Charles Fefferman's Multiplier Problem for the Ball

Charles Feﬀerman’s Multiplier Problem for the Ball

Decemeber, 2017

Prepared in partial completion of a reading course on Harmonic Analysis at the University of Mas- sachusetts, Amherst. Very little of the below is original work, the main contributions are making omitted details explicit. This was the exercise intended in the study.

The material and exposition below was discussed extensively with Professor Andrea Nahmod.

1 Contents

1 Motivation 2 1.1 The Setting ...... 2 1.2 Basic Results for the Fourier Transform ...... 3

2 The Lemma of Y. Meyer. 4 2.1 Proof of the Lemma ...... 4

3 Proof of Theorem 1.1 5 3.1 The Contradiction with Lemma 3.1 ...... 5 3.2 Proof of Lemma 3.1 ...... 6

4 Explicit Counterexample to the Disk Conjecture. 8

Appendix 14

A Vector Valued Inequalities 14 A.1 `2-Valued extensions of Linear Operators...... 14

B Lower Bound on χH (dχR)(x). 17 B.1 The calculation...... 17

C Mikhlin Multiplier Theorem 20 C.1 Motivation...... 20 C.2 Some Lp bounds for multipliers ...... 20

D Littlewood-Paley square-function estimate. 24 D.1 Probabilistic Preliminaries ...... 24 D.2 The Littlewood-Paley Theorem ...... 25

1 1 Motivation

1.1 The Setting 1 Consider the operator T deﬁned by ˆ Tc f(x) = χ|x|≤1(x)f(x). (1.1.1)

All functions have fourier transforms, denoted by the hat notation as above. It is a fundamental problem to recover a function from its fourier transform fˆ. In particular, deﬁne

Z ∨ ˆ 2πiξ·x ˆ SRf(x) := χ|x|≤R(x) f(ξ)e dξ = χ|x|≤R(x) f(ξ) (x), (1.1.2)

which corresponds to a partial sum in the fourier series of f, noting that this spatial indicator commutes with the inverse fourier transform. Worded diﬀerently, SRf(x) is the image, in space, of the set of frequen- cies associated to f restricted to the ball of radius R . p The question is then whether, for f ∈ L , SRf → f in || · ||Lp(Rd) as R → ∞. The choice of p turns out to be crucial. Suppose that p = 2. Then, since L2 is a Hilbert Space, we can utilize Plancheral’s Theorem. ˆ (SR\f − f)(ξ) = −χ|x|>R(ξ)f(ξ), (1.1.3)

so using Plancheral’s,

ˆ lim ||SRf − f||2 = lim ||(SR\f − f)(ξ)||2 = lim ||χ|x|>R(ξ)f(ξ)||2 = 0 R→∞ R→∞ R→∞

since χ|x|>R → 0 as R diverges. If f is smooth with compact support, the result is also aﬃrmative. p d If f is only assumed to be in L (R ), then use the density of smooth functions of compact support on d ∞ d p d ∞ d ∞ R C0 (R ) in L (R ) to ﬁnd g ∈ C0 (R ) such that ||f − g||p < /2. g ∈ C0 means that SRg → g in Lp for any p, so

lim sup ||SRf − f||p ≤ lim sup (||SRf − g||p + ||f − g||p) = lim sup (||SR(f − g)||p + ||f − g||p) ≤ R→∞ R→∞ R→∞

≤ sup ||SR||op||f − g||p + /2 < sup ||SR||op(/2) + /2. R R

This shows that if SR were bounded, then we would have convergence of SRf to f. If SR were not bounded, p then there is a function on L such that ||SRf||p = ∞ for each R. This means that ||SRf − f||p ≥ ||SRf||p + ||f||p = ∞, and so,

Result 1.1. Boundedness of SR for all R, which is equivalent to boundedness of S1 =: T above, is equivalent p d to the convergence of SRf to f in L (R ). It is with this in mind that Charles Feﬀerman studied the operator (1.1). Although Some progress was made towards a proof that T is bounded, ultimately Feﬀerman was able to provide a proof that T is only bounded on L2 (d > 1).

Theorem 1.2. T is bounded only on L2 (d > 1).

1Credit to https://lewko.wordpress.com/2009/08/03/lp-convergence-of-fourier-transforms/ for the example.

2 By assuming that T is bounded, a certain vector valued inequality on sequences of operators related to T , due to Y. Meyer, can be proven (Lemma 2.1). Fefferman is able to construct a counterexample to Lemma 2.1 via a geometrical result on Besicovitch sets (Lemma 3.1). p 2 The first reduction we need is that it suffices to disprove L , p > 2, boundedness on R . de Leeuw proved the following Theorem p j+i Theorem 1.3. Let m(x, y) be a Fourier multiplier for L (R ). Then for almost every x, mx(y) = m(x, y) p j is a Fourier multiplier for L (R ) and the multiplier norm of mx does not exceed that of m. In particular, the restriction is possible for each x such that (x, y) is a Lebesgue point of m for almost all y ∈ Rj.

proof. see [4]. p d d−1 Using this theorem, we know that L boundedness of T on R implies boundedness on R , and the case p > 2 yields p < 2 by duality2. It will be useful to record some basics of the Fourier Transform and Fourier Series.

1.2 Basic Results for the Fourier Transform We deﬁne the Fourier transform of f via fˆ(ξ) = R f(x)e−2πix·ξdx with inverse formally given as f(x) = Rd R fˆ(ξ)e2πi·ξdξ. This is motivated by the expansion of f in the orthogonal basis e2πikx: Rd ∞ X 2πikx f(x) = cke . (1.2.1) −∞ −2πimx Assume uniform convergence of 1.2.1. Then, term by term integration against e returns cm = R −2πimx ˆ P∞ ˆ 2πikx3 f(x)e dx. Hence, with our deﬁnition of f, the fourier series of f is given by −∞ f(k)e . We now record some basic facts for the fourier transform that will be useful in manipulations. f[∗ g = fˆ· g.ˆ (1.2.2)

f(\x + h)(ξ) = fˆ(ξ)e2πih·ξ (1.2.3)

fe\2πih·x(ξ) = fˆ(ξ − h) (1.2.4)

λ[−df(λ−1ξ) = fˆ(λξ) (1.2.5) d Deﬁne the Schwartz class S(R ) as the set of functions f that are smooth and such that for each n n α ∂ β d multi-Index α ∈ N and β ∈ N , the function x ∂x f is bounded on R . Via the Lemma, d 2 d Lemma 1.4. S(R ) is dense in L (R ). it can be shown that

d 2 d Theorem 1.5. The Fourier transform on S(R ) has a unique extension to a unitary mapping of L (R ) to itself such that ˆ ||f||L2 = ||f||L2 . (1.2.6) Equation 1.2.6 is konwn as Plancheral’s Theorem, a very useful and important fact when passing between a function and its fourier transform. 2This is a typical application of the TT ∗ duality lemma, an appropriate version of which can be found in [5] 3It is a fundamental question of fourier analysis to ask when this sum is equal to f itself.

