Convex sumsets and sum-product estimates

Thanks to Junxian Li for carefully reading and providing useful suggestions.

Here we advertise a conjecture that would have consequences to two major problems in additive combinatorics and provide a proof of the state of the art progress towards them. In this post, we say {b \gtrsim a} if {a = O(b \log^c |A|)} for some {c >0}. Let {A, B \subset \mathbb{R}} be finite. We define the sumset and product set via

\displaystyle A+B := \{a + b : a \in A , b \in B\} , \ \ \ \ \ \ AB := \{ab : a \in A , b \in B\}.

We have the beautiful sum–product conjecture of Erdős and Szemerédi.

Conjecture (ESconj): Let {\delta < 1}. Then for any finite {A \subset \mathbb{Z}} one has

\displaystyle |A+A| + |A A| \gtrsim |A|^{1 + \delta}. \ \ \spadesuit

We say a finite {A = \{a_1 < \ldots < a_k\} \subset \mathbb{R}} is convex if

\displaystyle a_{i+1} - a_i > a_i - a_{i-1}, \ \ \ 2 \leq i \leq k-1.

The following conjecture is due to Erdős.

Conjecture (Convex) Suppose {A \subset \mathbb{R}} is convex. Then

\displaystyle |A+A| \gtrsim |A|^2. \ \ \spadesuit

Study of the quantity {d^+(A)} has played a crucial role in recent progress on both conjectures, which we define now. For {A \subset \mathbb{R}} finite, we define

\displaystyle d^{+}(A) := \sup_{B \neq \emptyset}\frac{ E_3^+(A,B) }{|A| |B|^2},


\displaystyle E_3^+(A,B) = \sum_x r_{A- B}(x)^3 , \ \ \ \ r_{A-B}(x) = \#\{(a, b ) \in A \times B : x = a - b\}.

See my recent paper on Conjecture (ESconj) for an introduction to {d^+(A)}. We mention here that {1 \leq d^+(A) \leq |A|} and intuitively the larger {d^+(A)} is the more additive structure {A} has. Indeed {d^+(A)} is a higher order analog of the more common additive energy of a set. An application of Cauchy–Schwarz implies, as explain in equation (2) of this paper, reveals

\displaystyle |A|^{3/2} \leq |A+A| d^+(A)^{1/2}.\ \ \ \ \ (1)

Thus if {d^+(A) \lesssim 1}, we have that {|A+A| \gtrsim |A|^{3/2}}. We conjecture that this can be improved.

Conjecture (dplus): Let {A \subset \mathbb{R}} be finite. Then

\displaystyle |A+A| \gtrsim |A|^2 d^+(A)^{-1}. \ \ \ \ \spadesuit

This would imply Conjecture (Convex) and Conjecture (ESconj) for {\delta = 1/2}. The goal of the rest of this post is to prove the current state of the art progress due to Shkredov. We remark that the rest of the post is technical in nature. Please see this survey of de Zeeuw and the introduction of this paper of mine for background. Also, one can see this paper of Murphy, Rudnev, Shkredov, and Shteinikov for another perspective (thanks to Sophie Stevens for pointing this reference out to me).

Theorem 1: (Shkredov) Let {A \subset \mathbb{R}} be finite. Then

\displaystyle |A+A| \gtrsim |A|^{58/37} d^+(A)^{-21/37} . \ \ \spadesuit

His idea is to analyze spectral and combinatorial properties of an associated matrix: the spectral properties depend on {|A+A|}, while the combinatorial properties depend on {d^+(A)}. One of the simplest versions of this idea is that the number of {k}–walks in a graph is the sum of the the {k^{\rm th}} powers of the eigenvalues of the graph, which plays a crucial role in the theory of expander graphs.

For a function {g : \mathbb{R} \rightarrow \mathbb{C}}, we define the following {|A| \times |A|} matrices

\displaystyle T_A^g(x,y) : = 1_A(x)1_A(y) g(x-y), \ \ S_A^g(x,y) : = 1_A(x)1_A(y) g(x+y).

We set

\displaystyle E^+(A , B) := \sum_x r_{A-B}(x)^2 , \ \ \ E^+(A) : = E^+(A,A),

\displaystyle E_3^+(A, B , C) := \sum_x r_{A-A}(x) r_{B-B}(x) r_{C-C}(x) , \ \ E_3^+(A) : = E_3^+(A, A , A) .

Theorem 1 follows from the following two propositions.

