[proofplan] We prove both assertions by the same Fourier-analytic argument. First we build a normalized convolution whose positivity at a point is equivalent to the existence of a representation of that point as a sum of two elements of $A$ minus two elements of $A$. Fourier inversion expresses this convolution as a character expansion with non-negative fourth-power Fourier weights. We then separate the large Fourier spectrum from the small Fourier spectrum: in $\mathbb{F}_p^n$ we annihilate the large spectrum on a common subspace, while in $\mathbb{Z}/N\mathbb{Z}$ we make the large spectrum close to $1$ on a Bohr set, and [Parseval's identity](/theorems/434) controls the remaining error. [/proofplan]

[step:Fix the Fourier normalization and identify the convolution that detects $2A - 2A$]Let $\widehat{G}$ denote the dual group of all characters $\gamma: G \to \mathbb{C}$, and normalize counting measure on $G$ by \begin{align*} \mathbb{E}_{x \in G} f(x) := \frac{1}{|G|}\sum_{x \in G} f(x) \end{align*} for every function $f: G \to \mathbb{C}$. For functions $f,g: G \to \mathbb{C}$, define their normalized convolution \begin{align*} f * g: G &\to \mathbb{C} \\ x &\mapsto \mathbb{E}_{y \in G} f(y)g(x-y). \end{align*} For a function $f: G \to \mathbb{C}$ and a character $\gamma \in \widehat{G}$, define the Fourier coefficient \begin{align*} \widehat{f}(\gamma) := \mathbb{E}_{x \in G} f(x)\overline{\gamma(x)}. \end{align*} Let $\mathbb{1}_A: G \to \{0,1\}$ be the indicator function of $A$, and let $-A := \{-a : a \in A\}$. Define \begin{align*} F: G &\to [0,\infty) \\ x &\mapsto (\mathbb{1}_A * \mathbb{1}_A * \mathbb{1}_{-A} * \mathbb{1}_{-A})(x). \end{align*} By the definition of normalized convolution, \begin{align*} F(x) = \frac{1}{|G|^3} \#\{(a_1,a_2,a_3,a_4) \in A^4 : a_1+a_2-a_3-a_4=x\}. \end{align*} Therefore $F(x)>0$ implies $x \in 2A-2A$.[/step]

[guided]The function $F$ is chosen because it is exactly the representation-counting function for $2A-2A$, up to the harmless normalization factor $|G|^{-3}$. More explicitly, each convolution variable records one choice from $A$, $A$, $-A$, and $-A$. Thus \begin{align*} F(x) = \frac{1}{|G|^3} \#\{(a_1,a_2,a_3,a_4) \in A^4 : a_1+a_2-a_3-a_4=x\}. \end{align*} The factor $|G|^{-3}$ is positive, so it does not affect whether $F(x)$ vanishes. Hence, to prove that a set $S \subseteq G$ lies in $2A-2A$, it is enough to prove $F(x)>0$ for every $x \in S$.[/guided]

[step:Express the detecting convolution by Fourier inversion]Since $G$ is finite and all functions under consideration are complex-valued maps on $G$, the finite abelian group [Fourier inversion formula](/theorems/528), Parseval's identity, and the convolution identity for normalized convolution all apply to these functions. For every $\gamma \in \widehat{G}$, \begin{align*} \widehat{\mathbb{1}_{-A}}(\gamma) = \mathbb{E}_{x \in