Lectures in Harmonic Analysis

Thomas H. Wol↵

Revised version, March 2002

n n If f L1(R )thenitsFouriertransformisfˆ : R C defined by 2 ! 2⇡ix ⇠ fˆ(⇠)= e · f(x)dx. Z n n More generally, let M(R )bethespaceoffinitecomplex-valuedmeasuresonR with the norm n µ = µ (R ), k k | | n n where µ is the total variation. Thus L1(R )iscontainedinM(R )viatheidentification | | f µ, dµ = fdx.WecangeneralizethedefinitionofFouriertransformvia ! 2⇡ix ⇠ µˆ(⇠)= e · dµ(x). Z n Example 1 Let a R and let a be the Dirac measure at a, a(E)=1ifa E and 2 2⇡ia ⇠ 2 (E)=0ifa E.Then (⇠)=e · . a 62 a ⇡ x 2 Example 2 Let (x)=eb | | .Then

⇡ ⇠ 2 ˆ(⇠)=e | | . (1)

Proof The integral in question is

2⇡ix ⇠ ⇡ x 2 ˆ(⇠)= e · e | | dx. Z Notice that this factors as a product of one variable integrals. So it suces to prove (1) ⇡x2 when n = 1. For this we use the formula for the integral of a Gaussian: 1 e dx =1. It follows that 1 R 1 2⇡ix⇠ ⇡x2 1 ⇡(x i⇠)2 ⇡⇠2 e e dx = e dx e · Z1 Z1 i⇠ 1 ⇡x2 ⇡⇠2 = e dx e i⇠ · Z1 1 ⇡x2 ⇡⇠2 = e dx e · Z1 ⇡⇠2 = e ,

1 where we used contour integration at the next to last line. ⇤

There are some basic estimates for the L1 Fourier transform, which we state as Propo- sitions 1 and 2 below. Consideration of Example 1 above shows that in complete generality not that much more can be said.

n Proposition 1.1 If µ M(R )thenˆµ is a bounded function, indeed 2 µˆ µ n . (2) k k1 k kM(R )

Proof For any ⇠,

n Proposition 1.2 If µ M(R )thenˆµ is a continuous function. 2 Proof Fix ⇠ and consider

2⇡ix (⇠+h) µˆ(⇠ + h)= e · dµ(x). Z 2⇡ix ⇠ As h 0theintegrandsconvergepointwisetoe · .Sincealltheintegrandshave ! n absolute value 1 and µ (R ) < , the result follows from the dominated convergence | | 1 theorem. ⇤

We now list some basic formulas for the Fourier transform; the ones listed here are roughly speaking those that do not involve any di↵erentiations. They can all be proved by n using the formula ea+b = eaeb and appropriate changes of variables. Let f L1, ⌧ R , n n 2 2 and let T be an invertible linear map from R to R .

1. Let f (x)=f(x ⌧). Then ⌧ 2⇡i⌧ ⇠ ˆ f⌧ (⇠)=e · f(⇠). (3)

2⇡ix ⌧ 2. Let e⌧ (x)=e · .Then b

e f(⇠)=fˆ(⇠ ⌧). (4) ⌧ d 2 t 3. Let T be the inverse transpose of T .Then

1 t f[T = det(T ) fˆ T . (5) | | 4. Deﬁne f˜(x)=f( x). Then fˆ˜ = f.ˆ (6)

We note some special cases of 3. If T is an orthogonal transformation (i.e. TTt is the identity map) then f[T = fˆ T ,sincedet(T )= 1. In particular, this implies that if f ± is radial then so is fˆ, since orthogonal transformations act transitively on spheres. If T is adilation,i.e. Tx = r x for some r>0, then 3. says that the Fourier transform of the · n 1 1 function f(rx)isthefunctionr fˆ(r ⇠). Replacing r with r and multiplying through n by r , we see that the reverse formula also holds: the Fourier transform of the function n 1 r f(r x)isthefunctionfˆ(r⇠).

There is a general principle that if f is localized in space, then fˆ should be smooth, and conversely if f is smooth then fˆ should be localized. We now discuss some simple n manifestations of this. Let D(x, r)= y R : y x

We are using multiindex notation here and will do so below as well. Namely, a multi- n index is a vector ↵ R whose components are nonnegative integers. If ↵ is a multiindex 2 then by deﬁnition @↵1 @↵n D↵ = ... , ↵1 ↵n @x1 @xn ↵ n ↵j x =⇧j=1xj . The length of ↵,denoted ↵ ,is ↵ .Onedeﬁnesapartialorderonmultiindicesvia | | j j ↵ P ↵ for each i,  , i  i ↵< ↵ and ↵ = . ,  6

Proof of Proposition 1.3 Notice that (8) follows from (7) and Proposition 1 since the ↵ ↵ norm of the measure (2⇡ix) µ is (2⇡R)| | µ .  k k

3 Furthermore, for any ↵ the measure (2⇡ix)↵µ is again a ﬁnite measure with compact support. Accordingly, if we can prove thatµ ˆ is C1 and that (7) holds when ↵ =1,then the lemma will follow by a straightforward induction. | | Fix then a value j 1,...,n ,andletej be the jth standard basis vector. Also ﬁx n 2{ } ⇠ R ,andconsiderthedi↵erencequotient 2 µˆ(⇠ + he ) µˆ(⇠) (h)= j . (9) h This is equal to 2⇡ihxj e 1 2⇡i⇠ x e · dµ(x). (10) h Z As h 0, the quantity ! 2⇡ihxj e 1 h 2⇡ihx e j 1 converges pointwise to 2⇡ixj.Furthermore, h 2⇡ xj for each h.Accordingly, the integrands in (10) are dominated by 2⇡x |,whichisaboundedfunctiononthesupport| | | | j| of µ. It follows by the dominated convergence theorem that

2⇡ihxj e 1 2⇡i⇠ x lim (h)= lim e · dµ(x), h 0 h 0 h ! Z ! which is equal to 2⇡i⇠ x 2⇡ix e · dµ(x). j Z This proves the formula (7) when ↵ = 1. Formula (7) and Proposition 2 imply thatµ ˆ is | | C1. ⇤

Remark The estimate (8) is tied to the support of µ.However,thefactthatˆµ is C1 and the formula (7) are still valid whenever µ has enough decay to justify the di↵erentiations under the integral sign. For example, they are valid if µ has moments of all orders, i.e. x N d µ (x) < for all N. | | | | 1 R The estimate (2) can be seen as justiﬁcation of the idea that if µ is localized thenµ ˆ should be smooth. We now consider the converse statement, µ smooth impliesµ ˆ localized.

4 The proof is based on an integration by parts which is most easily justiﬁed when f has compact support. Accordingly, we include the following lemma before giving the proof.

n Let : R R be a C1 function with the following properties (4. is actually ! irrelevant for present purposes):

1. (x)=1if x 1 | | 2. (x)=0if x 2 3. 0 1. | |   4. is radial.

x Deﬁne k(x)=( k ); thus k is similar to but lives on scale k instead of 1. If ↵ C↵ ↵ is a multiindex, then there is a constant C↵ such that D k ↵ uniformly in k. k| | Furthermore, if ↵ =0thenthesupportofD↵ is contained| in the| region k x 2k. 6 | | N ↵ 1 Lemma 1.5 If f is C , D f L for all ↵ with ↵ N and if we let fk = kf then ↵ ↵ 2 | | limk D fk D f 1 =0forall↵ with ↵ N. !1 k k | | Proof It is obvious that

↵ ↵ lim kD f D f 1 =0, k !1 k k so it suces to show that

↵ ↵ lim D (kf) kD f 1 =0. (13) k !1 k k However, by the Leibniz rule

↵ ↵ ↵ D ( f) D f = c D fD , k k k 0< ↵ X where the c’s are certain constants. Thus

↵ ↵ ↵ D (kf) kD f 1 C D k D f L1( x: x k ) k k  k k1k k { | | } 0< ↵ X 1 ↵ Ck D f L1( x: x k )  k k { | | } 0< ↵ X The last line clearly goes to zero as k . There are two reasons for this (either would 1 !1 1 suce): the factor k ,andthefactthattheL norms are taken only over the region x k. | | ⇤

5 Proof of Proposition 1.4 If f is C1 with compact support, then by integration by parts we have @f 2⇡ix ⇠ 2⇡ix ⇠ (x)e · dx =2⇡i⇠ e · f(x)dx, @x j Z j Z i.e. (11) holds when ↵ = 1. An easy induction then proves (11) for all ↵ provided that | | f is CN with compact support. To remove the compact support assumption, let fk be as in Lemma 1.5. Then (11) ↵ holds for fk.Nowwepasstothelimitask .OntheonehandD[fk converges ↵ !1 uniformly to D f as k by Lemma 1.5 and Proposition 1.1. On the other hand fk !1 ↵ ↵ converges uniformly to fˆ,so(2⇡i⇠) fk converges to (2⇡i⇠) fˆ pointwise. This proves (11) in general. d b ↵ To prove (12), observe that (11)b and Proposition 1 imply that ⇠ fˆ L1 if ↵ N. On the other hand, it is easy to estimate 2 | |

1 N ↵ N C (1 + ⇠ ) ⇠ C (1 + ⇠ ) , (14) N | |  | | N | | ↵ N | X| so (12) follows. ⇤

Together with (14), let us note the inequality

2. Schwartz space

n The Schwartz space is the space of functions f : R C such that: S !

1. f is C1, 2. x↵Df is a bounded function for each pair of multiindices ↵ and .

For f we deﬁne 2S ↵ f ↵ = x D f . k k k k1 It is possible to see that with the family of norms ↵ is a Frechet space, but we don’t discuss such questionsS here (see [27]). However,k we·k deﬁne a notion of sequential convergence in : S

Asequence fk converges in to f if limk fk f ↵ =0foreachpair { }⇢S S 2S !1 k k of multiindices ↵ and .

Examples 1. Let C1 be the C1 functions with compact support. Then C1 . 0 0 ⇢S

6 ↵ Namely, to prove that x D f is bounded, just note that if f C01 then D f is a continuous function with compact support, hence bounded, and that2 x↵ is a bounded function on the support of Df.

⇡ x 2 2. Let f(x)=e | | .Thenf . 2S For the proof, notice that if p(x)isapolynomial,thenanyﬁrstpartialderivative @ ⇡ x 2 ⇡ x 2 (p(x)e | | )isagainoftheformq(x)e | | for some polynomial q.Itfollowsby @xj induction that each Df is a polynomial times f for each .Hencex↵Df is a polynomial times f for each ↵ and .ThisimpliesusingL’Hospital’srulethatx↵Df is bounded for each ↵ and .

2 N 3. The following functions are not in : fN (x)=(1+x ) for any given N, ⇡ x 2 ⇡ x 2 S | | and g(x)=e | | sin(e | | ). Roughly, although fN decays rapidly at ,itdoesnot decay rapidly enough, whereas g decays rapidly enough but its derivatives1 do not decay. Detailed veriﬁcation is left to reader.

We now discuss some simple properties of ,thensomewhichareslightlylesssimple. S I. is closed under di↵erentiations and under multiplication by polynomials. Fur- thermore,S these operations are continuous on in the sense that they preserve sequential S convergence. Also f,g implies fg . 2S 2S Proof Let f .If is a multiindex, then x↵D(D )f = x↵D+f,whichisbounded 2S since f .SoDf . By the2S Leibniz rule,2Sx↵D(x f)isafinitesumoftermseachofwhichisaconstant multiple of ↵ x D (x )D f for some .Furthermore,D (x )isaconstantmultipleofx if ,andotherwise   is zero. Thus x↵D(x f) is a linear combination of monomials times derivatives of f,and is therefore bounded. So x f . 2S The continuity statements follow from the proofs of the closure statements; we will normally omit these arguments. As an indication of how they are done, let us show that if is a multiindex then f Df is continuous. Suppose that f f in .Fixapair ! n ! S of multiindices ↵ and .Applyingthedefinitionofconvergencewiththemultiindices↵ and + ,wehave ↵ + lim x D (fn f) =0. n 1 !1 k k Equivalently, ↵ lim x D (D fn D f) =0 n 1 !1 k k which says that D fn converges to D f. The last statement (that is an algebra) follows readily from the product rule and S the definitions. ⇤

7 II. The following alternate definitions of are often useful. S f (1 + x )N Df is bounded for each N and ,(16) 2S, | | f lim x↵Df =0foreach↵ and .(17) x 2S, !1 Indeed, (16) follows from the definition and (14). The backward implication in (17) is trivial, while the forward implication follows by applying the definition with ↵ replaced by appropriate larger multiindices, e.g. ↵ + e for arbitrary j 1,...,n . j 2{ }

Proposition 2.1 C1 is dense in ,i.e.foranyf there is a sequence f C1 0 S 2S { k}⇢ 0 with f f in . k ! S

Proof This is almost the same as the proof of Lemma 1.5. Namely, deﬁne k as there and consider fk = kf,whichisevidentlyinC01.Wemustshowthat

x↵D( f) x↵Df k ! uniformly as k .Forthis,weestimate !1 ↵ ↵ ↵ ↵ ↵ x D (kf) x D f kx D f x D f + x (D (kf) kD f) . k k1 k k1 k k1 ↵ The ﬁrst term is bounded by sup x k x D f and therefore goes to zero as k by (17). The second term is estimated| | using| the Leibniz| rule by !1

↵ C x D f D k . (18) k k1k k1 X< C Since f and D ,theexpression(18)goestozeroask . 2S k kk k !1 ⇤

There is a stronger density statement which is sometimes needed. Deﬁne a C01 tensor n function to be a function f : R C of the form !

f(x)= j(xj), j Y where each j C01(R). 2

Proposition 2.10 Linear combinations of C1 tensor functions are dense in . 0 S

Proof In view of Proposition 2.1 it suces to show that if f C1 then there is a 2 0 sequence g such that: { k}

1. Each gk is a C01 tensor function.

8 2. The supports of the gk are contained in a ﬁxed compact set E which is independent of k. ↵ ↵ 3. D gk converges uniformly to D f for each ↵.

To construct gk , we use the fact (a basic fact about Fourier series) that if f is a C1 n { } function in R which is 2⇡-periodic in each variable then f can be expanded in a series

i⌫ ✓ f(✓)= a⌫e · n ⌫ X2Z where the a satisfy { ⌫} (1 + ⌫ )N a < | | | ⌫| 1 ⌫ X for each N.ConsideringpartialsumsoftheFourierseries,wethereforeobtainasequence ↵ ↵ of trigonometric polynomials pk such that D pk converges uniformly to D f for each ↵. In constructing gk we can assume that x suppf implies xj 1, say, for each j - otherwise we work with{ } f(Rx) for suitable fixed2R instead and undo| | the rescaling at the end. Let be a C01 function of one variable which is equal to 1 on [ 1, 1] and vanishes outside [ 2, 2]. Let f˜ be the function which is equal to f on [ ⇡,⇡] ... [ ⇡,⇡]and ⇥ ⇥ is 2⇡-periodic in each variable. Then we have a sequence of trigonometric polynomials pk such that D↵p converges uniformly to D↵f˜ for each ↵. Let g (x)=⇧n (x ) p (x). k k j=1 j · k Then gk clearly satisfies 1. and 2, and an argument with the product rule as in Lemma 1.5 and Proposition 2.2 will show that g satisfies 3. The proof is complete. { k} ⇤ 1 The next proposition is an alternate definition of using L instead of L1 norms. S

Proposition 2.2 A C1 function f is in i↵the norms S x↵Df k k1 are ﬁnite for each ↵ and .Furthermore,asequence f converges in to f i↵ { k}⇢S S 2S ↵ lim x D (fk f) 1 =0 k !1 k k for each ↵ and .

Proof We only prove the ﬁrst part; the equivalence of the two notions of convergence follows from the proof and is left to the reader. First suppose that f .Fix↵ and . Let N = ↵ + n +1.Thenweknowthat (1 + x )N Df is bounded.2S Accordingly, | | | | ↵ N ↵ N x D f 1 (1 + x ) D f x (1 + x ) 1 k k k | | k1k | | k < 1

9 n 1 using that the function (1 + x ) is integrable. | | n For the converse, we first make a definition and state a lemma. If f : R is Ck n !1 and if x R then 2 f def (x) = D↵f(x) . k | | ↵ =k |X| We denote D(x, r)= y : x y r .Wealsonowstarttousethenotationx . y to mean that x Cy where{ C| is a fixed| } but unspecified constant. 

Lemma 2.2’ Suppose f is a C1 function. Then for any x

f f(x) 1 . | | . k j kL (D(x,1)) 0 j n+1 X This is contained in Lemma A2 which is stated and proved at the end of the section. To ﬁnish the proof of Proposition 2.2, we apply the preceding lemma to Df.This gives Df(x) Df(y) dy, | | . | | +n+1 ZD(x,1) | |X| | therefore

(1 + x )N Df(x) (1 + x )N Df(y) dy | | | | . | | | | +n+1 D(x,1) | |X| | Z (1 + y )N Df(y) dy, . | | | | +n+1 ZD(x,1) | |X| | where we used the elementary inequality

1+ x 2min(1 + y ). | | y D(x,1) | | 2 It follows that N N (1 + x ) D f . (1 + x ) D f 1, k | | | k1 k | | k +n+1 | |X| | and then Proposition 2.2 follows from (14). ⇤

Theorem 2.3 If f then fˆ .Furthermore,themapf fˆ is continuous from to . 2S 2S ! S S Proof As usual we explicitly prove only the ﬁrst statement. If f then f L1,sofˆ is bounded. Thus if f ,thenD\↵xf is bounded for any given ↵2andS ,since2D↵xf is again in .However,Propositions1.3and1.4implythat2S S ↵ ↵ ↵ D\x f(⇠)=(2⇡i)| |( 2⇡i)| |⇠ D fˆ(⇠) 10 so ⇠↵Dfˆ is again bounded, which means that fˆ . 2S ⇤

Appendix: pointwise Poincare inequalities

This is a little more technical than the preceding and we will omit some details. We prove a frequently used pointwise estimate for a function in terms of integrals of its gradient, which plays a similar role that the mean value inequality plays in calculus. Then we prove a generalization involving higher derivatives which includes Lemma 2.3. Let ! be the volume of the unit ball.

Lemma A1 Suppose that f is C1.Then 1 f(y) f(x) f(y)dy |r | dy. | ! | . x y n 1 ZD(x,1) ZD(x,1) | | Proof Applying the fundamental theorem of calculus to the function

t f(x + t(y x)) ! shows that 1 f(y) f(x) x y f(x + t(y x)) dt. | || | |r | Z0 Integrate this with respect to y over D(x, 1) and divide by !.Thus 1 1 f(x) f(y)dy f(x) f(y) dy | ! |! | | ZD(x,1) ZD(x,1) 1 x y f(x + t(y x)) dtdy . | | |r | ZD(x,1) Z0 1 = x y f(x + t(y x)) dydt. (19) | ||r | Z0 ZD(x,1) Make the change of variables z = x + t(y x), and then reverse the order of integration again. This leads to

1 1 dz (19) = t z x f(z) dt | ||r | tn Zt=0 ZD(x,t) 1 (n+1) = z x f(z) t dtdz D(x,1) | ||r | t= z x Z Z | | (n 1) x z f(z) dz . | | |r | ZD(x,1) as claimed.