3 2 The Lemma of Y. Meyer.

2.1 Proof of the Lemma We know that the ball multiplier (1.1) is bounded on L2. Charles Feﬀerman proved that T is bounded only on L2 for spatial dimension n > 1. The ﬁrst step of the proof of this result is to show the following lemma, which can proven if we assume that ||T f||p ≤ C||f||p for p > 2.

2 2 Lemma 2.1. Let v1, v2, ··· be a sequence of unit vectors in R , and let Hj = {x ∈ R x · vj ≥ 0} p 2 be the upper half-plane deﬁned by vj. Deﬁne a sequence of operators T1,T2, ··· on L (R ) by setting ˆ Tdjf(x) = χHj (x)f(x). Then, for any sequence of functions f1, f2, ··· ,

 1/2  1/2

X 2 X 2  |Tjfj|  ≤ C  |fj|  . (2.1.1) j j p p

r r r proof. Replace T via the following. Define T by T f(x) = χ r (x)fˆ(x). With D the disk of radius j j dj Dj j r 2 ∞ 2 r and center rvj. limr→∞ Tj f(x) = Tjf(x) for x ∈ R . For f ∈ C0 (R ), this holds via estimates on r Tdjf − Tdj f . To see this, fix j and write Z Z r −2πix·ξ −2πi·xξ T f − T f = χ r (x) f(x)e dx − χ (x) f(x)e dx ≤ dj dj Dj Hj Z −2πix·ξ ≤ χDr (x) − χH (x) f(x)e dx. j j ∞ Now, we have assumed that f ∈ C , and so eventually as r → ∞, χ r (x) − χ (x) 6= 0 in a set A that 0 Dj Hj r r has A ∩ suppf = ∅. I.e., Dj has moving center a scalar multiplie of vj, so in particular the center of Dj lies r in H , r eventually is large enough that D does not intersect suppf. Hence, either χ r (x) − χ (x) = 0, j j Dj Hj or, when r is large enough that the difference of these indicators is nonzero for a significant set but far from the support of f, f = 0 for large enough r to move the support of χDr (x) − χH (x) sufficiently far j j into Hj. Now, we will use this to show that

Z Z r ˆ 2πix·ξ ˆ 2πix·ξ Tjf − Tj f = f(ξ)e dξ − f(ξ)e dξ → 0. r Hj Dj Consider Z Z Z Z r −2πix·ξ 2πiξ·x −2πix·ξ 2πiξ·x Tif − T f = χH (ξ) f(x)e dx e dξ − χDr (ξ) f(x)e dx e dξ = j j j Z Z Z 2πiξ·x r 2πiξ·x r = Tdjfe dξ − Tdfe dξ ≤ Tdjf − Tdf dξ. j j PN 2 Then, Fatou’s Lemma applied to the nonnegative partial sums SN (T f) = j |Tjfj| now shows that

 1/2  1/2 X 2 X r 2 |Tjfj| ≤ lim inf T fj . (2.1.2)   r→∞  j  j j p p

r ˆ ˆ 1 2 Now, Tjf − T f ≤ χDr (ξ) − χH (ξ) |f|, and χDr (ξ) − χH (ξ) |f| ∈ L ( ) for r large enough since d dj i i i i R

r f was assumed to be smooth. Thus, we can use the DCT to say that Tdjf − Tdj f → 0, then apply Fatou’s Lemma as stated above. This shows via the inequality in (2.1.2), (2.1.1) holds from the corresponding

21/2 P r bound on j Tj fj . p

21/2 1/2 P r P 2 It thus suﬃces to show that T fj ≤ C |fj| . j j j p p r This requires connecting the operators Tj to the T , where we can use the main assumption above. 2 1 2πivj ·x −2πivj ·y Dilating R we can choose r = 1. Furthermore, Tj f(x) = e T (e f(y)) Where we have just shifted the center of the ball of radius 1 from the origin to vj .

21/2 1/2 P r P iv ·y 2 So, T fj = T (e j fj(y)) . Considering now the assumption that ||T f||p ≤ j j j p p C||f||p, an application of A.1.1 shows that

 1/2  1/2  1/2

X ivj·y 2 X ivj·y 2 X 2  T (e fj(y))  ≤ C  e fj(y)  = C  |fj(y)|  j j j p p p which is what is we wanted to conclude.

3 Proof of Theorem 1.1

The next step in the proof that T is unbounded is showing the following Lemma.

2 Lemma 3.1. Fix a small η > 0. There is a set E in R and a collection R = {Rj} of pairwise disjoint rectangles with the following properties.

(1) At least one-tenth of the area of each R˜j lies in E P (2) m(E) ≤ η j m(Rj) Note the following diagram (ﬁgure 1) for Rj and its neighbors R˜j.

3.1 The Contradiction with Lemma 3.1

It can be shown, deﬁning T f = χ χ , that |T f | is strictly positive (see Appendix B, B.0.10). The dj j Hj dRj j j consequence of this lower bound is a contradiction that proves the desired conclusion: that T is an unbounded operator. The contradiction is with Lemma 2.1, and the proof of Lemma 2.1 relies on T being bounded. Hence, the contradiction will show that T cannot be bounded.

Using (3.1.10) and Lemma 3.1 part (1) for fj = χRj ,   Z Z X 2 X 2 2 X ˜  |Tjfj(x)|  dx = |Tjfj| dx ≥ δ m(E ∩ Rj) E j j E j

5 Figure 1: R˜j from Rj and vj.

j j p

1/2 2 2/p (p−2)/p X 2 (p−2)/p X (p−2)/p X ≤ Cm(E) |fj(x)| = Cm(E) m(Rj) ≤ Cη m(Rj). (3.1.2)

j p j We can choose η small enough in 3.1.12 to contradict 3.1.11, and so we would contradict Lemma 3.1 part (1). Thus, once we’ve proven Lemma 3.1 we can show that Lemma 2.1 does not hold, and thus that T is unbounded. It thus remains to prove Lemma 3.1.