Proposition 1: Let

\displaystyle S = \{x \in A+A : r_{A+A}(x) \geq \frac{ |A|^2 }{2|A+A|}\},

be a set of popular sums of {A} and {g = 1_S}. Let {\mu_1 , \ldots , \mu_{|A|}} be the eigenvalues of {S_A^g} with associated orthonormal eigenvectors {f_j : A \rightarrow \mathbb{R}}. Then

\displaystyle \frac{|A|^5}{2^5 |S|} \leq \frac{\mu_1^5}{||g||_2^2 ||g||_{\infty}} \leq \mu_1^2 \langle T_A^{r_{A-A}} f_1 , f_1 \rangle \leq \sum_{j=1}^{|A|} \mu_j^2 \langle T_A^{r_{A-A}} f_j , f_j \rangle \leq d^+(A)^{1/2} |A|^{3/2}E_3^+(A,A,S)^{1/2} . \ \ \ \spadesuit

The second and third inequality hold for general {g}. The right hand side arises from combinatorial properties of {T_A^{r_{A-A}}}, while the left hand side comes from spectral properties. Before proving Proposition 1, we note that in order to apply Proposition 1, we need a bound for {E_3^+(A,A,S)}, which is slightly different than what is considered in the definition of {d^+(A)}. This is handled by the following.

Proposition 2: Let {A \subset \mathbb{R}} be finite and

\displaystyle S = \{x \in A+A : r_{A+A}(x) \geq \frac{ |A|^2 }{2|A+A|}\}.


\displaystyle E_3^+(A,A,S) \lesssim d^+(A)^{11/10} |A|^{6/5} |A+A|^{17/10}. \ \ \ \spadesuit

Theorem 1 follows immediately from combining Proposition 1 and Proposition 2. We start with the proof of Proposition 2.

Proof of Proposition 2: The idea is to bound {E_3^+(A,A,S)} by various second moment estimates and then use the fact that {a - a' = b - b'} if and only if {a + b' = b + a'}. By a dyadic decomposition, there is a set {Q \subset A-A} such that

\displaystyle E_3^+(A,A,S) \gtrsim \sum_{x \in Q} r_{A-A}(x)^2 r_{S-S}(x) , \ \ Q = \{x : q \leq r_{A-A}(x) \leq 2q \}.

We have

\displaystyle E_3^+(A,A,S) \lesssim q E(A,S).\ \ \ \ \ (2)

On the other hand,

\displaystyle E_3^+(A,A,S) \lesssim q^2 \sum_x 1_Q(x) \sum_y 1_S(y) 1_S(x+ y)

\displaystyle \lesssim \frac{|A+A|}{|A|^2} q^2 \sum_y r_{A+A}(y)\sum_x 1_Q(x) 1_S(x+y)

\displaystyle = \frac{|A+A|}{|A|^2} q^2 \sum_y r_{A+A}(y)r_{S-Q}(y) \leq \frac{|A+A|}{|A|^2} q^2 E^+(A,S)^{1/2} E^+(A,Q)^{1/2}.\ \ \ \ \ (3)

By the asymmetric version of (1), we have

\displaystyle E^+(A,Q) \leq d^+(A)^{1/2} |A| |Q|^{3/2},

and we also have

\displaystyle |Q| \leq \#\{x : r_{A-A} (x) \geq q\} \leq E_3^+(A) q^{-3} \leq d^+(A) |A|^3 q^{-3}.

Combining these two, we find

\displaystyle E^+(A,Q) \leq \frac{d^+(A)^2 |A|^{11/2}}{q^{9/2}}.

Using (3), we find

\displaystyle E_3^+(A,A,S) \lesssim E^+(A,S)^{1/2} |A+A| d^+(A) |A|^{3/4} q^{-1/4} \ \ \ \ \ (4)

Combining (2) and (4) yields

\displaystyle E_3^+(A,A,S) \lesssim E^+(A,S)^{1/2} \min \{ q E^+(A,S)^{1/2} , |A+A| d^+(A) |A|^{3/4} q^{-1/4}. \}.

This quantity is minimized when

\displaystyle q^{5/4} = \frac{|A+A| d^+(A) |A|^{3/4}}{E^+(A,S)^{1/2}},

and so

\displaystyle E_3^+(A,A,S) \lesssim E^+(A,S)^{1/10} \left( |A+A| d^+(A) |A|^{3/4} \right)^{4/5}

The result follows from

\displaystyle E^+(A,S) \leq d^+(A)^{1/2} |A| |S|^{3/2}, \ \ |S| \leq |A+A|,

and simplification.

Proof of Proposition 1: Note that {T_A^{r_{A-A}}} is a symmetric, positive–semidefinite matrix. To see the second point, note

\displaystyle T_A^{r_{A-A}}(x,y) = \langle 1_{A - x} , 1_{A - y} \rangle.

