2.1. Lower bound for t_n

In view of proposition 2.1, the lower bound for t_n in theorem 1.1 is equivalent to the following proposition.

Proposition 2.2. Let ɛ > 0 and let n be sufficiently large in terms of ɛ. Let

\begin{equation*} q = \frac{4}{3 - (1+\varepsilon)\log_n\frac{3\sqrt{3}}{4}}. \end{equation*}

There exists a function $f: \mathbb{Z}\rightarrow \mathbb{R}$, which is supported on an interval of length n such that $\|\widehat{f}\|_4 \gt \|f\|_q$.

Our motivation for the construction of f in proposition 2.2 comes from the Babenko–Beckner inequality [Reference Babenko1, 2.3], a sharpened form of the Hausdorff–Young inequality for functions on $\mathbb{R}$ (and more generally on $\mathbb{R}^d$). It asserts that for any function $g: \mathbb{R}\rightarrow \mathbb{R}$, we have

(2.3)\begin{equation} \|\widehat{g}\|_4 \leq \left(\frac{4\sqrt{3}}{9}\right)^{1/4} \|g\|_{4/3}. \end{equation}

Moreover, equality is achieved when g is the Gaussian function $g(x) = e^{-x^2}$. In other words, Gaussian functions (and similarly their dilated versions) maximize the $\widehat{L}_4$-norm if we hold the $\ell^{4/3}$-norm fixed. If we take $g(x) = e^{-x^2/A}$ with $A \approx n^2$ (so that g is essentially supported on an interval of length ≈ n), then direct computations show that

\begin{equation*} \frac{\|g\|_{4/3}}{\|g\|_q} = c A^{\frac{1}{2}(\frac{3}{4}-\frac{1}{q})}, \end{equation*}

where c is an explicit constant depending on q and c ≈ 1 when $q \approx 4/3$. By our choice of A and q, we have

\begin{equation*} A^{\frac{1}{2}(\frac{3}{4}-\frac{1}{q})} \approx n^{\frac{3}{4}-\frac{1}{q}} = n^{\frac{1}{4}(1+\varepsilon)\log_n\frac{3\sqrt{3}}{4}} \approx \left(\frac{3\sqrt{3}}{4}+c\right)^{1/4} \end{equation*}

for some constant $c=c(\varepsilon) \gt 0$. Hence, this function g(x) satisfies

\begin{equation*} \|\widehat{g}\|_4 = \left(\frac{4\sqrt{3}}{9}\right)^{1/4} \|g\|_{4/3} \approx \left(\frac{4\sqrt{3}}{9}\right)^{1/4} \left(\frac{3\sqrt{3}}{4} + c\right)^{1/4}\|g\|_q \gt \|g\|_q. \end{equation*}

If we define $f: \mathbb{Z}\rightarrow \mathbb{R}$ by sampling the values of g(x) at integral points, then we may expect that

\begin{equation*} \|\widehat{f}\|_4 \approx \|\widehat{g}\|_4, \ \ \|f\|_q \approx \|g\|_q, \end{equation*}

and thus, we should also have $\|\widehat{f}\|_4 \gt \|f\|_q$. The details are worked out in § 3.

2.2. Upper bound for t_n

In view of proposition 2.1, the upper bound for t_n in theorem 1.1 is equivalent to the following proposition.

Proposition 2.3. Let $n \geq 2$ be a positive integer and let $f: \mathbb{Z}\rightarrow \mathbb{R}$ be a function, which is supported on a set of size n. Let

\begin{equation*} q = \frac{4}{3 - \log_n(1+c)} \end{equation*}

for some sufficiently small absolute constant c > 0. Then, $\|\widehat{f}\|_4 \leq \|f\|_q$.