G}\mathbb{1}_{-A}(x)\overline{\gamma(x)} = \mathbb{E}_{a \in G}\mathbb{1}_A(a)\overline{\gamma(-a)} = \mathbb{E}_{a \in G}\mathbb{1}_A(a)\gamma(a) = \overline{\widehat{\mathbb{1}_A}(\gamma)}. \end{align*} The convolution identity for the normalized convolution convention states that, for functions $f,g: G \to \mathbb{C}$, \begin{align*} \widehat{f*g}(\gamma)=\widehat{f}(\gamma)\widehat{g}(\gamma). \end{align*} Applying this identity three times to the fourfold convolution defining $F$ gives \begin{align*} \widehat{F}(\gamma) = |\widehat{\mathbb{1}_A}(\gamma)|^4. \end{align*} Fourier inversion on the finite abelian group $G$ gives, for every $x \in G$, \begin{align*} F(x) = \sum_{\gamma \in \widehat{G}} |\widehat{\mathbb{1}_A}(\gamma)|^4\gamma(x). \end{align*} The identity character $\gamma_0: G \to \mathbb{C}$ is defined by $\gamma_0(x)=1$ for every $x \in G$, and \begin{align*} \widehat{\mathbb{1}_A}(\gamma_0)=\mathbb{E}_{x \in G}\mathbb{1}_A(x)=\alpha. \end{align*} Thus \begin{align*} F(0) = \sum_{\gamma \in \widehat{G}} |\widehat{\mathbb{1}_A}(\gamma)|^4 \geq \alpha^4. \end{align*}[/step]

[guided]We now translate the representation-counting function into Fourier language. The normalized average on $G$ is the map assigning to each function $f: G \to \mathbb{C}$ the scalar \begin{align*} \mathbb{E}_{x \in G} f(x) := \frac{1}{|G|}\sum_{x \in G} f(x). \end{align*} For functions $f,g: G \to \mathbb{C}$, the normalized convolution is the function \begin{align*} f*g: G &\to \mathbb{C} \\ x &\mapsto \mathbb{E}_{y \in G} f(y)g(x-y), \end{align*} and for a character $\gamma \in \widehat{G}$ the Fourier coefficient of $f$ at $\gamma$ is \begin{align*} \widehat{f}(\gamma) := \mathbb{E}_{x \in G} f(x)\overline{\gamma(x)}. \end{align*} These definitions are the normalization under which the convolution identity takes the form \begin{align*} \widehat{f*g}(\gamma)=\widehat{f}(\gamma)\widehat{g}(\gamma). \end{align*} Because $G$ is finite abelian, the finite abelian group Fourier inversion formula, Parseval's identity, and this convolution identity apply to all complex-valued functions on $G$. First compute the [Fourier transform](/page/Fourier%20Transform) of the reflected indicator. For every $\gamma \in \widehat{G}$, \begin{align*} \widehat{\mathbb{1}_{-A}}(\gamma) &= \mathbb{E}_{x \in G}\mathbb{1}_{-A}(x)\overline{\gamma(x)} \\ &= \mathbb{E}_{a \in G}\mathbb{1}_A(a)\overline{\gamma(-a)} \\ &= \mathbb{E}_{a \in G}\mathbb{1}_A(a)\gamma(a) \\ &= \overline{\widehat{\mathbb{1}_A}(\gamma)}. \end{align*} The second equality is the substitution $x=-a$, which is a bijection of the finite group $G$ and therefore preserves the normalized average. The third equality uses the character identity $\gamma(-a)=\gamma(a)^{-1}=\overline{\gamma(a)}$, so $\overline{\gamma(-a)}=\gamma(a)$. Now apply the normalized convolution identity three times to \begin{align*} F=\mathbb{1}_A*\mathbb{1}_A*\mathbb{1}_{-A}*\mathbb{1}_{-A}. \end{align*} For