LECTURES IN HARMONIC ANALYSIS Thomas H. Wol↵ Revised version, March 2002 1. L1 Fourier transform 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 δa be the Dirac measure at a, δa(E)=1ifa E and 2 2⇡ia ⇠ 2 δ (E)=0ifa E.Thenδ (⇠)=e− · . a 62 a ⇡ x 2 Example 2 Let Γ(x)=eb− | | .Then ⇡ ⇠ 2 Γˆ(⇠)=e− | | . (1) Proof The integral in question is 2⇡ix ⇠ ⇡ x 2 Γˆ(⇠)= e− · e− | | dx. Z Notice that this factors as a product of one variable integrals. So it suffices 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 ⇠, 2⇡ix ⇠ µˆ(⇠) = e− · dµ(x) | | | | Z 2⇡ix ⇠ e− · d µ (x) | | | | Z = µ . k k ⇤ 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 <r . { 2 | − | } n Proposition 1.3 Suppose that µ M(R )andsuppµ is compact. Thenµ ˆ is C1 and 2 D↵µˆ =(( 2⇡ix)↵µ) . (7) − Further, if supp µ D(0,R)then ⇢ b ↵ ↵ D µˆ (2⇡R)| | µ . (8) k k1 k k 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 β ↵ β for each i, , i i ↵<β ↵ β and ↵ = β. , 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↵erentia- tions 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. Proposition 1.4 Suppose that f is CN and that D↵f L1 for all ↵ with 0 ↵ N. Then 2 | | D↵f(⇠)=(2⇡i⇠)↵fˆ(⇠)(11) when ↵ N and furthermore | | d N fˆ(⇠) C(1 + ⇠ )− (12) | | | | for a suitable constant C. 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 φ : R R be a C1 function with the following properties (4. is actually ! irrelevant for present purposes): 1. φ(x)=1if x 1 | | 2. φ(x)=0if x 2 3. 0 φ 1. | |≥ 4. φ is radial. x Define φk(x)=φ( k ); thus φk is similar to φ but lives on scale k instead of 1. If ↵ C↵ ↵ is a multiindex, then there is a constant C↵ such that D φk ↵ uniformly in k. k| | Furthermore, if ↵ =0thenthesupportofD↵φ 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 = φkf then ↵ ↵ 2 | | limk D fk D f 1 =0forall↵ with ↵ N. !1 k − k | | Proof It is obvious that ↵ ↵ lim φkD f D f 1 =0, k !1 k − k so it suffices to show that ↵ ↵ lim D (φkf) φkD f 1 =0. (13) k !1 k − k However, by the Leibniz rule ↵ ↵ ↵ β β D (φ f) φ D f = c D − fD φ , k − k β k 0<β ↵ X where the cβ’s are certain constants. Thus ↵ ↵ β ↵ β D (φkf) φkD f 1 C D φk D − f L1( x: x k ) k − k k k1k k { | |≥ } 0<β ↵ X 1 ↵ β Ck− D − f L1( x: x k ) k k { | |≥ } 0<β ↵ X The last line clearly goes to zero as k . There are two reasons for this (either would 1 !1 1 suffice): 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 n 1+ x (1 + y )(1 + x y ),x,y R (15) | | | | | − | 2 which will be used several times below. 2. Schwartz space n The Schwartz space is the space of functions f : R C such that: S ! 1.