The starting point of our proof of proposition 2.3 is the inequality

(2.4)\begin{equation} \|\widehat{f}\|_4 \leq \|f\|_{4/3}, \end{equation}

which follows from the Hausdorff–Young inequality or Young’s convolution inequality. By Hölder’s inequality (see (2.2)) and the definition of q, we have

\begin{equation*} \|f\|_{4/3} \leq n^{3/4 - 1/q} \|f\|_q = (1+c)^{1/4} \|f\|_q. \end{equation*}

Thus, the proof is already complete unless

\begin{equation*} \|\widehat{f}\|_4 \geq (1+c)^{-1/4} \|f\|_{4/3}, \end{equation*}

and thus, a key part of our argument is to analyse when equality almost holds in (2.4). Note that equality holds exactly in (2.4) when f is supported on a singleton set. We prove in proposition 4.5 that if equality almost holds in (2.4), then f is well approximated by a function f ₀, which is supported on a singleton set, up to an error g, which is small in $\ell^{4/3}$-norm. Clearly, the function f ₀ satisfies $\|\widehat{f_0}\|_4 = \|f_0\|_q$. The remaining task is then to show that the error g can only swing the inequality in the desired direction. The details are carried out in § 4.

We remark that proposition 4.5 is not new. In fact, it is a special case of [Reference Charalambides and Christ4, theorem 1.2] (see also [Reference Christ5] for an analogous result in Euclidean spaces) and of [Reference Eisner and Tao7, proposition 5.4]. As it turns out, our proof idea is the same as that in [Reference Eisner and Tao7], which, in turn, has its origin from [Reference Fournier8]. For completeness, we still give a self-contained proof of it in § 4.

2.3. Questions and speculations

Our proof of the lower bounds for t_n is not constructive, which motivates the question of constructing explicit subsets of $\{0,1,\cdots,n-1\}^d$ with large additive energies.

Questiona 2.4.

For sufficiently large n, construct a subset $A \subset \{0,1,\cdots,n-1\}^d$ for some d such that $E(A) \geq |A|^t$, where

\begin{equation*} t = 3 - (1+o_{n\rightarrow\infty}(1)) \log_n \frac{3\sqrt{3}}{4}. \end{equation*}

A possible candidate for such a set A is the set of lattice points in a d-dimensional ball $B_d \subset \mathbb{R}^d$ (with an appropriate choice of d and an appropriate centre and radius). This choice is motivated by results in [Reference Shao12], which implies, roughly speaking, that such a set A maximizes the additive energy among all genuinely d-dimensional subsets of $\mathbb{Z}^d$ of a given cardinality. Moreover, $E(A) \approx E(B_d)$, and it follows from the computations in [Reference Mazur11, § 3.1] that

\begin{equation*} E(B_d) = \left(\frac{4\sqrt{3}}{9} + o_{d\rightarrow\infty}(1)\right)^d |B_d|^3, \end{equation*}

where $|B_d|$ denotes the Lebesgue measure of B_d.

Next we speculate the asymptotic behaviour of t_n as $n\rightarrow\infty$. Note that if $g: \mathbb{R}\rightarrow \mathbb{R}$ is a (continuous) function supported on an interval of length n and

\begin{equation*} q = \frac{4}{3-\log_n\frac{3\sqrt{3}}{4}}, \end{equation*}

then

\begin{equation*} \|\widehat{g}\|_4 \leq \left(\frac{4\sqrt{3}}{9}\right)^{1/4} \|g\|_{4/3} \leq \left(\frac{4\sqrt{3}}{9}\right)^{1/4} n^{3/4-1/q} \|g\|_q = \|g\|_q, \end{equation*}

where the first inequality follows from the Babenko–Beckner inequality (2.3) and the second inequality follows from Hölder’s inequality (a continuous version of (2.2)). Based on this, it is perhaps reasonable to conjecture that a similar bound holds for discrete functions.

Conjecture 2.5.

Let ɛ > 0 and let n be sufficiently large in terms of ɛ. Let

\begin{equation*} q = \frac{4}{3 - (1-\varepsilon)\log_n\frac{3\sqrt{3}}{4}}. \end{equation*}

Then, for any function $f: \mathbb{Z}\rightarrow \mathbb{R}$, which is supported on an interval of length n, we have $\|\widehat{f}\|_4 \leq \|f\|_q$.

