THE 2D CASE OF THE BOURGAIN-DEMETER-GUTH ARGUMENT

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1 THE 2D CASE OF THE BOURGAIN-DEMETER-GUTH ARGUMENT ZANE LI Let e(z) := e 2πiz and for g : [0, ] C and J [0, ], define the extension oerator E J g(x) := g(t)e(tx + t 2 x 2 ) dt. J For a ositive weight ν : R 2 [0, ), define the weighted L norm by f L (ν) := ( f(x) ν(x) dx) /. Furthermore for a ball B R 2 centered at c R 2 B, let w B (x) := ( + x c B R ) 200. We think of w B as a smoothed out version of B and so L (w B ) can be thought of as L (B). Next, let f L # (w B) := ( B R 2 f(x) w B (x) dx) / and so thinking w B to be basically B gives that f L # (w B) is essentially the L norm of f with resect to the uniform robability measure on B. Finally, let geom x i := (x x 2 ) /2 and so geom x i is the geometric mean of x i. The goal of this note is to sketch the roof of the following theorem, concentrating on the iteration scheme from [4] in the two dimensional case. This justifies (9) of [4] without an aeal to []. We also include some routine alications of Hölder, Minkowski, etc. that are omitted from [4] for comleteness. We will rove Theorem. Let 0 < δ. For each ball B R 2 of radius δ 2 and each g : [0, ] C, we have E [0,] g L 6 (w B ) ε δ ε ( E J g 2 L 6 (w B ) )/2 J [0,] J =δ where the imlicit constant is indeendent of δ, B, and g. Here and throughout, δ = N Z and J [0,], J =δ is a sum when J ranges over the intervals [jδ, (j + )δ), 0 j N. Other sums over intervals are defined similarly. Denote by V (δ) = V,2 (δ) the smallest constant such that E [0,] g L (w B ) V (δ)( E J g 2 L (w B ) )/2. J [0,] J =δ It is useful to think of V (δ) as a ower of δ. Our goal will be to show that V 6,2 (δ) ε δ ε. We defer alications of estimates of this tye to Vinogradov s Mean Value Theorem to [4] and [5].

2 2 ZANE LI. Main Tools We mention the main tools used in the iteration scheme to rove Theorem. We will only at most briefly mention how these results are roven and refer the reader to [4], [5], and [6]. In a few instances, the tool follows from a routine alication Hölder or Minkowski and its roof was omitted from [4], we will write out the full details in those instances for comleteness. For some K 4, let I, I 2 be intervals of length /K in [0, ] which are searated by at least /K... Ball inflation and L 2 decouling. We record the two dimensional analogue of the key ball inflation theorem. Ball inflation allows us to inflate from L /2 # (or L #) norms over the smaller balls ( below) to L/2 # (or L #) norms over bigger balls (B below). Theorem 2 (Ball inflation, Theorem 5.6 in [4]). Let ρ < /K and fix 2. Let B be an arbitrary ball in R 2 with radius ρ 2 and let B be a finitely overlaing cover of B with balls of radius ρ. Then for each g : [0, ] C, () geom( E Ji g 2 B L /2 # (w ) )/2 B J i I i J i =ρ ε,k ρ ε geom( J i I i where the imlicit constant is indeendent of g, ρ, and B. E Ji g 2 L /2 # (w B) )/2 Proof. The roof will use the uncertainty rincile, bilinear Kakeya, and dyadic igeonholing. It will be useful to observe that the Fourier transform of E J g is suorted on {(t, t 2 ) : t J}. We refer the reader to [4] or Section 3 of [6]. We also record the L analogue of the above inequality which is just equation () in [4]. We have that if B is an arbitrary ball in R 2 with radius R and B is a finitely overlaing cover of B with balls of radius r < R/2, then for each g : [0, ] C and 2, we have (2) geom( E Ji g 2 L B # (w ) )/2 B J i I i geom( E Ji g 2 L # (w B) )/2. J i I i We shall refer to both () and (2) as ball inflation. Since the roof of (2) is only briefly mentioned in [4], we include the full details at the end of this section. We will also use the following result that will allow us to decoule at a scale equal to the inverse of the radius of the ball we are analyzing. Proosition 3 (L 2 decouling, Lemma 7. of [4]). Let {J i } i be arbitrary collections of airwise disjoint intervals J with length equal to an integer multile of R. Then

