[proofplan] We use Fourier analysis on the finite abelian group $G=\mathbb{F}_p^n$. The normalized fourfold convolution $\mathbb{1}_A * \mathbb{1}_A * \mathbb{1}_{-A} * \mathbb{1}_{-A}$ is positive exactly on elements of $2A-2A$, so it suffices to find a large subspace on which this convolution is positive. Fourier inversion writes this convolution as a positive main term $\alpha^4$ plus nonzero frequency terms. We define the large spectrum of $A$, take its common annihilator, bound its codimension by Parseval, and show that the remaining small spectrum cannot cancel the main term. [/proofplan] [step:Define the finite Fourier transform and record the identities used] Let $G := \mathbb{F}_p^n$, and write $N := |G| = p^n$. Fix the primitive $p$-th root of unity $\omega := e^{2\pi i/p}$. For each $\xi \in G$, define the character \begin{align*} \chi_\xi: G &\to \mathbb{C} \\ x &\mapsto \omega^{\xi \cdot x}, \end{align*} where $\xi \cdot x := \sum_{i=1}^n \xi_i x_i \in \mathbb{F}_p$ is represented by its standard element of $\{0,1,\dots,p-1\}$ in the exponent. For a function $f: G \to \mathbb{C}$, define its normalized [Fourier transform](/page/Fourier%20Transform) \begin{align*} \widehat f: G &\to \mathbb{C} \\ \xi &\mapsto \frac{1}{N}\sum_{x \in G} f(x)\overline{\chi_\xi(x)}. \end{align*} For functions $f,g: G \to \mathbb{C}$, define their normalized convolution \begin{align*} f*g: G &\to \mathbb{C} \\ x &\mapsto \frac{1}{N}\sum_{y \in G} f(y)g(x-y). \end{align*} The characters satisfy the orthogonality relation \begin{align*} \frac{1}{N}\sum_{x \in G} \chi_\xi(x)\overline{\chi_\eta(x)} = \begin{cases} 1, & \xi=\eta,\\ 0, & \xi\neq \eta. \end{cases} \end{align*} Indeed, if $\xi=\eta$, every summand is $1$. If $\xi\neq\eta$, choose $v \in G$ with $(\xi-\eta)\cdot v\neq 0$; translation by $v$ multiplies the sum by the nontrivial scalar $\chi_{\xi-\eta}(v)\neq 1$, so the sum must be $0$. Therefore Fourier inversion and [Parseval's identity](/theorems/434) hold: \begin{align*} f(x) &= \sum_{\xi \in G} \widehat f(\xi)\chi_\xi(x),\\ \frac{1}{N}\sum_{x \in G}|f(x)|^2 &= \sum_{\xi \in G}|\widehat f(\xi)|^2. \end{align*} Also, \begin{align*} \widehat{f*g}(\xi)=\widehat f(\xi)\widehat g(\xi) \end{align*} for every $\xi \in G$, obtained by substituting the definition of convolution and changing variables $z=x-y$ in the finite sum. [guided] We first set up the Fourier normalization because the constants in the argument depend on it. The group is the finite [vector space](/page/Vector%20Space) $G=\mathbb{F}_p^n$, with $N=|G|=p^n$. For each frequency $\xi\in G$, the associated character is the map \begin{align*} \chi_\xi: G &\to \mathbb{C} \\ x &\mapsto \omega^{\xi \cdot x}, \end{align*} where $\omega=e^{2\pi i/p}$ and $\xi\cdot x=\sum_{i=1}^n \xi_i x_i$ is computed in $\mathbb{F}_p$. For a function $f:G\to\mathbb{C}$, define \begin{align*} \widehat