interpolation theorem, is strong type . Contradiction!
(b) Lemma: Let be a finite collection of open rectangles in , whose eccentricity is . Then there exists a finite subcollection of pairwise disjoint rectangles such that Proof of lemma: First notice that if two rectangles with eccentricity intersect, we claim that the smaller one is contained in the bigger one scaled times.
We place the center of at , the longer edge being parallel to the -axis. Let the length of the edge of be , where .
So the abscissa of points in is no more than , and the ordinate of points in is no more than .
This implies .
Then follows the proof of Lemma 2.1.5 to prove the lemma. Similar to the proof of Theorem 2.1.6, we see that (c) Notice that for rectangles satisfying the assumption above, if and intersect, , then . The coefficient does not depend on the .
Then we can deduce the similar conclusion as in part (b).

2.1.10

(a) Define , then for ,

We claim

Note that So we have

Let , we have proved our assertion, which implies that .

(b) DefineBy Minkowski’s inequality, we have So we have
When , by Minkowski’s inequality, when , by part (c) of Exercise 1.1.5, Note that is bounded, we have So is of weak type , and the norm is less equal to , which implies is of weak type . For continuous, , and is dense in . Then we finish the proof by part (a).
(c) Set , then . Apply part (b) to , we have , a.e. on every , therefore a.e. on .

2.1.11

(a) It’s easy to see is open.
Write . For , let The set is nonempty, and . Otherwise, there exists s.t. . Thus , namely . But as , for all . Take , then . This leads to , whence , contradiction! Let , then such that . If , from the definition of , there exists , such that .
Let , contradictory to the supremum. So .
Let . If , then , a contradiction. Thus,which implies that Similar for .

(b)Here we use Hölder’s inequality for the last inequality. Then we get(c) Consider
We have SoLet , then
(d)Let , then , so . Similarly we have .

We deduce from the three inequalities that .
(e)(f) Use part (e), we further getHere we used Hölder’s inequality. Cancel the , then we obtain the required inequality.

2.1.12

(a) Firstly notice that , defined to be the set of all dyadic cubes, is countable. So any cover of dyadic cubes has a countable subcover, which is pairwise disjoint.
Let . , containing , such that .
So , where is countable, are mutually disjoint.
(b) Apply the result of part (a), we have

2.1.13

Use the result of Exercise 1.1.12 (a), we deduceTaking the supremum in yields

2.1.14

It suffices to prove the inequality for , as we may replace by . Define as in Theorem 2.1.10, and let Then for any ,

2.2 The Schwartz Class and the Fourier Transform

2.2.1

(a)

(b)

(c) Take some such that is odd, real–valued and compactly supported.

For instance, where and is defined as in part (b).

Then takes purely imaginary values, so takes real values.

Letthenis nonnegative and compactly supported.

2.2.2

For all multi–indices ,The last step is because in implies in , and subsequently in , hence in by Proposition 2.2.6. And gives .

2.2.3

If there exists nonzero functions such that , thenso , then . Now verify that is exactly the spectrum of the Fourier transform indeed:shows that is an eigenvalue;shows that is an eigenvalue;shows that is an eigenvalue;shows that is an eigenvalue.

2.2.4

Without loss of generality, assume , , then the proposition becomes

This is because

2.2.5

Let .

When , let where , then is a smooth approximate identity all the supports of which are contained in . For , we have by the Lebesgue dominated convergence theorem, and by Theorem 1.2.19 (a), which shows that is dense in , hence dense in .

When , by the case is dense in with respect to the norm and therefore also with respect to the norm because for we haveMoreover, is dense in since for we have , hence dense in .

When , any sequence of functions converges with respect to the norm to a continuous function, so is not dense in .

2.2.6

(a)

by Proposition 2.2.11 (6)

as by the Lebesgue dominated convergence theorem.

(b)by Fubini’s theorem.

(c) Use (2.2.15). But now the left–hand side of (2.2.15) may not converge to pointwise; only is known that the left–hand side converges to with respect to the norm. Yet by the Riesz–Fischer theorem, there is a sequence of as such that the left–hand side converges to a.e. if is taken to be , and the right–hand side still converges to after the restriction of , so we obtain a.e.

2.2.7

(a) Let in Exercise 2.2.6 (b), then , which is an approximate identity, so because is continuous at and is an approximate identity.

(b) Suppose that , then for some bounded area , so contradicting the result of part (a). So , and it follows that by the Lebesgue dominated convergence theorem.