3 THE 2D CASE OF THE BOURGAIN-DEMETER-GUTH ARGUMENT 3 for each integer M and each ball B R R 2 of radius R we have geom( E J g 2 L 2 # (w B R ) )/2 geom( E g 2 L 2 # (w B R ) )/2 J J i =/R, J for some J J i Proof. The roof of this roosition essentially just uses that the Fourier transform of E J g is suorted on {(t, t 2 ) : t J} and L 2 orthogonality, see [4]. In the terminology that will be introduced in Section.2, the above roosition will allow us to increase q in D 2 (q, B r ) if q < r. Proof of (2). By how our weight is defined, note that B w w B. Then geom( E Ji g 2 L (w ) )/2 geom ( E Ji g 2 L (w ) )/2 B B J i I i geom( J i I i J i I i E Ji g 2 L (w B ) )/2 where the first inequality is by Hölder and the second inequality is by Minkowski (in the sense of i ( k f k 2 L (w i ) )/2 ( k f k 2 L ( i w i) )/2 ) and that w w B (this last inequality is one of the reasons why we work with w B instead of B since is not necessarily B, for examle consider balls that cover the boundary of B). To change from L to L #, multily everything by ( B ) and use that this is B. This comletes the roof of (2)..2. Hölder s inequality and averaging. Given a function g : [0, ] C, a ball B r with radius δ r in R n, t, and q s r, define D t (q, B r ) = geom( E Ji,q g 2 L t # (w B r ))/2 J i,q I i J i,q =δ q and given a finitely overlaing cover B s (B r ) of B r with balls B s, define ( A (q, B r, s) = B s (B r ) B s B s (B r ) D 2 (q, B s ) ) /. At times to emhasize the deendence of A (q, B r, s) on g and δ, we will sometimes instead write A,δ (q, B r, s) or A (q, B r, s, g). We note that A (q, B r, r) = D 2 (q, B r ). We also note that A (q, B r, s) will also deend on B s (B r ). We first will rove a Hölder tye inequality for D t, more secifically, (3) D t (q, B r ) D (q, B r ) α D 2 (q, B r ) α for t = α + α 2. For simlicity, we will ignore the weights w B and simly relace them with B. Note that the factor / B in the definition of D t balances out by

4 4 ZANE LI how α is defined and hence we may relace L t #, L #, and L 2 # with Lt, L, and L 2, resectively. Next, it suffices to rove that ( 2 E J g L t (B r ) J I J I J =δ q J =δ q E J g 2 L (B r ) ) α ( E J g J I J =δ q ) α 2 L 2 (B r ) for I = I and I = I 2. The roof of the above equation will just use Hölder s inequality twice and so we will write J in lace of J I, J =δ and L in lace of q L (B). Let a, b be such that ar =, bs = 2, a + b = t and /r + /s =. Then E J g 2 L = (E t J g) a (E J g) b 2/t L (E J g) a 2/t L (E Jg) b 2/t r L s J J J = J E J g 2/(tr) L E J g 22/(ts) L 2. Now we aly Holder s inequality to the sum. Since /( α) + /α =, 2 J E J g 2 /(tr) L E J g 2 2/(ts) L 2 ( J E J g tr( α) L ) α ( J E J g 2 2 tsα L 2 ) α. Thus it remains to check that tr( α) = and tsα = 2. But this follows from solving the following three equations r + s =, α + α 2 = t, r + 2 s = t. Finally, we record a simle averaging result that follows from the definition of A. Fix arbitrary ositive reals r < s < t. Suose B r (B s ) is a finitely overlaing cover of a ball B s of radius δ s by balls B r of radius δ r and similarly define B s (B t ). Finally, define B r (B t ) := {B r : B r B r (B s ), B s B s (B t )}. Then we have the following averaging result. Proosition 4. B s (B t ) B s B s(b t ) A (r, B s, r) = A (r, B t, r) Proof. This will follow immediately from the definition of B s (B t ) and A. Exanding the left hand side, we have B s (B t A (r, B s, r) ) B s B s(b t ) = = B s (B t ) B r (B t ) B s B s (B t ) B r B r(b t ) This comletes the roof of Proosition 4. B r (B s ) B r B r (B s ) D 2 (r, B r ) D 2 (r, B r ) = A (r, B t, r). We also remark that ball inflation () and (2) will be useful in bounding averages of D t (q, B r ) when B r runs over some finitely overlaing cover B.

