--- title: '18.156 notes 16: bourgain''s projection theorem iii' description: In Subsection 16.1, we will sketch how the robust lemma, Lemma 16.2, implies Lemma 16.1. date: '2025-04-15' tags: - course - mit - ocw - lecture-notes cssclasses: - course-notes license: Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International license_url: https://creativecommons.org/licenses/by-nc-sa/4.0/ latex: source: static_resources/mit18_156_s25_lec_full.tex lines: - 3001 - 3126 section: 16 unit: 16 pdf: mit18_156_s25_lec_full_pdf pages: - 87 - 90 created: '2025-04-15' modified: '2025-04-15' published: '2025-04-15' pageLayout: default slug: courses/18-156-projection-theory-spring-2025/notes/16 permalink: https://aarnphm.xyz/courses/18-156-projection-theory-spring-2025/notes/16.md generator: quartz: v4.6.0 hostedProvider: Cloudflare baseUrl: aarnphm.xyz full: https://aarnphm.xyz/llms-full.txt --- Let us review the problem where we left off last time. Suppose that $A$ is a $(\delta,s,C)$-set. Lemma [[courses/18-156-projection-theory-spring-2025/notes/14#^lemma-14-8|14.8]] tells us that there is a fixed polynomial $Q$ so that $|Q(A)|_\delta \ge \delta^{-s \epsilon}$. We would like to iterate this lemma to prove a stronger lemma, which we now state. > \[!lemma|latex\] Lemma 16.1. > > For each $s > 0$ and each $\epsilon >0$, there is a polynomial $P = P_{s, \epsilon}$ so that, if $A$ is a $(\delta, s, C)$ set, then $|P(A)|_{\delta} \geq \delta^{-1 + \epsilon}$. > > ^lemma-16-1 However, we cannot prove Lemma [[courses/18-156-projection-theory-spring-2025/notes/16#^lemma-16-1|16.1]] just by iterating Lemma [[courses/18-156-projection-theory-spring-2025/notes/14#^lemma-14-8|14.8]], because we don’t yet know whether $Q(A)$ contains a $(\delta, s+\epsilon, C')$ set. Using the uniformization lemma, we can reduce to the case that $A$ is uniform. In this case, we know that $A$ is a $(\rho, s, C)$ set for all $\rho \ge \delta$. Now, by Lemma [[courses/18-156-projection-theory-spring-2025/notes/14#^lemma-14-8|14.8]], we know that $|Q(A)|_\rho \geq \rho^{-s-\varepsilon}$ for all $\rho \ge \delta$. If we knew that $Q(A)$ was uniform, then it would follow that $Q(A)$ is a $(\delta, s+ \epsilon, C')$ set with a reasonable $C'$. However, just because $A$ is uniform, it does not tells us that $Q(A)$ is uniform. The main enemy here is that $|Q(A)|_\rho$ may be large, and $|Q(A)|_\delta$ may be large, but it could still happen that there is a subset $B \subset Q(A)$ so that $|B|_\rho \ll |Q(A)|_\rho$ and yet $|Q(A) \setminus B|_\delta \ll |Q(A)|_\delta$ . (It’s a good exercise to draw a picture of this scenario.) Recall that the map $Q$ is a polynomial map from $\mathbb{R}^k$ to $\mathbb{R}$ for some $k$. And recall that $Q(A)$ is shorthand for $Q(A^k)$. In Lemma [[courses/18-156-projection-theory-spring-2025/notes/14#^lemma-14-8|14.8]], we showed that the entire image $Q(A^k)$ is large: $|Q(A^k)|_\delta \ge \delta^{-s - \varepsilon}$. To deal with this technical problem, it is very helpful to have a more robust estimate. > \[!lemma|latex\] Lemma 16.2. > > There is a polynomial $Q: \mathbb{R}^k \rightarrow \mathbb{R}$ so that the following holds. If $A$ is a $(\delta, s, C)$ set and $X \subset A^k$ with $|X|_\delta \gtrapprox |A^k|_\delta$, then $|Q(X)|_\delta \ge \delta^{- s - \varepsilon}$. > > ^lemma-16-2 In Subsection 16.1, we will sketch how the robust lemma, Lemma [[courses/18-156-projection-theory-spring-2025/notes/16#^lemma-16-2|16.2]], implies Lemma [[courses/18-156-projection-theory-spring-2025/notes/16#^lemma-16-1|16.1]]. Then in Subsection 16.2, we will sketch the proof of Lemma [[courses/18-156-projection-theory-spring-2025/notes/16#^lemma-16-2|16.2]]. In these sketches, we will deal with an important technical issue in the theory : formulating theorems in a robust way. We will see that more robust estimates are more useful – for instance because they work better in iteration. So having a more robust estimate is really useful. But on the other hand, we will see that the more robust estimate in Lemma [[courses/18-156-projection-theory-spring-2025/notes/16#^lemma-16-2|16.2]] does not follow from simple tweaks to our previous Lemma [[courses/18-156-projection-theory-spring-2025/notes/14#^lemma-14-8|14.8]]. It requires a really new input – the Balog-Szemeredi-Gowers theorem. This part further develops the ideas from Lecture [[courses/18-156-projection-theory-spring-2025/notes/12|12]] where we introduced BSG. ## 16.1 why robust estimates are useful Let us begin on the positive side and discuss how to use Lemma [[courses/18-156-projection-theory-spring-2025/notes/16#^lemma-16-2|16.2]]. Suppose that $A$ is uniform and $A$ is $(\delta, s, C)$. We will use Lemma [[courses/18-156-projection-theory-spring-2025/notes/16#^lemma-16-2|16.2]] to show that $Q(A)$ contains a $(\delta, s + \epsilon/2, C')$ set. Such a result can then be iterated to prove Lemma [[courses/18-156-projection-theory-spring-2025/notes/16#^lemma-16-1|16.1]]. We are going to build a $(\delta, s+\epsilon/2, C)$ subset of $Q(A)$. Let us recall the definition of a $(\delta, s, C)$ set. A set $S$ is $(\delta, s, C)$ if, for every ball $B(x,r)$ we have $$ |S \cap B(x,r)|_\delta \le C r^s |S|_\delta. $$ We are going to build a set which is $(\rho, s, C)$ for every $\rho \in [\delta, 1]$. So for every $\rho \in [\delta, 1]$, and every ball $B(x,r)$, out set will obey $$ |S \cap B(x,r)|_\rho \le C r^s |S|_\rho. \tag{27} $$ Consider a sequence of scales $1 > \rho_1 > \rho_2 > ... > \rho_N = \delta$. Assume these scales are very close together. First consider $| Q(A) |_{\rho_1}$. Since $A$ is uniform, we know that $A$ is $(\rho_1, s , C)$ and so $| Q(A) |_{\rho_1} \ge \rho_1^{-s - \epsilon}$. Cover $Q(A)$ with disjoint intervals $I_1$ of length $\rho_1$. We will pick some of these intervals $I_1$ to include in $B$. Initially, we include all of them, but as we continue through the construction, we will remove bad intervals. We pick a small parameter $\eta > 0$ with $\eta < \epsilon$. Next we consider scale $\rho_2$. We know that $A$ is $(\rho_2, s, C)$ and so $|Q(A)|_{\rho_2} \ge \rho_2^{-s - \epsilon}$. Cover $Q(A)$ with disjoint intervals $I_2$ of length $\rho_2$. Now we notice how many intervals $I_2$ lie in each interval $I_1$. We say that an interval $I_1$ is bad if $$ |Q(A) \cap I_1|_{\rho_2} > \rho_1^{s + \epsilon - \eta} |Q(A)|_{\rho_2}. $$ (Notice that a bad interval $I_1$ is a ball $B(x,r)$ that violates ([[courses/18-156-projection-theory-spring-2025/notes/16#eq-27|27]]) with $\rho = \rho_2$. ) The