11 Lemma A.2 Suppose that f is Ck.Then

Proof The case k = 1 follows immediately from Lemma A.1, so it has already been proved. To pass to general k we use induction based on the inequalities (here a>0,b> 0, z x constant) | | (n a b) C x z if a + bn, and : (n 1) 1 x y log dy C (22) y D(x,C) | | z y  Z 2 | | In fact, (21) may be proved by subdividing the region of integration in the three 1 1 regimes y x 2 z x , y z 2 z x and “the rest”, and noting that on the third | | | | | | | | (2n a b) regime the integrand is comparable to y x . (22) may be proved similarly. We now prove (20) by induction on| k.Wehavedonethecase| k =1.Supposethat 2 k n 1andthatthecasesuptoandincludingk 1havebeenproved.Then  

f (n k+1) f f(x) . j L1(D(x,1)) + x y k 1(y)dy | | k k | | j k 2 D(x,1) X Z f (n k+1) f . j L1(D(x,1)) + x y k 1(z)dzdy k k | | j k 2 D(x,1) D(y,1) X Z Z (n k+1) (n 1) f + x y y z (z)dzdy | | | | k ZD(x,1) ZD(y,1) f (n k) f 1 + x z (z)dz. . k j kL (D(x,2)) | | k j k 1 D(x,2) X Z For the ﬁrst two inequalities we used (20) with k replaced by k 1and1respectively, and for the last inequality we reversed the order of integration and used (21). The disc D(x, 2) can be replaced by D(x, 1) using rescaling, so we have proved (20) for k n 1.  To pass from k = n 1tok = n we argue similarly using the second case of (21), and to pass from k = n to k = n + 1 we argue similarly using (22). ⇤

3. Fourier inversion and Plancherel

12 Convolution of and f is deﬁned as follows:

f(x)= (y)f(x y)dy (23) ⇤ Z We assume the reader has seen this deﬁnition before but will summarize some facts, mostly without giving the proofs. There is an issue of the appropriate conditions on and f under which the integral (23) makes sense. We recall the following.

1. If L1 and f Lp,1 p , then the integral (23) is an absolutely convergent Lebesgue integral2 for a.e.2 x and 1 f f . (24) k ⇤ kp k k1k kp 1 2. If is a continuous function with compact support and f Lloc,thentheintegral (23) is an absolutely convergent Lebesgue integral for every x and2 f is continuous. ⇤ 3. If Lp and f Lp0 , 1 + 1 = 1, then the integral (23) is an absolutely convergent p p0 Lebesgue integral2 for every2 x,and f is continuous. Furthermore, ⇤ f p f p . (25) k ⇤ k1 k k k k 0 The absolute convergence of (23) in 2. is trivial, and (25) follows from Cauchy- Schwarz. Continuity then follows from the dominated convergence theorem. 1. is obvious when p = .Itisalsotrueforp =1,byFubini’stheoremandachangeofvariables. 1 The general case follows by interpolation, see the Riesz-Thorin theorem in Section 4.

In any of the above situations convolution is commutative: the integral deﬁning f is again convergent for the same values of x,andf = f.Thisfollowsbymaking⇤ the change of variables y x y.Noticealsothat⇤ ⇤ ! supp( f) supp +suppf, ⇤ ⇢ where the sum E + F means x + y : x E,y F . In many applications the function{ 2is ﬁxed2 and} very “nice”, and one considers convolution as an operator f f. ! ⇤

1 Lemma 3.1 If C1 and f L then f is C1 and 2 0 2 loc ⇤ D↵( f)=(D↵) f. (26) ⇤ ⇤

Proof It is enough to prove that f is C1 and (26) holds for multiindices of length ⇤ 1, since one can then use induction. Fix j and consider di↵erence quotients 1 d(h)= ( f)(x + he ) ( f)(x) . h ⇤ j ⇤ ⇣ ⌘ 13 Using (23) and commutativity of convolution, we can rewrite this as 1 d(h)= ((x + he y) (x y))f(y)dy. h j Z The quotients 1 A (y)= ((x + he y) (x y)) h h j are bounded by @ by the mean value theorem. For ﬁxed x and for h 1, the k @xj k1 | | support of Ah is contained in the ﬁxed compact set E = D(x, 1)+ supp .Thustheinte- 1 @ grands Ahf are dominated by the L function E f .Thedominatedconvergence k @xj k1 | | theorem implies @ lim d(h)= f(y)limAh(y)dy = f(y) (x y)dy. h 0 h 0 @x ! Z ! Z j This proves (26) (when ↵ = 1). The continuity of the partials then follows from 2. | | above. ⇤

Corollary 3.1’ If f,g then f g . 2S ⇤ 2S Proof By Lemma 3.1 it suces to show that if (1 + x )N f(x)and(1+ x )N g(x) are bounded for every N then so is (1 + x )N f g(x). This| | follows by writing| | out the | | ⇤ deﬁnitions and using (15); the details are left to the reader. ⇤

Convolution interacts with the Fourier transform as follows: Fourier transform con- verts convolution to ordinary pointwise multiplication. Thus we have the following formulas: f[g = fˆg,ˆ f,g L1, (27) ⇤ 2 fg = fˆ g,ˆ f,g . (28) ⇤ 2S (27) follows from Fubini’s theorem and is in many textbooks; the proof is left to the c reader. (28) then follows easily from the inversion theorem, so we defer the proof until after Theorem 3.4. ✏ n 1 Let ,andassumethat =1.Deﬁne (x)=✏ (✏ x). The family of functions 2✏S is called an approximate identity.Noticethat ✏ =1forall✏.Thus R one can regard{ } the ✏ as roughly convergent to the Dirac mass as ✏ 0. Indeed, the R 0 ! following fact is basic but quite standard; see any reasonable book on real analysis for the proof.

Lemma 3.2 Let and =1.Then: 1. If f is a continuous2S function which goes to zero at then ✏ f f uniformly as R 1 ⇤ ! ✏ 0. !2. If f Lp,1 p< then ✏ f f in Lp as ✏ 0. 2  1 ⇤ ! ! 14 Let us note the following corollary:

1 Lemma 3.3 Suppose f Lloc.Thenthereisaﬁxedsequence gk C01 such that if p [1, )andf Lp,then2 g f in Lp.Iff is continuous and{ goes} to⇢ zero at ,then 2 1 2 k ! 1 g f uniformly. k ! The reason for stating the lemma in this way is that one sometimes has to deal with several notions of convergence simultaneously, e.g., L1 and L2 convergence, and it is convenient to be able to approximate f in both norms simultaneously.

Proof Let C01, =1, 0, and let be as in Lemma 1.5. Fix a sequence 2 x ✏k ✏k 0. Let gk(x)=( k ) ( f). # p R· ⇤ ✏k If f L ,thenforlargek the quantity f Lp( x k) is bounded by f Lp( x k 1) 2 k ⇤ k | | k k | | using (24) and that supp ✏k is contained in D(0, 1). Accordingly, g ✏k f 0as k k ⇤ kp ! k .Ontheotherhand, ✏k f f 0 by Lemma 3.2. If f is continuous and !1 k ⇤ kp ! goes to zero at ,thenonecanarguethesamewayusingtheﬁrstpartofLemma3.2. 1 Smoothness of gk follows from Lemma 3.1, so the proof is complete. ⇤

Theorem 3.4 (Fourier inversion) Suppose that f L1,andassumethatfˆ is also in L1. Then for a.e. x, 2 2⇡i⇠ x f(x)= fˆ(⇠)e · d⇠. (29) Z Equivalently, fˆ(x)=f( x)fora.e.x. (30) b The proof uses Lemma 3.2 and also the following facts:

⇡ x 2 A. The gaussian (x)=e | | satisﬁes ˆ =, and therefore also satisﬁes (30). So at any rate there is one function f for which Theorem 3.4 is true. In fact this implies that there are many such functions. Indeed, if we form the functions

⇡✏2 x 2 ✏(x)=e | | , then we have ⇠ 2 n ⇡ | |2 ✏(⇠)=✏ e ✏ . (31) 1 Applying this again with ✏ replaced by ✏ ,onecanverifythat✏ satisﬁes (30). See the c discussion after formula (5).

B. The duality relation for the Fourier transform, i.e., the following lemma.

15 n n Lemma 3.5 Suppose that µ M(R )and⌫ M(R ). Then 2 2 µdˆ ⌫ = ⌫ˆdµ. (32) Z Z In particular, if f,g L1,then 2 fˆ(x)g(x)dx = f(x)ˆg(x)dx. Z Z Proof This follows from Fubini’s theorem:

2⇡i⇠ x µdˆ ⌫ = e · dµ(x)d⌫(⇠) Z ZZ 2⇡i⇠ x = e · d⌫(⇠)dµ(x) ZZ = ⌫ˆdµ. Z ⇤

Proof of Theorem 3.4 Consider the integral in (29) with a damping factor included:

⇡✏2 ⇠ 2 2⇡i⇠ x I✏(x)= fˆ(⇠)e | | e · d⇠. (33) Z We evaluate the limit as ✏ 0intwodi↵erentways. ! ˆ 2⇡i⇠ x 1. As ✏ 0, I✏(x) f(⇠)e · d⇠ for each fixed x. This follows from the dominated ! ! convergence theorem, since fˆ L1. R 2 ⇡✏2 ⇠ 2 2⇡i⇠ x 2. With x and ✏ fixed, define g(⇠)=e | | e · .Thus

I✏(x)= f(y)ˆg(y)dy Z by Lemma 3.5. On the other hand, we can evaluateg ˆ using the fact that g(⇠)=ex(⇠)✏(⇠) and (4), (31). Thus ✏ gˆ(y)=✏(y x)= (x y), ✏ n y where (y)=✏ (✏) is an approximate identity as in Lemma 3.2, and we have used that is even. c Accordingly, ✏ I✏ = f , ⇤ and we conclude by Lemma 3.2 that

I✏ f ! 16 in L1 as ✏ 0. ! ˆ 2⇡ix ⇠ Summing up, we have seen that the functions I✏ converge pointwise to f(⇠)e · d⇠, and converge in L1 to f.Thisisonlypossiblewhen(29)holds. R ⇤ Corollaries of the inversion theorem The first corollary below is not really a corollary, but a reformulation of the proof without the assumption that fˆ L1.Thisistheformtheinversiontheoremtakes 2 for general f. Notice that the integrals I✏ are well defined for any f L1,sincethe ⇡✏2 x 2 2 Gaussian e | | is integrable for each fixed ✏.Corollary3.6isoftenstatedas“the Fourier transform of f is Gauss-Weierstrass summable to f”, and can be compared to the theorem on Cesaro summability for Fourier series.

Corollary 3.6 1. Suppose f L1 and deﬁne I✏(x) via (33). Then I✏ f in L1 as 2 ! ✏ 0. ! 2. If 1

Corollary 3.7 If f L1 and fˆ =0,thenf =0. 2 This is immediate from Theorem 3.4. ⇤

Theorem 3.8 The Fourier transform maps onto . S S Proof Given f ,letF (x)=f( x)andletg = Fˆ.Theng by Theorem 2.3, and (30) implies 2S 2S ˆ gˆ(x)=Fˆ(x)=F ( x)=f(x) ⇤

Let us also prove formula (28). Let f,g .Then 2S ˆ f[ˆ gˆ( x)=fˆ( x)gˆ( x) ⇤ = f(x)g(x) by (27) and Theorem 3.4. Using Theorem 3.4 again, it follows that fˆ gˆ = fg. ⇤ ⇤ Theorem 3.9 (Plancherel theorem, ﬁrst version) If u, v then c 2S uˆvˆ = uv. (34) Z Z

17 Proof By the inversion theorem,

u(x)v(x)dx = uˆ( x)v(x)dx = uˆ(x)v( x)dx Z Z Z i.e., uˆvˆ = uˆv.˜ Z Z Applying the duality relation to the right-hand side we obtain

uv = uˆv,ˆ˜ Z Z and now (34) follows from (6). ⇤

Theorem 3.9 says that the Fourier transform restricted to Schwartz functions is an isometry in the L2 norm. Since is dense in L2 (e.g., by Lemma 3.3) this suggests a way S of extending the Fourier transform to L2.

Theorem 3.10 (Plancherel theorem, second version) There is a unique bounded operator : L2 L2 such that f = fˆ when f . has the following additional properties:F ! F 2S F

1. is a unitary operator. 2. Ff = fˆ if f L1 L2. F 2 \ Proof The existence and uniqueness statement is immediate from Theorem 3.9, as is the fact that f 2 = f 2.Inviewofthisisometryproperty,therangeof must be closed, and unitaritykF k ofk kwill follow if we show that the range is dense. However,F the latter statement is immediateF from Theorem 3.8 and Lemma 3.3. It remains to prove 2. For f ,2.istruebydefinition.Supposenowthatf L1 L2. 2S 1 2 \ 2 By Lemma 3.3, there is a sequence gk which converges to f both in L and in L . {ˆ }⇢S By Proposition 1.1, gk converges to f uniformly. On the other hand, gk converges to f in L2 by boundedness of the operator . It follows that f = fˆ. F F F ⇤ b b Statement 2. allows us to use the notation fˆ for f if f L2 without any possible ambiguity. We may therefore extend the definition ofF the Fourier2 transform to L1 + L2 n (in fact to -finite measures of the form µ + fdx, µ M(R ),f L2)viaf[+ g = fˆ+ˆg. 2 2 Corollary 3.12 The following form of the duality relation is valid:

⌫ˆ = ˆd⌫, , 2S Z Z

18 n if ⌫ = µ + fdx, µ M(R ), f L2. 2 2 Proof We have already proved this in Lemma 3.5 if f =0,soitsucestoproveit when µ =0,i.e.,toshowthat fˆ = f ˆ Z Z if f L2, . This is true by Lemma 3.5 if f L1 L2. Therefore it is also true for f 2L2,sinceforﬁxed2S both sides depend continuously2 \ on f (in the case of the left-hand 2 side this follows from the Plancherel theorem). ⇤

n Theorem 3.13 If µ M(R ), f L2 and 2 2 fˆ+ˆµ =0,

n then µ = fdx. In particular, if µ M(R )andˆµ L2,thenµ is absolutely continuous 2 2 with respect to the Lebesgue measure with an L2 density.

Proof By the Riesz representation theorem for measures on compact sets, the measure µ + fdx will be zero provided dµ + fdx =0 (35) Z for continuous with compact support.

If C01 then (35) follows from Corollary 3.12. In general, we choose (e.g., by 2 2 Proposition 3.2) a sequence k in C01 which converges to uniformly and in L .We write down (35) for the k’s and pass to the limit. This proves (35). To prove the last statement, suppose thatµ ˆ L2, and choose (by Theorem 3.10) a 2 function g L2 withg ˆ =ˆµ.Thendµ gdx has Fourier transform zero, so by the ﬁrst 2 part of the proof dµ = gdx. ⇤

All the basic formulas for the L1 Fourier transform extend to the L1 + L2-Fourier transform by approximation arguments. This was done above in the case of the duality relation. Let us note in particular that the transformation formulas in Section 1 extend. For example, in the case of (5), one has

1 t f[T = det (T ) fˆ T (36) | | if f L1 + L2.Sincewealreadyknowthiswhenf L1,itsucestoproveitwhen 2 2 f L2.Choose f L1 L2, f f in L2.CompositionwithT is continuous on L2, 2 { k}⇢ \ k ! as is Fourier transform, so we can write down (36) for the fk and pass to the limit. The L1 + L2 domain for the Fourier transform is wide enough{ } to include many natural examples. Note in particular that Lp L1 + L2 if p (1, 2). Furthermore, certain homogeneous functions belong to L1 + L2⇢although none of2 them can belong to Lp for any a 1 2 n 1 ﬁxed p. For example, x belongs to L + L if 2

19 most natural way to proceed would be to develop the idea of tempered distributions, but we don’t want to do this explicitly. Instead, we further broaden the deﬁnition of Fourier n transform as follows: a tempered function is a function f L1 (R )suchthat 2 loc

N (1 + x ) f(x) dx < | | | | 1 Z for some constant N. Roughly, f has at most polynomial growth in the sense of L1 averages. It is clear that if f is tempered and ,then f < .Furthermore,themap 2S | | 1 f is continuous on . It follows that f is well deﬁned if and f is R tempered,! and a simple estimationS shows that f⇤ is again tempered. 2S R If f and g are tempered functions, we say that⇤g is the distributional Fourier transform of f if g = fˆ (37) Z Z for all .Forgivenf,suchafunctiong is unique using the density properties of as in several2S previous arguments. We denote g by fˆ.Noticealsothatiff L1 + LS2, 2 then its L1 + L2-Fourier transform coincides with its distributional Fourier transform by Proposition 3.12. All the basic formulas, in particular (3), (4), (5), (6), extend to the case of distributional Fourier transforms, e.g., if g is the distributional Fourier transform of f,then 1 t det T fˆ T is the distributional Fourier transform of f T .Thismaybeseenby | | making appropriate changes of variable in the integrals in (37). We indicate how these arguments are carried out by proving the extended version of formula (27). Namely, if f is tempered, ,andf has a distributional Fourier transform, then so does f and 2S ⇤ [f = ˆfˆ (38) ⇤ The proof is as follows. Let be another Schwartz function. Then

( ˆfˆ) = fˆ( ˆ) Z Z = f ˆ Z ˆ = f(c ˆ ˆ) ⇤ Z = f(x) ( (x y))ˆ(y)dydx Z Z = f(y)ˆ(y). ⇤ Z The second line followed from the deﬁnition of distributional Fourier transform, the third line from (28), and the next to last line used the inversion theorem for .Comparing with the deﬁnition (37), we see that we have proved (38).

20 Let us note also that the inversion theorem is true for distributional Fourier transforms: if f is tempered and has a distributional Fourier transform fˆ,thenfˆ has the distributional Fourier transform f( x). Here is the proof. If ,then 2S f( x)(x)dx = f(x)( x)dx Z Z ˆ = f(x)ˆ(x)dx Z = fˆ(x)ˆ(x)dx. Z We used a change of variables, the inversion theorem for , and the deﬁnition (37) of fˆ. Comparing again with (37), we have the stated result. ⇤

4. Some speciﬁcs, and Lp for p<2

We ﬁrst discuss a couple of basic examples where the Fourier transform can be calculated, namely powers of the distance to the origin and complex Gaussians.