5 THE 2D CASE OF THE BOURGAIN-DEMETER-GUTH ARGUMENT 5.3. Rescaling and multilinear equivalence. We will also need to make use of two revious results from two different Bourgain-Demeter aers ([2], [3]). Towards the end of our iteration scheme (secifically the roof of Theorem 7), we will need to use the following rescaling lemma. Proosition 5 (Section 7 of [2]). Let 0 < ρ. For each interval I [0, ] of length δ ρ and each ball B R 2 of radius δ 2, E I g L (w B ) V (δ ρ )( J I J =δ E J g 2 L (w B ) )/2. It will haen in Section 2 that to analyze V (δ), it will be easier to analyze a multilinear version of V. With I, I 2, and K as defined at the beginning of this section, let V (δ, K) denote the smallest constant such that the inequality (4) geom E Ii g L (w B ) V (δ, K) geom( J I i J =δ E J g 2 L (w B ) )/2 holds for each ball B R 2 of radius δ 2 and each g : [0, ] C. By Hölder s inequality, V (δ, K) V (δ) (without using rescaling, just extend g by 0). We will make use that the reverse inequality is essentially true. Proosition 6 (Theorem 4. of [3]). Fix arbitrary 0 < δ. For each K, there exists ε (K) with lim K ε (K) = 0 such that V (δ),k δ ε(k) V (δ, K). With all our tools now set u, we next will rove Theorem given we know a certain inequality is true regarding A (u, B 2, u). The roof of this inequality will use all the tools in this section. 2. Proving V 6,2 (δ) ε δ ε Using the results from Section, we can rove the following theorem which is the two dimensional analogue of Theorem 7.3 and Theorem 8. in [4]. Theorem 7. Let < 6 be sufficiently close to 6. For each W > 0, for each sufficiently small u > 0, the following inequality holds for each g : [0, ] C, 0 < δ and for each ball B 2 in R 2 of radius δ 2 A (u, B 2, u, g) ε,k,w δ ε V,2 (δ) uw D (, B 2, g). We will rove this theorem in the next two sections. We first see how this theorem with Proosition 6 work together to rove V 6,2 (δ) ε δ ε. The roof will be exactly as in Section 8 in [4] and so we will not dwell too much on the details here. Essentially an alication of Hölder s inequality will show that V 6,2 (δ) ε δ ε if V,2 (δ) ε δ ε for all < 6 sufficiently close to 6 (see the roof of Lemma 8.2 in [4]). Fix < 6 sufficiently close to 6 so that Theorem 7 holds. Let η 0 be the such that (5) and (6) lim V,2(δ)δ η+σ = 0 for each σ > 0 δ 0 lim V,2(δ)δ η σ = for each σ > 0. δ 0

6 6 ZANE LI Note that (5) combined with (6) show that log V,2 (δ)/ log δ η as δ 0. Such an η exists since V,2 (δδ ) V,2 (δ)v,2 (δ ) and so log V,2 is subadditive. We claim that η = 0. We will use the multilinear equivalence. Let B 2 be an arbitrary ball of radius δ 2 in R 2. By Cauchy-Schwarz, geom E Ii g L # (w B 2 ) δ u/2 geom( J i,u I i J i,u =δ u E Ji,u g 2 ) /2 L # (w B 2 ). Exanding the integral and using that B u (B 2 ) B u B 2 and Minkowski s inequality the right hand side of the above is dominated by δ u/2 ( B u (B 2 D (u, B u, g) ) /. ) B u B u (B 2 ) Since D (u, B u ) D 2 (u, B u ), the above is dominated by A (u, B 2, u) and so geom E Ii g L # (w B 2 ) δ u/2 A (u, B 2, u, g). Now we are in a osition to use Theorem 7 and Proosition 6. above equation with Theorem 7 and (5) gives that for each σ > 0, Combining the geom E Ii g L # (w B 2 ) ε,k,w,σ δ (ε+u/2+(η+σ)( uw )) D (, B 2, g). But now since D (, B 2, g) is just B 2 times the right hand side of (4) and noticing that we have a L # on the left hand side of the above equation, we have geom E Ii g L (w B 2 ) ε,k,w,σ δ (ε+u/2+(η+σ)( uw )) geom( J I i J =δ E J g 2 L (w B 2 ) )/2. By using that V (δ, K) is the best constant for the multilinear inequality (4), Proosition 6 and (6), and the fact that the above inequality holds uniformly for all g : [0, ] C and balls B 2, we have δ η+σ+ε (K) ε,k,w,σ δ (ε+u/2+(η+σ)( uw )). Since this is true for arbitrarily small δ, the exonent of δ on the left hand side must be the exonent of δ on the right hand side. Rearranging gives η 2W + ε + ε (K) + σ(2 uw ) uw and using that uw < and the above inequality is true for all ε, σ > 0 and K 4, it follows that η /(2W ). Letting W gives that η = 0. Since η = 0, log V,2 (δ)/ log δ 0 as δ 0. Therefore V,2 (δ) ε δ ε for < 6 sufficiently close to 6 which roves that V 6,2 (δ) ε δ ε. It now remains to rove Theorem The iteration for dyadic balls To analyze A (u, B 2, u) with u sufficiently small, we will first analyze A (, B, ) where B is a ball of radius δ 2r, r =, 2,... and then by redefining δ we can trade large r for small u. For each n = 0,, 2,..., choose a finitely overlaing cover B 2 n(b 2n+ ) of a ball B 2n+ of radius δ 2n+ by balls B 2n of radius δ 2n. With these

