LECTURES IN HARMONIC ANALYSIS Thomas H. Wolff Revised version, March 2002 1. L1 Fourier transform n n If f L1(R ) then its Fourier transform is fˆ : R C defined by ∈ → 2πix ξ fˆ(ξ)= e− · f(x)dx. Z n n More generally, let M(R ) be the space of finite complex-valued measures on R with the norm n µ = µ (R ), k k n| | n where µ is the total variation. Thus L1(R ) is contained in M(R ) via the identification f µ,| dµ| = fdx. We can generalize the definition of Fourier transform via → 2πix ξ µˆ(ξ)= e− · dµ(x). Z n Example 1 Let a R and let δa be the Dirac measure at a, δa(E)=1ifa E and ∈ 2πia ξ ∈ δa(E)=0ifa E.Thenδa(ξ)=e− · . 6∈ π x 2 Example 2 Let Γ(x)=eb− | | .Then π ξ 2 Γ(ˆ ξ)=e− | | . (1) Proof The integral in question is 2πix ξ π x 2 Γ(ˆ ξ)= e− · e− | | dx. Z Notice that this factors as a product of one variable integrals. So it suffices to prove (1) πx2 when n = 1. For this we use the formula for the integral of a Gaussian: ∞ e− dx =1. It follows that −∞ R ∞ 2πixξ πx2 ∞ π(x iξ)2 πξ2 e− e− dx = e− − dx e− · −∞ −∞ Z Z iξ ∞− πx2 πξ2 = e− dx e− iξ · Z−∞− ∞ πx2 πξ2 = e− dx e− · Z−∞ πξ2 = 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 ∈ µˆ µ n . (2) k k∞ ≤k kM(R ) Proof For any ξ, 2πix ξ µˆ(ξ) = e− · dµ(x) | | | | Z 2πix ξ e− · d µ (x) ≤ | | | | Z = µ . k k n Proposition 1.2 If µ M(R )thenˆµ is a continuous function. ∈ Proof Fix ξ and consider 2πix (ξ+h) µˆ(ξ + h)= e− · dµ(x). Z 2πix ξ As h 0 the integrands converge pointwise to e− · . Since all the integrands have → n absolute value 1 and µ (R ) < , the result follows from the dominated convergence | | ∞ theorem. We now list some basic formulas for the Fourier transform; the ones listed here are roughly speaking those that do not involve any differentiations. They can all be proved by n using the formula ea+b = eaeb and appropriate changes of variables. Let f L1, τ R , n n ∈ ∈ and let T be an invertible linear map from R to R . 1. Let fτ (x)=f(x τ). Then − 2πiτ ξ ˆ fτ (ξ)=e− · f(ξ). (3) 2πix τ 2. Let eτ (x)=e · .Then b ˆ eτ f(ξ)=f(ξ τ). (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 a dilation, i.e. Tx = r x for some r>0, then 3. says that the Fourier transform of the · n 1 1 function f(rx) is the function r− fˆ(r− ξ). 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) is the function fˆ(rξ). 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 . { ∈ | − | } n Proposition 1.3 Suppose that µ M(R ) and supp µ is compact. Thenµ ˆ is C∞ and ∈ Dαµˆ =(( 2πix)αµ) . (7) − Further, if supp µ D(0,R)then ⊂ b α α D µˆ (2πR)| | µ . (8) k k∞ ≤ k k We are using multiindex notation here and will do so below as well. Namely, a multi- n index is a vector α R whose components are nonnegative integers. If α is a multiindex then by definition ∈ ∂α1 ∂αn Dα = ... , α1 αn ∂x1 ∂xn α n αj x =Πj=1xj . The length of α, denoted α ,is αj. One defines a partial order on multiindices via | | j α P β αi βi for each i, ≤ ⇔ ≤ α<β α β and α = β. ⇔ ≤ 6 ProofofProposition1.3Notice that (8) follows from (7) and Proposition 1 since the α α norm of the measure (2πix) µ is (2πR)| | µ . ≤ k k 3 Furthermore, for any α the measure (2πix)αµ is again a finite measure with compact support. Accordingly, if we can prove thatµ ˆ is C1 and that (7) holds when α =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 ∈{ } ξ R , and consider the difference quotient ∈ µˆ(ξ + hej) µˆ(ξ) ∆(h)= − . (9) h This is equal to 2πihxj e− 1 2πiξ x − e− · dµ(x). (10) h Z As h 0, the quantity → 2πihxj e− 1 − h 2πihx e− j 1 converges pointwise to 2πixj. Furthermore, − 2π xj for each h. Accordingly, − | h |≤ | | the integrands in (10) are dominated by 2πxj , which is a bounded function on the support of µ. It follows by the dominated convergence| | theorem that 2πihxj e− 1 2πiξ x lim ∆(h)= lim − e− · dµ(x), h 0 h 0 h → Z → which is equal to 2πiξ x 2πixje− · dµ(x). − Z This proves the formula (7) when α = 1. Formula (7) and Proposition 2 imply thatµ ˆ is 1 | | C . Remark The estimate (8) is tied to the support of µ. However, the fact thatµ ˆ is C∞ and the formula (7) are still valid whenever µ has enough decay to justify the differentia- 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. | | | | ∞ TheR 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 Dαf L1 for all α with 0 α N. Then ∈ ≤| |≤ Dαf(ξ)=(2πiξ)αfˆ(ξ) (11) when α N and furthermore | |≤ d N fˆ(ξ) C(1 + ξ )− (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 C∞ 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 α Cα α is a multiindex, then there is a constant Cα such that D φk α uniformly in k. k| | Furthermore, if α = 0 then the support of Dαφ is contained| in the|≤ region k x 2k. 6 ≤| |≤ N α 1 Lemma 1.5 If f is C , D f L for all α with α N and if we let fk = φkf then α α ∈ | |≤ limk D fk D f 1 = 0 for all α with α N. →∞ k − k | |≤ Proof It is obvious that α α lim φkD f D f 1 =0, k →∞ k − k so it suffices to show that α α lim D (φkf) φkD f 1 =0. (13) k →∞ k − k However, by the Leibniz rule α α α β β D (φkf) φkD f = cβD − fD φk, − 0<β α X≤ where the cβ’s are certain constants. Thus α α β α β D (φkf) φkD f 1 C D φk D − f L1( x: x k ) k − k ≤ k k∞k k { | |≥ } 0<β α X≤ 1 α β Ck− D − f L1( x: x k ) ≤ k k { | |≥ } 0<β α X≤ The last line clearly goes to zero as k . There are two reasons for this (either would 1 →∞ 1 suffice): the factor k− , and the fact that the L norms are taken only over the region x k. | |≥ 5 ProofofProposition1.4If f is C1 with compact support, then by integration by parts we have ∂f (x)e 2πix ξdx =2πiξ e 2πix ξf(x)dx, ∂x − · j − · Z j Z i.e. (11) holds when α = 1. An easy induction then proves (11) for all α provided that f is CN with compact| support.| To remove the compact support assumption, let fk be as in Lemma 1.5. Then (11) α holds for fk. Now we pass to the limit as k . On the one hand D[fk converges α →∞ uniformly to D f as k by Lemma 1.5 and Proposition 1.1. On the other hand fk →∞ˆ α α ˆ converges uniformly to f,so(2πiξ) fk converges to (2πiξ) f pointwise.