8.8 DECOUPLIG THEORY HW2 DOGHAO WAG DATE:OCT. 3 207 Problem We shall start by reformulatng the problem. Denote by δ S n the delta functon that s evenly dstrbuted at the n ) dmensonal unt sphere. As a temporal dstrbuton, ts acton on a Schwartz functon s gven by δ S n ψ) = ψω)dvol S n S n Smlarly, we can defne the unevenly dstrbuted delta functon φδ S n where φω) C S n ) s the weght functon. As a temporal dstrbuton, ts acton s defned as φδ S n ψ) = ψω)φω)dvol S n S n The goal of study s to understand the nverse Fourer transformaton of φδ. We need to show for certan p and each n > 0 we can fnd some contnous functons φ n C 0 S n ) such that Eφ n L p n φ n We can strengthen ths goal a lttle bt and try to prove for some M > 0, Eφ n L p B M ) n φ n In what follows, we wll take M = R and construct a functon φ R such that Eφ R L p B R ) R γ φ R for some γ > 0. From now on, we wll focus on the constructon of φ R. Take L = R 2 and choose an /L separated 0/L net on S n. We use ω,, ω m to record ther drectons. For each drecton ω, let θ be a /L cap on S n centered at ω. Snce the net s /L separated, these caps are dsjont. For each θ, we choose a smooth bump functon
2 DOGHAO WAG DATE:OCT. 3 207 f θ that s almost on the whole cap θ and vanshes outsde. The functon to be consdered has the form φ R = θ f θ e ω x Each φ θ = f θ ω)e ω x s supported at a n dm cap θ of sze R R 2 R 2 and ts nverse Fourer transform concentrates on the translated dual rectangle θ +x. Ths dual rectangle has sze R R 2 R 2 R 2 wth drecton ω ow by the counterexample of Kakeya problem, we can choose the net and x carefully, such that Ω = θ + x ) CL γ R n 2 n+γ R 2 The second factor comes from rescalng the pcture. ow to estmate Eφ R L p B R ), we proceed as follows and keep n mnd that energy s concentrated on Ω, Eφ R p L p B R ) Eφ R p L p Ω) we assume Ω B R ) Eφ R p L 2 Ω) Ω p 2 Holder Inequalty) Eφ R p L 2 B R ) Ω p 2 Because energy concentrates) At the last step, we replace the regon Ω by B R. It s temptng to replace t by the whole space R n and use Plancherel s dentty. Snce φ R δ S n s not an L 2 functon at all, ths attempt won t work. The lesson s even though the energy concentrates, you stll get a dvergent term f ntegratng over a non-compact regon. Thus, t s only feasble to estmate over B R. To do ths, we need a cut off functon. Choose a postve bump functon η supported on B wth ntegral R η =. We requre η n on B /2. Then ts nverse Fourer transformaton η s almost a bump functon of heght on B. ow consder η R = η ) and η R R = η ). R n R ow Eφ R L 2 B R ) η R Eφ R L 2 R n ) = δ S n φ η R L 2 R n ) We need to understand the last convoluton. It s a smooth bump functon convolved wth a tempered dstrbuton. The clam s that t
8.8 DECOUPLIG THEORY HW2 3 s a smooth functon supported n R neghborhood of the sphere wth heght R. To be specfc, we wrte down the expresson, δ S n φ R η R ξ) = φ R ω)η R ξ ω)dω S n If ξ s out of /R neghborhood of the unt sphere, then the ntegral s smply zero. If t s wthn /2R neghborhood and ξ s near the support of a cap θ then ths θ s unque by separateness), then the ntegral s taken over S n B /R ξ) whose n ) dmensonal area s roughly R n ). On that regon, η R has heght R n and φ R s constant. Ths shows the outcome s roughly R. Fnally, to compute the L 2 norm, snce the doman s of sze R whle the heght s roughly R, the result s R /2. ow combnng all these together, we have, Eφ R L p B R ) R 2 Ω p 2 R 2 + p 2 ) γ 2 + n 2 ) The condton γ = 2 + p 2 ) γ 2 + n 2 ) > 0 gves Problem2 2n n <) p < 2n + γ) n + γ) 2 We shall use nducton on n, the dmenson of ambent Eucldean space and each tme try to reduce t by one untl t reaches k, the number of dfferent drectons. Suppose the drectons are along the frst k coordnates, say x,, x k. Consder the projecton Π : R n R n, whch forgets the last component. Let Q S be the mage of Q S under Π and that s a n ) dmensonal cube of sde length S. Let l j,a be the mage of l j,a and T j,a be the characterstc functon of the -neghborhood of l j,a. We shall prove j j Q S Q S T j,a) To do ths, the frst step s to use Fubn s theorem and rewrte LHS as j j = Q S Q S [ S,S]