In particular, the conjecture would imply that

\begin{equation*} t_n = 3 - (1+o_{n\rightarrow\infty}(1)) \log_n \frac{3\sqrt{3}}{4}. \end{equation*}

So perhaps the lower bound in theorem 1.1 is sharp up to the error in o(1).

4.1. Near equality in Hölder’s inequality

In this section, all implied constants are allowed to depend on the exponents p,q, and r.

Lemma 4.1. Let $p,q \in (1,+\infty)$ be exponents with $1/p + 1/q = 1$. Let $a, b$ be non-negative reals. Suppose that

\begin{equation*} \frac{a^p}{p} + \frac{b^q}{q} \leq (1 + \delta)ab \end{equation*}

for some sufficiently small constant δ > 0. Then, $a^p = (1 + O(\delta^{1/2}))b^q$.

Proof. If ab = 0, then the conclusion holds trivially. Henceforth, assume that $a,b \gt 0$. By Taylor’s theorem applied to the function $\psi(x) = \log x$ at the point $x_0 = a^p/p + b^q/q$, we have

\begin{equation*} \psi(a^p) = \psi(x_0) + (a^p - x_0) \psi'(x_0) + \frac{1}{2} (a^p-x_0)^2 \psi''(\xi_1) \end{equation*}

and

\begin{equation*} \psi(b^q) = \psi(x_0) + (b^q - x_0) \psi'(x_0) + \frac{1}{2} (b^q-x_0)^2 \psi''(\xi_2) \end{equation*}

for some $\xi_1,\xi_2$ lying between a^p and b^q. Since

\begin{equation*} a^p - x_0 = \frac{a^p - b^q}{q}, \ \ b^q-x_0 = \frac{b^q-a^p}{p}, \end{equation*}

it follows that

\begin{equation*} \frac{1}{p}\psi(a^p) + \frac{1}{q}\psi(b^q) = \psi(x_0) + \frac{(a^p-b^q)^2}{2pq^2} \psi''(\xi_1) + \frac{(a^p-b^q)^2}{2p^2q}\psi''(\xi_2). \end{equation*}

Since $\psi''(x) = -1/x^2$, we have

\begin{equation*} \psi''(\xi_i) \leq -\min\left(\frac{1}{a^{2p}}, \frac{1}{b^{2q}}\right). \end{equation*}

From hypothesis, we have

\begin{equation*} \frac{1}{p}\psi(a^p) + \frac{1}{q}\psi(b^q) - \psi(x_0) = \log a + \log b - \log\left(\frac{a^p}{p} + \frac{b^q}{q}\right) \geq -\log(1+\delta) \geq - \delta. \end{equation*}

Hence, it follows that

\begin{equation*} -\delta \leq -(a^p-b^q)^2\left(\frac{1}{2pq^2} + \frac{1}{2p^2q}\right) \min\left(\frac{1}{a^{2p}}, \frac{1}{b^{2q}}\right) = -\frac{(a^p-b^q)^2}{2pq} \min\left(\frac{1}{a^{2p}}, \frac{1}{b^{2q}}\right), \end{equation*}

and thus,

\begin{equation*} (a^p - b^q)^2 \ll \delta \max(a^{2p}, b^{2q}). \end{equation*}

The desired conclusion follows immediately.

Lemma 4.2. Let $p,q,r \in (1,+\infty)$ be exponents with $1/p + 1/q + 1/r = 1$. Let a, b, and c be non-negative reals. Suppose that

\begin{equation*} \frac{a^p}{p} + \frac{b^q}{q} + \frac{c^r}{r} \leq (1 + \delta)abc \end{equation*}

for some sufficiently small constant δ > 0. Then, $a^p = (1 + O(\delta^{1/2}))b^q = (1 + O(\delta^{1/2})c^r$.