number of bad intervals $I_1$ is at most $\rho_1^{-(s + \epsilon - \eta)}$. Next define $X_{1, bad} \subset A^k$ by $$ X_{1,bad} = \{ (a_1, ..., a_k) \in A^k: Q(a_1, ..., a_k) \text{ lies in a bad interval } I_1 \}. $$ Our robust estimate Lemma [[courses/18-156-projection-theory-spring-2025/notes/16#^lemma-16-2|16.2]] tells us that $|X_{1,bad}|_{\rho_1} \ll |A^k|_{\rho_1}$. Since $A$ is uniform, this also tells us that for every $\rho \le \rho_1$, $$ |X_{1, bad}|_\rho \ll |A^k|_\rho. $$ Define $X_1 = A^k \setminus X_{1,bad}$. Applying Lemma [[courses/18-156-projection-theory-spring-2025/notes/16#^lemma-16-2|16.2]], we also see that $$ |Q(X_1)|_{\rho_1} \ge \rho_1^{-s -\epsilon} \tag{28} $$ $$ |Q(X_1)|_{\rho_2} \ge \rho_2^{-s -\epsilon} \tag{29} $$ Typically, we have $|Q(X_1)|_{\rho_2} \approx |Q(A)|_{\rho_2}$. We will focus on that special case in this sketch. (If $|Q(X_1)|_{\rho_2} \ll |Q(A)|_{\rho_2}$, then we redefine bad intervals and repeat the argument above.) We claim that $Q(X_1)$ obeys ([[courses/18-156-projection-theory-spring-2025/notes/16#eq-27|27]])with dimension $s = s + \epsilon - \eta$, in the special case where $r$ and $\rho$ are either 1 or $\rho_1$ or $\rho_2$. There are three cases here. If $r = 1$ and $\rho = \rho_1$, ([[courses/18-156-projection-theory-spring-2025/notes/16#eq-27|27]]) boils down to ([[courses/18-156-projection-theory-spring-2025/notes/16#eq-28|28]]. If $r=1$ and $\rho = \rho_2$, then ([[courses/18-156-projection-theory-spring-2025/notes/16#eq-27|27]]) boils down to ([[courses/18-156-projection-theory-spring-2025/notes/16#eq-29|29]]). And if $r = \rho_1$ and $\rho = \rho_2$, then ([[courses/18-156-projection-theory-spring-2025/notes/16#eq-27|27]]) boils down to the definition of a good interval: $$ |Q(X_1) \cap I_1 |_{\rho_2} = |Q(A) \cap I_1|_{\rho_2} \le \rho_1^{s + \epsilon - \eta} |Q(A)|_{\rho_2} \approx \rho_1^{s + \epsilon - \eta} |Q(X_1)|_{\rho_2}. $$ Now we continue by the same method working through all the scales $\rho_j$. In this way, we will find a subset $X= X_N \subset A^k$ so that $Q(X_N)$ obeys ([[courses/18-156-projection-theory-spring-2025/notes/16#eq-27|27]]) at all the scales $r, \rho$ of the form $\rho_j$. Since these cover essentially all scales, this finishes our proof sketch that $Q(A)$ contains a $(\delta, s + \epsilon/2, C')$ set. ## 16.2 how to prove robust estimates In this Subsection, we will outline the proof of Lemma [[courses/18-156-projection-theory-spring-2025/notes/16#^lemma-16-2|16.2]]. We first encountered the issue of robust estimates in the proof of the Bourgain-Katz-Tao projection theorem in Lecture [[courses/18-156-projection-theory-spring-2025/notes/12|12]]. Recall that in the previous lecture, we had proven that if $A \subset \mathbb{F}_p$ with $|A| = p^{s_A}$ and $D \subset \mathbb{F}_p$ with $|D| = p^{s_D}$ with $0 < s_A, s_D < 1$, then there exists $t \in D$ so that $| \pi_t(A \times A) | \ge p^{s_A + \epsilon}$ for $\epsilon = \epsilon(s_A, s_D) > 0$. We wanted to replace the product set $A \times A$ by a general set $X \subset \mathbb{F}_p^2$ and to prove that there exists $t \in D$ so that $| \pi_t (X) | \ge p^\epsilon |X|^{1/2}$. By changing variables we could assume that our direction set $D$ included horizontal and vertical