4 DOGHAO WAG DATE:OCT. 3 207 It s enough to show, j Π k [ S,S] Π k T j,a) for any x = x,, x n ) R n. From now on, we regard T j,a as a functon on {x } [ S, S]. If T j,a s dentcally zero, then T j,ax ) = 0. But even though t s not zero, t s a characterstc functon of compact nterval of length 2. Ths s because {x } [ S, S] and l j,a are almost perpendcular to each other and -neghborhood of l j,a has dameter 2. ow by Holder nequalty, j j Π k Π k T j,a ) k/) ) /k [ S,S] [ S,S] and t suffces to show j T j,a ) k/) [ S,S] T j,a) k/) WLOG, we assume each T j,a on RHS s not zero, otherwse we can smply dscard that term. Agan by Holder nequalty, we can use L and L norm to control LHS, T j,a L k/) [ S,S]) T j,a )/k T j,a /k Snce each T j,a, T j,a. Also, snce [ S,S] T j,a 2, T j,a 2. Therefore, RHS. Ths completes the nducton step. Fnally, we use mult-lnear Kekaya theorem n dmenson k to conclude. The last step s not very enlghtenng, though t s vald as a proof. The pcture s along {x } R, these tubes may ntersects at m dfferent ponts and for each m, there are m,,, m,k tubes comng from each drecton. Therefore, the number of ntersectons at -th pont s estmated as, 2,k The nequalty s sayng a dscrete verson), 2,k ) /) j,j ) /)
8.8 DECOUPLIG THEORY HW2 5 The ntuton s the quantty on LHS acheves ts maxmum when ntersectons happen at a sngle pont, say, when m =. Ths dscrete verson s proven smlarly: frst, use Holder nequalty,, 2,k ) /) k/),j ) /k j and for each j, we need to show k/),j,j ) k/) we can smply repeat the L and L argument. Problem3 a) Let Σ,, Σ k be hyperplanes nsde R n wth normal vectors parallel to x -axs,, x k -axs respectvely. For each j, choose a small rectangle θ j of sze R R α R α and a smooth bump functon f j supported on θ j. Here, the power α s allowed to vary ths s made possble because Σ j doesn t have curvature) and the rectangle concde wth our old frend, cap when α = 2. Then f j concentrates on θ j and s roughly a constant functon on ths dual rectangle. WLOG, assume f j on θ j. If k 2, then the product Π f j concentrates on B R α. We want to make a guess about e n the expresson, ow, LHS s roughly, whle RHS s f j /k /) L avg B R ) Re+ɛ f L 2 avg ω BR )) /k B R α B R ))/ = R n α) R e+ɛ f L 2 avg B R ) = R e+ɛ Rn )α R ) α)n ) R n 2 = R e+ɛ 2 Therefore, e has be greater than n α) n α) k 2 = n k α) To make t large, we put α = 0. Ths s the best we can do, because each Σ j s only part of a plane and the radus s controlled by. b) Before we start, we rewrte multlnear Kakeya n a proper way. By what s proven n the last problem, f projected to R k {0}, we
6 DOGHAO WAG DATE:OCT. 3 207 have j Q n S j Q k S T k) j,a ) By multlnear Kakeya n k dmenson, RHS s controlled by ɛ /) j ɛ S j ɛ S k/) Q n S Q n S We transform both sdes to the average: S n Q n S j ɛ S k/) S nk/) Q n S Fnally, we get j ɛ S n k)/) Q n S Q n S Ths non-trval power of S n front of the product wll explan the appearance of e. In below, we wll take S to be R /2. To prove multlnear restrcton theorem n ths case, we use nducton on the scale
8.8 DECOUPLIG THEORY HW2 7 and proceed as follows, f j 2/) Average BR /2 B R B R B R 2 By nducton) Average R ɛ+e) 2 BR /2 Local Orthogonalty) Average BR /2 Rɛ+e) k Local Constant) Average BR /2 Rɛ+e) k = R ɛ+e) k B R θ Multlnear Kakeya) R n k)/2) R ɛ+e) k Local Orthogonalty) R ɛ+e) k +n k)/2) f j 2/) f L 2 avg ω BR /2 ) ) 2/) B R /2 f,θ 2 ) /) θ B R /2 θ f,θ 2 ) /) B R θ f j 2/) L avgb R ) f,θ 2 ) /) f,θ 2 ) /) Therefore, the correct power should result n ek + n k)/ ) = k k e Ths s precsely e = n k. I should say a few words about why we used Multlnear Kakeya for R /2, nstead of R. Ths s because each f,θ s a tube of sze R R /2 R /2. But what we proved for multlnear Kakeya s for tubes of radus, so we need to rescale the whole pcture. Snce we used mean ntegral, whch s nvarant under scalng, no other powers of R wll appear due to ths procedure.