(a/2) a 1 2 Proposition 4.1 Let ha(x)= ⇡a/2 x .Thenha = hn a in the sense of L + L n | | Fourier transforms if 2 < Re (a)

Here is the gamma function, i.e.,

1 t s 1 (s)= e t dt. Z0

n 1 2 Proof Suppose that a is real and 2

n n 1. f is radial, i.e., f ⇢ = f for all linear ⇢ : R R with ⇢t⇢ =identity. 2. f is homogeneous of degree a,i.e., ! a f(✏x)=✏ f(x)(39) for each ✏>0.

We will use the notation f✏(x)=f(✏x), (40) ✏ n x f (x)=✏ f( ). (41) ✏

21 a n Let f(x)= x ,

a ⇡ x 2 (n a) ⇡ x 2 x e | | dx = c x e | | dx. (42) | | | | Z Z To evaluate the left hand side, change to polar coordinates and then make the change of variable t = ⇡r2.Thus,if is the area of the unit sphere, we get

a ⇡ x 2 1 ⇡r2 n a dr x e | | dx = e r | | r Z Z0 n a 1 t t dt = e ( ) 2 ⇡ 2t Z0 n a ( ) n a = ⇡ 2 ( ), 2 2

a ( 2 ) a and similarly the right hand side of (42) is c 2 ⇡ ( 2 ). Hence

a 2 n a ⇡ ( 2 ) c = n a , 2 a ⇡ ( 2 )

n and the proposition is proved in the case 2

A(z)= hzˆ, Z

B(z)= hn z. Z Both A and B may be seen to be analytic in z in the indicated regime: since is analytic, this reduces to showing that z x (x)dx | | Z is analytic when ,whichmaybedonebyusingthedominatedconvergencetheorem to justify complex2 di↵Serentiation under the integral sign. n By Proposition 3.14, A and B agree for z in ( 2 ,n). So they agree everywhere by the uniqueness theorem. This proves that the distributional Fourier transform of ha exists n 1 2 1 2 and is hn a.IfRea> ,thenha L + L ,sothatitsL + L and distributional Fourier 2 2 transforms coincide. ⇤

22 Let T be an invertible n n real symmetric matrix. The signature of T is the quantity ⇥ k+ k where k+ and k are the numbers of positive and negative eigenvalues of T , counted with multiplicity. We also deﬁne

⇡i Tx,x GT (x)=e h i, and observe that GT has absolute value 1 and is therefore tempered.

Proposition 4.2 Let T be an invertible n n real symmetric matrix with signature . ⇥ Then GT has a distributional Fourier transform, which is equal to

⇡i 1 e 4 det T 2 G T 1 | |

Remark This can easily be generalized to complex symmetric T with nonnegative imaginary part (the latter condition is needed, else GT is not tempered). See [17], Theorem 7.6.1. If n = 1, we do this case in the course of the proof.

Proof We need to show that

⇡i 1 1 ⇡i Tx,x ⇡i T x,x e h iˆ(x)dx = e 4 det T 2 e h i(x)dx (43) | | Z Z if and T is invertible real symmetric. 2S First consider the n = 1 case. Let pz be the branch of the square root deﬁned on the complement of the nonpositive real numbers and positive on the positive real axis. Thus ⇡i p i = e± 4 .Accordingly,(43)withn =1isequivalentto ± 2 ⇡zx2 1 ⇡ x e ˆ(x)dx =(pz) e z (x)dx (44) Z Z if and z is pure imaginary and not equal to zero. We prove this formula by analytic 2S continuation from the real case. Namely, if z =1then(44)isExample2inSection1,andthecaseofz real and ✏ positive then follows from scaling, i.e., the fact that the Fourier transform of f✏ is fˆ ,see (5). Both sides of (44) are easily seen to be analytic in z when Re z>0andcontinuous in z when Re z 0,z = 0, so (44) is proved. 6 Now consider the n 2case.Observethatif(43)istrueforagivenT (and all ), 1 it is true also when T is replaced by UTU for any U SO(n). This follows from the 2 fact that f[U = fˆ U.However,sincewedidnotgiveanexplicitproofofthelatter fact for distributional Fourier transforms, we will now exhibit the necessary calculations. 1 Let S = UTU .ThusS and T have the same determinant and the same signature. Accordingly, if (43) holds for T then

⇡i Sx,x ⇡i TU 1x,U 1x e h iˆ(x)dx = e h iˆ(x)dx Z Z 23 ⇡i Tx,x = e h iˆ(Ux)dx Z ⇡i Tx,x = e h i[U(x)dx Z ⇡i 1 1 ⇡i T x,x = e 4 det T 2 e h i U(x)dx | | Z ⇡i 1 1 1 1 ⇡i T U x,U x = e 4 det T 2 e h i(x)dx | | Z ⇡i 1 1 ⇡i S x,x = e 4 det S 2 e h i(x)dx. | | Z We used that [U = ˆ U for Schwartz functions , see the comments after formula (5) in Section 1. It therefore suces to prove (43) when T is diagonal. If T is diagonal and is a tensor function, then the integrals in (43) factor as products of one variable integrals and (43) follows immediately from (44). The general case then follows from Proposition 2.1’ and the fact that integration against a tempered function deﬁnes a continuous linear functional on . S ⇤ We now brieﬂy discuss the Lp Fourier transform, 1

Riesz-Thorin interpolation theorem. Let T be a linear operator with domain Lp0 +Lp1 , 1 p

For the proof see [20], [34], or numerous other textbooks.

24 We will adopt the convention that when indices p and p0 are used we must have 1 1 + =1. p p0

Proposition 4.3 (Hausdor↵-Young) If 1 p 2then   fˆ f . (49) k kp0 k kp

Proof We interpolate between the cases p =1and2,whichwealreadyknow.Namely, apply the Riesz-Thorin theorem with p =1,q = , p = q =2,A = A =1.The 0 0 1 1 1 0 1 hypotheses (45) and (46) follow from Proposition 1.1 and Theorem 3.10 respectively. For given p, q,existenceof✓ (0, 1) for which (47) and (48) hold is equivalent to 1

For later reference we insert here another basic result which follows from Riesz-Thorin, although this one (in contrast to Hausdor↵-Young) could also be proved by elementary manipulation of inequalities.

Proposition 4.4(Young’s inequality) Let Lp, Lr,where1 p, r and 2 2  1 1 + 1 1. Let 1 = 1 1 .Thentheintegraldeﬁning is absolutely convergent for p r q p r0 a.e. x and ⇤ . k ⇤ kq k kpk kr

Proof View as ﬁxed, i.e., deﬁne

T = . ⇤ Inequalities (24) and (25) imply that

1 p p T : L + L 0 L + L1 ! with T k kp k kpk k1 T p p . k k1 k k k k 0 If 1 = 1 1 then there is ✓ [0, 1] with q p r0 2 1 1 ✓ ✓ = + r 1 p0 1 1 ✓ ✓ = + . q p 1 25 The result now follows from Riesz-Thorin. ⇤

Remarks 1. Unless p = 1 or 2, the constant 1 in the Hausdor↵-Young inequality is not the best possible; indeed the best constant is found by testing the Gaussian function . This is much deeper and is due to Babenko when p0 is an even integer and to Beckner [1] in general. There are some related considerations in connection with Proposition 4.4, due also to Beckner.

2. Except in the case p =2theinequality(49)isnotreversible,inthesensethat there is no constant C such that f f when f .Equivalently(inviewofthe k kp0 k kp 2S inversion theorem) the result does not extend to the case p>2. This is not at all dicult to show, but we discuss it at some length in order to illustrate a few di↵erent techniques used for constructing examples in connection with the Lp Fourier transform. Here is the most elementary argument.

Exercise Using translation and multiplication by characters, construct a sequence of Schwartz functions n so that { } p 1. Each n has the same L norm. p0 2. Each n has the same L norm. 3. The supports of the n are disjoint. 4. The supportsc of the n are “essentially disjoint” meaning that c N N p p ( N) k nkp ⇡ k nkp ⇡ n=1 n=1 X X uniformly in N. Use this to disprove the converse of Hausdor↵-Young.

Here is a second argument based on Proposition 4.2. This argument can readily be adapted to show that there are functions f Lp for any p>2whichdonothavea 2 distributional Fourier transform in our sense. See [17], Theorem 7.6.6. ⇡ix2 Take n =1andf (x)=(x)e ,where C1 is ﬁxed. Here is a large positive 2 0 number. Then f is independent of for any p.BythePlanchereltheorem, f k kp k k2 ˆ 1 is also independent of .Ontheotherhand,f is the convolution of ,whichisinL , 1 1 ⇡i x2 1 c with (pi) e ,whichhasL1 norm 2 .Accordingly,ifp<2then c 2 1 2 p0 p0 f p f f k k 0 kk2 k k1 ( 1 1 ) 2 p0 . c . c c Since f is independent of ,thisshowsthatwhenp<2thereisnoconstantC k kp such that C f f for all f . k kp0 k kp 2S ⇤

26 Here now is another important technique (“randomization”) and a third disproof of the converse of Hausdor↵-Young. Let ! N be independent random variables taking values 1withequalprobability. { n}n=1 ± Denote expectation (a.k.a. integral over the probability space in question) by E,and probability (a.k.a. measure) by Prob. Let a N be complex numbers. { n}n=1 Proposition 4.5 (Khinchin’s inequality)

N N p 2 p E( an!n ) ( an ) 2 (50) | | ⇡ | | n=1 n=1 X X for any 0 1. There are three steps.

(i) When p =2itissimpletoseefromindependencethat(50)istruewithequality: expand out the left side and observe that the cross terms cancel.

(ii) The upper bound. This is best obtained as a consequence of a stronger (“subgaus- sian”) estimate. One can clearly assume the an are real and (52) below is for real an . Let t>0. We have { } { }

t an!n tan!n 1 tan tan E(e n )= E(e )= (e + e ), 2 n n Y Y where the ﬁrst equality follows from independence and the fact that ex+y = exey.Usethe numerical inequality 2 1 x x x (e + e ) e 2 (51) 2  to conclude that 2 t a ! t a2 E(e n n n ) e 2 n n ,  therefore 2 t+ t a2 Prob( a ! ) e 2 n n n n  n X for any t>0and>0. Taking t = 2 gives n an 2 2 Prob( a ! ) e 2 n an , (52) n n  n X hence 2 2 Prob( a ! ) 2e 2 n an . | n n|  n X 27 From this and the formula for the Lp norm in terms of the distribution function,

p p 1 E( f )=p Prob( f )d | | | | Z one gets 2 p p p 1 2 p p 2 an 2+ 2 2 2 E( an!n ) 2p e n d =2 p( )( an) . | |  2 n n X Z X This proves the upper bound.

(iii) The lower bound. This follows from (i) and (ii) by duality. Namely

To apply this in connection with the converse of Hausdor↵-Young, let be a C01 def function, and let k N be such that the functions = ( k ) have disjoint support. { j}j=1 j · j 2⇡i⇠ kn ˆ p Thus n(⇠)=e · (⇠). The L norm of n N !nn is independent of ! in view of the disjoint support, indeed  P 1 c ! = CN p , (53) k n nkp n N X where C = p. Now considerk k the corresponding Fourier side norms, more precisely the expectation of their p0 powers: p0 E( !nn . (54) k kp0 n N X We have by Fubini’s theorem c

2⇡i⇠ kn ˆ p0 (54) = E( !ne · (⇠)) p k kL 0 (d⇠) n N X ˆ p0 2⇡i⇠ kn p0 = (⇠) E( !ne · ) n | | | | R n N Z X p0 N 2 , ⇡ 28 where at the last step we used Khinchin. 1 2 It follows that we can make a choice of ! so that n N !nn p0 . N .Ifp<2and if N is large, this is much smaller than the right handk side of (53),k so we are done. P c

5. Uncertainty Principle

The uncertainty principle is1 the heuristic statement that if a measure µ is supported on R,thenformanypurposesˆµ may be regarded as being constant on any dual ellipsoid R⇤. The simplest rigorous statement is

Proposition 5.1 (L2 Bernstein inequality) Assume that f L2 and fˆ is supported in 2 D(0,R). Then f is C1 and there is an estimate

↵ ↵ D f (2⇡R)| | f . (55) k k2  k k2 Proof Essentially this is an immediate consequence of the Plancherel theorem, see the calculation at the end of the proof. However, there are some details to take care of if one wants to be rigorous. The Fourier inversion formula

2⇡ix ⇠ f(x)= fˆ(⇠)e · d⇠ (56) Z is valid (in the naive sense). Namely, note that the support assumption implies fˆ L1, 12 and choose a sequence of Schwartz functions k which converges to fˆ both in L and 2 in L . Let k satisfy k = k. The formula (56) is valid for k.Ask ,the left sides converge2S in L2 to f by Theorem 3.10 and the right sides converge uniformly!1 to 2⇡ix ⇠ fˆ(⇠)e · d⇠,whichproves(56)forc f. Proposition 1.3 applied to fˆ now implies that f is C and that D↵f is obtained by R 1 di↵erentiation under the integral sign in (56). The estimate (55) holds since

↵ ↵ ↵ ↵ ↵ D f = D f = (2⇡i⇠) fˆ (2⇡R)| | fˆ =(2⇡R)| | f . k k2 k k2 k k2  k k2 k k2 ⇤ d AcorrespondingstatementisalsotrueinLp norms, but proving this and other related results needs a di↵erent argument since there is no Plancherel theorem.

Lemma 5.2 There is a ﬁxed Schwartz function such that if f L1 + L2 and f is 2 supported in D(0,R), then 1 f = R f. b ⇤ 1This should be qualiﬁed by adding “as far as we are concerned”. There are various more sophisticated related statements which are also called uncertainty principle; see for example [14], [15] and references there.

29 R 1 1 Proof Take so that is equal to 1 on D(0, 1). Thus [ (⇠)=(R ⇠)isequal 2S 1 1 to 1 on D(0,R), so (R f f) vanishes identically. Hence R f = f. ⇤ ⇤ ⇤ b b Proposition 5.3 (Bernstein’s inequality for a disc) Suppose that f L1 + L2 and fˆ is b 2 supported in D(0,R). Then

1. For any ↵ and p [1, ], 2 1 ↵ ↵ D f (CR)| | f . k kp  k kp 2. For any 1 p q   1 n( 1 1 ) f CR p q f . k kq  k kp

1 Proof The function = R satisﬁes

n = CRr0 (57) k kr for any r [1, ], where C = .Also,usingthechainrule 2 1 k kr = R . (58) kr k1 k k1 We know that f = f.Inthecaseofﬁrstderivatives,1.thereforefollowsfrom(57) ⇤ and (24). The general case of 1. then follows by induction. For 2., let r satisfy 1 = 1 1 .ApplyYoung’sinequalityobtaining q p r0 f = f k kq k ⇤ kq r f p knk k k R r0 f . k kp n( 1 1 ) = R p q f . k kp ⇤

We now extend the Lp Lq bound to ellipsoids instead of balls, using change of n ! variable. An ellipsoid in R is a set of the form

2 n (x a) ej E = x R : | · | 1 (59) { 2 r2  } j j X n for some a R (called the center of E), some choice of orthonormal basis ej (the axes) 2 { } and some choice of positive numbers rj (the axis lengths). If E and E⇤ are two ellipsoids,

30 then we say that E⇤ is dual to E if E⇤ has the same axes as E and reciprocal axis lengths, i.e., if E is given by (59) then E⇤ should be of the form

n 2 2 x R : rj (x b) ej 1 { 2 | · |  j X for some choice of the center point b.

Proposition 5.4 (Bernstein’s inequality for an ellipsoid) Suppose that f L1 + L2 and 2 fˆ is supported in an ellipsoid E.Then

( 1 1 ) f E p q f k kq . | | k kp if 1 p q .   1 One could similarly extend the ﬁrst part of Proposition 5.3 to ellipsoids centered at the origin, but the statement is awkward since one has to weight di↵erent directions di↵erently, so we ignore this.

Proof Let k be the center of E. Let T be a linear map taking the unit ball onto E k. t t 2⇡ik x Let S = T ;thusT = S also. Let f (x)=e · f(x)andg = f S,sothat 1 1 1 t gˆ(⇠)= det S f1(S (⇠)) | | 1 t = det S fˆ(S (⇠ + k)) | | = det T fˆ(Tb(⇠)+k)). | | Thusg ˆ is supported in the unit ball, so by Proposition 5.3

g g . k kq . k kp On the other hand,

1 1 1 g = det S q f = det T q f = E q f k kq | | k kq | | k kq | | k kq and likewise with q replaced by p.So

1 1 E q f E p f | | k kq . | | k kp as claimed. ⇤

For some purposes one needs a related “pointwise statement”, roughly that if suppfˆ ⇢ E, then for any dual ellipsoid E⇤ the values on E⇤ are controlled by the average over E⇤. 2 N To formulate this precisely, let N be a large number and let (x)=(1+x ) . | | Suppose an ellipsoid R⇤ is given. Deﬁne (x)=(T (x k)), where k is the center E⇤ of E⇤ and T is a selfadjoint linear map taking E⇤ k onto the unit ball. If T and T 1 2 31 1 are two such maps, then T T is an orthogonal transformation, so is well deﬁned. 1 2 E⇤ Essentially, E⇤ is roughly equal to 1 on E⇤ and decays rapidly as one moves away from E⇤.Wecouldalsowritemoreexplicitly

(x k) e 2 N (x)= 1+ | · j| . E⇤ r2 j j ⇣ X ⌘ Proposition 5.5 Suppose that f L1 + L2 and fˆ is supported in an ellipsoid E.Then 2 for any dual ellipsoid E⇤ and any z E⇤, 2 1 f(z) C f(x) dx. (60) | | N E | | E⇤ | ⇤| Z

Proof Assume ﬁrst that E is the unit ball, and E⇤ is also the unit ball. Then f is the convolution of itself with a ﬁxed Schwartz function .Accordingly

f(z) f(x) (z x) dx | | | || | Z 2 N C f(x) (1 + z x )  N | | | | Z 2 N C f(x) (1 + x ) .  N | | | | Z We used the Schwartz space bounds for and that 1 + z x 2 1+ x 2 uniformly in x | | & | | when z 1. This proves (60) when E = E⇤ =unit ball. | | Suppose next that E is centered at zero but E and E⇤ are otherwise arbitrary. Let k and T be as above, and consider

1 g(x)=f(T x + k)).