Proof. We may assume that abc > 0, since otherwise the conclusion holds trivially. Choose exponent $p' \in (1,+\infty)$ such that $1/p + 1/p' = 1$. Let

\begin{equation*} d = \left(\frac{p'}{q} b^q + \frac{p'}{r} c^r\right)^{1/p'}. \end{equation*}

Then,

\begin{equation*} \frac{a^p}{p} + \frac{b^q}{q} + \frac{c^r}{r} = \frac{a^p}{p} + \frac{d^{p'}}{p'} \geq ad. \end{equation*}

From hypothesis, it follows that $d \leq (1+\delta)bc$, which can be rewritten as

\begin{equation*} \frac{x^{q'}}{q'} + \frac{y^{r'}}{r'} \leq (1+\delta)^{p'} xy, \end{equation*}

where

\begin{equation*} q' = \frac{q}{p'}, \ \ r' = \frac{r}{p'}, \ \ x = b^{p'}, \text{and} \ \ y = c^{q'}. \end{equation*}

Note that $1/q' + 1/r' = 1$. Hence, by lemma 4.1, it follows that

\begin{equation*} x^{q'} = (1 + O(\delta^{1/2})) y^{r'}, \end{equation*}

which implies that

\begin{equation*} b^q = (1 + O(\delta^{1/2})) c^r. \end{equation*}

Similarly, one can also prove that $a^p = (1 + O(\delta^{1/2})) c^r$.

Lemma 4.3. Let $p,q,r \in (1,+\infty)$ be exponents with $1/p + 1/q + 1/r = 1$. Let $a_1,\cdots,a_n$, $b_1,\cdots,b_n$, $c_1,\cdots,c_n$ be non-negative reals such that

\begin{equation*} \sum_{i=1}^n a_i^p = \sum_{i=1}^n b_i^q = \sum_{i=1}^n c_i^r = 1. \end{equation*}

Suppose that

\begin{equation*} \sum_{i=1}^n a_ib_ic_i \geq 1 - \delta \end{equation*}

for some sufficiently small constant δ > 0. Then, we have

\begin{equation*} a_i^p = (1 + O(\delta^{1/4})) b_i^q = (1 + O(\delta^{1/4})) c_i^r \end{equation*}

for each i outside an exceptional set E satisfying

\begin{equation*} \sum_{i \in E} (a_i^p + b_i^q + c_i^r) \ll \delta^{1/2}. \end{equation*}

Proof. For each i, we have

\begin{equation*} a_ib_ic_i \leq \frac{a_i^p}{p} + \frac{b_i^q}{q} + \frac{c_i^r}{r}. \end{equation*}

Let $E \subset \{1,2,\cdots,n\}$ be the exceptional set of indices i such that

\begin{equation*} \frac{a_i^p}{p} + \frac{b_i^q}{q} + \frac{c_i^r}{r} \geq (1+\delta^{1/2}) a_ib_ic_i. \end{equation*}

Then,

\begin{equation*} \delta \geq \sum_{i=1}^n \left(\frac{a_i^p}{p} + \frac{b_i^q}{q} + \frac{c_i^r}{r} - a_ib_ic_i\right) \gg \delta^{1/2} \sum_{i \in E} \left(\frac{a_i^p}{p} + \frac{b_i^q}{q} + \frac{c_i^r}{r}\right), \end{equation*}

and hence, $\sum_{i \in E} (a_i^p + b_i^q + c_i^r) \ll \delta^{1/2}$. For $i \notin E$, lemma 4.2 implies that

\begin{equation*} a_i^p = (1 + O(\delta^{1/4})) b_i^q = (1 + O(\delta^{1/4})) c_i^r. \end{equation*}

This concludes the proof.

4.2. Near equality in Young’s inequality

In this section, all implied constants are allowed to depend on the exponents p, q, and r. Before proving the approximate inverse of Young’s inequality, we need the following standard result in additive combinatorics.