projections, and then we could reduce to the case that $X \subset A_1 \times A_2$ with $|X| \ge p^{- 2 \epsilon} |A_1| |A_2|$. So we only needed to make our previous estimates a little more robust, extending from the case when $X$ is an honest product $A \times A$ to the case when $X$ is a large subset of a product $A_1 \times A_2$. But we found that this extension was not straightforward. It required a signficant new idea. The key idea to make this extension work is the Balog-Szemeredi-Gowers theorem. The BSG theorem can be used in a similar way in the proof of Lemma [[courses/18-156-projection-theory-spring-2025/notes/16#^lemma-16-2|16.2]]. To prove the more robust estimate Lemma [[courses/18-156-projection-theory-spring-2025/notes/16#^lemma-16-2|16.2]], we use the BSG theorem and follow some of the ideas from Lecture [[courses/18-156-projection-theory-spring-2025/notes/12|12]]. We will ultimately prove Lemma [[courses/18-156-projection-theory-spring-2025/notes/16#^lemma-16-2|16.2]] with $k=3$ and with polynomial $\tilde Q(a_1, a_2, a_3) = a_1 + a_2 a_3$. We sketch the steps of this argument. Each step is similar to proofs we have done in the last lectures. It is a good exercise to fill in the details of these arguments. The first step is to prove that if $A$ is a $(\delta, s, C)$ set, then there is an $a \in A$ so that $$ |A + a A|_\delta \ge \delta^{-s - \varepsilon} \tag{30} $$ By Lemma [[courses/18-156-projection-theory-spring-2025/notes/14#^lemma-14-8|14.8]], we know that there is a polynomial $Q$ so that $|Q(A)|_\delta \ge \delta^{-s - \varepsilon}$, and it’s not hard to show that $Q(A)$ is a $(\delta, s, C)$ set. Using a careful double counting argument, we can then show that there exists $b \in Q(A)$ so that $|A + b A|_\delta \ge \delta^{-s - \epsilon}$. And then using the contagious structure argument, based on Plunnecke-Ruzsa, we can find $a \in A$ so that $|A + a A|_\delta \ge \delta^{-s - \epsilon}$. This argument is similar to Lecture [[courses/18-156-projection-theory-spring-2025/notes/11|11]]. The second step is to upgrade this estimate by proving that there is some $a \in A$ so that if $X \subset A\times A$ is a large subset, then $|\pi_a(X)|_\delta \gtrapprox \delta^{-s - \epsilon}$. More precisely, we would prove that there is some $\eta > 0$ so that if $|X|_\delta \ge \delta^\eta |A \times A|_\delta$, then $|\pi_a(X)|_\delta \gtrapprox \delta^{-s - \epsilon}$. This upgrade is based on Balog-Szemeredi-Gowers and a symmetry argument, as in Lecture [[courses/18-156-projection-theory-spring-2025/notes/12|12]]. With just a little more work, we can prove that almost all $a \in A$ have the good property in the second step. To prove this upgrade, we set $A_{good} \subset A$ to be the set of $a \in A$ with the good property in the second step, and we set $A_{bad} = A \setminus A_{good}$. If $A_{bad}$ is a large subset of $A$, then we can get a contradiction by applying our previous results to $A_{bad}$. All together, we see that if $\tilde Q(a_1, a_2, a_3) = a_1 + a_3 a_2$ and $\tilde X \subset A \times A \times A$ is a large subset, then $| \tilde Q(\tilde X) |_\delta \ge \delta^{-s - \epsilon}$. This finishes our proof sketch for Lemma [[courses/18-156-projection-theory-spring-2025/notes/16#^lemma-16-2|16.2]].