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Apr 04, 2016

🖉 18.099 Transcript: Bourgain's Theorem

As part of the 18.099 Discrete Analysis reading group at MIT, I presented section 4.7 of Tao-Vu’s Additive Combinatorics textbook. Here were the notes I used for the second half of my presentation.

1. Synopsis

We aim to prove the following result.

Theorem 1 (Bourgain)

Assume N2N \ge 2 is prime and A,BZ=ZNA, B \subseteq Z = \mathbb Z_N. Assume that δ(loglogN)3logN\delta \gg \sqrt{\frac{(\log \log N)^3}{\log N}} is such that min{PZA,PZB}δ\min\left\{ \mathbf P_ZA, \mathbf P_ZB \right\} \ge \delta. Then A+BA+B contains a proper arithmetic progression of length at least exp(Cδ2logN3)\exp\left( C\sqrt[3]{\delta^2 \log N} \right) for some absolute constant C>1C > 1.

The methods that we used with Bohr sets fail here, because in the previous half of yesterday’s lecture we took advantage of Parseval’s identity in order to handle large convolutions, always keeping two 1^\widehat 1_\ast term’s inside the \sum sign. When we work with A+BA+B this causes us to be stuck. So, we instead use the technology of Λ(p)\Lambda(p) constants and dissociated sets.

2. Previous results

As usual, let ZZ denote a finite abelian group. Recall that

Definition 2. Let SZS \subseteq Z and 2p2 \le p \le \infty. The Λ(p)\Lambda(p) constant of SS, denoted SΛ(p)\left\lVert S \right\rVert_{\Lambda(p)}, is defined as

SΛ(p)=supc:SCc≢0ξSc(ξ)e(ξx)Lp(Z)c2(S). \left\lVert S \right\rVert_{\Lambda(p)} = \sup_{\substack{c : S \rightarrow \mathbb C \\ c \not\equiv 0}} \frac{\left\lVert \displaystyle\sum_{\xi \in S} c(\xi) e(\xi \cdot x) \right\rVert_{L^p(Z)}} {\left\lVert c \right\rVert_{\ell^2(S)}}.

Definition 3. If SZS \subseteq Z, we say SS is a dissociated set if all 2S2^{|S|} subset sums of SS are distinct.

For such sets we have the Rudin’s inequality (yes, Walter) which states that

Lemma 4 (Rudin’s inequality)

If SS is dissociated then SΛ(p)p.\left\lVert S \right\rVert_{\Lambda(p)} \ll \sqrt p.

Disassociated sets come up via the so-called “cube covering lemma”:

Lemma 5 (Cube covering lemma)

Let SZS \subseteq Z and d1d \ge 1. Then we can partition S=D1D2DkRS = D_1 \sqcup D_2 \sqcup \dots \sqcup D_k \sqcup R such that

  • Each DiD_i is dissociated of size d+1d+1,
  • There exists η1\eta_1, \dots, ηd\eta_d such that RR is contained in a dd-cube, i.e. it’s covered by c1η1++cdηdc_1\eta_1 + \dots + c_d\eta_d, where ci{1,0,1}c_i \in \{-1,0,1\}.

Finally, we remind the reader that

Lemma 6 (Parseval)

We have fL2Z=f^2Z.\left\lVert f \right\rVert_{L^2Z} = \left\lVert \widehat f \right\rVert_{\ell^2Z}.

Since we don’t have Bohr sets anymore, the way we detect progressions is to use the pigeonhole principle. In what follows, let TnfT^n f be the shift of xx by nn, id est Tnf(x)=f(xn)T^nf(x) = f(x-n).

Proposition 7 (Pigeonhole gives arithmetic progressions)

Let f:ZR0f : Z \rightarrow \mathbb R_{\ge 0}, J1J \ge 1 and suppose rZr \in \mathbb Z is such that EZmax1jJTjrff<EZf.\mathbf E_Z \max_{1 \le j \le J} \left\lvert T^{jr}f - f \right\rvert < \mathbf E_Z f. Then supp(f)\operatorname{supp}(f) contains an arithmetic of length jj and spacing rr.