f: G &\to \mathbb{C} \\ \xi &\mapsto \frac{1}{N}\sum_{x \in G} f(x)\overline{\chi_\xi(x)}. \end{align*} The normalization by $1/N$ is chosen so that densities appear directly as Fourier coefficients at frequency $0$. For example, if $f=\mathbb{1}_A$, then \begin{align*} \widehat{\mathbb{1}_A}(0)=\frac{1}{N}\sum_{x\in G}\mathbb{1}_A(x)=\frac{|A|}{|G|}=\alpha. \end{align*} For functions $f,g:G\to\mathbb{C}$, define normalized convolution by \begin{align*} f*g: G &\to \mathbb{C} \\ x &\mapsto \frac{1}{N}\sum_{y \in G} f(y)g(x-y). \end{align*} The key algebraic fact is character orthogonality: \begin{align*} \frac{1}{N}\sum_{x \in G} \chi_\xi(x)\overline{\chi_\eta(x)} = \begin{cases} 1, & \xi=\eta,\\ 0, & \xi\neq \eta. \end{cases} \end{align*} When $\xi=\eta$, the summand is identically $1$. When $\xi\neq\eta$, the character $\chi_{\xi-\eta}$ is nontrivial. Choose $v\in G$ such that $(\xi-\eta)\cdot v\neq 0$. If \begin{align*} S:=\sum_{x\in G}\chi_{\xi-\eta}(x), \end{align*} then translation by $v$ gives \begin{align*} S=\sum_{x\in G}\chi_{\xi-\eta}(x+v) =\chi_{\xi-\eta}(v)\sum_{x\in G}\chi_{\xi-\eta}(x) =\chi_{\xi-\eta}(v)S. \end{align*} Since $\chi_{\xi-\eta}(v)\neq 1$, this forces $S=0$. This orthogonality gives Fourier inversion: \begin{align*} f(x)=\sum_{\xi\in G}\widehat f(\xi)\chi_\xi(x), \end{align*} and Parseval's identity: \begin{align*} \frac{1}{N}\sum_{x\in G}|f(x)|^2=\sum_{\xi\in G}|\widehat f(\xi)|^2. \end{align*} Finally, the normalized convolution was chosen so that Fourier transform turns convolution into multiplication: \begin{align*} \widehat{f*g}(\xi)=\widehat f(\xi)\widehat g(\xi). \end{align*} This follows by inserting the definition of convolution and using the finite change of variables $z=x-y$. [/guided] [/step] [step:Express the fourfold convolution by its Fourier expansion] Let \begin{align*} f: G &\to \{0,1\}\\ x &\mapsto \mathbb{1}_A(x) \end{align*} be the indicator function of $A$, and define \begin{align*} \tilde f: G &\to \{0,1\}\\ x &\mapsto \mathbb{1}_{-A}(x). \end{align*} Define the fourfold convolution \begin{align*} F: G &\to [0,\infty)\\ x &\mapsto (f*f*\tilde f*\tilde f)(x). \end{align*} Since $\tilde f(x)=f(-x)$, we have \begin{align*} \widehat{\tilde f}(\xi) &= \frac{1}{N}\sum_{x\in G}f(-x)\overline{\chi_\xi(x)} = \frac{1}{N}\sum_{y\in G}f(y)\overline{\chi_\xi(-y)} = \frac{1}{N}\sum_{y\in G}f(y)\chi_\xi(y) = \overline{\widehat f(\xi)}. \end{align*} Thus, by the convolution identity, \begin{align*} \widehat F(\xi)=|\widehat f(\xi)|^4. \end{align*} Fourier inversion gives, for every $x\in G$, \begin{align*} F(x)=\sum_{\xi\in G}|\widehat f(\xi)|^4\chi_\xi(x). \end{align*} Since $\widehat f(0)=\alpha$, this becomes \begin{align*} F(x)=\alpha^4+\sum_{\xi\in G\setminus\{0\}}|\widehat f(\xi)|^4\chi_\xi(x). \end{align*} [guided] The object whose positivity detects membership in $2A-2A$ is the fourfold