3.2 Proof of Lemma 3.1 We will use the following process on a triangle in the plane. Consider some triangle with vertices a, b, c. Let h be the height of the original triangle T . Extend the lines ac and bc to endpoints a0 and b0 such that they are at height h0 > h. Create two triangles with vertices a0, (a + b)/2, a and b0(a + b)/2, b, and call these ”T sprouts from height h to height h0”.

6 Figure 2: Sprouting from a single triangle to 2 triangles with base heights halved

This process will allow us to construct the besicovitch set E from which we will be able to get our disjoint rectangles Rj. √ o proof of Lemma 3.1. Let T be the equilateral triangle with base [0, 1]. The initial height is h0 = 3/2. o 0 Pick an increasing sequence of numbers h1, h2, ··· , hk. Sprout T from h0 to h1 to obtain 2 triangles T 00 1 0 00 and T with bases 2 wide. Then Sprout both T and T from height h1 to h2 to obtain four triangles 1 2 3 4 1 1 T ,T ,T ,T with bases 4 wide. Continue to obtain k triangles Tj with base widths 2k and heights hk. 2k Let E = ∪i=1Ti If √ √ √ 3 1 3 1 1 3 1 1 1 h = (1 + ), h = (1 + + ), h = (1 + + + ),..., 1 2 2 2 2 2 3 3 2 2 3 4 the area of E was computed by Busemann and Feller to be no more than 17. Choosing heights hj to be strictly less than these, we can be sure that the area of E is no more than 17. −k Note that each dyadic interval I ⊂ [0, 1] of length 2 is the base of exactly one Tj. From each T (I), triangle with base I, construct a rectangle R(I) as in ﬁgure 3.

S The important thing is that R(I) remains in ∆P (I)AB. Deﬁne R = I R(I) where the union is over all dyadic intervals I of [0, 1] with length 2−k. Clearly one-tenth of the area of each R˜(I) lies in E, since T (I) lies in E for all I, and tildeR(I) must intersect this triangle -see Figure 3. This proves (1). −k P −k k Each R(I) has area 2 ∗ ln(k + 1), and so m ( I R(I)|) = ln(k + 1) ∗ 2 ∗ 2 = ln(k + 1). Hence, take k large enough such that ln(k + 1) > 17/η.

To show that the R(I) are disjoint, we need only note that for dyadic intervals I1 and I2, if I1 lies to the left of I2, then P (I1) lies to the left of P (I2). This then implies that R(I1) and R(I2) are disjoint as follows.

7 Figure 3: R˜(I) from R

Figure 4: Disjointness of two diﬀerent R(Ik)

With this , we have proven Lemma 3.1, and so the proof that T is unbounded is complete.

4 Explicit Counterexample to the Disk Conjecture.

We now present a way to construct an explicit counterexample for the disk conjecture. Using the same construction as in Lemma 3.1, generate a collection of disjoint rectangles {Rj} which satisfy

(1) At least one-tenth of the area of each R˜j lies in E. P (2) m(E) ≤ η j=1 m(Rj). 1/2 We can choose the rectangles Rj to have sidelengths approximately N × N for any N we want. Furthermore, we can assume that E is made up of squares with sidelengths N 1/2 without loosing properties 1/2 (1) and (2). To do this, for every triangle Tl that makes up E, take one square of sidelength at least N

8 Figure 5: Disjointness of several diﬀerent R(Ik)

with base the same as the base of Tl. Then, we will maintain both properties guaranteed by the Lemma: For k ≥ 2, the new set E is strictly smaller in area than the old set, being made up of sets of area ≈ N 1 3/2 vs sets of area 2 N so (2) holds. Additionally, the area of each square being ≈ N means that the ratio of the areas of R˜ and the respective square is ≈ N = 1 . If N > 100, then use multiple squares per j N 3/2 N 1/2 1 3/2 triangle, and note that N << 2 N for large enough N (N = 100 corresponds to a diﬀerence of 100 vs. 50 ∗ 10 = 500), so that we can add additional squares to E for each triangle Tj to achieve (1) if necessary.. Thus, assme that E is a union of squares of side lengths N 1/2.

Calculate the fourier transform of χ[0,1]: Z 1 −i2πx·ξ 1 i2πξ χˆ[0,1](ξ) = e dx = e − 1 . (4.0.1) 0 i2πξ

Once |ξ| > 1/2, 1 < 1 < 1 , and decreases rapidly as |ξ| increases . This leads us to viewχ ˆ (ξ) i2πξ π 3 [0,1] R as approximately lying in the interval [−1/2, 1/2]. Importantly, we’d like to know if χˆ[0,1](ξ) dξ ≈ R χ[−1/2,1/2](ξ)dξ = 0.5. Numerically, we can show that this holds, but we can calculate Z 1 Z 1 ei2πξ − 1 dξ = cos2(2πξ) − 2 cos(2πξ) + 1 + sin2(2πξ) dξ ξ 2πξ Z 1 = 2 (1 − cos(2πξ))dξ. 2πξ cos x 1 x2 x4 x6 1 x x3 x5 = (1 − + − + ··· ) = − + − + ··· , x x 2 4! 6! x 2 4! 6! so Z cos x x2 x4 x6 dx = ln x − + − + ··· . x 4 4 ∗ 4! 6 ∗ 6!

9 Figure 6: Constructing E from N (1/2) sided squares

Using a change of variable, redeﬁne ξ := 2πξ, so that we have. Z 2 4 6 ∞ 2 4 6 8 ∞ 1 cos ξ ξ ξ ξ ξ ξ ξ ξ − dξ = ln ξ − ln ξ − + − + ··· = − + − + ··· ξ ξ 4 4 ∗ 4! 6 ∗ 6! −∞ 4 4 ∗ 4! 6 ∗ 6! 8 ∗ 8! −∞ 2 4 6 8 ∞ ξ ξ ξ ξ = 2 − + − + ··· . 4 4 ∗ 4! 6 ∗ 6! 8 ∗ 8! 0 This is the sum ∞ X ξ2k (−1)k−1 , 2k(2k!) k=1 which, by the ratio test, which involves taking the limit k 1 lim = 0 < 1, k→∞ k + 1 (2k + 1)(2k + 1) converges to a ﬁnite number. This doesn’t really give us an actual estimate on that number, but we note that numerical integration ofχ ˆ[0,1](ξ) returns ≈ 0.5. Up to scaling and since if χR = χ[0,a]×[0,b], thenχ ˆR =χ ˆ[0,a]χˆ[0,b], this leads us to conclude that if f is supported in a rectangular region [0,N 1/2] × [0,N] (in particular if it is the indicator of that region), then fˆ is supported (nearly) in [− 1 , 1 ] × [− 1 , 1 ], and 2N 1/2 2N 1/2 2N 2N ˆ ||f||1 ≈ ||χ[− 1 , 1 ]×[− 1 , 1 ]||L1 (4.0.2) 2N1/2 2N1/2 2N 2N