every $\gamma \in \widehat{G}$, \begin{align*} \widehat{F}(\gamma) &= \widehat{\mathbb{1}_A}(\gamma)^2\widehat{\mathbb{1}_{-A}}(\gamma)^2 \\ &= \widehat{\mathbb{1}_A}(\gamma)^2\overline{\widehat{\mathbb{1}_A}(\gamma)}^2 \\ &= |\widehat{\mathbb{1}_A}(\gamma)|^4. \end{align*} Fourier inversion on the finite abelian group $G$ then gives, for every $x \in G$, \begin{align*} F(x) = \sum_{\gamma \in \widehat{G}} \widehat{F}(\gamma)\gamma(x) = \sum_{\gamma \in \widehat{G}} |\widehat{\mathbb{1}_A}(\gamma)|^4\gamma(x). \end{align*} This formula is the bridge between additive structure and the Fourier spectrum. Let $\gamma_0: G \to \mathbb{C}$ denote the identity character, defined by $\gamma_0(x)=1$ for every $x \in G$. Since $A$ has density $\alpha$, we have \begin{align*} \widehat{\mathbb{1}_A}(\gamma_0) = \mathbb{E}_{x \in G}\mathbb{1}_A(x) = \alpha. \end{align*} Evaluating the inversion formula at $0 \in G$ gives \begin{align*} F(0) = \sum_{\gamma \in \widehat{G}} |\widehat{\mathbb{1}_A}(\gamma)|^4 \geq |\widehat{\mathbb{1}_A}(\gamma_0)|^4 = \alpha^4. \end{align*}[/guided]

[step:Separate the large Fourier spectrum from the small Fourier spectrum]Define the threshold \begin{align*} \delta := \frac{\alpha^{3/2}}{4}, \end{align*} and define the large spectrum \begin{align*} \Gamma := \{\gamma \in \widehat{G} : |\widehat{\mathbb{1}_A}(\gamma)| \geq \delta\}. \end{align*} Parseval's identity gives \begin{align*} \sum_{\gamma \in \widehat{G}} |\widehat{\mathbb{1}_A}(\gamma)|^2 = \mathbb{E}_{x \in G} |\mathbb{1}_A(x)|^2 = \alpha. \end{align*} Since each $\gamma \in \Gamma$ contributes at least $\delta^2$ to the left-hand side, \begin{align*} |\Gamma|\delta^2 \leq \alpha. \end{align*} Therefore \begin{align*} |\Gamma| \leq \frac{\alpha}{\delta^2}=16\alpha^{-2}. \end{align*} For the small spectrum $\widehat{G}\setminus\Gamma$, we have $|\widehat{\mathbb{1}_A}(\gamma)|<\delta$, and hence \begin{align*} \sum_{\gamma \notin \Gamma} |\widehat{\mathbb{1}_A}(\gamma)|^4 \leq \delta^2 \sum_{\gamma \notin \Gamma} |\widehat{\mathbb{1}_A}(\gamma)|^2 \leq \delta^2\alpha = \frac{\alpha^4}{16}. \end{align*}[/step]

[guided]The large spectrum is the set of characters whose Fourier coefficients are too large to be treated as an error term. We choose \begin{align*} \delta := \frac{\alpha^{3/2}}{4} \end{align*} because then the total fourth-power contribution of all coefficients smaller than $\delta$ is bounded by a fixed small multiple of $\alpha^4$, which is the size of the identity-character contribution. Parseval's identity says \begin{align*} \sum_{\gamma \in \widehat{G}} |\widehat{\mathbb{1}_A}(\gamma)|^2 = \mathbb{E}_{x \in G} |\mathbb{1}_A(x)|^2 = \alpha. \end{align*} Since every character in $\Gamma$ has squared Fourier coefficient at least $\delta^2$, the number of such characters satisfies \begin{align*} |\Gamma|\delta^2 \leq \alpha. \end{align*} Substituting $\delta^2=\alpha^3/16$ gives \begin{align*} |\Gamma| \leq 16\alpha^{-2}. \end{align*} Now consider the complementary set $\widehat{G}\setminus\Gamma$. For every $\gamma \notin \Gamma$, \begin{align*} |\widehat{\mathbb{1}_A}(\gamma)|^2 < \delta^2. \end{align*} Multiplying this pointwise bound by $|\widehat{\mathbb{1}_A}(\gamma)|^2$ and summing gives \begin{align*} \sum_{\gamma \notin \Gamma} |\widehat{\mathbb{1}_A}(\gamma)|^4 \leq \delta^2 \sum_{\gamma \notin \Gamma} |\widehat{\mathbb{1}_A}(\gamma)|^2 \leq \delta^2\alpha = \frac{\alpha^4}{16}. \end{align*} This is the error estimate that will leave the positive main term intact.[/guided]

[step:Force the large spectrum to vanish in $\mathbb{F}_p^n$ and obtain a subspace]Assume now that $G=\mathbb{F}_p^n$. Define \begin{align*} V := \{x \in \mathbb{F}_p^n : \gamma(x)=1 \text{ for every } \gamma \in \Gamma\}. \end{align*} Each set $\ker \gamma := \{x \in \mathbb{F}_p^n : \gamma(x)=1\}$ is an $\mathbb{F}_p$-linear subspace, so $V$ is an $\mathbb{F}_p$-linear subspace. Since $V$ is the intersection of at most $|\Gamma|$ kernels of characters, \begin{align*} \operatorname{codim}_{\mathbb{F}_p} V \leq |\Gamma| \leq 16\alpha^{-2}. \end{align*} Let $x \in V$. Then $\gamma(x)=1$ for every $\gamma \in \Gamma$, and Fourier inversion gives \begin{align*} F(x) &= \sum_{\gamma \in \Gamma} |\widehat{\mathbb{1}_A}(\gamma)|^4 + \sum_{\gamma \notin \Gamma} |\widehat{\mathbb{1}_A}(\gamma)|^4\gamma(x) \\ &= F(0) + \sum_{\gamma \notin \Gamma} |\widehat{\mathbb{1}_A}(\gamma)|^4(\gamma(x)-1). \end{align*} Define the complex number \begin{align*} z_x := \sum_{\gamma \notin \Gamma} |\widehat{\mathbb{1}_A}(\gamma)|^4(\gamma(x)-1). \end{align*} The displayed identity says $F(x)=F(0)+z_x$. Since $F(x)$ and $F(0)$ are [real numbers](/page/Real%20Numbers), taking real parts and then applying the triangle inequality gives $F(x)=F(0)+\operatorname{Re} z_x \geq F(0)-|z_x|$. Using $|\gamma(x)-1|\leq 2$ for every character value $\gamma(x)$ on the unit circle, \begin{align*} F(x) &\geq F(0) - 2\sum_{\gamma \notin \Gamma} |\widehat{\mathbb{1}_A}(\gamma)|^4 \\ &\geq \alpha^4 - 2\cdot \frac{\alpha^4}{16} = \frac{7\alpha^4}{8} >0. \end{align*} Therefore $x \in 2A-2A$. Since $x \in V$ was arbitrary, $V \subseteq 2A-2A$.[/step]

[guided]In the vector-space case, we can make every large Fourier character contribute exactly as it does at $0$. The common kernel \begin{align*} V := \{x \in \mathbb{F}_p^n : \gamma(x)=1 \text{ for every } \gamma \in \Gamma\} \end{align*} is a subspace because each character kernel is a subspace and intersections of subspaces are subspaces. Intersecting $m$ kernels can increase codimension by at most $m$, so \begin{align*} \operatorname{codim}_{\mathbb{F}_p} V \leq |\Gamma| \leq 16\alpha^{-2}. \end{align*} Now fix $x \in V$. For every large-spectrum character $\gamma \in \Gamma$, we