Lemma 4.4. Let G be an abelian group and let $X, Y \subset G$ be finite subsets with $|X| = |Y| = N$. Let $\varepsilon \in (0, 1/20)$ and let δ > 0 be sufficiently small in terms of ɛ. Let $M \subset X \times Y$ be a subset with $|M| \geq (1-\delta) N^2$. Suppose that the restricted sumset

\begin{equation*} X+_MY := \{x+y: (x,y) \in M\} \end{equation*}

has size at most $(1+\varepsilon)N$. Then, there exists a coset x + H of a subgroup $H \subset G$ such that $|X \setminus (x+H)| \leq \varepsilon N$ and $|(x+H) \setminus X| \leq 3\varepsilon N$.

Proof. By an almost-all version of the Balog–Szemeredi–Gowers theorem as in [Reference Shao13, theorem 1.1] (see also [Reference Shao and Xu14, theorem 1.1] for a version with $G = \mathbb{Z}$ and [Reference Campos, Coulson, Serra and Wötzel3, theorem 3.3] for an asymmetric version), one can find subsets $X' \subset X$ and $Y' \subset Y$ such that

\begin{equation*} |X'| \geq (1-\varepsilon)N, \ \ |Y'| \geq (1-\varepsilon)N, \ \ |X'+Y'| \leq |X+_MY| + \varepsilon N \leq (1+2\varepsilon)N. \end{equation*}

By Kneser’s theorem [Reference Kneser10] (see [Reference Tao and Vu15, theorem 5.5]), we have

\begin{equation*} |X'+Y'| \geq |X'| + |Y'| - |H|, \end{equation*}

where $H \subset G$ is the subgroup defined by

\begin{equation*} H = \{h \in G: X'+Y'+h = X'+Y'\}. \end{equation*}

It follows that $|H| \geq (1-4\varepsilon)N$ and hence $|X'+Y'| \lt 2|H|$. Since $X'+Y'$ is the union of cosets of H, it must be a single coset of H, and thus X ^ʹ is contained in a single coset x + H of H. Hence,

\begin{equation*} |X\setminus (x+H)| \leq |X\setminus X'| \leq \varepsilon N \end{equation*}

and

\begin{equation*} |(x+H) \setminus X| \leq |(x+H) \setminus X'| = |X'+Y'| - |X'| \leq 3\varepsilon N. \end{equation*}

Proposition 4.5. Let $p,q,r \in (1,+\infty)$ be exponents with $1/p + 1/q = 1 + 1/r$. Let $f,g: \mathbb{Z}\rightarrow\mathbb{C}$ be finitely supported functions such that $\|f\|_p = \|g\|_q = 1$. Suppose that

\begin{equation*} \|f*g\|_r \geq 1-\delta \end{equation*}

for some sufficiently small constant δ > 0. Then, there exists a singleton set $\{x_0\}$ for some $x_0 \in \mathbb{Z}$ such that

\begin{equation*} \|f - f(x_0)1_{\{x_0\}}\|_p^p \ll \delta^{1/8}. \end{equation*}

Proof. By replacing $f,g$ by $|f|, |g|$, we may assume that $f,g$ take non-negative real values. For every $x \in \mathbb{Z}$, we have

\begin{equation*} (f*g)(x) = \sum_{y\in\mathbb{Z}} f(x-y) g(y) = \sum_{y \in \mathbb{Z}} f(x-y)^{p/r} g(y)^{q/r} \cdot f(x-y)^{(r-p)/r} \cdot g(y)^{(r-q)/r}. \end{equation*}

By Hölder’s inequality, we have

(4.1)\begin{equation} (f*g)(x) \leq \left(\sum_{y \in \mathbb{Z}} f(x-y)^p g(y)^q \right)^{\frac{1}{r}} \left(\sum_{y \in \mathbb{Z}} f(x-y)^p\right)^{\frac{r-p}{pr}} \left(\sum_{y \in \mathbb{Z}} g(y)^q\right)^{\frac{r-q}{qr}}. \end{equation}

Since $\|f\|_p = \|g\|_q = 1$, it follows that

\begin{equation*} (f*g)(x)^r \leq \sum_{y \in \mathbb{Z}} f(x-y)^p g(y)^q. \end{equation*}