Deﬁne the operators T fj by

T f = χ eivj ·kχ (4.0.3) dj B bRj

10 Figure 7: Fourier Transform of χ[0,1] and χ[0,10] showing the rapid decay of amplitudes as we get away from 1 1 1 1 [− 2 , 2 ] and [− 20 , 20 ] respectively

for all Rj in our collection of rectangles. We want to show that {T fj} form an essentially orthogonal collection on our set E. Hence, calculate |hT fj, T, fli| (Translations /dilations allow us to assume Rj are centered at the origin.): Z Z Z |hT f ,T , f i| = χ eivj ·kχ dξ χ eivl·kχ (xi)dξ dx j j l l B bRj B bRl E Z Z ivj ·k ivl·k ≤ χBe χR χBe χR dx = χR 1 ||χR || 1 dx b j 1 b l 1 b j L (B) b l L (B) E L L E Z 2

≈ χ[− 1 , 1 ]×[− 1 , 1 ] dx. E 2N1/2 2N1/2 2N 2N L1(B) E is made up of nearly disjoint squares of side length N 1/2, so this is a ﬁnite sum of the integrals

M Z 2 X |hT f , T f i| ≤ χ 1 1 1 1 dx, j l h 1/2 1/2 i [− , ]×[− , ] N N 2N1/2 2N1/2 2N 2N 1 j − 2 , 2 L (B)

so choose N large enough that the rectangle [− 1 , 1 ] × [− 1 , 1 ] is contained in the unit ball. 2N 1/2 2N 1/2 2N 2N Then, this is all equal to

M 2 X Z 1 1 1 = dx = M ∗ N ∗ = M . (4.0.4) h N1/2 N1/2 i N 3/2 N 3 N 2 j − 2 , 2

11 M is some power of 2, say 2p, so we just need to choose N such that 2p << N, which we can do using translations and dilations. Therefore, this inner product is small, and the collection {Tjfj} is essentially orthogonal. Figure 9 shows the inner product of two T fj. They are, essentially, orthogonal plane waves. vj were generated randomly and k was set to 3.

Figure 8:

P With the fact that T fj are essentially orthogonal, and deﬁning f = j fj, we have an f such that 2 Z Z Z 2 X X 2 |T f(x)| dx = T fj(x) dx ≈ |T fj(x)| dx. (4.0.5)

E E j j E Arguing as in the proof of Theorem 1 from Lemma 2, we know that X Z 1 X 1 |T f (x)|2dx ≥ m(R ) = MN 3/2. (4.0.6) j 20 j 20 j E j

If we assume that ||T f||p ≤ C||f||p, then the following is true Z 2 (p−2)/p X 2 (p−2)/p 2 2 |T f| ≤ m(E) || T fj||Lp ≤ m(E) C ||f||Lp E j   (p−2)/p 2 X 2/p 2 (p−2)/p X = m(E) C  m(Rj)  ≤ C η m(Rj). (4.0.7) j j

12 For some η, which was ﬁxed at the beginning of the construction, we can choose N large enough such that 4.0.6 contradicts 4.0.7.

13 Appendix

A Vector Valued Inequalities

A.1 `2-Valued extensions of Linear Operators. The following is taken from Classical Fourier Analysis by Gafakos:

Theorem A.1. Let 0 < p, q < ∞ and let (X, µ) and (Y, ν) be two measure spaces. Then (a) If T is a bounded linear operator from Lp(X) to Lq(Y ) with norm A, then T has an `2- valued p extension, such that for all complex valued fj in L (X) we have

1 1   2   2

X 2 X 2  |T (fj)|  ≤ Cp,qA  |fj|  (A.1.1) j j Lq Lp

for some constant Cp,q depending only on p and q, and Cp,q = 1 if p ≤ q. (b) If T is a bounded operator from Lp(X) to Lq,∞(Y ) with norm A, then T has an `2-valued extension. That is,

1 1   2   2

X 2 X 2  |T (fj)|  ≤ Dp,qA  |fj|  (A.1.2) j j Lq,∞ Lp

for some constant Dp,q depending only on p and q.

The following Lemma is needed.

Lemma A.2. For 0 < r < ∞, deﬁne 1 r+1 ! r Γ( 2 ) Ar = r+1 π 2 and 1 r r Γ( 2 + 1) Br = r . π 2

Then, for any λ1, λ2, . . . , λn ∈ R we have

1 Z r r −π|x|2 2 2 1/2 |λ1x1 + ··· + λnxn| e dx = Br(λ1 + ··· λn) , (A.1.3) R and for all w1, . . . , wn ∈ C, we have

1 Z r r −π|z|2 2 2 1/2 |w1x1 + ··· + wnxn| e dz = Ar(w1 + ··· wn) . (A.1.4) R

14 2 2 1/2 proof. Divide both inequalities by (λ1 + ··· + λn) and the corresponding sum of wis to reduce the 2 2 t problem to the situation when λ1 + ··· + λn = 1 and likewise for wis. Let e1 = (1, 0,..., 0) , and ﬁnd an n −1 t orthogonal matrix in O(R ) such that A e1 = (λ1, . . . , λn) . Then, the ﬁrst coordinate of Ax is

t (Ax)1 = Ax · e1 = x · A e1 = λ1x1 + ··· + λnxn.