have $\gamma(x)=1$. Thus the large-spectrum part of $F(x)$ is exactly the same as the large-spectrum part of $F(0)$. Only the small spectrum can change: \begin{align*} F(x) = F(0) + \sum_{\gamma \notin \Gamma} |\widehat{\mathbb{1}_A}(\gamma)|^4(\gamma(x)-1). \end{align*} Define the complex error term \begin{align*} z_x := \sum_{\gamma \notin \Gamma} |\widehat{\mathbb{1}_A}(\gamma)|^4(\gamma(x)-1). \end{align*} Then $F(x)=F(0)+z_x$. The values $F(x)$ and $F(0)$ are real because $F$ is a nonnegative real-valued convolution of indicator functions, so \begin{align*} F(x)=F(0)+\operatorname{Re} z_x \geq F(0)-|z_x|. \end{align*} Since character values lie on the unit circle, $|\gamma(x)-1|\leq 2$. Therefore \begin{align*} F(x) \geq F(0) - 2\sum_{\gamma \notin \Gamma} |\widehat{\mathbb{1}_A}(\gamma)|^4. \end{align*} The preceding small-spectrum estimate and the lower bound $F(0)\geq\alpha^4$ give \begin{align*} F(x) \geq \alpha^4 - 2\cdot \frac{\alpha^4}{16} = \frac{7\alpha^4}{8} >0. \end{align*} Since positivity of $F(x)$ means that $x$ has a representation $x=a_1+a_2-a_3-a_4$ with all $a_i \in A$, every $x \in V$ lies in $2A-2A$.[/guided]

[step:Force the large spectrum to be close to one in $\mathbb{Z}/N\mathbb{Z}$ and obtain a Bohr set]Assume now that $G=\mathbb{Z}/N\mathbb{Z}$. For each $k \in \mathbb{Z}/N\mathbb{Z}$, define the character \begin{align*} \gamma_k: \mathbb{Z}/N\mathbb{Z} &\to \mathbb{C} \\ x &\mapsto \exp\left(\frac{2\pi i kx}{N}\right), \end{align*} where representatives of $k$ and $x$ in $\{0,1,\dots,N-1\}$ are used in the exponent. Let \begin{align*} \Lambda := \{k \in \mathbb{Z}/N\mathbb{Z} : \gamma_k \in \Gamma\}. \end{align*} Then $|\Lambda|=|\Gamma| \leq 16\alpha^{-2}$. Define the Bohr set \begin{align*} B(\Lambda,\rho) := \left\{x \in \mathbb{Z}/N\mathbb{Z} : \left\|\frac{kx}{N}\right\|_{\mathbb{R}/\mathbb{Z}} \leq \rho \text{ for every } k \in \Lambda \right\}, \end{align*} with $\rho := 1/(8\pi)$. If $x \in B(\Lambda,\rho)$ and $\gamma_k \in \Gamma$, then \begin{align*} |\gamma_k(x)-1| = \left|\exp\left(2\pi i \frac{kx}{N}\right)-1\right| \leq 2\pi \left\|\frac{kx}{N}\right\|_{\mathbb{R}/\mathbb{Z}} \leq \frac{1}{4}. \end{align*} Therefore, for $x \in B(\Lambda,\rho)$, define the complex number \begin{align*} w_x := \sum_{\gamma \in \widehat{G}} |\widehat{\mathbb{1}_A}(\gamma)|^4(\gamma(x)-1). \end{align*} Fourier inversion gives $F(x)=F(0)+w_x$. Since $F(x)$ and $F(0)$ are real, $F(x)=F(0)+\operatorname{Re}w_x \geq F(0)-|w_x|$. Splitting the sum defining $w_x$ over $\Gamma$ and $\widehat{G}\setminus\Gamma$, and applying the triangle inequality with the bounds above, gives \begin{align*} F(x) &\geq F(0) - \frac{1}{4}\sum_{\gamma \in \Gamma}|\widehat{\mathbb{1}_A}(\gamma)|^4 - 2\sum_{\gamma \notin \Gamma}|\widehat{\mathbb{1}_A}(\gamma)|^4. \end{align*} Since \begin{align*} \sum_{\gamma \in \Gamma}|\widehat{\mathbb{1}_A}(\gamma)|^4 \leq F(0), \end{align*} we obtain \begin{align*} F(x) &\geq \frac{3}{4}F(0) - 2\cdot \frac{\alpha^4}{16} \\ &\geq \frac{3\alpha^4}{4} - \frac{\alpha^4}{8} = \frac{5\alpha^4}{8} >0. \end{align*} Thus every $x \in B(\Lambda,\rho)$ belongs to $2A-2A$. Hence \begin{align*} B(\Lambda,1/(8\pi)) \subseteq 2A-2A, \end{align*} with $|\Lambda|\leq 16\alpha^{-2}$.[/step]