convolution. Let \begin{align*} f: G &\to \{0,1\}\\ x &\mapsto \mathbb{1}_A(x) \end{align*} and let \begin{align*} \tilde f: G &\to \{0,1\}\\ x &\mapsto \mathbb{1}_{-A}(x). \end{align*} Define \begin{align*} F: G &\to [0,\infty)\\ x &\mapsto (f*f*\tilde f*\tilde f)(x). \end{align*} We compute the Fourier transform of $\tilde f$. Since $\tilde f(x)=f(-x)$, the finite change of variables $y=-x$ gives \begin{align*} \widehat{\tilde f}(\xi) &= \frac{1}{N}\sum_{x\in G}f(-x)\overline{\chi_\xi(x)}\\ &= \frac{1}{N}\sum_{y\in G}f(y)\overline{\chi_\xi(-y)}\\ &= \frac{1}{N}\sum_{y\in G}f(y)\chi_\xi(y)\\ &= \overline{\widehat f(\xi)}. \end{align*} Therefore the Fourier transform of the fourfold convolution is \begin{align*} \widehat F(\xi) = \widehat f(\xi)^2\widehat{\tilde f}(\xi)^2 = |\widehat f(\xi)|^4. \end{align*} Fourier inversion then gives \begin{align*} F(x)=\sum_{\xi\in G}|\widehat f(\xi)|^4\chi_\xi(x). \end{align*} The frequency $\xi=0$ contributes the main term. Since $\chi_0(x)=1$ and \begin{align*} \widehat f(0)=\frac{1}{N}\sum_{x\in G}\mathbb{1}_A(x)=\alpha, \end{align*} we obtain \begin{align*} F(x)=\alpha^4+\sum_{\xi\in G\setminus\{0\}}|\widehat f(\xi)|^4\chi_\xi(x). \end{align*} This is the analytic form of the argument: the main term is positive, and the rest must be controlled. [/guided] [/step] [step:Relate positivity of the convolution to membership in $2A-2A$] For every $x\in G$, expanding the convolution gives \begin{align*} F(x) = \frac{1}{N^3} \left| \left\{ (a_1,a_2,a_3,a_4)\in A^4: a_1+a_2-a_3-a_4=x \right\} \right|. \end{align*} Indeed, the three convolution sums choose $a_1,a_2\in A$ and $b_1,b_2\in -A$, with $a_1+a_2+b_1+b_2=x$; writing $b_1=-a_3$ and $b_2=-a_4$ gives the displayed count. Hence $F(x)>0$ implies $x\in 2A-2A$. [guided] We now justify why positivity of $F$ gives membership in the sum-difference set. By the definition of normalized convolution, expanding $F=f*f*\tilde f*\tilde f$ gives \begin{align*} F(x) &= \frac{1}{N^3} \sum_{u\in G}\sum_{v\in G}\sum_{w\in G} f(u)f(v-u)\tilde f(w-v)\tilde f(x-w). \end{align*} Each nonzero summand has value $1$ and corresponds exactly to choices $a_1,a_2\in A$ and $b_1,b_2\in -A$ satisfying \begin{align*} a_1+a_2+b_1+b_2=x. \end{align*} Writing $b_1=-a_3$ and $b_2=-a_4$ with $a_3,a_4\in A$, this condition becomes \begin{align*} a_1+a_2-a_3-a_4=x. \end{align*} Therefore \begin{align*} F(x) = \frac{1}{N^3} \left| \left\{ (a_1,a_2,a_3,a_4)\in A^4: a_1+a_2-a_3-a_4=x \right\} \right|. \end{align*} The factor $N^{-3}$ is positive, so $F(x)>0$ is equivalent to the existence of at least one quadruple $(a_1,a_2,a_3,a_4)\in A^4$ with $x=a_1+a_2-a_3-a_4$. This is precisely the assertion that $x\in 2A-2A$. [/guided] [/step] [step:Build the annihilator of the large spectrum and bound its codimension] Define the large spectrum \begin{align*} \Gamma := \left\{\xi\in