7 THE 2D CASE OF THE BOURGAIN-DEMETER-GUTH ARGUMENT 7 covers, define a finitely overlaing cover B (B 2n ) of B 2n by balls B as follows (this is analogous to how B (B 3 ) is defined in Proosition 7.5 of [4]) Then (7) B (B 2n ) = {B : B 2i B 2 i(b 2i+ ) for i = 0,,..., n }. For 4, let α be defined by /2 = α + α. 2 A (, B 2, ) ε,k δ ε D /2 (, B 2 ) ε,k δ ε D (, B 2 ) α A (2, B 2, 2) α where the first inequality is by ball inflation () combined with the following alication of Hölder (here we use 4) E Ji, g L 2 # (w B ) E Ji, g L /2 # (w B ) and the second inequality is by Hölder s inequality for D t, the observation that A (q, B r, r) = D 2 (q, B r ), and L 2 decouling (Proosition 3) to increase from q = to q = 2. Raise (7) to the ower and then average the resulting inequality over balls B 2 B 2 (B 4 ). By Hölder s inequality, B 2 (B 4 ) B 2 B 2(B 4 ) A (, B 2, ) ( ε,k δ ε B 2 (B 4 ) ( B 2 (B 4 ) B 2 B 2(B 4 ) B 2 B 2 (B 4 ) D (, B 2 ) ) α A (2, B 2, 2) ) α. Alying Lemma 4 for the left hand side of the above equation (here we have also have used how B (B 2n ) is defined) and ball inflation (2) for the first term on the right hand side (note that the second term on the right hand side does not need Lemma 4, however we will need this Lemma at the next ste of the iteration to rove (8)), we obtain Iterating this r times, we have (8) A (, B 4, ) ε,k δ ε D (, B 4 ) α A (2, B 4, 2) α. A (, B 2r, ) ε,k,r δ ε D (, B 2r ) α A (2, B 2r, 2) α. 4. Finishing the iteration By raising (8) to the ower and summing over finitely overlaing families of balls, the above inequality holds for all balls B of radius δ 2r (once again using ball inflation and Lemma 4). Then (9) A (, B, ) ε,k,r δ ε D (, B) α A (2, B, 2) α. To emhasize the deendence of A on δ, we will write A,δ to emhasize this deendence.

8 8 ZANE LI Fix a small δ > 0. We will rove an inequality about A,δ (u, B 2, u) for small u deending on δ. Let δ := δ u. Then A,δ (u, B 2, u) = A,δ (, B 2/u, ) since A,δ (, B 2/u, ) = D B (B 2/u 2 (, B ) ) = B (B 2/u ) B B (B 2/u ) B B (B 2/u ) geom( J i, I i J i, =δ E Ji, g 2 L 2 # (w B ))/2 where B (B 2/u ) is a finitely overlaing cover of B 2/u (a ball of radius δ 2/u ) by balls of radius δ and A,δ (u, B 2, u) = B u (B 2 geom( E Ji,u g 2 L ) 2 # (w B u ))/2 B u B u (B 2 ) J i,u I i J i,u =δ u where B u (B 2 ) is a finitely overlaing cover of B 2 (a ball of radius δ 2 ) by balls of radius δ u. Similarly, A,δ (2u, B 2, 2u) = A,δ (2, B 2/u, 2) and D,δ (u, B 2 ) = D,δ (, B 2/u ). Since (9) is true for all balls B of radius δ 2r, if u > 0 is such that 2 r 2/u, then and hence A,δ (, B 2/u, ) ε,k,r δ ε D,δ (, B 2/u ) α A,δ (2, B 2/u, 2) α A,δ (u, B 2, u) ε,k,r δ ε D,δ (u, B 2 ) α A,δ (2u, B 2, 2u) α. We summarize this result in the following roosition. Proosition 8. Let 4 and u > 0 such that 2 r u. Then for each ball B 2 of radius δ 2, we have A (u, B 2, u) ε,k,r δ ε D (u, B 2 ) α A (2u, B 2, 2u) α. We now iterate Proosition 8. If 2 r u, then u = 2u satisfies the condition 2 r u and hence A (2u, B 2, 2u) ε,k,r δ ε D (2u, B 2 ) α A (4u, B 2, 4u) α (note that key feature is that the ball B 2 stays fixed so we can now iterate) and hence A (u, B 2, u) ε,k,r δ ε D (u, B 2 ) α D (2u, B 2 ) α( α) A (4u, B 2, 4u) ( α)2 Iterating this rocess M times gives the following result. Proosition 9. Let 4. Given integers r, M and let u > 0 be such that 2 r +M u. Then for each ball of radius δ 2, we have M A (u, B 2, u) ε,k,r,m δ ε D (2 i u, B 2 ) α( α)i A (2 M+ u, B 2, 2 M+ u) ( α)m+ (0) By how V (δ) is defined, i=0 D (2 i u, B 2 ) V (δ)d (, B 2 )