Let Ax = y and note that |Ax| = |x| such that

1 1 1 Z r Z r Z ∞ r r −π|x|2 r −π|y|2 r −πt2 |λ1x1 + ··· + λnxn| e dx = |y1| e dy = 2 t e dt = Rn Rn 0

1 1 ∞ r+1 ! r Z r−1 r Γ( ) 2 −πs 2 = s e ds = r+1 . 0 π 2 We used a change of variables ﬁrst to radial coordinates and again to s = t2 in two of the inequalities.

n −1 t Repeating the argument on C , using a unitary matrix A such that A e1 = (w ¯1,..., w¯n) , so that (Az)1 = w1z1 + ··· + wnzn and |Az| = |z|. Let y = Az and rewrite the intetral as

1 1 Z r Z ∞ r r −π|y|2 r −πt2 |y1| e dy = 2π t e dt = Cn 0 1 Z ∞ r r −πs = π s 2 e ds = Br. 0

proof of Theorem. In general, T maps complex valued functions to complex valued functions, so the second conclusion of the Lemma (A.1.4) is used. (a) Suppose that q ≤ p. The following line of computations utilizes the identity of the Lemma referring 0 to Br, the boundedness of T , Hodlers inequality with p/q and (p/q) = 1 − p/q with respect to the measure −π|z|2 e dz, and the identity for Br again.

1/2 q  n  Z 1 q X Bq |T (f )|2 = |T f |2 + |T f |2 + ··· + |T f |2 2 dν =  j  B 1 2 n j=1 Y q Lq(Y ) Z Z −q q −π|z|2 = (Bq) |z1T (f1) + ··· + znT (fn)| e dzdν Y C Z Z q/p −q q p −π|z|2 ≤ (Bq) A |z1T (f1) + ··· + znT (fn)| dµ e dz C X Z Z q/p −q q p −π|z|2 ≤ (Bq) A |z1T (f1) + ··· + znT (fn)| dµe dz C X  q/p q  p/2  1/2 Z −q q  p X 2  −1 q q X 2 = (Bq) A Bp  |fj|  dµ = (Bq Bp) A  |fj|  . X j j Lp(X) −1 Letting n → ∞ and setting Cp,q = BpBq , we have the desired result with Cp,q = 1 if p = q.

15 If q > p, then make a similar calculation. 1/2 q  n  Z Z 2 X 2 −q q −π|z|  |T (fj)|  = (Bq) |z1T (f1) + ··· + znT (fn)| dνe dz j=1 Y C Lq(Y ) Z Z q/p −1 q p −π|z|2 ≤ (ABq ) |z1f1 + ··· + znfn| dµ e dz C X Z q/p −1 q p = (ABq ) |z1f1 + ··· + znfn| dµ 2 X Lq/p(Cn,e−π|z| dz) Z q/p −1 q p ≤ (ABq ) |||z1f1 + ··· + znfn| ||Lq/p( n,e−π|z|2dz) dµ X C Z Z p/q !q/p −1 q q −π|z|2 = (ABq ) |z1f1 + ··· + znfn| e dz dµ X Cn  q/p q  p/2  1/2 Z n n −1 q  q X 2  q X 2 = (ABq )  (Bq)  |fj|  dµ = A  |fj|  . X j j=1 Lp(X) (b) This is a consequence of part (a) and the following (an exc. in Gafakos’s book).

1 1 Z r r 1 − 1 r q ||g||Lq,∞ ≤ sup ν(E) q r |g| dν ≤ ||g||Lq,∞ . (A.1.5) 0<ν(E)<∞ E q − r q  1/2 n X 2  |T (fj)|  j=1 Lq,∞(Y ) 1 r     r Z 2 1 − 1 X q r  2  ≤ sup ν(E)   |T (fj)|  dν 0<ν(E)<∞ E j

1 r     r Z 2 1 − 1 X q r  2  = sup ν(E)   |χET (fj)|  dν 0<ν(E)<∞ Y j

p 1     p Z 2 1 − 1 X q r  2  ≤ sup ν(E) ||χET ||Lp→Lr Cp,r   |fj|  dµ . 0<ν(E)<∞ X j Using the above quoted result: 1 1 1 − 1 q r q r ν(E) q r ||χ T (f)|| r ≤ ||T (f)|| q,∞ ≤ A||f|| p , E L q − r L q − r L we conclude that 1 1 − 1 q r ν(E) q r ||χ T || q r ≤ A. E L →L q − r

16 B Lower Bound on χH (dχR)(x). B.1 The calculation.

Let Ti be deﬁned as in the proof of lemma 1, and let fj = χRj where Rj are as in Lemma 2. Fix j, some half plane relative to a unit vector vj, and drop the ”j” notation. Furthermore, up to translation, we can assume that R is centered at the origin, and that v = (0, 1).

Figure 9: An Rj translatd/dilated to be centered at the origin with vj = (0 1)

We wish to show that |T f| > 0. Z Z ∨ −2πix·ξ 2πiξ·x T f = (χH χcR) = χH (ξ) χR(x)e dx e dξ.

2 v = (0, 1) means that H = R × [0, ∞) = {r = (x1, x2) ∈ R x2 ≥ 0}. Hence, rewrite the double integral and write out the dot products as follows,

Z Z ! ∨ −2πi(x1ξ1+x2ξ2) 2πi(ξ1x1+ξ2x2) (χH χcR) = e dx1dx2 e dξ. (−∞,∞)×[0,∞) [−a,a]×[−b,b]

Now apply Fubini’s twice to obtain

Z Z ! −2πi(x1ξ1+x2ξ2) 2πi(ξ1x1+ξ2x2) e dx1dx2 e dξ (−∞,∞)×[0,∞) [−a,a]×[−b,b] Z Z Z ! Z ! −2πiξ1x1 −2πix2ξ2 2πiξ1x1 2πiξ2x2 = e dx1 e dx2 e e dξ1dξ2 (−∞,∞) [0,∞) [−a,a] [−b,b]

17 Z Z ! ! Z Z ! ! −2πiξ1x1 2πiξ1x1 −2πix2ξ2 2πiξ2x2 = e dx1 e dξ1 e dx2 e dξ2 . (B.1.1) (−∞,∞) [−a,a] [0,∞) [−b,b]

Simplfying these integrals,

Z Z ! Z Z ! −2πiξ1x1 2πiξ1x1 −2πiξ1x1 2πiξ1x1 e dx1 e dξ1 = χ[−a,a](x1)e dx1 e dξ1 (−∞,∞) [−a,a] (−∞,∞) (−∞,∞)

∨ = (χ\[−a,a](x1)) = χ[−a,a](x1), and

Z Z ! Z Z ! −2πix2ξ2 2πiξ2x2 −2πix2ξ2 2πiξ2x2 e dx2 e dξ2 = χ[0,∞)(ξ2) e dx2 e dξ2 [0,∞) [−b,b] [−b,b] ∨ = χ\[−b,b]χ[0,∞) (x2).