1 Its Fourier transform is supported on T E,andifT maps E⇤ onto the unit ball, then 1 T maps E onto the unit ball. Accordingly,

g(y) (x) g(x) dx | | | | Z if y D(0, 1), so that 2

32 Remarks 1. Proposition 5.5 is an example of an estimate “with Schwartz tails”. It is not possible to make the stronger conclusion that, say, f(x) is bounded by the average | | of f over the double of E⇤ when x E⇤, even in the one dimensional case with E = E⇤ = 2 unit interval. For this, consider a ﬁxed Schwartz function g whose Fourier transform is supported in the unit interval [ 1, 1]. Consider also the functions x2 f (x)=(1 )N g(x). N 4

Since fˆN are linear combinations ofg ˆ and its derivatives, they have the same support as gˆ.Moreover,theyconvergepointwiseboundedlytozeroon[ 2, 2], except at the origin. It follows that there can be no estimate of the value of fN at the origin by its average over [ 2, 2]. 2. All the estimates related to Bernstein’s inequality are sharp except for the values of the constants. For example, if E is an ellipsoid, E⇤ adualellipsoid,N< ,thenthere 1 is a function f with suppfˆ E⇤ and with ⇢ f E , (61) k k1 | | f(x) C (x), (62) | | E (N) where E = E was deﬁned above. In the case E = E⇤ =unit ball this is obvious: take f to be any Schwartz function with Fourier support in the unit ball and with the appropriate L1 norm. The general case then follows as above by making changes of variable. 1 The estimates (61) and (62) imply that f p E p for any p, so it follows that Proposition 5.4 is also sharp. k k ⇡| |

6. Stationary phase

Let be a real valued C1 function, let a be a C01 function, and deﬁne

⇡i(x) I()= e a(x)dx. Z Here is a parameter, which we always assume to be positive. The issue is the behavior of the integral I()as + . ! 1 Some general remarks 1. I() is clearly bounded by a constant depending on a only. | | One may expect decay as ,sincewhen is large the integral will involve a lot of cancellation. !1

2. On the other hand, if is constant then I() is independent of .Sooneneedsto put nondegeneracy hypotheses on . As it turns| out,| properties of a are less important.

33 Note also that one can always cut up a with a partition of unity, which means that the question of how fast I()decayscanbe“localized”toasmallneighborhoodofapoint.

3. Suppose that = G where G is a smooth di↵eomorphism. Then 1 2

1 ⇡i2(x) ⇡i1(G x) e a(x)dx = e a(x)dx Z Z ⇡i1(y) = e a(Gy)d(Gy) Z ⇡i1(y) = e a(Gy) J (y) dy | G | Z where J is the Jacobian determinant. The function y a(Gy) J (y) is again C1,sowe G ! | G | 0 see that any bound for I()whichisindependentofchoiceofa will be “di↵eomorphism invariant”.

4. Recall from advanced calculus [23] the normal forms for a function near a regular point or a nondegenerate critical point:

n Straightening Lemma Suppose ⌦ R is open, f :⌦ R is C1, p ⌦and ⇢ ! 2 f(p) =0.ThenthereareneighborhoodsU and V of 0 and p respectively and a C1 r 6 di↵eomorphism G : U V with G(0) = p and ! f G(x)=f(p)+x . n

n Morse Lemma Suppose ⌦ R is open, f :⌦ R is C1, p ⌦, f(p)=0,and ⇢ 2 ! 2 r suppose that the Hessian matrix H (p)= @ f (p) is invertible. Then, for a unique k f @xi@xj (= number of positive eigenvalues of Hf ;seeLemma6.3below)thereareneighborhoods⇣ ⌘ U and V of 0 and p respectively and a C1 di↵eomorphism G : U V with G(0) = p and ! k n f G(x)=f(p)+ x2 x2. j j j=1 X j=Xk+1 We consider now I()ﬁrstwhena is supported near a regular point, and then when a is supported near a nondegenerate critical point. Degenerate critical points are easy to deal with if n =1,see[33],chapter8,butinhigherdimensionstheyaremuchmore complicated and only the two-dimensional case has been worked out, see [36].

n Proposition 6.1 (Nonstationary phase) Suppose ⌦ R is open, :⌦ R is C1, p ⇢ ! 2 ⌦and (p) =0.Supposea C1 has its support in a suciently small neighborhood r 6 2 0 of p.Then N N C : I() C , 8 9 N | | N 34 and furthermore CN depends only on bounds for ﬁnitely many derivatives of and a and alowerboundfor (p) (and on N). |r | Proof The straightening lemma and the calculation in 3. above reduce this to the case (x)=xn + c.Inthiscase,lettingen =(0,...,0, 1) we have

⇡ic I()=e aˆ( e ), 2 n and this has the requisite decay by Proposition 2.3. ⇤

Now we consider the nondegenerate critical point case, and as in the preceding proof we ﬁrst consider the normal form.

Proposition 6.2 Let T be a real symmetric invertible matrix with signature ,leta be C1 (or just in ), and deﬁne 0 S ⇡i Tx,x I()= e h ia(x)dx. Z Then, for any N, N ⇡i 1 n j (N+1) I()=e 4 det T 2 2 a(0) + a(0) + ( ) . | | Dj O j=1 ! X Here j are certain explicit homogeneous constant coecient di↵erential operators of orderD 2j,dependingonT only, and the implicit constant depends only on T and on bounds for ﬁnitely many Schwartz space seminorms of a.

Proof Essentially this is just another way of looking at the formula for the Fourier transform of an imaginary Gaussian. By Proposition 4.2, the deﬁnition of distributional Fourier transform, and the Fourier inversion theorem for a we have

1 n 1 1 ⇡i ⇡i T ⇠,⇠ I()=e 4 2 det T 2 aˆ( ⇠)e h id⇠. | | Z We can replacea ˆ( ⇠)withˆa( ⇠) by making a change of variables, since the Gaussian 1 is even. To understand the resulting integral, use that 0as ,sotheGaussian term is approaching 1. To make this quantitative, use Taylor’s! theorem!1 for eix: N 1 1 1 j 2N+2 1 (⇡i T ⇠,⇠ ) ⇠ ⇡i T ⇠,⇠ e h i = h i + (| | ) j! O N+1 j=0 X uniformly in ⇠ and .Accordingly, N 1 1 1 j 1 (⇡i T ⇠,⇠ ) ⇡i T ⇠,⇠ aˆ(⇠)e h id⇠ = aˆ(⇠)(1 + h i )d⇠ j! j=1 Z Z X ⇠ 2N+2 + aˆ(⇠) | | . O | | N+1 ⇣ Z ⌘ 35 Now observe that aˆ(⇠)d⇠ = a(0) by the inversion theorem, and similarly

1 j R (⇡i T ⇠,⇠ ) aˆ(⇠) h i d⇠ j! Z is the value at zero of a for an appropriate di↵erential operator .Thisgivesthe Dj Dj result, since aˆ(⇠) ⇠ 2N+2d⇠ | | | Z is bounded in terms of Schwartz space seminorms ofa ˆ,andthereforeintermsofderivatives of a. ⇤

Now we consider the case of a general phase function with a nondegenerate critical point. It is clear that this should be reducible to the Gaussian case using the Morse lemma and remark 3. above. However, there is more calculation involved than in the proof of Proposition 5.1, since we need to obtain the correct form for the asymptotic expansion. We recall the following formula which follows from the chain rule:

Lemma 6.3 Suppose that is smooth, (p)=0andG is a smooth di↵eomorphism, G(0) = p.Then r t H G(0) = DG(0) H(p)DG(0). Thus H(p)andH G(0) have the same signature and 2 det (H G(0)) = JG(0) det (H(p)).

Proposition 6.4 Let be C1 and assume that (p)=0andH(p)isinvertible.Let n r be the signature of H(p), and let = 2 det (H(p) . Let a be C01 and supported in asucientlysmallneighborhoodofp.Deﬁne| |

⇡i(x)dx I()= e a(x). Z Then, for any N,

N ⇡i(p) ⇡i 1 n j (N+1) I()=e e 4 2 2 a(p)+ a(p)+ ( ) . Dj O j=1 ! X 2 Here j are certain di↵erential operators of order 2j,withcoecientsdependingon ,andtheimplicitconstantdependsonD and on bounds for ﬁnitely many derivatives of a.

2Actually the order is exactly 2j but we have no need to know that.

36 Proof We can assume that (p)=0;elsewereplace with (p). Choose a C1 di↵eomorphism G by the Morse lemma and apply remark 3. Thus

⇡i Ty,y I = e h ia(Gy) J (y) dy, | G | Z where T is a diagonal matrix with diagonal entries 1andwithsignature.Also 1 ± J (0) = 2 by Lemma 6.3 and an obvious calculation of the Hessian determinant | G | of the function y Ty,y . Let j be associated to this T as in Proposition 5.2 and let b(y)=a(Gy) J (y!h) .Theni D | G | N ⇡i n j (N+1) I()=e 4 2 b(0) + b(0) + ( ) Dj O j=1 ! X 1 by Proposition 6.2. Now b(0) = J (0) a(p)= 2 a(p), so we can write this as | G | N ⇡i 1 n j 1 (N+1) I()=e 4 2 2 a(p)+ 2 b(0) + ( ) . Dj O j=1 ! X Further, it is clear from the chain rule and product rule that any 2j-th order derivative of b at the origin can be expressed as a linear combination of derivatives of a at p of order 1 2j with coecients depending on G,i.e.,on.Otherwisestated,theterm2 b(0)  Dj can be expressed in the form ˜ a(p), where ˜ is a new di↵erential operator of order 2j Dj Dj  with coecients depending on .Thisgivestheresult. ⇤

In practice, it is often more useful to have estimates for I()insteadofanasymptotic n expansion. Clearly an estimate I() 2 could be derived from Proposition 6.4, but | | . one also sometimes needs estimates for the derivatives of I()withrespecttosuitable parameters. For now we just consider the technically easiest case where the parameter is itself.

Proposition 6.5 (i) Assume that (p) = 0. Then for a supported in a small neigh- r 6 dj I() N borhood of p, j CjN for any N. d  (ii)Assume that (p)=0,andH(p)isinvertible.Then,fora supported in a small r neighborhood of p, k d ⇡i(p) ( n +k) (e I()) C 2 . k  k d Proof We only prove (ii), since (i) follows easily from Proposition 6.1 after di↵erenti- ating under the integral sign as in the proof below. For (ii) we need the following.

M Claim. Let i i=1 be real valued smooth functions and assume that i(p)=0, { } M i(p) = 0. Let =⇧i=1i.Thenallpartialderivativesofoforderlessthan2M also vanishr at p.

37 Proof By the product rule any partial D↵is a linear combination of terms of the form M i D i i=1 Y with = ↵.If ↵ < 2M,thensome must be less than 2, so by hypothesis all such i i | | i terms vanish at p. P To prove the proposition, di↵erentiate I()undertheintegralsignobtaining

k ⇡i(p) d (e I()) k k ⇡i((x) (p) =( ⇡i) ((x) (p)) a(x)e dx. dk Z Let b(x)=((x) (p))ka(x). By the above claim all partials of b of order less than 2k vanish at p. Now look at the expansion in Proposition 6.4 replacing a with b and setting N = k 1. By the claim the terms jb(p)mustvanishwhenj0. Put local coordinates on the sphere as follows: the ﬁrst “local coordinate” is the map

x (x, 1 x 2), ! | | n 1 p1 n 1 R D(0, ) S . 2 ! The second is the map x (x, 1 x 2), ! | | n 1 p1 n 1 R D(0, ) S , 2 ! and the remaining ones map onto sets whose closures do not contain e . Let q be {± n} { k} asuitablepartitionofunitysubordinatetothiscoveringbycharts.Deﬁne(x)=en x, n · : R R. Thus the gradient of is en and is normal to the sphere at en only. ! ± 38 Now

2⇡ien x ˆ(en)= e · d(x)

Zk 2⇡ien x = e · qk(x)d(x) j=1 X Z 2⇡ip1 x 2 q1(x) 2⇡ip1 x 2 q2(x) = e | | dx + e | | dx 2 2 D(0, 1 ) 1 x D(0, 1 ) 1 x Z 2 | | Z 2 | | 2⇡ik(x) + e apk(x)dx, p (63) k 3 X Z n 1 where the dx integrals are in R ,andthephasefunctionsk for k 3 have no critical points in the support of a . The Hessian of 2 1 x 2 at the origin is 2timesthe k | | identity matrix, and in particular is invertible. It is also clear that the ﬁrst and second p terms are complex conjugates. We conclude from Proposition 6.5 that

2⇡i ˆ(en)=Re(a()e )+y() with j n 1 d a() j . 2 , (64) dj j d y() N . (65) dj for any N.Infact is real and even and therefore ˆ must be real valued. Multiplying y 2⇡i by e does not a↵ect the estimate (65), so we can absorb y into a and rewrite this as

2⇡i ˆ(en)=Re(a()e ), where a satisﬁes (64). Since ˆ is radial, we have proved the following.

Corollary 6.6 The function ˆ (is a C1 function and) satisﬁes

2⇡i x ˆ(x)=Re(a( x )e | |), (66) | | where for large r j n 1 d a ( +j) C r 2 . (67) |drj | j

Furthermore, looking at the ﬁrst term in (63), and using the expansion of Proposition 6.4, with N = 0, we can obtain the leading behavior at .Namely,fortheﬁrsttermin (63) at its critical point x =0wehaveaphasefunction21 1 x 2 with (0) = 2, = 1 | | p

39 2 1/2 and signature (n 1), and an amplitude q2(x)(1 x ) which is 1 at the critical point. By Proposition 6.4 the integral is | |

⇡i n 1 n+1 2⇡i (n 1) e e 4 2 + ( 2 ). O N The second term is the complex conjugate and the others are ( )foranyN.Hence the quantity (63) is O n 1 n+1 n 1 2 2 cos(2⇡( )) + ( 2 ) 8 O and we have proved

Corollary 6.7 For large x | |

n 1 n+1 n 1 ˆ(x)=2x 2 cos(2⇡( x )) + ( x 2 ). | | | | 8 O | |

Remarks Of course it is possible to consider surfaces other than the sphere, see for example [17], Theorem 7.7.14. The main point in regard to the latter is that the nondegeneracy of the critical points of the phase function which arises when calculating the Fourier transform is equivalent to nonzero Gaussian curvature, so a hypersurface with nonzero Gaussian curvature everywhere behaves essentially the same as the sphere, whereas if there are ﬂat directions the decay becomes weaker. Obtaining derivative bounds like Corollary 6.6 in the above manner requires a somewhat more complicated version of Proposition 6.5 with and a depending on an auxiliary parameter z,whichwenowexplainwithout giving the proofs. n k Suppose that (x, z)isaC1 function of x and z,wherex R ,andz R should be regarded a parameter. Assume that for a certain p and z we2 have (p,2 z )=0and 0 rx 0 that the matrix of second x-partials of at (p, z0)isinvertible.

1. Prove that there are neighborhoods U of z0 and V of p and a smooth function  : U V with the following property: if z V ,then (x, z)=0ifandonlyif ! 2 rx x = (z).

2. Let a(x, z)beC01 and supported in a small enough neighborhood of (p, z0). Deﬁne

⇡i(x,z) I(, z)= e a(x, z)dx. Z Prove the following:

j+k ⇡i((z),z) d (e I(, z)) n ( 2 +j) j Cjk . d dzk  This is the analogue of Proposition 6.5 for general parameters.

40 7. Restriction problem

n 1 We now suppose given a function f : C and consider the Fourier transform S !

2⇡ix ⇠ fd(⇠)= f(x)e · d(x). (68) n 1 ZS d If f is smooth, then one can use stationary phase to evaluate fd to any desired degree of precision, just as with Corollary 6.6. In particular this leads to the bound

n 1 d fd(⇠) C f 2 (1 + ⇠ ) 2 | | k kC | | ↵ 2 say, where f C = 0 ↵ 2 dD f L1 . On thek otherk hand there| | cank bek no similar decay estimate for functions f which are just bounded. The reasonP for this is that then there is no distinguished reference point in 2⇡ik x Fourier space. Thus, if we let fk(x)=e · and set ⇠ = k,wehave

n 1 f[d(⇠) = (S ) 1. | k | ⇡ 2 Taking a sum of the form f = j f ,where k suciently rapidly, we obtain j kj | j|!1 a continuous function f such that there is no estimate P ✏ fd(⇠) C(1 + ⇠ ) | | | | for any ✏>0. d On the other hand, if we consider instead Lq norms then the issue of a distinguished origin is no longer relevant. The following is a long standing open problem in the area.

n 1 Restriction conjecture (Stein) Prove that if f L1(S )then 2

fd q Cq f (69) k k  k k1 2n for all q>n 1 . d 2n The example of a constant function shows that the regime q>n 1 would be best q n 1 possible. Namely, Corollary 6.7 implies that ˆ L if and only if q 2 >n. The corresponding problem for L2 densities2f was solved in the· 1970’s:

2 n 1 Theorem 7.1 (P. Tomas-Stein) If f L (S )then 2

fd C f 2 n 1 (70) k kq  k kL (S ) 2n+2 for q n 1 , and this range of q isd best possible. 41 Remarks 1. Notice that the assumptions on q in (70) and (69) are of the form q>q0 or q q . The reason for this is that there is an obvious estimate 0

fd f 1 k k1 k k by Proposition 1.1, and it follows by thed Riesz-Thorin theorem that if (69) or (70) holds for a given q, then it also holds for any larger q.

2. The restriction conjecture (69) is known to be true when n =2;thisisduetoC. Fe↵erman and Stein, early 1970’s. See [12] and [33].

3. Of course there is a di↵erence in the Lq exponent in (70) and the one which is conjectured for L1 densities. Until fairly recently it was unknown (in three or more dimensions) whether the estimate (69) was true even for some q less than the Stein-Tomas 2n+2 exponent n 1 .ThiswasﬁrstshownbyBourgain[3],apaperwhichhasbeenthestarting point for a lot of recent work.