G\setminus\{0\}: |\widehat f(\xi)| \geq \frac{\alpha^{3/2}}{\sqrt{2}}\right\}. \end{align*} By Parseval's identity applied to $f=\mathbb{1}_A$, \begin{align*} \sum_{\xi\in G}|\widehat f(\xi)|^2 = \frac{1}{N}\sum_{x\in G}|f(x)|^2 = \alpha. \end{align*} Therefore \begin{align*} |\Gamma|\cdot \frac{\alpha^3}{2} \leq \sum_{\xi\in\Gamma}|\widehat f(\xi)|^2 \leq \alpha, \end{align*} so \begin{align*} |\Gamma|\leq 2\alpha^{-2}. \end{align*} Define the common annihilator \begin{align*} V:=\{x\in G:\xi\cdot x=0 \text{ for every } \xi\in\Gamma\}. \end{align*} This is an $\mathbb{F}_p$-linear subspace of $G$. Each condition $\xi\cdot x=0$ is one homogeneous linear equation over $\mathbb{F}_p$, so \begin{align*} \operatorname{codim}_{\mathbb{F}_p} V\leq |\Gamma|\leq 2\alpha^{-2}. \end{align*} [guided] We now isolate the frequencies that might be too large to control by a crude estimate. Define \begin{align*} \Gamma := \left\{\xi\in G\setminus\{0\}: |\widehat f(\xi)| \geq \frac{\alpha^{3/2}}{\sqrt{2}}\right\}. \end{align*} The threshold $\alpha^{3/2}/\sqrt{2}$ is chosen so that the total contribution of all frequencies outside $\Gamma$ will be at most $\alpha^4/2$. We bound the size of $\Gamma$ using Parseval. Since $f=\mathbb{1}_A$, we have $|f(x)|^2=f(x)$ for every $x\in G$, and hence \begin{align*} \sum_{\xi\in G}|\widehat f(\xi)|^2 = \frac{1}{N}\sum_{x\in G}|f(x)|^2 = \frac{|A|}{|G|} = \alpha. \end{align*} Every $\xi\in\Gamma$ contributes at least $\alpha^3/2$ to this sum, so \begin{align*} |\Gamma|\cdot \frac{\alpha^3}{2} \leq \sum_{\xi\in\Gamma}|\widehat f(\xi)|^2 \leq \alpha. \end{align*} Dividing by $\alpha^3/2$ gives \begin{align*} |\Gamma|\leq 2\alpha^{-2}. \end{align*} Now define the subspace that kills every large frequency: \begin{align*} V:=\{x\in G:\xi\cdot x=0 \text{ for every } \xi\in\Gamma\}. \end{align*} This is an $\mathbb{F}_p$-linear subspace because it is the intersection of kernels of homogeneous linear maps $x\mapsto \xi\cdot x$. Each such equation can increase codimension by at most $1$, and there are $|\Gamma|$ equations. Therefore \begin{align*} \operatorname{codim}_{\mathbb{F}_p}V\leq |\Gamma|\leq 2\alpha^{-2}. \end{align*} [/guided] [/step] [step:Show the convolution is positive on the annihilator] Let $x\in V$. For every $\xi\in\Gamma$, the definition of $V$ gives $\chi_\xi(x)=1$. Therefore \begin{align*} F(x) &= \alpha^4 + \sum_{\xi\in\Gamma}|\widehat f(\xi)|^4 + \sum_{\xi\in (G\setminus\{0\})\setminus\Gamma}|\widehat f(\xi)|^4\chi_\xi(x). \end{align*} The middle sum is nonnegative. For the remaining frequencies, by the definition of $\Gamma$, \begin{align*} |\widehat f(\xi)|^2 < \frac{\alpha^3}{2} \end{align*} for every $\xi\in (G\setminus\{0\})\setminus\Gamma$. Hence \begin{align*} \left| \sum_{\xi\in (G\setminus\{0\})\setminus\Gamma}|\widehat f(\xi)|^4\chi_\xi(x) \right| &\leq \sum_{\xi\in (G\setminus\{0\})\setminus\Gamma}|\widehat