Hence, letting x1 = x and x2 = y,

∨ ∨ T f = (χH χcR) (x, y) = χ[−a,a](x) χ\[−b,b]χ[0,∞) (y). (B.1.2) Recall that the Hilbert Transform H is a multiplier operator with symbol m = −isgn(ξ) such that ∨ Hf = −isgn(ξ)fˆ(ξ) , (B.1.3)

Hdf = −isgn(ξ)fˆ(ξ). (B.1.4)

Recall Duoandikoetxea 3.9 ([3]): i S = (M HM − M HM ) (B.1.5) a,b 2 a −a b −b

2πibx where Mbf(x) = e f(x). Furthermore, the scaling properties of the Fourier transform imply that iMbHM−b = sgn(ξ − b). Using this,

i 1 1 χ[0,∞)(x) = (M0HM0 − lim MtHM−t) = sgn(ξ) − lim sgn(ξ − t) = (sgn(ξ) − (−1)) . 2 t→∞ 2 t→∞ 2 Thus, 1 ∨ chi (x) χ χ ∨ (y) = χ (x) χ (1 + sgn(ξ)) (y) [−a,a] \[−b,b] [0,∞) [−a,a] \[−b,b] 2

1 = χ (x) (I + iH) χ (y). (B.1.6) [−a.a] 2 [−b,b] Thus, 1 |T f| ≥ χ[−a,a](x) H(χ[−b,b])(y) . 2

We now will calculate H(χ[−b,b])(y).

18 1 Z 1 1 Z b 1 H(χ[a,b])(x) = p.v. χ[a,b](x − y) dy = p.v. dz, (B.1.7) π y π a x − z where z = x − y.

1 Z b 1 1 |x − a| 1 |x − a| p.v. dz = lim ln + ln = ln . (B.1.8) π a x − z π →0 |x − b| π |x − b| Hence,

1 |y + b| |T f| ≥ χ[−a,a](x) ln . (B.1.9) 2π |y − b| ˜ Let x ∈ R, as depicted in the diagram after Lemma 2. Then, χ[−a,a](x) = 1, and b < |y| < 3b. If y > 0, |y+b| y+b |y+b| −y+b ˜ then |y−b| = y−b > 2, and if y < 0, then |y−b| = −y−b > 2 as well. Therefore, on R we have that 1 |T f| ≥ ln 2 = δ > 0. (B.1.10) 2π as desired.

19 C Mikhlin Multiplier Theorem

C.1 Motivation. Multiplier operators are operators T deﬁned by the relation of the form

Z 2πix·ξ ∨ (Tmf)(x) = e m(ξ)fˆ(ξ)dξ = (mfˆ) (C.1.1) Rd d 0 for some bounded m : R → C. Over a distribution K ∈ S , we may write Tmf = K ∗ f, K =m ˇ We are interested in bounds for T as an operator on Lp, 1 < p < ∞. Note ﬁrst a couple of cases where 1 1 the bounds are clear. Tm is bounded on L if and only if the kernel K is itself bounded on L , and in this case ||T || = ||K|| 1 . This is reliant on the following result. opL1→L1 L Proposition C.1. f ∈ Lp, g ∈ L1, X = d, and f ∗ g = R f(x − y)g(y)dy. Then, if f ∈ Lp, 1 ≤ p ≤ ∞ R Rd and g ∈ L1, then f(x − y)g(y) is integrable a.e. in y. Furthermore, f ∗ g ∈ Lp with

||f ∗ g||Lp ≤ ||f||Lp ||g||L1 . (C.1.2)

R proof. Let p = ∞. Then, we know that |f(y)g(x − y)| ≤ |g(x − y)| · ||f||L∞ , so | (f ∗ g)(x)dy| ≤ R R R R 1 |(f ∗ g)(x)|dy = |f(x − y)g(y)|dy = |f(y)g(x − y)|dy ≤ ||f||L∞ |g(x − y)|dy < ∞ since g ∈ L . For p < ∞, Using The Minkowski inequality: Z Z

f(x, y)dy ≤ ||f(x, y)||Lp(X)dy Lp(X) for measure spaces X and Y, by translation invariance of the integral/change of variable, and Fubini’s R R R p 1/p Theorem, we can say that ||(f ∗ g)(x)||Lp ≤ ||f(x − y)g(y)||Lp dx = |f(x − y)g(y)| dy dx = R R p 1/p R R p 1/p R p 1/p R |f(x − y)| dx |g(y)|dy = |f(x)| dx |g(y)|dy = |f(x)| dx |g(y)|dy = ||f||Lp ||g||L1 . R p 1 d R p Thus, f(x − y)g(y)dy is in L (R ), so f(x − y)g(y)dy is ﬁnite a.e. This means, additionally, R that f(x − y)g(y)dy is ﬁnite a.e., so f(x − y)g(y) is integrable in y a.e. x.

∞ 2 d 2 If the multipler m ∈ L , then if f is in L (R ), Tm is a bounded operator on L . This follows from the Plancherel theorem and the triangle inequality,

||Tmf||2 ≤ ||m||∞||f||2.

C.2 Some Lp bounds for multipliers To generalize to Lp functions, we must make additional assumptions on m. Furthermore, a geometric partition of unity must be utilized.

∞ d d Lemma C.2. There is ψ ∈ C (R ) such that supp(ψ) ⊂ R \{0} and

∞ X ψ(2−jx) = 1, x 6= 0. (C.2.1) j=−∞

20 proof. Deﬁne χ ∈ C∞ to satisfy χ(x) = 1 for all |x| ≤ 1 and χ(x) = 0 for all |x| ≥ 2. Then, deﬁne ψ(x) = χ(x) − χ(2x). For any N > 0, we have

N X ψ(2−jx) = χ(2−N x) − χ(2−N+1x) + χ(2−N+1x) − χ(2−N−2x) + χ(2−N−2x) − · · · − χ(2N+1x) = j=−N

= χ(2−N x) − χ(2N+1x). Given some non-zero x, we can make N large enough that 2−N |x| ≤ 1, so that χ(2−N x) = 1, and such that 2N+1|x| ≥ 2, so that χ(2N+1x) = 0. The function ψ deﬁned in the Lemma is said to give rise to a dyadic partition of unity.