Proof: Apply the pigeonhole principle to find an xx such that max1jJTjrf(x)f(x)<f(x).\max_{1 \le j \le J} \left\lvert T^{jr}f(x) - f(x) \right\rvert < f(x). Then the claim follows. \Box

3. Periodicity

Proposition 8 (Estimate for maxhHThf\max_{h \in H} |T^hf| for supp(f^)\operatorname{supp}(\widehat f) dissociated)

Let f:ZRf : Z \rightarrow \mathbb R, supp(f^)SZ\operatorname{supp}(\widehat f) \subseteq S \subseteq Z with SS dissociated. Then for any set HH with H>1|H| > 1 we have

maxhHThfL2ZlogHfL2Z. \left\lVert \max_{h \in H} \left\lvert T^h f \right\rvert \right\rVert_{L^2Z} \ll \sqrt{\log|H|} \left\lVert f \right\rVert_{L^2Z}.

Proof: Let p>2p > 2 be large and note

maxhHThfL2ZmaxhHThfLpZ(hHThfp)1/pLpZ=(EZ(hHThfp))1/p=(hHEZThfp)1/p=(hHEZfp)1/p=H1/pξf^(ξ)e(ξx)LpZH1/pSΛ(p)f^2Z \begin{aligned} \left\lVert \max_{h \in H} \left\lvert T^h f \right\rvert \right\rVert_{L^2Z} &\le \left\lVert \max_{h \in H} \left\lvert T^h f \right\rvert \right\rVert_{L^pZ} \\ &\le \left\lVert \left( \sum_{h \in H} \left\lvert T^h f \right\rvert^p \right)^{1/p} \right\rVert_{L^pZ} \\ &= \left( \mathbf E_Z \left( \sum_{h \in H} \left\lvert T^h f \right\rvert^p \right) \right)^{1/p} \\ &= \left( \sum_{h \in H} \mathbf E_Z \left\lvert T^h f \right\rvert^p \right)^{1/p} \\ &= \left( \sum_{h \in H} \mathbf E_Z \left\lvert f \right\rvert^p \right)^{1/p} \\ &= \left\lvert H \right\rvert^{1/p} \left\lVert \sum_\xi \widehat f(\xi) e(\xi \cdot x) \right\rVert_{L^pZ} \\ &\le \left\lvert H \right\rvert^{1/p} \left\lVert S \right\rVert_{\Lambda(p)} \left\lVert \widehat f \right\rVert_{\ell^2Z} \\ \end{aligned}

Then by Parseval and Rudin,

maxhHThfL2ZH1/pSΛ(p)fL2ZH1/ppfL2Z. \begin{aligned} \left\lVert \max_{h \in H} \left\lvert T^h f \right\rvert \right\rVert_{L^2Z} &\le \left\lvert H \right\rvert^{1/p} \left\lVert S \right\rVert_{\Lambda(p)} \left\lVert f \right\rVert_{L^2Z} \\ &\ll \left\lvert H \right\rvert^{1/p} \sqrt p \left\lVert f \right\rVert_{L^2Z}. \end{aligned}

We may then take plogHp \ll \log H. \Box

We combine these two propositions into the following lemma which applies if f^\widehat f has nonzero values of “uniform” size.

Lemma 9 (Uniformity estimate for shifts)

Let f:ZRf : Z \rightarrow \mathbb R and J,d>1J, d > 1. Suppose that f^\widehat f is “uniform in size” across its support, in the sense that

supξsupp(f^)f^(ξ)infξsupp(f^)f^(ξ)2016. \frac {\sup_{\xi \in \operatorname{supp}(\widehat f)} \left\lvert \widehat f(\xi) \right\rvert} {\inf_{\xi \in \operatorname{supp}(\widehat f)} \left\lvert \widehat f(\xi) \right\rvert} \le 2016.

Then one can find SZS \subseteq Z such that S=d|S| = d and for all rZr \in Z,

EZmax1jJTjrff(ξf^(ξ))(logJd+JdmaxηSηrR/Z). \mathbf E_Z \max_{1 \le j \le J} \left\lvert T^{jr}f - f \right\rvert \ll \left( \sum_\xi \left\lvert \widehat f(\xi) \right\rvert \right) \left( \sqrt{\frac{\log J}{d}} + Jd\max_{\eta \in S} \left\lVert \eta \cdot r \right\rVert_{\mathbb R/\mathbb Z} \right).