Let $E_1 \subset \mathbb{Z}$ be the exceptional set of $x \in \mathbb{Z}$ such that

\begin{equation*} (f*g)(x)^r \leq (1 - \delta^{1/2}) \sum_{y \in \mathbb{Z}} f(x-y)^p g(y)^q. \end{equation*}

From hypothesis, we have

\begin{align*} 1 - (1-\delta)^r & \geq \sum_{x \in \mathbb{Z}} \left(\sum_{y \in \mathbb{Z}} f(x-y)^p g(y)^q - (f*g)(x)^r\right) \\ & \geq \delta^{1/2} \sum_{x \in E_1} \sum_{y \in \mathbb{Z}} f(x-y)^p g(y)^q, \end{align*}

and hence,

(4.2)\begin{equation} \sum_{(x,y) \in E_1\times \mathbb{Z}} f(x-y)^p g(y)^q \ll \delta^{1/2}. \end{equation}

For each $x \notin E_1$, we have almost equality in (4.1), and hence, by lemma 4.3 applied to the three sequences

\begin{equation*} a_x(y) = \frac{f(x-y)^{p/r} g(y)^{q/r}}{h(x)^{1/r}}, \ \ b_x(y) = f(x-y)^{(r-p)/r}, \ \ c_x(y) = g(y)^{(r-q)/r}, \end{equation*}

where

\begin{equation*} h(x) = \sum_{z \in \mathbb{Z}} f(x-z)^p g(z)^q, \end{equation*}

we conclude that

(4.3)\begin{equation} \frac{f(x-y)^pg(y)^q}{h(x)} = (1 + O(\delta^{1/8})) f(x-y)^p = (1 + O(\delta^{1/8})) g(y)^q \end{equation}

for each y outside an exceptional set $E_2(x)$ satisfying

(4.4)\begin{equation} \sum_{y \in E_2(x)} f(x-y)^p g(y)^q \ll \delta^{1/4}h(x). \end{equation}

Define

\begin{equation*} E = (E_1 \times \mathbb{Z}) \cup \{(x, y) \in \mathbb{Z} \times \mathbb{Z}: x \notin E_1, y \in E_2(x)\}. \end{equation*}

Then, from (4.2) and (4.4), it follows that

\begin{equation*} \sum_{(x,y) \in E} f(x-y)^p g(y)^q \ll \delta^{1/2} + \delta^{1/4} \sum_{x \in \mathbb{Z}}h(x) \ll \delta^{1/4}, \end{equation*}

and (4.3) holds for every $(x,y) \notin E$.

Now make a change of variables and consider

\begin{equation*} E' = \{(x,y) \in \mathbb{Z} \times \mathbb{Z}: (x+y, y) \in E\}. \end{equation*}

Then,

(4.5)\begin{equation} \sum_{(x,y) \in E'} f(x)^p g(y)^q = \sum_{(x,y) \in E} f(x-y)^pg(y)^q \ll \delta^{1/4}, \end{equation}

and we have

(4.6)\begin{equation} \frac{f(x)^p g(y)^q}{h(x+y)} = (1 + O(\delta^{1/8}))f(x)^p = (1 + O(\delta^{1/8})) g(y)^q \end{equation}

for every $(x,y) \notin E'$. Let $X \subset \mathbb{Z}$ be the set of $x \in \mathbb{Z}$ such that

\begin{equation*} \sum_{y: (x,y) \in E'} g(y)^q \leq \delta^{1/8}. \end{equation*}

Then, from (4.5), it follows that

\begin{equation*} \delta^{1/4} \gg \sum_{x \notin X} f(x)^p \sum_{y: (x,y) \in E'} g(y)^q \geq \delta^{1/8} \sum_{x \notin X} f(x)^p, \end{equation*}

and hence,

(4.7)\begin{equation} \sum_{x\notin X} f(x)^p \ll \delta^{1/8}. \end{equation}

For every $x_1, x_2 \in X$, since

\begin{equation*} \sum_{y: (x_1,y) \in E'} g(y)^q + \sum_{y: (x_2,y) \in E'} g(y)^q \ll \delta^{1/8}, \end{equation*}

there exists $y \in \mathbb{Z}$ such that $(x_1,y) \notin E'$ and $(x_2,y)\notin E'$. By (4.6), we have

\begin{equation*} f(x_1)^p = (1 + O(\delta^{1/8})) g(y)^q, \ \ f(x_2)^p = (1 + O(\delta^{1/8})) g(y)^q. \end{equation*}