[guided]In the cyclic case we cannot usually force a non-identity character to be exactly $1$ on a large subgroup, because $\mathbb{Z}/N\mathbb{Z}$ may have too few subgroups. Instead, we force the large characters to be close to $1$ by taking a Bohr set. The characters of $\mathbb{Z}/N\mathbb{Z}$ are \begin{align*} \gamma_k: \mathbb{Z}/N\mathbb{Z} &\to \mathbb{C} \\ x &\mapsto \exp\left(\frac{2\pi i kx}{N}\right), \end{align*} with $k \in \mathbb{Z}/N\mathbb{Z}$. Let $\Lambda$ be the set of frequencies whose characters lie in the large spectrum: \begin{align*} \Lambda := \{k \in \mathbb{Z}/N\mathbb{Z} : \gamma_k \in \Gamma\}. \end{align*} Then the rank of the Bohr set is bounded by the large-spectrum estimate: \begin{align*} |\Lambda|=|\Gamma| \leq 16\alpha^{-2}. \end{align*} Choose \begin{align*} \rho := \frac{1}{8\pi}. \end{align*} If $x \in B(\Lambda,\rho)$, then for each $k \in \Lambda$, \begin{align*} \left\|\frac{kx}{N}\right\|_{\mathbb{R}/\mathbb{Z}} \leq \rho. \end{align*} The elementary estimate $|e^{2\pi i t}-1|\leq 2\pi\|t\|_{\mathbb{R}/\mathbb{Z}}$ gives \begin{align*} |\gamma_k(x)-1| \leq 2\pi\rho = \frac{1}{4}. \end{align*} Thus every large-spectrum character changes its contribution by at most one quarter of its Fourier weight. Using Fourier inversion and comparing $F(x)$ with $F(0)$, define the complex error term \begin{align*} w_x := \sum_{\gamma \in \widehat{G}} |\widehat{\mathbb{1}_A}(\gamma)|^4(\gamma(x)-1). \end{align*} Then \begin{align*} F(x)=F(0)+w_x. \end{align*} The quantities $F(x)$ and $F(0)$ are real because $F$ is a nonnegative real-valued convolution of indicator functions. Hence \begin{align*} F(x)=F(0)+\operatorname{Re}w_x \geq F(0)-|w_x|. \end{align*} For $\gamma \in \Gamma$ the change is bounded by $\frac14|\widehat{\mathbb{1}_A}(\gamma)|^4$, while for $\gamma \notin \Gamma$ the unit-circle bound gives $|\gamma(x)-1|\leq 2$. Splitting $w_x$ over $\Gamma$ and $\widehat{G}\setminus\Gamma$ and applying the triangle inequality gives \begin{align*} F(x) \geq F(0) - \frac{1}{4}\sum_{\gamma \in \Gamma}|\widehat{\mathbb{1}_A}(\gamma)|^4 - 2\sum_{\gamma \notin \Gamma}|\widehat{\mathbb{1}_A}(\gamma)|^4. \end{align*} The first sum is at most $F(0)$, and the second sum is at most $\alpha^4/16$. Therefore \begin{align*} F(x) \geq \frac{3}{4}F(0) - 2\cdot\frac{\alpha^4}{16} \geq \frac{3\alpha^4}{4} - \frac{\alpha^4}{8} = \frac{5\alpha^4}{8} >0. \end{align*} Since $F(x)>0$ implies $x \in 2A-2A$, the entire Bohr set $B(\Lambda,1/(8\pi))$ is contained in $2A-2A$.[/guided]