Proof: Use the cube covering lemma to put supp(f^)=D1DkR\operatorname{supp}(\widehat f) = D_1 \sqcup \dots \sqcup D_k \sqcup R where RR is contained in the cube of S={η1,,ηd}S = \left\{ \eta_1, \dots, \eta_d \right\} and Di=d+1|D_i| = d+1 for 1ik1 \le i \le k. Accordingly, we decompose ff over its Fourier transform as f=f1++fk+gf = f_1 + \dots + f_k + g by letting fif_i be supported on DiD_i and g(x)g(x) supported on RR.

First, we can bound the “leftover” bits in RR:

EZmax1jJTjrgg=EZmax0jJξRf^(ξ)(e(ξ(x+jr))e(ξx))EZmax0jJξRf^(ξ)(e(ξ(x+jr))e(ξx))(ξRf^(ξ))max0jJξR(e(ξ(x+jr))e(ξx))(ξRf^(ξ))max0jJξRe(ξjr)1(ξRf^(ξ))2πmax0jJξRξjrR/Z \begin{aligned} \mathbf E_Z \max_{1 \le j \le J} \left\lvert T^{jr} g - g \right\rvert &= \mathbf E_Z \max_{0 \le j \le J} \sum_{\xi \in R} \left\lvert \widehat f(\xi) \cdot (e(\xi \cdot (x+jr)) - e(\xi \cdot x)) \right\rvert \\ &\le \mathbf E_Z \max_{0 \le j \le J} \sum_{\xi \in R} \left\lvert \widehat f(\xi) \right\rvert \left\lvert (e(\xi \cdot (x+jr)) - e(\xi \cdot x)) \right\rvert \\ &\le \left( \sum_{\xi \in R} \left\lvert \widehat f(\xi) \right\rvert \right) \max_{\substack{0 \le j \le J \\ \xi \in R}} \left\lvert (e(\xi \cdot (x+jr)) - e(\xi \cdot x)) \right\rvert \\ &\le \left( \sum_{\xi \in R} \left\lvert \widehat f(\xi) \right\rvert \right) \max_{\substack{0 \le j \le J \\ \xi \in R}} \left\lvert e(\xi \cdot jr) - 1 \right\rvert \\ &\le \left( \sum_{\xi \in R} \left\lvert \widehat f(\xi) \right\rvert \right) 2\pi \max_{\substack{0 \le j \le J \\ \xi \in R}} \left\lVert \xi \cdot jr \right\rVert_{\mathbb R/\mathbb Z} \end{aligned}

Since the ξR\xi \in R are covered by a cube of S={η1,,ηd}S = \left\{ \eta_1, \dots, \eta_d \right\}, we get

EZmax1jJTjrgg(ξRf^(ξ))2πJdmax0jJηSηjrR/Z. \mathbf E_Z \max_{1 \le j \le J} \left\lvert T^{jr} g - g \right\rvert \le \left( \sum_{\xi \in R} \left\lvert \widehat f(\xi) \right\rvert \right) 2\pi Jd \max_{\substack{0 \le j \le J \\ \eta \in S}} \left\lVert \eta \cdot jr \right\rVert_{\mathbb R/\mathbb Z}.

Let’s then bound the contribution over each dissociated set. We’ll need both the assumption of uniformity and the proposition we proved for dissociated sets.

EZmax1jJTjrfifi2EZmax0jJTjrfi2max0jJTjrfiL2Z.log(J)fiL2Z=log(J)ξDif^(ξ)2logJDξDif^(ξ) \begin{aligned} \mathbf E_Z \max_{1 \le j \le J} \left\lvert T^{jr} f_i - f_i \right\rvert &\le 2\mathbf E_Z \max_{0 \le j \le J} \left\lvert T^{jr} f_i \right\rvert \\ &\le 2\left\lVert \max_{0 \le j \le J} \left\lvert T^{jr} f_i \right\rvert \right\rVert_{L^2Z}. \\ &\ll \sqrt{\log(J)} \left\lVert f_i \right\rVert_{L^2Z} \\ &= \sqrt{\log(J)} \sqrt{\sum_{\xi \in D_i} \left\lvert \widehat f(\xi) \right\rvert^2 } \\ &\ll \sqrt{\frac{\log J}{D}} \sum_{\xi \in D_i} \left\lvert \widehat f(\xi) \right\rvert \end{aligned}

where the last step is by uniformity of ξ^\widehat \xi. Now combine everything with triangle inequality. \Box

4. Proof of main theorem

Without loss of generality PZA=PZB=δ\mathbf P_ZA = \mathbf P_ZB = \delta. Of course, we let f=1A1Bf = 1_A \ast 1_B so EZf=δ2\mathbf E_Z f = \delta^2. We will have parameters d1d \ge 1, M1M \ge 1, and Jexp(Cδ2logN3)J \ge \exp(C\sqrt[3]{\delta^2 \log N}) which we will select at the end.