2n+2 4. The fact that q n 1 is best possible for (70) is due I believe to A. Knapp. We 2 now discuss the construction. Notice that in order to distinguish between L and L1 norms, one should use a function f which is highly localized. Next, in view of the nice behavior of rectangles under the Fourier transform discussed e.g. in our section 5, it is n 1 natural to take the support of f to be the intersection of S with a small rectangle. Now we set up the proof. Let n 1 2 C = x S :1 x e , { 2 · n  } where e =(0,...,0, 1). Since x e 2 =2(1 x e ), it is easy to show that n | n| · n 1 x e C x C x e C (71) | n| ) 2 )| n| for an appropriate constant C.Nowletf = f be the indicator function of C.We 2 n 1 calculate f L (S ) and fd q.Allconstantsareofcourseindependentof. In thek ﬁrstk place, f k is thek square root of the measure of C , so by (71) and the k k2 dimensionality of the sphered we have

n 1 f 2 . (72) k k2 ⇡ 2 The support of fd is contained in the rectangle centered at en with length about in the en direction and length about in the orthogonal directions. We look at fd on the 1 2 1 1 dual rectangle centered at 0. Suppose then that ⇠ C and that ⇠ C | n| 1 | j| 1 when j

2⇡ix ⇠ fd(⇠) = e · d(x) | | C Z d 42 2⇡i(x en) ⇠ = e · d(x) C Z cos(2⇡(x e ) ⇠)d(x). n · ZC Our conditions on ⇠ imply if C is large enough that (x e ) ⇠ ⇡ ,say,forallx C . 1 | n · | 3 2 Accordingly, 1 n 1 fd(⇠) C . | |2| |⇡ (n+1) Our set of ⇠ has volume about d , so we conclude that

n 1 n+1 fd q . k kq & Comparing this estimate with (72) wed ﬁnd that if (70) holds then

n 1 n+1 n 1 q . 2

n+1 n 1 2n+2 uniformly in (0, 1]. Hence n 1 q 2 ,i.e.q n 1 . 2 For future reference we record the following variant on the above example: if f is as 2⇡ix ⌘ n n 1 above and g = e · Tf,where⌘ R and T is a rotation mapping en to v S ,then g is supported on 2 2 n 1 2 x S :1 x v , { 2 ·  } and n 1 gd | | & 1 2 1 1 on a cylinder of length C and cross-section radius C ,centeredat⌘ and with 1 d 1 the axis parallel to v. Before giving the proof of Theorem 7.1 we need to discuss convolution of a Schwartz n function with a measure, since this wasn’t previously considered. Let µ M(R ); assume µ has compact support for simplicity, although this assumption is not2 really needed. Deﬁne µ(x)= (x y)dµ(y). ⇤ Z Observe that µ is C1,sincedi↵erentiationundertheintegralsignisjustiﬁedasin Lemma 3.1. ⇤ It is convenient to use the notationµ ˇ forµ ˆ( x). We need to extend some of our formulas to the present context. In particular the following extends (28) since if µ then the Fourier transform ofµ ˇ is µ by Theorem 3.4: 2S

µˇ = ˆ µ when , (73) ⇤ 2S µ = ˆ µˆ when . (74) c ⇤ 2S

c 43 Notice that (73) can be interpreted naively: Proposition 1.3 and the product rule imply that µˇ is a Schwartz function. To prove (73), by uniqueness of distributional Fourier transforms it suces to show that

ˆµˇ = (ˆ µ) ⇤ Z Z if is another Schwartz function. This is done as follows. Denote Tx = x,then ˆ(x)(x)ˆµ( x)dx = ˆ( x)( x)ˆµ(x)dx Z Z = (( ˆ) T )ˆµ Z = (( ˆ) T )ˆdµ by the duality relation Z = ( \ˆ T ) (\ T )dµ ⇤ Z = (ˆ T )dµ ⇤ Z = (y)ˆ( x + y)dydµ(x) ZZ = (y)ˆ µ(y)dy. ⇤ Z For (74), again let be another Schwartz function. Then

µ dx = ˆdµ by the duality relation Z Z c = ˇ[ dµ ⇤ Z = (ˇ )ˆµdx by the duality relation ⇤ Z = (ˆ µˆ) dx. ⇤ Z The last line may be seen by writing out the deﬁnition of the convolution and using Fubini’s theorem. Since this worked for all ,weget(74). 2S Lemma 7.2 Let f,g ,andletµ be a (say) compactly supported measure. Then 2S fˆgdµˆ = (ˆµ g) fdx. (75) ⇤ · Z Z Proof Recall that gˆ˜ = g,ˆ

44 so that gˆ = gˆ˜ = g by the inversion theorem. Now apply the duality relation and (74), obtaining

fˆgdµˆ = f (gµˆ )ˆdx · Z Z = f (g µˆ)dx · ⇤ Z as claimed. ⇤

Lemma 7.3 Let µ be a ﬁnite positive measure. The following are equivalent for any q and any C. 1. fdµ C f , f L2(dµ). k kq  k k2 2 2. gˆ 2 C g , g . k kL (dµ)  k kq0 2S 3. µˆdf C2 f , f . k ⇤ kq  k kq0 2S Proof Let g , f L2(dµ). By the duality relation 2S 2 gfdµˆ = fdµ gdx. (76) · Z Z

d 2 If 1. holds, then the right side of (76) is g q0 fdµ q C g q0 f L (dµ) for any f L2(dµ). Hence so is the left side. This proveskk 2.k by duality.k  Ifk 2.k holdsk k then the left 2 side is gˆ 2 f 2 C g f 2 for g d.Hencesoistherightside.Since k kL (dµ)k kL (dµ)  k kq0 k kL (dµ) 2S is dense in Lq0 ,thisproves1.byduality. S If 3. holds, then the right side of (75) is C2 f 2 when f = g .Hencesois q0 the left side, which proves 2. If 2. holds then, for anyk kf,g , using2 alsoS the Schwartz 2 2S inequality the left side of (75) is C f q0 g q0 . Hence the right side of (75) is also C2 f g ,whichproves3.byduality. k k k k  k kq0 k kq0 ⇤ Remark One can ﬁt lemma 7.3 into the abstract setup

2 q q 2 q q T : L L T ⇤ : L 0 L TT⇤ : L 0 L . ! , ! , ! This is the standard way to think about the lemma, although it is technically a bit easier to present the proof in the above ad hoc manner. Namely, if T is the operator f fd ! then T ⇤ is the operator f fˆ, where we regard fˆ as being deﬁned on the measure space ! associated to µ,andTT⇤ is convolution withµ ˆ. d

Proof of Theorem 7.1 We will not give a complete proof; we only prove (70) when 2n+2 q> n 1 instead of . For the endpoint, see for example [35], [9], [32], [33]. 2n+2 We will show that if q> n 1 ,then ˆ f C f , (77) k ⇤ kq  qk kq0 45 which suces by Lemma 7.3. The relevant properties of will be

n 1 (D(x, r)) . r , (78) which reﬂects the n 1-dimensionality of the sphere, and the bound n 1 ˆ(⇠) (1 + ⇠ ) 2 (79) | | . | | from Corollary 6.6. Let be a C1 function with the following properties: 1 supp x : x 1 , ⇢{ 4   }

j if x 1then (2 x)=1. | | j 0 X Such a function may be obtained as follows: let be a C1 function which is equal to 1 1when x 1andto0when x 2 ,andlet(x)=(2x) (x). We now| | cut up ˆ as follows:| |

1 ˆ = K + Kj, 1 j=0 X where j Kj(x)=(2 x)ˆ(x),

1 j K (x)=(1 (2 x))ˆ. 1 j=0 X Then K is a C01 function, so 1

K f q . f p k 1 ⇤ k k k by Young’s inequality, provided q p.Inparticular,sinceq>2wemaytakep = q0. We now consider the terms in the sum. The logic will be that we estimate convolution 1 2 2 with Kj as an operator from L to L1 and from L to L ,andthenweuseRiesz-Thorin. We have n 1 j Kj . 2 2 k k1 1 by (79). Using the trivial bound Kj f Kj f 1 we conclude our L L1 1 1 bound, k ⇤ k k k k k ! n 1 j Kj f . 2 2 f 1. (80) k ⇤ k1 k k On the other hand, we can use (78) to estimate Kj.Namely,let = ˆ.Notealsothat ˆ =ˇ,since and therefore ˆ are invariant under the reﬂection x x.Accordingly, ! we have c j K = 2 , j ⇤ c 46 ✏ using (73) and the fact that ✏ = ˆ .Since , it follows that 2S

jn j N K (⇠c) C 2 (1 + 2 ⇠ ⌘ ) d(⌘) | j | N | | Z for any ﬁxed N< . Thereforec 1

jn j N Kj(⇠) CN 2 (1 + 2 ⇠ ⌘ ) d(⌘) | | j | | ✓ZD(⇠,2 ) c j N + (1 + 2 ⇠ ⌘ ) d(⌘) k+1 j k j | | k 0 D(⇠,2 ) D(⇠,2 ) ! X Z \ jn j Nk k+1 j k j C 2 (D(⇠,2 )) + 2 (D(⇠,2 ) D(⇠,2 ))  N \ k 0 ! X jn j(n 1) Nk (n 1)(k j) . 2 2 + 2 2 k 0 ! j X . 2 , where we used (78) at the next to last line, and at the last line we ﬁxed N to be equal to n and summed a geometric series. Thus

j Kj . 2 . (81) k k1 Now we mention the trivial but importantc fact that

K f 2 Kˆ f 2 (82) k ⇤ k k k1k k if say K and f are in .Thisfollowssince S K f = K\f k ⇤ k2 k ⇤ k2 = Kˆ fˆ k k2 Kˆ fˆ 2 kk1k k = Kˆ f 2. k k1k k Combining (81) and (82) we conclude

K f 2j f . (83) k j ⇤ k2 . k k2 Accordingly, by (80), (83) and Riesz-Thorin we have

n 1 j✓ j (1 ✓) K f 2 2 2 f k j ⇤ kq . k kq0 ✓ 1 ✓ 1 if 2 + = q . This works out to 1 n+1 n 1 j( ) K f 2 q 2 f (84) k j ⇤ kq . k kq0 47 2n+2 n+1 n 1 for any q [2, ]. If q> n 1 then the exponent q 2 is negative, so we conclude that 2 1 K f f . k j ⇤ kq . k kq0 j X Since f is a Schwartz function, the sum

K f + Kj f 1 ⇤ ⇤ j X is easily seen to converge pointwise to ˆ f.WeconcludeusingFatou’slemmathat ⇤ ˆ f f ,asclaimed. k ⇤ kq . k kq0 ⇤ Further remarks 1. Notice that the L2 estimate in the preceding argument was based only on dimensionality considerations. This suggests that there should be an L2 bound for fd valid under very general conditions.

Theoremd 7.4 Let ⌫ be a positive ﬁnite measure satisfying the estimate

⌫(D(x, r)) Cr↵. (85)  Then there is a bound n ↵ fd⌫ 2 CR 2 f 2 . (86) k kL (D(0,R))  k kL (d⌫)

The proof uses the followingd “generic” test for L2 boundedness.

Lemma 7.5 (Schur’s test) Let (X, µ)and(Y,⌫)bemeasurespaces,andletK(x, y)be ameasurablefunctiononX Y with ⇥ K(x, y) dµ(x) A for each y, (87) | |  ZX K(x, y) d⌫(y) B for each x. (88) | |  ZY Deﬁne T f(x)= K(x, y)f(y)d⌫(y). Then for f L2(d⌫)theintegraldeﬁningT f K 2 K converges a.e. (dµ(x)) and there is an estimate R

T f 2 pAB f 2 . (89) k K kL (dµ)  k kL (d⌫)

Proof It is possible to use Riesz-Thorin here, since (88) implies TKf B f k k1  k k1 and (87) implies TKf 1 A f 1. Amore“elementary”argumentgoesasfollows.Ifk k  k k a and b are positive numbers then we have 1 1 pab =min (✏a + ✏ b), (90) ✏ (0, ) 2 2 1

48 since is the arithmetic-geometric mean inequality and follows by taking ✏ = b .  a To prove (89) it suces to show that if f L2(dµ) 1, g L2(d⌫) 1, then q k k  k k  K(x, y) f(x) g(y) dµ(x)d⌫(y) pAB. (91) | || || |  ZZ To show (91), we estimate K(x, y) f(x) g(y) dµ(x)d⌫(y) | || || | 1 RR = min ✏ K(x, y) g(y) 2dµ(x)d⌫(y) 2 ✏ | || | ✓ ZZ 1 2 +✏ K(x, y) f(x) d⌫(y)dµ(x) | || | ZZ ◆ 1 2 1 2 min ✏A g(y) d⌫(y)+✏ B f(x) dµ(x)  2 ✏ | | | | ✓ Z Z ◆ 1 1 min(✏A + ✏ B)  2 ✏ = pAB.

⇤

To prove Theorem 7.4, let be an even Schwartz function which is 1ontheunitdisc and whose Fourier transform has compact support. (Exercise: show that such a function 1 1 exists.) In the usual way deﬁne R (x)=(R x). Then

fd⌫ 2 1 (x)fd⌫( x) 2 k kL (D(0,R)) kR kL (dx) = [1 (fd⌫) d k R ⇤ d k2 by (73) and Plancherel. The last line is the L2(dx) norm of the function

Rnˆ(R(x y))f(y)d⌫(y). Z We have estimates Rnˆ(R(x y)) dx = ˆ < | | k k1 1 Z for each ﬁxed y,bychangeofvariables,and

n n ↵ R ˆ(R(x y)) d⌫(y) R | | . Z for each ﬁxed x, by (85) and the compact support of ˆ. By Lemma 7.5

n ↵ n R ˆ(R(x y))f(y)d⌫(y) 2 R 2 f 2 , k kL (dx) . k kL (d⌫) Z 49 and the proof is complete. ⇤

2. Another remark is that it is possible to base the whole proof of Theorem 7.1 on the stationary phase asymptotics in section 6, instead of explicitly using the dimensionality of .Thissortofargumenthastheobviousadvantagethatitismoreflexible,sinceit works also in other situations where the “convolution kernel” ˆ(x y)isreplacedbya kernel K(x, y) satisfing appropriate conditions. See for example [29], [32], [33]. On the other hand, it is more complicated and is not as relevant in connection with more delicate questions such as the restriction conjecture, which is known to be false in most of the more general situations (see [5], [26], [30]). We give a brief sketch omitting details. The basic result is the so-called variable coecient Plancherel theorem, due to Hörmander [16]. n n n n Let be a real valued C1 function defined on R R ,leta C01(R R )and ⇥ 2 ⇥ consider the “oscillatory integral operators”

⇡i(x,y) Tf(x)= e a(x, y)f(y)dy. (92) Z n n Since a has compact support, it is obvious that these map L2(R )toL2(R )witha norm bound independent of ,butwewanttoshowthatthenormdecaysinasuitable way as .Aswiththeoscillatoryintegralsofsection6,thiswillnotbethecaseif is too degenerate.!1 In the present situation, note that if depends on x only, then the ⇡i factor e in (92) may be taken outside the integral sign, so the norm is independent ⇡i of .Similarly,if depends on y only, then the factor e may be incorporated into f. We conclude in fact that if (x, y)=a(x)+b(y), then T L2 L2 is independent of . k k ! This strongly suggests that the appropriate nondegeneracy condition should involve the “mixed Hessian” @2 n H˜ = @x @y ✓ i j ◆i,j=1 since the mixed Hessian vanishes identically if (x, y)=a(x)+b(y).

Theorem A (H¨ormander) Assume that det (H˜ (x, y)) =0 6 at all points (x, y) supp a.Then 2 n T L2 L2 C 2 . k k ! 

Sketch of proof This is evidently related to stationary phase, but one cannot apply stationary phase directly to the integral (92), since f isn’t smooth. Instead, one looks at TT⇤ which is an integral operator TK with kernel

⇡i((x,z) (y,z)) K(x, y)= e a(x, z)a(y,z)dz. (93) Z 50 The assumption about the mixed Hessian guarantees that the phase function in (93) has no critical points if x and y are close together. Using a version3 of “nonstationary phase” one can obtain the estimate

n K(x, y) dx | | . Z n n for each y.ThenSchur’stestshowsthat T T ⇤ L2 L2 . ,so T L2 L2 . 2 .The k k ! k k ! small support assumption on a can then be removed using a partition of unity. ⇤. It is possible to generalize this to the case where the rank of H˜ is k,where n k k 1,...,n ;justreplacetheexponent 2 by 2 .Furthermore,thecompactsupport assumption2{ on} a may be replaced by “proper support” (see the statement below), and ﬁ- q q nally one can obtain L 0 L estimates by interpolating with the trivial T 1 1. L L1 Here then is the variable! coecient Plancherel, souped up in a mannerk whichk ! makes it applicable in connection with Stein-Tomas. See the references mentioned above.

Theorem B (H¨ormander) Assume that a is a C1 function supported on the set (x, y) n n { 2 R R : x y C whose all partial derivatives are bounded. Let be a real valued ⇥ | | } C1 function deﬁned on a neighborhood of supp a, all of whose partial derivatives are bounded, and such that the rank of H˜(x, y)isatleastk everywhere. Assume furthermore that the sum of the absolute values of the determinants of the k by k minors of H˜ is bounded away from zero. Then there is a bound

k q T Lq0 Lq C k k !  when 2 q .  1 Now look back at the proof we gave for Theorem 7.1. The main point was to obtain the bound (84). Now, Kj(x) is the real part of

def j 2⇡i x K˜ (x) = (2 x)a( x )e | |, j | | where a satisﬁes the estimates in Corollary 6.6. Accordingly, it suces to prove (84) with j Kj replaced by K˜j. Let Tj be convolution with K˜j,andrescaleby2;thusweconsider the operator f T (f j ) j . ! j 2 2 3One needs something a bit more quantitative than our Proposition 6.1; the necessary lemma is best proved by integration by parts. See for example [32].

51 This is an integral operator Sj whose kernel is

nj j 2⇡i2j x y 2 (x y)a(2 x y )e | |. | |

We want to apply Theorem B to Sj; toward this end we make the following observa- tions.

(i) From the estimates in Corollary 6.6, we see that the functions

n 1 j j 2 2 (x y)a(2 x y ) | | have derivative bounds which are independent of j,andclearlytheyaresupportedin 1 x y 1. 4 | | (ii) The mixed Hessian of the function x y has rank n 1. This is a calculation which we leave to the reader, just noting that| the| exceptional direction corresponds to the direction along the line segment xy.

n+1 j It follows that the operators 2 2 Sj satisfy the hypotheses of Theorem B with k = n 1, uniformly in j,ifwetake =2j+1.Accordingly, n+1 n 1 ( )j S f 2 2 q f , k j kq . k kq0 and therefore using change of variables

n n+1 n 1 n j ( )j j 2 q T f 2 2 q 2 q0 f , k j kq . k kq0 i.e. n+1 n 1 ( )j T f 2 q 2 f , k j kq . k kq0 which is (84). ⇤

Exercise: Use Theorem A for an appropriate phase function, and a rescaling argument of the preceding type, to prove the bound

fˆ C f . k k2  k k2 This explains the name “variable coecient Plancherel theorem”.