2.2.8

Let , then becauseby Theorem 1.2.12,where (5) and (4) stands for Proposition 2.2.11 (5) and (4), and is continuous at becauseExercise 2.2.7 (b) yields

2.2.9

(a) For where ,And for similar reasons, So

(b) is odd implies that , so by part (a)

(c) Suppose that such exists, then by part (b)a contradiction.

2.2.10

By substituting with which increases from 0 to then from to as ranges from to , we get

Then since increases from to twice as ranges from to ,

2.2.11

(a) Use Exercise 2.2.10 with to obtain that

(b) Set in part (a) to get that

For all , use a rectangle contour to lift the -axis and get thatwhere and , plugging into above

2.2.12

(a) For all ,

(b) For , write . We have

2.2.13

Let . Integrate by parts and apply the Cauchy–Schwarz inequality and Plancherel’s identity, we getThen replace by in the inequality above and the desired inequality follows from

2.2.14

Let be the unit ball and its complement. By the Cauchy–Schwarz inequality,Then repalce by for and get

2.3 The Class of Tempered Distributions

2.3.1

Such can be identified with the functionalwhich is a tempered distribution because the condition of Proposition 2.3.4 (b) is satisfied: in fact,where the last inequality is due to (2.2.3).

The Lebesgue measure as a tempered distribution coincides with since

2.3.2

Pick any nonzero vector , and it is easily shown that is times an approximate identity for any multi–index as . Then for all multi–indices ,In the last step and is hence uniformly continuous on .

2.3.3

We have

Or check that the two distributions have identical Fourier transforms:

2.3.4

(a)(b) To simplify notations, let . Sure enough, everything is essentially the same for .(c)(d)

2.3.5

(a) For all multi–indices and ,where the last step is because are both dominated by a finite sum of some ’s.

In part (b) and (c), assume is identically equal to in .

(b) For all multi–indices ,Noticing that for some when , we havewhere is the multi–index whose -th component is and the others , so as .

(c) By part (b) in since the Fourier transform is a homeomorphism (Corollary 2.2.15), which on one hand reduces the proposition to in , and on the other hand implies that for all and multi–indices ,

Then for all multi–indices ,

For the same reason as part (b),

so as .

2.3.6

The Fourier transform of a nonzero function, as a distribution with compact support, is a real analytic function by Theorem 2.3.21, which cannot be a function for it would otherwise be identically due to the uniqueness of real analytic functions (the uniqueness is obtained by showing that the interior of the zero set of a real analytic function is both open and closed).

2.3.7

For all ,

2.3.8

LetThen in because for all ,where as and in the last inequality (2.2.3) is used.

Since (see Example 2.3.9), for all ,

2.3.9

(a) The usual definition for functions (still denoted by ) homogeneous of degree is for all , which agrees with the one given in the problem because for functions ,

(b) This is because

(c) Since

(d) Since

The converse follows similarly.

2.3.10

(a) and converge to in and as as a consequence of Proposition 2.2.17, from which we know that the multiplication of distributions is not a continuous operation even when it is defined, for in yet .

(b) in .

This is because for all , by the Lebesgue dominated convergence theorem,

2.3.11

Let be a Schwartz function in whose Fourier transform is equal to on the ball and vanishes outside the ball (which can be constructed as the inverse Fourier transform of a Schwartz function with the properties that the desired Fourier transform of is supposed to have). Sincewe establish the identity . Then

2.3.12

(a) in for as because for all multi–indices ,and for the same reason as Exercise 2.3.5,

Thus for all ,

(b) Suppose that is equal to in , then is equal to in . Apply part (a), which is also true when is replaced by , and getnamely,

2.3.13

Assume the converse. Then for all , is locally integrable and hence the map is a well defined operator from to for all (i.e. for all , since is covered by a finite number of neighborhoods of on which is integrable).

Then we shall use the closed graph theorem to deduce that the operator is bounded, or in other words, .

Suppose in , then by Proposition 2.3.22, in , so in . Note that in implies convergence in , so we can conclude that is indeed once we check that: if two functions are identical as tempered distributions, then they are the same as functions.

In fact, if for all , , then so will be true for , which is justified by finding a sequence of Schwartz functions all bounded by converging pointwise to and making use of the Lebesgue dominated convergence theorem — such Schwartz functions can be taken as a sequence of step functions mollified at the endpoints approximating , where the ’s and ’s are each a finite combination of intervals contained in and with and . Then we get in the usual sense of functions.

To violate the previous inequality whenever , takewhere . Then for ,

Andusing Proposition 2.2.11 and Example 2.2.9, so is a constant. Therefore the inequality cannot be true, for as .