Our goal is to show there exists some integer rr such that EZmax1j<JTjrff<δ2.\mathbf E_Z \max_{1 \le j < J} \left\lvert T^{jr} f - f \right\rvert < \delta^2. Now we cannot apply the uniformity estimate directly since ff is probably not uniform, and therefore we impose a dyadic decomposition on the base group ZZ; let

Z0={ξZ:12δ2<f^(ξ)δ2}Z1={ξZ:14δ2<f^(ξ)12δ2}Z2={ξZ:18δ2<f^(ξ)14δ2}ZM1={ξZ:2Mδ2<f^(ξ)2M+1δ2}Zerr={ξZ:f^(ξ)<2Mδ2} \begin{aligned} Z_0 &= \left\{ \xi \in Z : \frac{1}{2} \delta^2 < \left\lvert \widehat f(\xi) \right\rvert \le \delta^2 \right\} \\ Z_1 &= \left\{ \xi \in Z : \frac14\delta^2 < \left\lvert \widehat f(\xi) \right\rvert \le \frac{1}{2}\delta^2 \right\} \\ Z_2 &= \left\{ \xi \in Z : \frac18\delta^2 < \left\lvert \widehat f(\xi) \right\rvert \le \frac14\delta^2 \right\} \\ &\vdots \\ Z_{M-1} &= \left\{ \xi \in Z : 2^{-M} \delta^2 < \left\lvert \widehat f(\xi) \right\rvert \le 2^{-M+1} \delta^2 \right\} \\ Z_{\mathrm{err}} &= \left\{ \xi \in Z : \left\lvert \widehat f(\xi) \right\rvert < 2^{-M} \delta^2 \right\} \\ \end{aligned}

Then as before we can decompose via Fourier transform to obtain f=f0+f1++fM1+ferrf = f_0 + f_1 + \dots + f_{M-1} + f_{\mathrm{err}} so that f^i\widehat f_i is supported on ZiZ_i.

Now we can apply the previous lemma to get for each 0m<M0 \le m < M:

EZmax1jJTjrfmfm(ξZmf^(ξ))(logJd+JdmaxηSmηrR/Z) \mathbf E_Z \max_{1 \le j \le J} \left\lvert T^{jr} f_m - f_m \right\rvert \ll \left( \sum_{\xi \in Z_m} \left\lvert \widehat f(\xi) \right\rvert \right) \left( \sqrt{\frac{\log J}{d}} + Jd\max_{\eta \in S_m} \left\lVert \eta \cdot r \right\rVert_{\mathbb R/\mathbb Z} \right)

for some SmS_m; hence by summing and using the fact that

ξZf^(ξ)=ξZ1^A(ξ)1^B(ξ)1^A2Z1^B2Z=1AL2Z1BL2Z=PZAPZB=δ \sum_{\xi \in Z} \left\lvert \widehat f(\xi) \right\rvert = \sum_{\xi \in Z} \left\lvert \widehat 1_A(\xi) \right\rvert \left\lvert \widehat 1_B(\xi) \right\rvert \le \left\lVert \widehat 1_A \right\rVert_{\ell^2Z} \left\lVert \widehat 1_B \right\rVert_{\ell^2Z} = \left\lVert 1_A \right\rVert_{L^2Z} \left\lVert 1_B \right\rVert_{L^2Z} = \sqrt{\mathbf P_ZA \mathbf P_ZB} = \delta

we obtain that

0m<MEZmax1jJTjrffδ(logJd+JdmaxηSmηrR/Z). \sum_{0 \le m < M} \mathbf E_Z \max_{1 \le j \le J} \left\lvert T^{jr} f - f \right\rvert \ll \delta \left( \sqrt{\frac{\log J}{d}} + Jd\max_{\eta \in \bigcup S_m} \left\lVert \eta \cdot r \right\rVert_{\mathbb R/\mathbb Z} \right).