Now, the main Theorem will be stated and proven. In the proof, we will need the Calderon-Zygmund Theorem, and to recall a few conditions. All this is recalled now.

d Deﬁnition C.3. A Calderon-Zygmund Kernel is K : R \{0} → C satisfying, for a ﬁnite constant B, −d d (i) |K(x)| ≤ B|x| , ∀x ∈ R R (ii) {|x|>2|y|} |K(x) − K(x − y)| dx ≤ B, y 6= 0, and R (iii) {r<|x|

Z T f(x) := lim K(x − y)f(y)dy (C.2.2) →0 {|x−y|>} .

Theorem C.4. (Calderon-Zygmund Theorem) For any singular operator T , for all 1 < p < ∞, we p can extend T to a bounded operator on L with ||T ||opLp→Lp < CB .

We will need the following alternative criterion to condition (ii) for being a CZ kernel.

d Lemma C.5. If K : R \{0} → C satisﬁes

|∇K(x)| ≤ B|x|−d−1, x 6= 0, (C.2.3) then K satisﬁes condition (ii) for some CB, C = C(d).

d d proof. For x, y ∈ R such that |x| > 2|y|, deﬁne the function θ : [0, 1] → R via

θ(t) = (1 − t)(x − y) + tx.

The image of θ is the line segment form x to x − y, and is contained in B(x, |x|/2). Hence,

Z Z 1

|K(x) − K(x − y)| = ∇Kdθ = (∇K)(x − (1 − t)y) · ydt 0 Z 1 ≤ |((∇K)(x − (1 − t)y)| |y|dt ≤ B2d+1|x|−d−1|y|. 0

21 Combine this with the inequality

Z 1 C∗(d) C∗(d) d+1 dy ≤ ≤ (C.2.4) |x|>2|y| |x| |x| |y| to conclude that Z Z |K − K(x − y)|dx ≤ B2d+1|y| |x|−d−1dx ≤ C∗(d)B2d+1 = C(d)B. |x|>2|y| |x|>2|y|

d Theorem C.6. (Mikhlin Multiplier Theorem) Let m : R \{0} → C satisfy, for any multi -index γ of length |γ| ≤ d + 2,

|∂γm(ξ)| ≤ B|ξ|−|γ| (C.2.5) for all ξ 6= 0. Then, for any 1 < p < ∞, there is a constant C = C(p, d) such that

∨ ||(mfˆ) ||p ≤ CB||f||p (C.2.6) for all f ∈ S.

proof. Let ψ deﬁne a dyadic partition of unity as in the Lemma. For each j, deﬁne

−j mj(ξ) = ψ(2 ξ)m(ξ)

PN and let Kj =m ˇj. Fix some N and set K(x) = j=−N Kj(x). γ −j|γ| By assumption, ||D mj||∞ ≤ CB2 , such that Z Z γ γ −j jd γ j −j|γ| jd ||D mj||1 = D ψ(2 ξ)m(ξ) dξ = 2 D ψ(ξ)m(2 ξ) dξ ≤ CB2 2

so long as |γ| ≤ d + 2. Similarly, Z γ γ −j −j(|γ|−1) jd ||D (ξjmj)||1 = D ξψ(2 ξ)m(ξ) dξ ≤ CB2 2

γ j(d−|γ|) γ j(1+d−|γ|) for some γ. Hence, ||x mˇj(x)||∞ ≤ CB2 , and ||x Dmˇj(x)||∞ ≤ CB2 . To see this, note the following.

k X γ X γ ∨ |x ||Kj(x)| ≤ C |x Kj| = C1 (∂ξ mj(ξ)) |γ|=k

and Z Z γ ∨ γ −j 2πiξ·x γ −j 2πiξ·x (∂ξ mj(ξ)) = ∂ξ m(ξ)ψ(2 ξ)e dξ = (2πix) m(ξ)ψ(2 ξ)e dξ,

γ so absorbing (2πi) into the constant C1 above gives the inequality, and justiﬁes the earlier calculation. Furthermore,

j(d−k) −k |mˇ j(x)| ≤ CB2 |x| (C.2.7)

22 and

j(d+1−k) −k |Dmˇ j(x)| ≤ CB2 |x| . (C.2.8)

For 0 ≤ k ≤ d + 2. Utilize this with k = 0 and k = d + 2, such that X X X |K(x)| ≤ |mˇj(x)| ≤ |mˇj(x)| + |mˇ j(x)| j 2j ≤|x|−1 2j >|x|−1 X X ≤ CB 2jd + CB 2jd(2j|x|)−(d+2) 2j ≤|x|−1 2j >|x|−1 ≤ CB|x|−d + CB|x|2|x−d−2 = 2CB|x|−d. Utilize B.2.8 to get the second statement. This all implies that K satisﬁes conditions (i) and (ii) for being a Calderon-Zygmund operator, which are the only conditions needed to satisfy the Claderon- Zygmund Theorem. Noting further that we only need to satisfy L2 boundedness and conditions (i) and ∨ (ii) in order to utilize the CZ theorem, the fact that ||m||∞ ≤ B, such that ||(mfˆ) ||2 ≤ B||f||2 allows us to conclude. That is, the C-Z Theorem then says that

∨ ||(mfˆ) ||p ≤ C(p, d)||f||p, f ∈ S, 1 < p < ∞ (C.2.9) as desired.

23 D Littlewood-Paley square-function estimate.

∨ −j Consider the multiplier operator Pjf = (ψjfˆ) , with ψj = ψ(2 ξ) where ψ is some chosen function, typically a nice function with compactly supported fourier transform. Deﬁne  1/2 X 2 Sf :=  |Pjf|  , (D.0.1) j

then we will show that

−1 C ||f||p ≤ ||Sf||p ≤ C||f||p. (D.0.2)

Plancherel’s Theorem immediately gives that

−1 2 2 2 C ||f||2 ≤ ||Sf||2 ≤ ||f||2, (D.0.3) 2 P 2 With ||Sf||2 = j ||Pjf||2, so the result we want to prove is, in eﬀect, a generalization.

D.1 Probabilistic Preliminaries

To show the generalization, we will use a set of numbers {rj} deﬁned by P ({rj = 1}) = P ({rj = −1}) = 1/2 for all j. The Khinchine inequality willl also be used, and to prove this we need the following result about the {rj} deﬁned above.