8. Hausdor↵measures

n Fix ↵>0, and let E R .For✏>0, one defines ⇢ 1 ✏ ↵ H↵(E)=inf( rj ), j=1 X 52 where the infimum is taken over all countable coverings of E by discs D(xj,rj)withrj <✏. ✏ It is clear that H↵(E)increasesas✏ decreases, and we define ✏ H↵(E)=limH↵(E). ✏ 0 ! It is also clear that H✏(E) H✏(E)if↵>and ✏ 1; thus ↵  

H↵(E)isanonincreasingfunctionof↵. (94) 1 Remarks 1. If H↵(E)=0,thenH↵(E) = 0. This follows readily from the deﬁnition, 1 1 ↵ since a covering showing that H↵(E) <will necessarily consist of discs of radius < .

n 2. It is also clear that H↵(E)=0forallE if ↵>n,sinceonecanthencoverR by ↵ discs D(xj,rj)with j rj arbitrarily small. P Lemma 8.1 There is a unique number ↵0,calledtheHausdor↵dimensionofE or dimE,suchthatH (E)= if ↵<↵ and H (E)=0if↵>↵. ↵ 1 0 ↵ 0 Proof Deﬁne ↵ to be the supremum of all ↵ such that H (E)= .ThusH (E)= 0 ↵ 1 ↵ 1 if ↵<↵,by(94).Suppose↵>↵. Let (↵ ,↵). Deﬁne M =1+H (E) < .If 0 0 2 0 1 ✏>0, then we have a covering by discs with r M and r <✏.So j j  j

↵ ↵ P ↵ r ✏ r ✏ M j  j  j j X X which goes to 0 as ✏ 0. Thus H (E)=0. . ! ↵ ⇤

Further remarks 1. The set function H↵ may be seen to be countably additive on Borel sets, i.e. deﬁnes a Borel measure. See standard references in the area like [6], [10], [25]. 1 This is part of the reason one considers H↵ instead of, say, H↵.Noticeinthisconnection that if E and F are disjoint compact sets, then evidently H↵(E F )=H↵(E)+H↵(F ). 1 [ This statement is already false for H↵.

1 2. The Borel measure Hn coincides with ! times Lebesgue measure, where ! is the volume of the unit ball. If ↵

1 Examples The canonical example is the usual 3 -Cantor set on the line. This has a n n covering by 2 intervals of length 3 , so it has ﬁnite H log 2 -measure. It is not dicult to log 3 show that in fact its H log 2 -measure is nonzero; this can be done geometrically, or one can log 3 apply Proposition 8.2 below to the Cantor measure. In particular, the dimension of the log 2 Cantor set is log 3 .

53 Now consider instead a Cantor set with variable dissection ratios ✏ ,i.e.onestarts { n} with the interval [0, 1], removes the middle ✏1 proportion, then removes the middle ✏2 proportion of each of the resulting intervals and so forth. If we assume that ✏ ✏ , n+1  n and let ✏ =limn ✏n, then it is not hard to show that the dimension of the resulting set log 2 !1 E will be log( 2 ) . In particular, if ✏n 0thendimE =1. 1 ✏ ! On the other hand, H (E)willbepositiveonlyif ✏ < ,sothisgivesexamples 1 n n 1 of sets with zero Lebesgue measure but “full” Hausdor↵dimension. P There are numerous other notions of dimension. We mention only one of them, the Minkowski dimension, which is only deﬁned for compact sets. Namely, if E is compact n then let E = x R :dist(x, E) < . Let ↵0 be the supremum of all numbers ↵ such that, for some{ constant2 C, } n ↵ E C | | for all (0, 1]. Then ↵ is called the lower Minkowski dimension and denoted d (E). 2 0 L Let ↵1 be the supremum of all numbers ↵ such that, for some constant C,

n ↵ E C | | for a sequence of ’s which converges to zero. Then ↵1 is called the upper Minkowski dimension and denoted dU (E). It would also be possible to deﬁne these like Hausdor↵dimension but restricting to coverings by discs all the same size, namely: deﬁne a set S to be -separated if any two distict points x, y S satisfy x y >. Let (E)(“-entropy on E”) be the maximal possible cardinality2 for a -separated| | subset4 ofE E. Then it is not hard to show that

log (E) dL(E)=liminf E 1 , 0 log ! log (E) dU (E)=limsup E 1 . 0 log ! Notice that a countable set may have positive lower Minkowski dimension; for example, 1 1 the set 1 0 has upper and lower Minkowski dimension . { n }n=1 [{ } 2 If E is a compact set, then let P (E)bethespaceoftheprobabilitymeasuressupported on E.

n Proposition 8.2 Suppose E R is compact. Assume that there is a µ P (E)with ⇢ 2 µ(D(x, r)) Cr↵ (95)  n for a suitable constant C and all x R , r>0. Then H↵(E) > 0. Conversely, if H (E) > 0, there is a µ P (E)suchthat(95)holds.2 ↵ 2 4Exercise: show that (E)iscomparabletotheminimumnumberof-discs required to cover E E 54 Proof The first part is easy: let D(x ,r ) be any covering of E by discs. Then { j j } 1=µ(E) µ(D(x ,r )) C r↵,  j j  j j j X X 1 which shows that H↵(E) C . The proof of the converse involves constructing a suitable measure, which is most easily k done using dyadic cubes. Thus we let k be all cubes of side length `(Q)=2 whose k n Q vertices are at points of 2 Z .Wecantakethesetobeclosedcubes,fordefiniteness. It is standard to work with these in such contexts because of their nice combinatorics: if Q k,thenthereisauniqueQ˜ k 1 with Q Q˜;furthermoreifwefixQ1 k 1, 2Q 2Q ⇢ 2Q then Q1 is the union of those Q k with Q˜ = Q1, and the union is disjoint except for edges. A dyadic cube is a cube which2Q is in for some k. Qk If Q is a dyadic cube, then clearly there is a disc D(x, r)withQ D(x, r)and r C`(Q). Likewise, if we fix D(x, r), then there are a bounded number of⇢ dyadic cubes Q...Q with `(Q ) Cr and whose union contains D(x, r). From these properties, it 1 C j  is easy to see that the definition of Hausdor↵measure and also the property (95) could equally well be given in terms of dyadic cubes. Thus, except for the values of the constants,

µ satisﬁes (95) µ(Q) C`(Q)↵ for all dyadic cubes Q. ,  Furthermore, if we deﬁne

h✏(E)=inf( `(Q)↵ : E Q), ↵ ⇢ Q Q X2F [2F where runs over all coverings of E by dyadic cubes of side length `(Q) <✏,and F ✏ h↵(E)=limh↵(E), ✏ 0 ! then we have 1 ✏ ✏ ✏ C H (E) h (E) CH (E), ↵  ↵  ↵ therefore h (E) > 0 H (E) > 0. ↵ , ↵ We return now to the proof of Proposition 8.2. We may assume that E is contained in the unit cube [0, 1] ... [0, 1]. By the preceding remarks and Remark 1. above we may assume that h1 (E⇥) > 0,⇥ and it suces to find µ P (E)sothatµ(Q) C`(Q)↵ for ↵ 2  all dyadic cubes Q with `(Q) 1. We now make a further reduction.  + Claim It suces to find, for each fixed m Z ,apositivemeasureµ with the following 2 properties:

µ is supported on the union of the cubes Q which intersect E;(96) 2Qm 55 1 µ C ;(97) k k ↵ m µ(Q) `(Q) for all dyadic cubes with `(Q) 2 . (98)  Here C is independent of m.

Namely, if this can be done, then denote the measures satisfying (96), (97), (98) by µm.(98)impliesaboundon µm ,sothereisaweak*limitpointµ.(96)thenshows that µ is supported on E,(98)showsthatk k µ(Q) `(Q)↵ for all dyadic cubes, and (97) 1  shows that µ C .Accordingly,asuitablescalarmultipleofµ gives us the necessary k k probability measure. There are a number of ways of constructing the measures satisfying (96), (97), (98). Roughly, the issue is that (97) and (98) are competing conditions, and one has to find a measure µ with the appropriate support and with total mass roughly as large as possible given that (97) holds. This can be done for example by using finite dimensional convexity theory (exercise!). We present a di↵erent (more constructive) argument taken from [6], Chapter 2. We fix m,andwillconstructafinitesequenceofmeasures⌫m,...,⌫0,inthatorder; ⌫0 will be the measure we want. Start by defining ⌫m to be the unique measure with the following properties.

1. On each Q , ⌫ is a scalar multiple of Lebesgue measure. 2Qm m 2. If Q m and Q E = ,then⌫m(Q)=0. 2Q \ ; m↵ 3. If Q and Q E = ,then⌫ (Q)=2 . 2Qm \ 6 ; m

If we set k = m,then⌫k has the following properties: it is absolutely continuous to Lebesgue measure, and

↵ j ⌫ (Q) `(Q) if Q is a dyadic cube with side 2 , k j m;(99) k   

k if Q1 is a dyadic cube of side 2 ,thenthereisacovering Q1 of Q1 E by F \ ↵ dyadic cubes contained in Q1 such that ⌫k(Q1) Q `(Q) .(100) 2F Q1 P Assume now that 1 k m and we have constructed an absolutely continuous   measure ⌫k with properties (96), (99) and (100). We will construct ⌫k 1 having these same properties, where in (99) and (100) k is replaced by k 1. Namely, to deﬁne ⌫k 1 it suces to deﬁne ⌫k 1(Y )whenY is contained in a cube Q k 1.FixQ k 1. 2Q 2Q Consider two cases.

↵ (i) ⌫k(Q) `(Q) .Inthiscasewelet⌫k 1 agree with ⌫k on subsets of Q.  ↵ (ii) ⌫k(Q) `(Q) .Inthiscasewelet⌫k 1 agree with c⌫k on subsets of Q,wherec is (k1)↵ the scalar 2 . ⌫k(Q)

56 ↵ Notice that ⌫k 1(Y ) ⌫k(Y )foranysetY , and furthermore ⌫k 1(Q) `(Q) if   Q k 1.Thesepropertiesand(99)for⌫k give (99) for ⌫k 1,and(96)for⌫k 1 follows 2Q trivially from (96) for ⌫k.Tosee(100)for⌫k 1,ﬁxQ k 1.IfQ is as in case (ii), then ↵ 2Q ⌫k 1(Q)=`(Q) ,sowecanusethecoveringbythesingleton Q .IfQ is as in case (i), then for each of the cubes Q whose union is Q we have{ } the covering of Q E j 2Qk j \ associated with (100) for ⌫k.Since⌫k and ⌫k 1 agree on subsets of Q,wecansimply put these coverings together to obtain a suitable covering of Q E.Thisconcludesthe \ inductive step from ⌫k to ⌫k 1. We therefore have constructed ⌫0.Ithasproperties(96),(98)(sincefor⌫0 this is 1 equivalent to (99)), and by (100) and the deﬁnition of h↵ it has property (99). ⇤

Let us now deﬁne the ↵-dimensional energy of a (positive) measure µ with compact support5 by the formula

↵ I (µ)= x y dµ(x)dµ(y). ↵ | | ZZ We always assume that 0 <↵

↵ ↵ V (y)= x y dµ(x). µ | | Z Thus ↵ I↵(µ)= Vµ dµ. (101) Z ↵ The “potential” Vµ is very important in other contexts (namely elliptic theory, since it is harmonic away from supp µ when ↵ = n 2) but less important than the energy here. Nevertheless we will use it in a technical way below. Notice that it is actually the ↵ convolution of the function x with the measure µ. | | Roughly, one expects a measure to have I↵(µ) < if and only if it satisﬁes (95); this precise statement is false, but we see below that nevertheless1 the Hausdor↵dimension of acompactsetcanbedeﬁnedintermsoftheenergiesofmeasuresinP (E).

Lemma 8.3 (i) If µ is a probability measure with compact support satisfying (95), then I (µ) < for all <↵. 1 (ii) Conversely, if µ is a probability measure with compact support and with I↵(µ) < ,thenthereisanotherprobabilitymeasure⌫ such that ⌫(X) 2µ(X)forallsetsX 1  and such that ⌫ satisﬁes (95).

Proof (i) We can assume that the diameter of the support of µ is 1. Then  1 j j Vµ (x) . 2 µ(D(x, 2 )). j=0 X 5The compact support assumption is not needed; it is included to simplify the presentation.

57 Accordingly, if µ satisﬁes (95), and <↵,then

1 j j↵ Vµ (x) . 2 2 j=0 X . 1.

It follows by (101) that I↵(µ) < . (ii) Let F be the set of points1x such that V ↵(x) 2I (µ). Then µ(F ) 1 by (101). µ  ↵ 2 Let F be the indicator function of F and let ⌫(X)=µ(X F )/µ(F ). We need to show that ⌫ satisfies (95). Suppose first that x F .Ifr>0then\ 2 ↵ ↵ ↵ r ⌫(D(x, r)) V (x) 2V (x) 4I (µ).  ⌫  µ  ↵ This verifies (95) when x F .Forgeneralx,considertwocases.Ifr is such that 2 D(x, r) F = then evidently ⌫(D(x, r)) = 0. If D(x, r) F = ,lety D(x, r) F . Then ⌫(\D(x, r)); ⌫(D(y,2r)) r↵ by the first part of the\ proof.6 ; 2 \  . ⇤ Proposition 8.4 If E is compact then the Hausdor↵dimension of E coincides with the number sup ↵ : µ P (E)withI (µ) < . { 9 2 ↵ 1}

Proof Denote the above supremum by s.If 0, so dim E. So s dim E.Conversely,if 0smallenough.ThenI (µ) < ,so s,which  1  shows that dim E s.  ⇤ The energy is a quadratic expression in µ and is therefore susceptible to Fourier transform arguments. Indeed, the following formula is essentially just Lemma 7.2 combined ↵ with the formula for the Fourier transform of x . | | Proposition 8.5 Let µ be a positive measure with compact support and 0 <↵

1 We sketch the proof as follows: one can easily reduce to the case where = D(0,R) D(0,R), | | and this case can be done by explicit calculation. ⇤

⇡ x 2 ✏ Now let (x)=e | | .Wehavethen µ ,so ⇤ 2S ↵ ✏ ✏ x y (x z) (y w)dxdy dµ(z)dµ(w) | | (104) 2 2 (n ↵) =RRc RRµˆ(⇠) ˆ(✏⇠) ⇠ d⇠. ↵ | | | | | | Now let ✏ 0.R On the left side of (104), the expression inside the parentheses ! ↵ converges pointwise to z w using a minor variant on Lemma 3.2. If I↵(µ) < then the convergence is dominated| | in view of the preceding lemma, so the integrals on1 the left side converge to I↵(µ). If I↵(µ)= ,thenthisremainstrueusingFatou’slemma.On the right hand side of (104) we can1 argue similarly: the integrands converge pointwise 2 (n ↵) 2 (n ↵) to µˆ(⇠) ⇠ .If µˆ(⇠) ⇠ d⇠< then the convergence is dominated since | | | | | | | | 1 2 (n ↵) the factors ˆ(✏⇠) are bounded by 1, so the integrals converge to µˆ(⇠) ⇠ d⇠.If 2 (n ↵) R | | | | µˆ(⇠) ⇠ d⇠ = then this remains true by Fatou’s lemma. Accordingly, we can R pass| to| the| | limit from1 (104) to obtain the proposition. R ⇤ n Corollary 8.7 Suppose µ is a compactly supported probability measure on R with µˆ(⇠) C ⇠ (105) | | | | for some 0 <

2 n 2 µˆ(⇠) d⇠ CN . (106) | |  ZD(0,N) Then the dimension of the support of µ is at least 2.

Proof It suces by Proposition 8.4 to show that if (106) holds then I↵(µ) < for all ↵<2.However, 1

1 2 (n ↵) j(n ↵) 2 µˆ(⇠) ⇠ d⇠ . 2 µˆ(⇠) d⇠ | | | | j j+1 | | ⇠ 1 j=0 2 ⇠ 2 Z| | X Z | | 1 j(n ↵) j(n 2) . 2 2 j=0 < X 1 59 if ↵<2 and (106) holds. Observe also that the integral over ⇠ 1isfinitesince µˆ(⇠) µ = 1. This completes the proof in view of Proposition| 8.5.| | |k k ⇤ One can ask the converse question, whether a compact set with dimension ↵ must support a measure µ with ↵ +✏ µˆ(⇠) C✏(1 + ⇠ ) 2 (107) | | | | for all ✏>0. The answer is (emphatically) no6. Indeed, there are many sets with positive dimension which do not support any measure whose Fourier transform goes to zero as 2 ⇠ .TheeasiestwaytoseethisistoconsiderthelinesegmentE =[0, 1] 0 R . | |!1 ⇥{ }⇢ E has dimension 1, but if µ is a measure supported on E thenµ ˆ(⇠)dependson⇠1 only, so it cannot go to zero at .Ifoneconsidersonlythecasen =1,thisquestionis related to the classical question1 of “sets of uniqueness”. See e.g. [28], [40]. One can show 1 for example that the standard 3 Cantor set does not support any measure such thatµ ˆ vanishes at . 1 Indeed, it is nontrivial to show that a “noncounterexample” exists, i.e. a set E with given dimension ↵ which supports a measure satisfying (107). We describe a construction of such a set due to R. Kaufman in the next section. As a typical application (which is also important in its own right) we now discuss a n n special case of Marstrand’s projection theorem. Let e be a unit vector in R and E R acompactset.TheprojectionP (E)istheset x e : x E .Wewanttorelatethe⇢ e { · 2 } dimensions of E and of its projections. Notice first of all that dim P E dim E;this e  follows from the definition of dimension and the fact that the projection Pe is a Lipschitz function. 2 Areasonableexample,althoughnotverytypical,isasmoothcurveinR .Thisis one-dimensional, and most of its projection will be also one-dimensional. However, if the curve is a line, then one of its projections will be just a point.

Theorem 8.8 (Marstrand’s projection theorem for one-dimensional projections) As- n sume that E R is compact and dim E = ↵.Then ⇢ n 1 (i) If ↵ 1 then for a.e. e S we have dimPeE = ↵.  2 n 1 (ii) If ↵>1 then for a.e. e S the projection PeE has positive one-dimensional Lebesgue measure. 2

n 1 Proof If µ is a measure supported on E, e S ,thentheprojectedmeasureµ is 2 e the measure on R deﬁned by fdµ = f(x e)dµ(x) e · Z Z for continuous f.Noticethatµe may readily be calculated from this deﬁnition:

2⇡ikx e µ (k)= e · dµ(x) be Z 6 2 On the other hand, if one interpretsb decay in an L averaged sense the answer becomes yes, because the calculation in the proof of the above corollary is reversible.

60 =ˆµ(ke).

Let ↵

2 1+↵ µˆ(ke) k dk < . | | | | 1 Z

It follows by Proposition 8.5 with n =1thatfora.e. e the projected measure µe has ﬁnite ↵-dimensional energy. This and Proposition 8.4 give part (i), since µe is supported on the projected set PeE. For part (ii), we note that if dim E>1wecantake↵ =1in 2 (108). Thus µe is in L for almost all e. By Theorem 3.13, this condition implies that 2 µe has an L density, and in particular is absolutely continuous with respect to Lebesgue measure. Accordinglyb PeE must have positive Lebesgue measure. ⇤

Remark Theorem 8.8 has a natural generalization to k-dimensional instead of 1- dimensional projections, which is proved in the same way. See [10].