We conclude that there exists a constant $a \in \mathbb{R}$ such that

\begin{equation*} f(x) = (1 + O(\delta^{1/8})) a \end{equation*}

for every $x \in X$. Moreover, since

\begin{equation*} 1 = \sum_{x \in \mathbb{Z}} f(x)^p = \sum_{x \in X} f(x)^p + O(\delta^{1/8}) = (1 + O(\delta^{1/8})) a^p |X| + O(\delta^{1/8}), \end{equation*}

we have

\begin{equation*} |X| = (1 + O(\delta^{1/8})a^{-p}. \end{equation*}

By symmetry, we may also conclude the existence of a constant $b \in \mathbb{R}$ such that

\begin{equation*} g(y) = (1 + O(\delta^{1/8})) b \end{equation*}

for every $y \in Y$, where $Y \subset \mathbb{Z}$ is a subset satisfying

\begin{equation*} |Y| = (1 + O(\delta^{1/8})b^{-q}. \end{equation*}

We now return to using the first part of (4.6) for $(x,y) \in M := (X \times Y)\setminus E'$. First note from (4.5) that

\begin{equation*} \delta^{1/4} \gg \sum_{(x,y) \in (X\times Y) \cap E'} f(x)^p g(y)^q \gg a^p b^q \cdot |(X\times Y) \cap E'| \gg |X|^{-1}|Y|^{-1} \cdot |(X\times Y) \cap E'|, \end{equation*}

and hence,

\begin{equation*} |M| \geq (1 - O(\delta^{1/4})) |X||Y|. \end{equation*}

For $(x,y) \in M$, (4.6) implies that

\begin{equation*} \frac{a^pb^q}{h(x+y)} = (1 + O(\delta^{1/8}) a^p = (1 + O(\delta^{1/8})) b^q. \end{equation*}

In particular, since M is non-empty, we have $a^p = (1 + O(\delta^{1/8})) b^q$, and hence, $|X| = (1 + O(\delta^{1/8})) |Y|$. Moreover, for $s \in X+_MY$, we have

\begin{equation*} h(s) = (1 + O(\delta^{1/8})) a^p. \end{equation*}

Since $\sum_{s \in \mathbb{Z}}h(s) = 1$, we have

\begin{equation*} 1 \geq \sum_{s \in X+_MY} h(s) = (1 - O(\delta^{1/8}))a^p \cdot |X+_MY|, \end{equation*}

and hence,

\begin{equation*} |X+_MY| \leq (1 + O(\delta^{1/8})) |X|. \end{equation*}

We now apply lemma 4.4 with $\varepsilon = 1/100$ (say), after possibly shrinking one of $X, Y$ slightly so that $|X| = |Y|$, to conclude that there exists a coset $x_0 + H$ of a subgroup $H \subset \mathbb{Z}$ such that

\begin{equation*} |X \setminus (x_0+H)| \leq \frac{1}{10}|X|, \ \ |(x_0+H) \setminus X| \leq \frac{1}{10}|X|. \end{equation*}

The only finite subgroup of $\mathbb{Z}$ is $H = \{0\}$, and hence, it must be that $X = \{x_0\}$. The desired conclusions follow immediately from (4.7).