As for the “error” term, we bound

EZmax1jJTjrferrferr2EZmax1jJTjrferr2EZ1jJTjrferr21jJEZTjrferr21jJEZferr2JEZferr2JferrL2Z=2Jf^err2Z=2JξZerrf^err(ξ)22JmaxξZerrf^err(ξ)ξZerrf^err(ξ)2J2Mδ2δ=2J2M/2δ3/22J2M/2δ. \begin{aligned} \mathbf E_Z \max_{1 \le j \le J} \left\lvert T^{jr} f_{\mathrm{err}} - f_{\mathrm{err}} \right\rvert &\le 2\mathbf E_Z \max_{1 \le j \le J} \left\lvert T^{jr} f_{\mathrm{err}} \right\rvert \\ &\le 2\mathbf E_Z \sum_{1 \le j \le J} \left\lvert T^{jr} f_{\mathrm{err}} \right\rvert \\ &\le 2\sum_{1 \le j \le J} \mathbf E_Z \left\lvert T^{jr} f_{\mathrm{err}} \right\rvert \\ &\le 2\sum_{1 \le j \le J} \mathbf E_Z \left\lvert f_{\mathrm{err}} \right\rvert \\ &\le 2J \mathbf E_Z \left\lvert f_{\mathrm{err}} \right\rvert \\ &\le 2J \left\lVert f_{\mathrm{err}} \right\rVert_{L^2Z} \\ &= 2J \left\lVert \widehat f_{\mathrm{err}} \right\rVert_{\ell^2 Z} \\ &= 2J \sqrt{\sum_{\xi \in Z_{\mathrm{err}}} \left\lvert \widehat f_{\mathrm{err}}(\xi) \right\rvert^2} \\ &\le 2J \sqrt{\max_{\xi \in Z_{\mathrm{err}}} \left\lvert \widehat f_{\mathrm{err}}(\xi) \right\rvert \sum_{\xi \in Z_{\mathrm{err}}} \left\lvert \widehat f_{\mathrm{err}}(\xi) \right\rvert} \\ &\le 2J \sqrt{2^{-M}\delta^2 \cdot \delta} \\ &= 2J 2^{-M/2} \delta^{3/2} \\ &\le 2J 2^{-M/2} \delta. \end{aligned}

Thus, putting these altogether we need to find R0R \neq 0 such that

logJd+JdmaxηSmηrR/Z+2J2M/2δ. \sqrt{\frac{\log J}{d}} + Jd \max_{\eta\in\bigcup S_m} \left\lVert \eta \cdot r \right\rVert_{\mathbb R/\mathbb Z} + 2J \cdot2^{-M/2} \ll \delta.

Now set MlogJM \asymp \log J and dδ2logJd \asymp \delta^{-2} \log J, so the first and third terms are less than 13cδ\frac13 c \delta, since by hypothesis δ(loglogN)3logN\delta \gg \sqrt{\frac{(\log \log N)^3}{\log N}} from which we deduce

Jexp(Cδ2logN3)=exp(CloglogN)(logN)Cδ1. J \gg \exp\left( C\sqrt[3]{\delta^2\log N} \right) = \exp\left( C\log \log N \right) \ge (\log N)^C \gg \delta^{-1}.

Thus it suffices that maxηSηrR/Zδ3JlogJ\max_{\eta\in S} \left\lVert \eta \cdot r \right\rVert_{\mathbb R/\mathbb Z} \ll \frac{\delta^3}{J \log J} where S=SmS = \bigcup S_m. Note SdM(logJδ)2\left\lvert S \right\rvert \le dM \ll \left( \frac{\log J}{\delta} \right)^2. Now we recall the result that Bohr(S,ρ)ZρS\operatorname{Bohr}(S, \rho) \ge |Z| \rho^{|S|} and so it suffices for us that N(c1δ3JlogJ)c2(δ1logJ)2>1N \cdot \left( \frac{c_1 \delta^3}{J \log J} \right) ^{c_2 \left( \delta^{-1} \log J \right)^2} > 1 for constants c1c_1 and c2c_2. Then J=exp(Cδ2logN3)J = \exp(C\sqrt[3]{\delta^2 \log N}) works now.