9. Sets with maximal Fourier dimension, and distance sets

A. Sets with maximal Fourier dimension Jarnik’s theorem is the following Proposition 9A1. Fix a number ↵>0, and let

a a (2+↵) E↵ = x R : inﬁnitely many rationals such that x q . { 2 9 q | q | }

2 Proposition 9A1 The Hausdor↵dimension of E↵ is equal to 2+↵ .

Proof We show only that dim E 2 .Theconverseinequalityisnotmuchharder ↵  2+↵ (see [10]) but we have no need to give a proof of it since it follows from Theorem 9A2 below using Corollary 8.7. It suces to prove the upper bound for E↵ [ N,N]. Consider the set of intervals a (2+↵) a (2+↵) \ I =( q , + q ), where 0 a Nq are integers. Then aq q q  

(2+↵) I q q , | aq| ⇡ · q>q a q>q X0 X X0 which is ﬁnite and goes to 0 as q if > 2 .Foranygivenq the set I : q>q 0 !1 2+↵ 0 { aq 0} covers E [ N,N], which therefore has H (E [ N,N]) = 0 when > 2 ,asclaimed. ↵ \ ↵ \ 2+↵ ⇤.

61 Theorem 9A2 (Kaufman [21]). For any ↵>0thereisapositivemeasureµ supported on a subset of E↵ such that 1 +✏ µˆ(⇠) C✏ ⇠ 2+↵ (109) | | | | for all ✏>0.

This shows then that Corollary 8.7 is best possible of its type. The proof is most naturally done using periodic functions, so we start with the following general remarks concerning “periodization” and “deperiodization”. Let Tn be the n-torus which we regard as [0, 1] ... [0, 1] with edges identiﬁed; thus a function on Tn n ⇥ ⇥ n is the same as a function on R periodic for the lattice Z . If f L1(Tn)thenonedeﬁnesitsFouriercoecientsby 2

2⇡ik x n fˆ(k)= f(x)e · dx, k Z n 2 Z and one also makes the analogous deﬁnition for measures. If f is smooth then one has ˆ N n ˆ 2⇡ik x f(k) CN k for all N and k f(k)e · = f(x). | | | | n Z Also, if f L1(R )onedeﬁnesitsperiodizationby2 2 P

fper(x)= f(x ⌫). n ⌫ X2Z 1 n Then fper L (T ), and we have 2 n ˆ Lemma 9A3 If k Z then fper(k)=f(k). 2 Proof d

2⇡ik x fˆ(k)= f(x)e · dx n ZR 2⇡ik x = f(x)e · dx n [0,1] ... [0,1]+⌫ ⌫ Z ⇥ ⇥ X2Z 2⇡ik (x ⌫) = f(x ⌫)e · dx n [0,1] ... [0,1] ⌫ Z ⇥ ⇥ X2Z 2⇡ik (x ⌫) = f(x ⌫)e · dx [0,1] ... [0,1] n Z ⇥ ⇥ ⌫ X2Z 2⇡ik x = fper(x)e · dx. [0,1] ... [0,1] Z ⇥ ⇥ 2⇡ik ⌫ At the last line we used that e · =1. ⇤

62 Suppose now that f is a smooth function on Tn; regard it as a periodic function on n R . Let and consider the function F (x)=(x)f(x). We have then 2S 2⇡i(⇠ ⌫) x Fˆ(⇠)= fˆ(⌫) e · (x)dx n ⌫ Z X2Z = fˆ(⌫)ˆ(⇠ ⌫). n ⌫ X2Z This formula extends by a limiting argument to the case where the smooth function f is replaced by a measure; we omit this argument7.Thuswehavethefollowing:letµ n be a measure on Tn,let ,anddeﬁneameasure⌫ on R by 2S d⌫(x)=(x)dµ( x ), (110) { } n where x is the fractional part of x. Then for ⇠ R { } 2 ⌫ˆ(⇠)= µˆ(k)ˆ(⇠ k). (111) n k X2Z Acorollaryofthisformulabysimpleestimateswithabsolutelyconvergentsums,using the Schwartz decay of ˆ,isthefollowing.

Lemma 9A4 If µ is a measure on Tn, satisfying ↵ µˆ(k) C(1 + k ) | | | | n for a certain ↵>0, and if ⌫ M(R ) is deﬁned by (110), then 2 ↵ ⌫ˆ(⇠) C0(1 + ⇠ ) . | | | |

This can be proved by using (111) and considering the range ⇠ k ⇠ /2andits complement separately. The details are left to the reader. | || | We now start to construct a measure on the 1-torus T,whichwillbeusedtoprove Theorem 9A2.

Let be a nonnegative C01 function on R supported in [ 1, 1] and with =1. ✏ 1 1 ✏ ✏ Deﬁne (x)=✏ (✏ x)andlet be the periodization of . M R Let (M)bethesetofprimenumberswhichlieintheinterval(2 ,M]. By the prime numberP theorem, (M) M for large M.Ifp (M)thenthefunction |P |⇡ log M 2P ✏ def ✏ 8 p(x) =(px)isagain1-periodic and we have ˆ(✏ k )ifp k, ✏(k)= p | (112) p 0otherwise. ⇢ 7It is based on the fact that everyc measure on Tn is the weak limit of a sequence of absolutely continuous measures with smooth densities, which is a corollary e.g.⇤ of the Stone-Weierstrass theorem. 8 1 Of course for ﬁxed p it is p -periodic

63 To see this, start from the formula

✏(k)=✏(k)=(✏k) which follows from Lemma 9A3.c Thus c b 2⇡ik x ✏(x)= (✏k)e · , Xk b ✏ 2⇡ikp x p(x)= (✏k)e · , Xk which is equivalent to (112). b Now deﬁne 1 F = ✏. (M) p |P | p (M) 2XP 1 Then F is smooth, 1-periodic, and 0 F =1(cf.(113)).OfcourseF depends on ✏ and M but we suppress this dependence. R Lemma The Fourier coecients of F behave as follows:

Fˆ(0) = 1, (113)

Fˆ(k)=0if0< k M , (114) | | 2 for any N there is CN such that log k ✏ k N Fˆ(k) C | | 1+ | | for all k =0. (115) | | N M M 6 ⇣ ⌘ Proof Both (113) and (114) are selfevident from (112). For (115) we use that a given log k integer k>0hasatmostC log M di↵erent prime divisors in the interval (M/2,M]. Hence, by (112) and the Schwartz decay of ˆ,

(1+↵) We now make up our mind to choose ✏ = M ,anddenotethefunctionF by FM . Thus we have the following

a (2+↵) suppF x : x p for some p ,a [0,p] , (116) M ⇢{ | p| 2PM 2 } log k k N F (k) C | | 1+ | | , (117) | M | N M M 2+↵ ⇣ ⌘ c 64 and furthermore FM is nonnegative and satisﬁes (113) and (114). It is easy to deduce from (117) with N =1that

1 2+ ↵ F (k) k | | log k (118) | M | . | | | | uniformly in M. In view of (116)c we could now try to prove the theorem by taking a weak limit of the measures FM dx and then using Lemma 9A4. However this would not be correct since the set E↵ is not closed, and ((116) notwithstanding) there is no reason why the weak limit should be supported on E↵. Indeed, (114) and (113) imply easily that the weak limit is the Lebesgue measure. The following is the standard way of getting around this kind of problem. It has something in common with the classical “Riesz product” construction; see [25].

I. First consider a ﬁxed smooth function on T. We claim that if M has been chosen large enough then

N log k l k l CN max | | 1+| | . (121) l: k l M/2 M M 2+↵ | | ⇣ ⌘ For any ﬁxed N the function log t t N f(t)= 1+ M M 2+↵ ⇣ ⌘ is decreasing for t M/10, provided that M is large enough. Thus (121) is bounded by N log(M/2) M/2 CN M 1+ M 2+↵ 100 . M⇣ , ⌘ which proves the second part of (119). To prove the ﬁrst part, we again use (120) and consider separately the range k l k /2anditscomplement.For k l k /2we | || | | || | 65 argue as above, replacing k l by k /2in(121).For k l k /2wehave l k /2, hence the estimate follows| easily | using| | the decay of the| Fourier || coe| cients of| |and| | the fact that F (k) 1. The details are left to the reader. | M | II. Let c 1 r 2+↵ log r when r r0, g(r)= 8 1 r 2+↵ log r when r r , < 0 0  0 where r > 1ischosensothatg(r) 1andg(r) is nonincreasing for all r.Thenforany 0 :  C1(T), ✏>0, and M0 > 10r0 we can choose N large enough and a rapidly increasing 2 sequence M0

G(k) (k) ✏g( k ), (122) | | | | 1 where G = N (FM1 + ...+ FdMN ). Thisb can be done as follows. Denote EM (k)= [F (k) (k) for M M . We will need the following consequences of (119): | M | 1 E (k) Cg( k )if k M/4, (123) b M  | | | | E (k) lim M =0foranyﬁxedM, (124) k g( k ) | |!1 | | 100 E (k) CM if k M/4, (125) M  | | with the constant in (123), (125) independent of k, M. C ✏ Fix N so that N < 100 .ObservethatforallM large enough

100 ✏ CM < g(M). (126) 100

We now choose M1,M2,...,MN inductively so that (126) holds, Mj+1 > 4Mj and

j 1 ✏ E (k) g( k )if k >M , (127) N i  100 | | | | j+1 i=1 X which is possible by (124). We claim that (122) holds for this choice of Mj.Toshowthis, we start with 1 N G(k) (k) E (k). (128) | |N i i=1 X Assume that M k M d(the casesb k M and k M are similar and are left j | | j+1 | | 1 | | N to the reader). By (127) we have

1 C C 100 ✏ C 100 E (k) g( k )+ M g( k )+ M , N j+1  N | | N j+1  100 | | N j+1

1 C 100 E (k) M for j +2 i N. N i  N i   Thus the right side of (128) is bounded by

N 3✏ C 100 g( k )+ M . 100 | | N i i=j+1 X By (126), we can estimate the last sum by

✏ N ✏ ✏ g(M ) g(M ) g( k ), 100N i  100 j+1  100 | | i=j+1 X which proves (122). We note that the support properties of G are similar to those of the F ’s. Namely, it follows from (116) that

a (2+↵) M1 supp G x : x p for some p ( ,M ),a [0,p] . (129) ⇢{ | p| 2 2 N 2 }

III. We now construct inductively the functions Gm and Hm, m =1, 2,...,asfol- lows. Let G0 1. If Gm has been constructed, we let Gm+1 be as in step II with ⌘ m 2 = G0G1 ...Gm, M0 10r0 + m,and✏ =2 .ThenthefunctionsHm = G1 ...Gm satisfy 1 3 H (0) 2  m  2 for each m and moreover the estimate (118)d holds also for the H’s, i.e. 1 H (k) k 2+↵ log k | m | . | | | | d IV. Now let µ be a weak limit point of the sequence H dx .Thesupportofµ will ⇤ { m } be contained in the intersection of the supports of the Gm , hence by (129) it will be a { } 1 compact subset of E .FromstepIII,wewillhave µˆ(k) k 2+↵ log k .Thetheorem ↵ | | . | | | | now follows by Lemma 9A4. ⇤

B. Distance sets

67 2 n If E is a compact set in R (or in R ), the distance set (E)isdeﬁnedas (E)= x y : x, y E . {| | 2 } One version of Falconer’s distance set problem is the conjecture that

2 E R , dim E>1 (E) > 0. ⇢ )| | One can think of this as a version of the Marstrand projection theorem where the nonlinear projection (x, y) x y replaces the linear ones. In fact, it is also possible to make the stronger conjecture!| that the| “pinned” distance sets

x y : y E {| | 2 } should have positive measure for some x E,orforasetofx E with large Hausdor↵ dimension. This would be analogous to Theorem2 8.8 with the nonlinear2 maps y x y !| | replacing the projections Pe. Alternately, one can consider this problem as a continuous analogue of a well known open problem in discrete geometry (Erd˝os’ distance set problem): prove that for finite sets 2 1 1 ✏ 2 F R there is a bound (F ) C✏ F , ✏>0. The example F = Z D(0,N), N ⇢ can be used to| show| that in| Erd˝| os’ problem one cannot take ✏ =0,anda\ !1 related example [11] shows that in Falconer’s problem it does not suce to assume that 1 H1(E) > 0. The current best result on Erd˝os’ problem is ✏ = 7 due to Solymosi and Tóth [31] (there were many previous contributions), and on Falconer’s problem the current best 4 result is dim E>3 due to myself [37] using previous work of Mattila [24] and Bourgain [4]. The strongest results on Falconer’s problem have been proved using Fourier transforms in a manner analogous to the proof of Theorem 8.8. We describe the basic strategy, which is due to Mattila [24]. Given a measure µ on E,thereisanaturalwaytoputameasure on (E), namely push forward the measure µ µ by the map :(x, y) x y .Ifone can show that the pushforward measure has an⇥ L2 Fourier transform, then!| (E| )must have positive measure by Theorem 3.13. In fact one proceeds slightly di↵erently for technical reasons. Let µ be a measure in 2 R ,then[24]oneassociatestoitthemeasure⌫ defined as follows. Let ⌫0 =(µ µ), i.e. ⇥ fd⌫ = f( x y )dµ(x)dµ(y). 0 | | Z Z Observe that 1 t 2 d⌫(t)=I 1 (µ). 2 Z Thus if I 1 (µ) < ,aswewillalwaysassume,thenthemeasurewenowdefinewillbein 2 1 M(R). Namely, let i ⇡ 1 i ⇡ 1 d⌫(t)=e 4 t 2 d⌫ (t)+e 4 t 2 d⌫ ( t). (130) 0 | | 0 Since ⌫ is supported on (E), ⌫ is supported on (E) (E). 0 [ 68 Proposition 9B1 (Mattila [24]) Assume that I↵(µ) < for some ↵>1. Then the following are equivalent: 1 (i)⌫ ˆ L2(R), (ii) the2 estimate 1 ( µˆ(Rei✓) 2d✓)2RdR < . (131) | | 1 ZR=1 Z Corollary 9B2 [24] Suppose that ↵>1isanumberwiththefollowingproperty:ifµ is a positive compactly supported measure with I (µ) < then ↵ 1

i✓ 2 (2 ↵) µˆ(Re ) d✓ C R . (132) | |  µ Z 2 Then any compact subset of R with dimension >↵must have a positive measure distance set.

2 Here and below we identify R with C in the obvious way.

Proof of the corollary Assume dim E>↵.ThenE supports a measure with I↵(µ) < .Wehave 1 1 i✓ 2 2 1 i✓ 2 (2 ↵) ( µˆ(Re ) d✓) RdR ( µˆ(Re ) d✓)R RdR | | . | | ZR=1 Z ZR=1 Z < . 1 On the ﬁrst line we used (132) to estimate one of the two angular integrals, and the last line then follows by recognizing that the resulting expression corresponds to the Fourier representation of the energy in Proposition 8.5. By Proposition 9B1 (E) (E) 2 [ supports a measure whose Fourier transform is in L ,whichsucesbyTheorem3.13.⇤.

3 At the end of the section we will prove (132) in the easy case ↵ = 2 where it follows from the uncertainty principle; we believe this is due to P. Sj¨olin. It is known [37] that 4 4 (132) holds when ↵> 3 , and this is essentially sharp since (132) fails when ↵< 3 .The negative result follows from a variant on the Knapp argument (Remark 4. in section 7); this is due to [24], and is presented also in several other places, e.g. [37]. The positive result requires more sophisticated Lp type arguments.

Before proving the proposition we record a few more formulas. Let R be the angular measure on the circle of radius R centered at zero; thus we are normalizing the arclength measure on this circle to have total mass 2⇡. Let µ be any measure with compact support. We then have µˆ(Rei✓) 2d✓ = µdµ. (133) | | R ⇤ Z Z c 69 This is just one more instance of the formula which ﬁrst appeared in Lemma 7.2 and was used in the proof of Proposition 8.5. This version is contained in Lemma 7.2 if µ has a Schwartz space density, and a limiting argument like the one in the proof of Proposition 8.4 shows that it holds for general µ. We also record the asymptotics for R which of course follow from those for 1 (Corollary 6.7) using dilations. Notice that the passage R from 1 to R preserves the total mass, i.e. essentially R =(1) . We concludec that b 1 1 3 (x)=2(R x ) 2 cos(2⇡(R x )) + ((R x ) 2 )(134) R | | | |8 O | | when R x 1, and is clearly also bounded independently of R. | | c| R| Proof of the propositionc From the deﬁnition of ⌫ we have

i ⇡ 1 2⇡ik x y ⌫ˆ(k)=e 4 x y 2 e | |dµ(x)dµ(y) | | Z i ⇡ 1 2⇡ik x y + e 4 x y 2 e | |dµ(x)dµ(y) | | Z 1 1 =2 x y 2 cos(2⇡( k x y ))dµ(x)dµ(y). | | | || |8 Z On the other hand, by (133) and (134) we have

i✓ 2 1 1 1 µˆ(ke ) d✓ = k 2 2 x y 2 cos(2⇡( k x y ))dµ(x)dµ(y) | | | | | | | || |8 Z Z 3 + ( k x y ) 2 dµ(x)dµ(y) O x y k 1 | || | ⇣ Z| || | ⌘ 1 + ( k x y ) 2 dµ(x)dµ(y) . O x y k 1 | || | ⇣ Z| || | ⌘ The last error term arises by comparing R,whichisbounded,tothemaintermonthe 1 right side of (134), which is ((R x ) 2 ), in the regime R x < 1. We may combine the O | | | | two error terms to obtain c

i✓ 2 1 1 1 µˆ(ke ) d✓ = k 2 2 x y 2 cos(2⇡( k x y ))dµ(x)dµ(y) | | | | | | | || |8 Z Z ↵ + ( ( k x y ) dµ(x)dµ(y)) O | || | Z for any ↵ [ 1 , 3 ]. Therefore 2 2 2

1 i✓ 2 1 ↵ ⌫ˆ(k)= k 2 µˆ(ke ) d✓ + ( k 2 I (µ)). | | | | O | | ↵ Z 1 2 ↵ 3 The error term here is evidently bounded by k I↵(µ)forany↵ (1, 2 ), and therefore belongs to L2( k 1). We conclude then| | that⌫ ˆ belongs to L2 2on k 1if | | | | 70 1 i✓ 2 and only if k 2 µˆ(ke ) d✓ does. This gives the proposition, since⌫ ˆ (being the Fourier transform of| a| measure)| clearly| belongs to L2 on [ 1, 1]. R ⇤ Proposition 9B3 If ↵ 1andifµ is a positive measure with compact support9 then

i✓ 2 (↵ 1) µˆ(Re ) d✓ I (µ)R , | | . ↵ Z where the implicit constant depends on a bound for the radius of a disc centered at 0 3 which contains the support of µ.Inparticular,(132)holdsif↵ = 2 .

3 Corollary 9B4 (originally due to Falconer [11] with a di↵erent proof) If dim E>2 then the distance set of E has positive measure.