Thus {\langle T_A^{r_{A-A}} h , h \rangle \geq 0} for any {h : A \rightarrow \mathbb{C}}. This proves the third inequality

Now, we show the first inequality. We have {||1_S||_2^2 = |S|} and by a popularity argument

\displaystyle \langle S_A^{1_S} 1 , 1 \rangle = \sum_{x \in S} r_{A+A}(x) \geq |A|^2 / 2,

and so {\mu_1 \geq |A| / 2}, since

\displaystyle \mu_1 \geq \sup_{||f||_2 = 1} \frac{\langle T_A^{r_{A-A}} f , f \rangle}{||f||_2}.

We now prove the fourth inequality. Since {f_1 , \ldots , f_{|A|}} are an orthonormal basis of eigenvectors for {S_A^{1_S}},

\displaystyle \sum_{j=1}^{|A|} \mu_j^2 \langle T_A^{r_{A-A}} f_j , f_j \rangle = \sum_{x,y , j } T_A^{r_{A-A}}(x,y) \mu_j f_j(x) \mu_j f_j(y) = \sum_{x,y , j , i} T_A^{r_{A-A}}(x,y) \mu_i f_i(x) \mu_j f_j(y) \sum_z f_j(z) f_i(z)

\displaystyle = \sum_{x,y , z } T_A^{r_{A-A}}(x,y) \sum_i \mu_i f_i(x) f_i(z) \sum_j \mu_j f_j(y) f_j(z) = \sum_{x, y , z} T_A^{r_{A-A}}(x,y) S_A^{1_S} (x,z) S_A^{1_S} (y , z),

where in the final equality, we used that

\displaystyle S_A^{1_S} = \sum_j \mu_j f_j f_j^T,

from the spectral theorem of self adjoint matrices. Now we have

\displaystyle \sum_{x, y , z} T_A^{r_{A-A}}(x,y) S_A^{1_S} (x,z) S_A^{1_S} (y , z) = \sum_{x, y , z \in A} T_A^{r_{A-A}}(x,y) 1_S(x+z) 1_S(y+z).

Using the change of variables {\alpha = x+z} and {\beta = y+z}, we have this is equal to

\displaystyle \sum_{z, \alpha , \beta} r_{A-A}(\alpha - \beta ) 1_S(\alpha) 1_S (\beta) \sum_z 1_A (z) 1_A(\alpha - z) 1_A( \beta - z).

By Cauchy–Schwarz and simplification, we get this is

\displaystyle \leq \left( \left( \sum_{\alpha , \beta} r_{A-A} (\alpha - \beta)^2 1_S(\alpha) 1_S(\beta) \right) \sum_{\alpha , \beta} |\sum_z 1_A(z) 1_A(\alpha - z) 1_A( \beta - z) |^2 \right)^{1/2}

\displaystyle = E_3^+(A)^{1/2} E_3^+(A, A , S)^{1/2}.

The inequality follows from

\displaystyle E_3^+(A) \leq d^+(A) |A|^3.

We now prove the second inequality. It is enough to show

\displaystyle \mu_1 \leq ||g||_{\infty}\left( \sum_z f_1(z) \right)^2, \ \ \ \left( \sum_z \mu_1 f_1(z) \right)^2 \leq ||g||_2^2 \langle T_A^{r_{A-A}} f_1 , f_1 \rangle .

We start with the first inequality. Since {f_1} is a unit vector and {S_A^g f_1 = \mu_1 f_1},

\displaystyle \mu_1 = \sum_x f_1(x) \mu_1 f_1(x) = \sum_{x,y} f_1(x) g(x +y) f_1(y) \leq ||g||_{\infty} \left( \sum_{x} f_1(x) \right)^2.

In the last inequality, we used that we may choose {f_1 \geq 0} entry–wise by the Perron–Frobenius theorem. We move onto the second inequality. Using the change of variables {w = x+y} we find

\displaystyle \sum_{x \in A} \mu_1 f_1(x) = \sum_{x,y} g(x+y) f_1(y) 1_A(x) = \sum_{w , x} g(w) f_1(w-x) 1_A(x) .

By Cauchy–Schwarz and a modest computation, we find this is

\displaystyle \leq ||g||_2 \left( \sum_w \left( \sum_x f_1(w-x) 1_A(x) \right)^2 \right)^{1/2} = ||g||_2 \langle T_A^{r_{A-A}} f_1 , f_1 \rangle^{1/2}. \ \ \spadesuit

We end by mentioning that

\displaystyle |A-A| \gtrsim |A|^{8/5} d^+(A)^{-3/5},

can be inferred from this paper of Schoen and Shkredov, which is stronger than the plus version discussed in this blog post (see this post). Any improvement to either bound would be of interest.

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