4.3. Proof of proposition 2.3

Let $f: \mathbb{Z} \rightarrow \mathbb{R}$ be a function that is supported on a set of size $n \geq 2$. By replacing f by $|f|$, we may assume that f takes non-negative real values. Let δ > 0 be a sufficiently small absolute constant and let

\begin{equation*} q = \frac{4}{3 - \log_n(1+\delta)}. \end{equation*}

First, consider the case when

\begin{equation*} \|\widehat{f}\|_4 \leq (1 - \delta) \|f\|_{4/3}. \end{equation*}

By Hölder’s inequality (see (2.2)), we have

\begin{equation*} \|f\|_{4/3} \leq n^{3/4-1/q} \|f\|_q = (1+\delta)^{1/4} \|f\|_q. \end{equation*}

It follows that

\begin{equation*} \|\widehat{f}\|_4 \leq (1-\delta) (1+\delta)^{1/4} \|f\|_q \leq \|f\|_q. \end{equation*}

Now suppose that

\begin{equation*} \|\widehat{f}\|_4 \geq (1 - \delta) \|f\|_{4/3}. \end{equation*}

By normalization, we may assume that $\|f\|_{4/3} = 1$, and thus, $\|f*f\|_2 = \|\widehat{f}\|_4^2 \geq 1-2\delta$. By proposition 4.5, there exists $x_0 \in \mathbb{Z}$ such that

(4.8)\begin{equation} \|f - f(x_0)1_{\{x_0\}}\|_{4/3} \ll \delta^{1/20}. \end{equation}

By translation, we may assume that $x_0=0$, and we may write f in the form $f = f(0)\delta_0 + g$, where δ ₀ is the Kronecker delta function and $g(0) = 0$. Let $x = f(0)$ and $y = \|g\|_{4/3}$. Since $\|f\|_{4/3}=1$, we have

\begin{equation*} x^{4/3} + y^{4/3} = 1. \end{equation*}

From (4.8), we have

\begin{equation*} y = O(\delta^{1/20}), \ \ x = 1 - O(\delta^{1/20}). \end{equation*}

In particular, we have $y/x \leq 0.01$. Since $f*f = x^2\delta_0 + 2xg + g*g$, we have

\begin{equation*} \|f*f\|_2 \leq x\|x\delta_0 + 2g\|_2 + \|g*g\|_2 = x \sqrt{\|x\delta_0\|_2^2 + \|2g\|_2^2} + \|g*g\|_2. \end{equation*}

Using the inequalities $\|g\|_2 \leq \|g\|_{4/3} = y$ and $\|g*g\|_2 \leq \|g\|_{4/3}^2 = y^2$, we obtain

\begin{equation*} \|f*f\|_2 \leq x\sqrt{x^2 + 4y^2} + y^2 = x^2 \sqrt{1 + \frac{4y^2}{x^2}} + y^2. \end{equation*}

Since $\sqrt{1+\lambda} \leq 1 + \lambda/2$ for $\lambda \geq 0$, it follows that

\begin{equation*} \|f*f\|_2 \leq x^2\left(1 + \frac{2y^2}{x^2}\right) + y^2 = x^2 + 3y^2. \end{equation*}

On the other hand, note that

\begin{equation*} \|f\|_q = (x^q + \|g\|_q^q)^{1/q}. \end{equation*}

Since g is supported on a set of size n, by Hölder’s inequality, we have

\begin{equation*} \|g\|_{4/3} \leq n^{3/4 - 1/q} \|g\|_q = (1+\delta)^{1/4} \|g\|_q. \end{equation*}

By choosing δ > 0 to be small enough, we have $\|g\|_q^q \geq 0.9 \|g\|_{4/3}^q = 0.9y^q$, and hence,

\begin{equation*} \|f\|_q^2 \geq (x^q + 0.9y^q)^{2/q} = x^2\left(1 + \frac{0.9y^q}{x^q}\right)^{2/q}. \end{equation*}

Since $4/3 \leq q \leq 3/2$, we have $(1+\lambda)^{2/q} \geq 1+\lambda \geq 1+4\lambda^{2/q}$ for $0 \leq \lambda \leq 1/64$. Hence,

\begin{equation*} \|f\|_q^2 \geq x^2\left(1 + 4 \cdot 0.9^{2/q} \cdot \frac{y^2}{x^2}\right) \geq x^2 + 3y^2. \end{equation*}

It follows that $\|f*f\|_2 \leq \|f\|_q^2$, as desired.