Proofs The corollary is immediate from the proposition and Corollary 9B2. The proof of the proposition is very similar to the proofs of Bernstein’s inequality and of Theorem

7.4. We can evidently assume that R is large. Let be a radial C01 function whose 1 Fourier transform is 1onthesupportofµ. Let d⌫(x)=((x)) dµ(x). Then it is obvious (from the deﬁnition, not the Fourier representation) that I (⌫) I (µ). Also ↵  ↵ µˆ = ⌫ˆ.Accordingly b ⇤ µˆ(Rei✓) 2d✓ = ⌫ˆ(Rei✓) 2d✓ | | | ⇤ | Z Z (Rei✓ x) ⌫ˆ(x) 2dxd✓ . | || | Z = ⌫ˆ(x) 2 (x Rei✓) d✓dx | | | | Z Z 1 2 . R ⌫ˆ(x) dx x R C | | Z|| | | 1+2 ↵ (2 ↵) 2 R x ⌫ˆ(x) dx . | | | | 1 ↵ Z R I (µ). ⇡ ↵ Here the second line follows by writing

⌫ˆ(Rei✓) (Rei✓ x) (Rei✓ x) ⌫ˆ(x) dx | ⇤ | | |· | || | Z p p and applying the Schwartz inequality. The fourth line follows since for ﬁxed x the set of i✓ 1 ✓ where (x Re ) =0hasmeasure R ,andisemptyif x R is large. The proof 6 . | | is complete. ⇤

9Here, as opposed to in some previous situations, the compact support is important.

71 ↵ Remark The exponent ↵ 1 is of course far from sharp; the sharp exponent is 2 if ↵>1, 1 if ↵ [ 1 , 1] and ↵ if↵< 1 . 2 2 2 2 Exercise Prove this in the case ↵ 1. (This is a fairly hard exercise.)  Exercise Carry out Mattila’s construction (formula (130) and the preceding discussion) n 2 in the case where µ is a measure in R instead of R , and prove analogues of Proposition n 9A1, Corollary 9A2, Proposition 9A3. Conclude that a set in R with dimension greater n+1 than 2 has a positive measure distance set. (See [24]. The dimension result is also due n originally to Falconer. The conjectured sharp exponent is 2 .)

10. The Kakeya Problem

n ABesicovitchset,oraKakeyaset,isacompactsetE R which contains a unit line segment in every direction, i.e. ⇢

n 1 n 1 1 e S x R : x + te E t [ , ]. (135) 8 2 9 2 2 8 2 2 2

n Theorem 10.1 (Besicovitch, 1920) If n 2, then there are Kakeya sets in R with measure zero.

There are many variants on Besicovitch’s construction in the literature, cf. [10], Chap- ter 7, or [39], Section 1. There is a basic open question about Besicovitch sets which can be stated vaguely as “How small can this really be?” This can be formulated more precisely in terms of fractal dimension. If one uses the Hausdor↵dimension, then the main question is the following.

n Open question (the Kakeya conjecture) If E R is a Kakeya set, does E necessarily have Hausdor↵dimension n? ⇢

If n =2thentheanswerisyes;thiswasprovedbyDavies[8]in1971.Forgeneraln, n+2 p what is known at present is that dim(E) min( 2 , (2 2)(n 4) + 3); the first bound which is better for n =3isduetomyself[38],andthesecondoneisduetoKatzand Tao [19]. Instead of the Hausdor↵dimension one can use other notions of dimension, for example the Minkowski dimension defined in Section 8. The current best results for the upper Minkowski dimension are due to Katz,Laba and Tao [18], [22], [19]. There is also a more quantitative formulation of the problem in terms of the Kakeya n 1 n maximal functions,whicharedefinedasfollows.Forany>0, e S and a R ,let 2 2 n 1 Te (a)= x R : (x a) e , (x a)? , { 2 | · |2 | | } where x? = x (x e)e.ThusTe (a)isessentiallythe-neighborhood of the unit line segment in thee direction· centered at a. Then the Kakeya maximal function of f 2 72 1 n n 1 L (R )isthefunctionf ⇤ : S R defined by loc ! 1 ⇤ f (e)= sup f . (136) n T (a) | | a e Te (a) 2R | | Z ✏ The issue is to prove a “ ”estimateforf⇤,i.e.anestimateoftheform " " C : f ⇤ p n 1 C f (137) 8 9 " k kL (S )  " k kp for some p< . 1 Remarks 1. It is clear from the definition that

f⇤ f , (138) k k1 k k1 (n 1) f⇤ f 1. (139) k k1  k k

2. If n 2andp< , there can be no bound of the form 1 f ⇤ C f , (140) k kq  k kp with C independent of .Thiscanbeseenasfollows.ConsiderazeromeasureKakeya set E. Let E be the -neighborhood of E,andletf = E .Thenf⇤(e)=1forall n 1 e S ,sothat f⇤ q 1. On the other hand, lim 0 E =0,hencelim 0 f p =0 ! ! for2 any p< . k k ⇡ | | k k 1 n 1 3. Let f = D(0,). Then for all e S the tube Te (0) contains D(0,), so that D(0,) 2 n/p f ⇤(e)= | | .Hence f ⇤ p .However, f p . This shows that (137) Te (0) & cannot hold| for| any p

Open problem (the Kakeya maximal function conjecture):provethat(137)holdswith p = n,i.e. " " C : f ⇤ n n 1 C f . (141) 8 9 " k kL (S )  " k kn

When n =2,thiswasprovedbyC´ordoba [7] in a somewhat di↵erent formulation and by Bourgain [3] as stated. These results are relatively easy; from one point of view, this is because (141) is then an L2 estimate. In higher dimensions the problem remains open. There are partial results on (141) which can be understood as follows. Interpolat- ing between (139), which is the best possible bound on L1, and (141) gives a family of conjectured inequalities

n +1 " f ⇤ C p f ,q= q(p). (142) k kq . " k kp Note that if (142) holds for some p > 1, it also holds for all 1 p p (again by 0   0 interpolating with (135)). The current best results in this direction are that (142) holds with p =min((n +2)/2, (4n +3)/7) and a suitable q [38], [19].

73 n Proposition 10.2 If (137) holds for some p< ,thenBesicovitchsetsinR have Hausdor↵dimension n. 1 Remark The inequality 1 " E C (143) | | " for any Kakeya set E follows immediately from (137) by the same argument that showed n (140). (143) says that Besicovitch sets in R have lower Minkowski dimension n.

Proof of the proposition Let E be a Besicovitch set. Fix a covering of E by discs D = D(x ,r ); we can assume that all r ’s are 1/100. Let j j j j  k (k 1) J = j :2 r 2 . k {  j  } n 1 For every e S , E contains a unit line segment I parallel to e. Let 2 e n 1 1 S = e S : I D . k { 2 | e \ j|100k2 } j J [2 k 1 n 1 Since 2 < 1and I D I = 1, it follows that 1 S = S . k 100k k e j Jk j e k=1 k Let | \ 2 || | P P S S f = Fk ,Fk = D(xj, 10rj). j J [2 k Then for e S we have 2 k k 1 k T 2 (a ) F T 2 (a ) , | e e \ k| & 100k2 | e e |

k 2 k where ae is the midpoint of Ie so that Te (ae) is a tube of radius 2 around Ie.Hence

2 1/p f2⇤ k p & k (Sk) . (144) k k On the other hand, (137) implies that

k" k" (k 1)np 1/p f ⇤ k p C"2 f p C"2 ( Jk 2 ) . (145) k 2 k  k k  | |· Comparing (144) and (145), we see that

kp" kn 2p k(n 2p") (S ) 2 k J 2 J . k . | k| . | k| Therefore n 2p" k(n 2p") r 2 J (S ) 1. j | k| & k & j X Xk Xk ↵ We have shown that rj & 1forany↵

74 Remark By the same argument as in the proof of Proposition 10.2, (142) implies that n the dimension of a Kakeya set in R is at least p.

A. The n = 2case

Theorem 10.3 If n =2,thenthereisabound

1 1/2 f ⇤ C(log ) f . k k2  k k2

We give two di↵erent proofs of the theorem. The ﬁrst one is due to Bourgain [3] and uses Fourier analysis. The second one is due to C´ordoba [7] and is based on geometric arguments.

e 1 Proof 1 (Bourgain) We can assume that f is nonnegative. Let ⇢(x)=(2) Te (0), then e f⇤(e)= sup(⇢ f)(a). 2 ⇤ a 2R Let be a nonnegative Schwartz function on R such that has compact support and 2 (x) 1when x 1. Deﬁne : R R by | | ! b 1 1 (x)=(x1) ( x2).

e Note that ⇢ when e = e ,sothatf ⇤(e ) sup ( f)(a). Similarly 1 1  a ⇤

f⇤(e) sup( e f)(a),  a ⇤ where = p for an appropriate rotation p .Hence e e e

f⇤(e) e f ef 1 = e(⇠) f(⇠) d⇠. (146) k ⇤ k1 k k | |·| | Z By H¨older’s inequality, c b c b

(⇠)) f(⇠) d⇠ | e || | 1/2 1/2 (147) R 2 e(⇠) | | c e(⇠b) f(⇠) (1 + ⇠ )d⇠ 1+ ⇠ d⇠ .  | || | | | | | ⇣ R ⌘ ⇣ R ⌘ Note that = p andc = b(x )(x ), so that 1and is supported on a e e 1 2 | e| . rectangle R of size about 1 1/.Accordingly, e ⇥ c b b b b c b 1/ e(⇠) d⇠ ds 1 | |d⇠ . =log( ). (148) 1+ ⇠ Re 1+ ⇠ ⇡ 1 s Z c | | Z | | Z 75 Using (146), (147) and (148) we obtain

1/2 2 1 2 f ⇤ log( ) (⇠) f(⇠) (1 + ⇠ )d⇠ k k2 . | e || | | | 1 2 ⌘ log( ) R 2 f(⇠) (1 + ⇠ ) (⇠) de d⇠ . Rc| |b | | S1 | e | 1 2 ⇣ ⌘ log R 2 fˆ(⇠) d⇠ R . R | b | c =log1 Rf 2. k k2 n 1 Here the third line follows since for ﬁxed ⇠ the set of e S where e(⇠) =0has measure 1/(1 + ⇠ ). The proof is complete. 2 6 2 . | | c Remark For n 3, the same argument shows that (n 2)/2 f ⇤ f , (149) k k2 . k k2 which is the best possible L2 bound.

Proof 2 (C´ordoba) The proof uses the following duality argument. 1 1 Lemma 10.4 Let 1

n 1 Proof. Let ek be a maximal -separated subset of S .Observethatif e e0 < { } | | then f⇤(e) Cf⇤(e0); this is because any Te (a)canbecoveredbyaboundednumberof  tubes T (a0). Therefore e0

p0 n 1 P for some sequence yk with k yk =1.Onthelastlineweusedthedualitybetween lp and lp .Hence 0 P p n 1 1 ⇤ f p . yk f T (a ) k k ek k Te (ak) | | Xk | | Z k

76 n 1 for some choice of ak .Since T (ak) , it follows that { } | ek |⇡

p ⇤ f p . ykTe (ak) f k k k ! | | Z Xk ykT (a ) p f p (H¨older’s inequality) k ek k k 0 ·k k Xk A f  k kp as claimed. 2

We continue with C´ordoba’s proof. In view of Lemma 10.4, it suces to prove that 2 1 for any sequence yk with yk =1andanymaximal-separated subset ek of S we have { } { } P 1 ykT (ak) . log . (150) ek 2 k r X The relevant geometric fact is 2 Te (a) Te (b) . . (151) | k \ l | e e + | k l| Using (151) we estimate

Observe that for ﬁxed k 1 1 . = log , el ek + l + l +1 ⇡ l l 1 l 1 X | | X X and similarly for ﬁxed l 1 . log . ek el + Xk | | Applying Schur’s test (Lemma 7.5) to the kernel /( e e + ) we obtain that | k l|

2 1 2 1 p ykT (ak) 2 . log ( yk) . log , (153) k ek k Xk Xk 77 which proves (150). 2

B. Kakeya Problem vs. Restriction Problem

Recall that the restriction conjecture states that 2n fd C f n 1 if p> . k kp  pk kL1(S ) n 1 In fact, the stronger estimated 2n fd C f p n 1 if p> (154) k kp  pk kL (S ) n 1 can be proved to be formallyd equivalent, see e.g. [3]. It is known that the restriction conjecture implies the Kakeya conjecture. This is due to Bourgain [3], although a related construction had appeared earlier in [2]; both constructions are variants on the argument in [13].

Proposition 10.5. (Fe↵erman, Bourgain) If (154) is true then the conjectured bound

" f ⇤ C f k kn  " k kn is also true.

Proof We will use Lemma 10.4. Accordingly, we choose a maximal -separated set n 1 (n 1) ej on S ;observethatsuchasethascardinality .Foreachj pick a tube { } ⇡ 2 Tej (aj), and let ⌧j be the cylinder obtained by dilating Tej (aj)byafactorof around 2 1 the origin. Thus ⌧j has length ,cross-sectionradius ,andaxisintheej direction. Also let n 1 1 2 S = e S :1 e.e C . j { 2 j  } 1 Then Sj is a spherical cap of radius approximately C ,centeredatej.Wechoosethe constant C large enough so that the Sj’s are disjoint. Knapp’s construction (see chapter n 1 7) gives a smooth function fj on S such that fj is supported on Sj and

f n 1 =1, k jkL1(S ) n 1 f d on ⌧ . | j | & j We consider functions of the form d

f! = !jyjfj, j X

78 where yj are nonnegative coecients and !j are independent random variables taking values 1withequalprobability.Sincef have disjoint supports, we have ± j q q f! q n 1 = yjfj q n 1 k kL (S ) k kL (S ) j X q n 1 y . (155) ⇡ j j X On the other hand,

q q E( f[!d n )= E( f[!d(x) )dx k kLq(R ) n | | ZR 2 2 q/2 ( y f[!d(x) ) dx (Khinchin’s inequality) ⇡ n j | | R j Z X q(n 1) 2 q/2 & yj ⌧j (x) dx. (156) n | | R j Z X 2n Assume now that (154) is true. Then for any q>n 1 it follows from (155) and (156) that q(n 1) 2 q/2 q n 1 yj ⌧j (x) dx . yj . n | | R j j Z X X 2 Let zj = yj and p0 = q/2, then the above inequality is equivalent to the statement

n 1 q/2 2(n 1) if z 1, then z (157) j  k j ⌧j kq/2 . j j X X n 2 for any p0 n 1 .Wenowrescalethisby to obtain n n 1 p0 2( (n 1)) if z 1, then z p0 . j  k j Tj kp0 . j j X X n n Observe that p (n 1) 0asp0 n 1 .Thusforany">0wehave 0 & &

n 1 p0 " if z 1, then z (158) j  k j Tj kp0 . j j X X n if p0 is close enough to n 1 . By Lemma 10.4, this implies that for any ">0 " f ⇤ f k kp . k kp provided that p

79 as claimed. 2

We proved that the restriction conjecture is stronger than the Kakeya conjecture. Bourgain [3] partially reversed this and obtained a restriction theorem beyond Stein- Tomas by using a Kakeya set estimate that is stronger than the L2 bound (153) used in the proof of (149). It is not known whether (either version of) the Kakeya conjecture implies the full restriction conjecture.

Theorem 10.6 (Bourgain [3]) Suppose that we have an estimate

n ( q 1+") T (a ) q C" (159) k ej j k 0  j X for any given ">0andforsomeﬁxedq>2. Then

fd C f n 1 (160) k kp  pk kL1(S ) 2n+2 for some p< n 1 . d n Remark The geometrical statement corresponding to (159) is that Kakeya sets in R have dimension at least q.

3 We will sketch the proof only for n =3.RecallthatinR we have the estimates

fd f 2 2 (161) k k4 . k kL (S ) from the Stein-Tomas theorem, and d

1/2 fd 2 R f 2 2 (162) k kL (D(0,R)) . k kL (S ) from Theorem 7.4 with ↵ = nd1=2.Interpolating(161)and(162)yieldsafamilyof estimates

2 1 fdˆ p R p 2 f 2 2 (163) k kL (D(0,R)) . k kL (S ) for 2 p 4. Below we sketch an argument showing that the exponent of R in (163)   2 can be lowered if the L norm on the right side is replaced by the L1 norm.

Proposition 10.7 Let n =3,22. Then ↵(p) fd p R f n 1 , k kL (D(0,R)) . k kL1(S ) where ↵(p) < 2 1 . p 2 d This of course implies (160) for all p such that ↵(p) 0; in particular, there are p<4 for which (160) holds. 

80 1 n 1 Heuristic proof of the proposition Assume that f L1(S ) =1,andlet = R .We cover S2 by spherical caps k k S = e S2 :1 e.e , j { 2 j  } where e is a maximal p-separated set on S2.Then { j}

f = fj, j X ˆ ˆ where each fj is supported on Sj. Let G = fd and Gj = fjd,sothatG = j Gj.Bythe uncertainty principle Gj is roughly constant on cylinders of length R and diameter pR | | P 10 pointing in the ej direction. To simplify the presentation, we now make the assumption that Gj is supported on only one such cylinder ⌧j. Next, we cover D(0,R)withdisjointcubesQ of side pR.ForeachQ we denote by N(Q)thenumberofcylinders⌧ which intersect it. Note that G = G ,wherewe j |Q j j|Q sum only over those j’s for which ⌧j intersects Q. Using this and (163), we can estimate P G p for 2 p 4: k kL (Q)   2 1 p 2 G Lp(Q) . pR fj k k L2(S2) j:⌧ Q= Xj \ 6 ; 2 1 pR p 2 (N(Q) S )1/2 . ·| i| 3 1 1/2 4 p N(Q) . (164) ⇡ Summing over Q,weobtain

p 3p 1 p/2 G 4 N(Q) k kLp(D(0,R)) . Q X 3p p+ 1 p/2 4 2 . ⇡ k ⌧j kp/2 j X (165) On the last line we used that

p/2 p/2 3/2 p/2 = N(Q) Q = N(Q) . k ⌧j kp/2 ·| | j X XQ XQ We now let p =2q0,whereq0 is the exponent in (159), and assume that p is suciently close to 4 (interpolate (159) with (149) if necessary). We have from (159)

( 3 1+") q p C p . T (a ) q0 " k ej j k  j X 10It is because of this assumption that our proof is merely heuristic. Of course the Fourier transform of a compactly supported measure cannot be compactly supported; the rigorous proof uses the Schwartz decay of Gj instead.

81 1 Rescaling this inequality by we obtain

3 ( 1+") 3 q 3/q 1 " p 0 = p . k ⌧j kq0 . j X We combine this with (164) and conclude that

p p 4 1 " G p , k kL (Q) . i.e., 1 1 " 1 1 +" fdˆ p 4 p = R p 4 , k kL (D(0,R)) . which proves the proposition since 1 1 < 2 1 if p<4. 2 p 4 p 2

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