9 THE 2D CASE OF THE BOURGAIN-DEMETER-GUTH ARGUMENT 9 for 0 i M. This inequality requires no rescaling. We illustrate how to rove such a result by just considering the case when i = M, the result for other values of i follow similarly. To show (0), it is enough to show that for all g : [0, ] C, ( E J g 2 L (w B 2 ) )/2 V (δ)( E J g 2 L (w B 2 ) )/2. J I i J =δ 2M u J I i J =δ Thus it is enough to show that for all J I i with J = δ 2M u we have () E J g L (w B 2 ) V (δ)( E J g 2 L (w B 2 ) )/2 for all g : [0, ] C. Now by how V (δ) is defined, we have (2) E [0,] h L (w B 2 ) V (δ)( J J, J =δ J [0,], J =δ E J h 2 L (w B 2 ) )/2 for all h : [0, ] C. In articular, this is true for all functions h which equal 0 outside the interval J. Since E [0,] (g J ) = E J g, () then follows from (2). Thus inserting (0) into Proosition 9 gives A (u, B 2, u) ε,k,r,m δ ε V (δ) ( α)m+ D (, B 2 ) ( α)m+ A (2 M+ u, B 2, 2 M+ u) ( α)m+. Since α = (4 )/(2 ), 2( α) > for < 6 and for each W > 0, there exists an M sufficiently large such that 2 M+ ( α) M+ > W. We can now rove Theorem 7. Proof of Theorem 7. Fix an arbitrary W > 0. Let u be so small and choose r and M (deending on W ) such that uw <, 2 M+ ( α) M+ > W and 2 r+m u. Hölder s inequality alied to B 2M+u B 2 M+ u (B 2 ) and using that B 2 M+ u(b 2 ) B 2M+u B 2 yields that A (2 M+ u, B 2, 2 M+ u) D (2 M+ u, B 2 ). Rescaling as in Proosition 5 gives D (2 M+ u, B 2 ) V,2 (δ 2M+u )D (, B 2 ) and thinking that V,2 (δ) as a ower of δ (strictly seaking we cannot do this by our definition of V,2 (δ), however, we just want to rove V,2 (δ) δ ε and to make this comletely rigorous one could go through the argument and relace V,2 (δ) by δ a,2 and analyze the a,2 instead) shows that and hence D (2 M+ u, B 2 ) V,2 (δ) 2M+u D (, B 2 ) A (u, B 2, u) ε,k,w δ ε V (δ) [2( α)]m+u D (, B 2 ) ε,k,w δ ε V,2 (δ) uw D (, B 2 ). This comletes the roof of Theorem 7.

10 0 ZANE LI References [] Jean Bourgain, Cirian Demeter, The roof of the l 2 decouling conjecture, Annals of Math. 82 (205), no., [2] Jean Bourgain, Cirian Demeter, Mean value estimates for Weyl sums in two dimensions, arxiv: [3] Jean Bourgain, Cirian Demeter, Decoulings for curves and hyersurfaces with nonzero Gaussian curvature, arxiv: [4] Jean Bourgain, Cirian Demeter, Larry Guth, Proof of the main conjecture in Vinogradov s mean value theorem for degrees higher than three, arxiv: [5] Terence Tao, htts://terrytao.wordress.com/205/2/0/decouling-and-the-bourgaindemeter-guth-roof-of-the-vinogradov-main-conjecture/, 205. [6] Terence Tao, htts://terrytao.wordress.com/205/2//the-two-dimensional-case-ofthe-bourgain-demeter-guth-roof-of-the-vinogradov-main-conjecture/, 205.