N Lemma D.1. For all N, sequences {aj}j=1 ⊆ C, and λ > 0,

N N 1/2 X X 2 −λ2/2 P rjaj > λ |aj| ≤ 4e . (D.1.1)

j=1 j=1

proof. If aj ∈ C, squaring shows that it suﬃces to prove the inequality for aj real. So, assume that {aj} ⊆ R. Let SN be the partial sum of rjaj. Then,

N N tSN Y trj aj Y Ee = Ee = cosh(taj). j=1 j=1

x2/2 It is a fact from calculus that cosh x ≤ e for all x ∈ R. Thus,

N  N  Y t2a2/2 X 1 etSN ≤ e j = exp t2 a2 . E  2 j  j=1 j=1

2 PN 2 Let σ = j=1 aj . Then t2σ2/2 −λtσ P SN > λσ ≤ e e ,

which when minimized over t is leqe−λ2/2. Apply the same arguement to {SN < −λσ} to say that −λ2/2 P SN < −λσ ≤ e

24 −λ2/2 such that P |SN | > λσ ≤ 2e .

We are now in a position to prove the Khinchine inequality.

Theorem D.2. (Khinchine’s Inequality) For all 1 ≤ p < ∞, there is C = C(p) such that

 p/2 p !p/2 N N N −1 X 2 X X 2 C  |aj|  ≤ E rjaj ≤ C |aj| . (D.1.2) j

PN 2 PN proof. WLOG let j=1 |aj| = 1 and denote SN = j=1 rjaj. Then, Z ∞ Z ∞ p p−1 −λ2/2 p−1 E|SN | = P({|SN | > λ})pλ dλ ≤ 4e pλ dλ = C(p) < ∞. 0 0 For the lower bound, it suﬃces to consider the case when p = 1. Use the upper bound, and Holders to write 2 E|SN | 2/3 4/3 2/3 41/3 = E|SN | E|SN | ≤ (E|SN |) E|SN | 2/3 22/3 ≤ C (E|SN |) E|SN | , 2 2 which implies that E|SN | ≤ C(E|SN |) .

D.2 The Littlewood-Paley Theorem Muscalu and Schlag ([5]) oﬀer a nice heuristic motivation for why it should be possible to achieve some estimate on ||Sf||p in terms of ||f||p (i.e., C.0.2). Take a probabilistic viewpoint of the exponentials n {exp(2 θ)}n≥1 on the interval [0, 1]. That is, we’ve restricted the typical basis components of the fourier series to those with freqences a power of 2. For some k, on [k2−n, (k + 1)2−n], sin(2nπθ) is either nonnegative or nonpositive. On this same interval however, if we doule the frequency, sin(2n+1πθ) has an equal probability of being positive or negative on this interval.

j Figure 10: Probabilisitic Similarity of the system {sin(2 πθ)}j∈Z to the Rademacher functions

25 This is much like the Rademacher functions described above. This leads us to guess that, utlizing j Khinechine’s Inequality, we can obtain estimates on ||Sf||p if we view ajexp(2 θ) as a random variable on [0, 1]. Taking this one step further would be viewing Pjf as a system of nearly independent random variables on [0, 1]. Doing this involves utilizing the Rademacher functions and Khinechine’s Inequality, and so the argument becomes, partially, a probabilistic one.

Theorem D.3. (Littlewood-Paley Theorem). For any 1 < p < ∞ there is a constant C = C(p, d) such that

−1 C ||f||p ≤ ||Sf||p ≤ C||f||p (D.2.1) for any f ∈ S.

PN proof. Claim: m(ξ) = j=−N rjψj(ξ) satisﬁes the conditions of the Mikhlin Theorem uniformly in N and uniformly in the realization of {rj}. For any γ,

N N γ X γ X −|γ| γ −j −|γ| |D m(ξ)| ≤ |D ψj(ξ)| ≤ C |ξ| |(D ψ)(2 ξ)| ≤ C|ξ j=−N j=−N where we used the fact only an absolutely bounded number of terms in the series is nonzero for ξ 6= 0, and we have absorbed the rest of the series into the constant C. To see the ﬁrst statement, note that ψ is assumed to be nonnegative, so

1 1 1 −1 1 0 1 −2 1 −1 |(∂ ψ1)(ξ)| + |(∂ ψ2)(ξ)| = ∂ χ(2 x) − ∂ χ(2 x) + ∂ χ(2 x) − ∂ χ(2 x), P etc. Hence, the terms will cancel in a telescoping way, just as with ψj. 1/2 p/2 p PN 2 PN 2 PN We continue by using Lemma 5.5 on −N |Pjf| , such that −N |Pjf| ≤ CE −N rj(Pjf)(x) , we get p Z Z N p X p |(Sf)(x)| dx ≤ C lim sup E rj(Pjf)(x) dx ≤ C||f||p N→∞ −N where the ﬁnal inequality is reliant on the Mikhlin Multiplier Theorem. To prove the lower bound, choose a function ψ˜ such that ψ˜ = 1 on supp(ψ) and ˜(ψ) is compactly ˜ d ˜ ˜ ˜ supported with supp(ψ) ⊂ R \{0}. Deﬁne Pj the same as Pj but replace ψ with ψ. Then PjPj = Pj by construction. As such, for any f, g ∈ S, and any 1 < p < ∞, use Plancherels and the fact that P∞ P∞ ˜ j=−∞ ψj(ξ) = 1 = j=−∞ ψj(ξ) to say that

Z Z ∞ ˆ X ˆ ˜ |hf, gi| = f(ξ)gˆ(ξ)dξ = ψj(ξ)f(ξ)gˆ(ξ)ψj(ξ)dξ .

j j k ˜ ≤ ||Sf||p||Sg||p0 ≤ C||Sf||p||g||p0 .

We used the argument for the upper bound on Sg˜ to get the last inequality. Furthermore, since ||f||p = sup |hf, gi|, we can conclude that ||f|| ≤ C||Sf|| . ||g||p0 =1 p p

26 References

[1] Duoandikoetxea, Javier . Fourier Analysis,American Mathematical Society, 2001

[2] Loukas Grafakos. Classical Fourier Analysis,Springer, Graduate Texts in Mathematics, 2008

[3] Feﬀerman,Charles . ”The Multiplier Problem for the Ball”. Annals of Mathematics, Vol. 94, No. 2, pp. 330-336, 1971, http://www.jstor.org/stable/1970864

[4] Jodeit Jr.,Max . ”A Note on Fourier Multipliers”. Proceeds of the American Mathematical Society Vol. 27 , No. 2, pp 423–424, 1971.

[5] Muscalu, Camil and Schlag, Wilhelm. Classical and Multilinear Harmonic Analysis, Volume I, Cam- bridge Studies in Advanced Mathematics. Cambridge University Press, 2013