f(\xi)|^4\\ &\leq \frac{\alpha^3}{2}\sum_{\xi\in G}|\widehat f(\xi)|^2\\ &= \frac{\alpha^3}{2}\alpha\\ &= \frac{\alpha^4}{2}. \end{align*} Thus \begin{align*} F(x)\geq \alpha^4-\frac{\alpha^4}{2}=\frac{\alpha^4}{2}>0. \end{align*} Since $F(x)>0$ implies $x\in 2A-2A$, every $x\in V$ lies in $2A-2A$. Therefore \begin{align*} V\subset 2A-2A. \end{align*} [guided] Take an arbitrary $x\in V$. The point of defining $V$ as an annihilator is that every large frequency becomes harmless on $V$: if $\xi\in\Gamma$, then $\xi\cdot x=0$, so \begin{align*} \chi_\xi(x)=\omega^{\xi\cdot x}=1. \end{align*} Substituting this into the Fourier expansion of $F$ gives \begin{align*} F(x) &= \alpha^4 + \sum_{\xi\in\Gamma}|\widehat f(\xi)|^4 + \sum_{\xi\in (G\setminus\{0\})\setminus\Gamma}|\widehat f(\xi)|^4\chi_\xi(x). \end{align*} The first term is the positive main term. The second term is nonnegative because every summand is a fourth power of an absolute value. The only possible cancellation can come from the small frequencies, namely those outside $\Gamma$. For every $\xi\in (G\setminus\{0\})\setminus\Gamma$, the definition of $\Gamma$ gives \begin{align*} |\widehat f(\xi)| < \frac{\alpha^{3/2}}{\sqrt{2}}, \end{align*} and hence \begin{align*} |\widehat f(\xi)|^2 < \frac{\alpha^3}{2}. \end{align*} Using $|\chi_\xi(x)|=1$ and the triangle inequality, \begin{align*} \left| \sum_{\xi\in (G\setminus\{0\})\setminus\Gamma}|\widehat f(\xi)|^4\chi_\xi(x) \right| &\leq \sum_{\xi\in (G\setminus\{0\})\setminus\Gamma}|\widehat f(\xi)|^4\\ &= \sum_{\xi\in (G\setminus\{0\})\setminus\Gamma}|\widehat f(\xi)|^2|\widehat f(\xi)|^2\\ &\leq \frac{\alpha^3}{2}\sum_{\xi\in G}|\widehat f(\xi)|^2. \end{align*} By Parseval applied to $f=\mathbb{1}_A$, the last sum is $\alpha$. Therefore \begin{align*} \left| \sum_{\xi\in (G\setminus\{0\})\setminus\Gamma}|\widehat f(\xi)|^4\chi_\xi(x) \right| \leq \frac{\alpha^4}{2}. \end{align*} Consequently, \begin{align*} F(x)\geq \alpha^4-\frac{\alpha^4}{2}=\frac{\alpha^4}{2}>0. \end{align*} The earlier counting interpretation of $F$ says that positivity of $F(x)$ means there exist $a_1,a_2,a_3,a_4\in A$ such that \begin{align*} x=a_1+a_2-a_3-a_4. \end{align*} Thus $x\in 2A-2A$. Since $x\in V$ was arbitrary, we have \begin{align*} V\subset 2A-2A. \end{align*} [/guided] [/step] [step:Conclude the codimension bound] We have constructed an $\mathbb{F}_p$-linear subspace $V\leq G$ such that \begin{align*} V\subset 2A-2A \end{align*} and \begin{align*} \operatorname{codim}_{\mathbb{F}_p}V\leq 2\alpha^{-2}. \end{align*} Thus the theorem holds with the explicit constant $2$. In particular, it holds in the stated form with $C_p=2$, which depends only on $p$. [guided] The construction has produced the exact object required by the theorem. Namely, $V$ is an $\mathbb{F}_p$-linear subspace of $G=\mathbb{F}_p^n$, and the preceding step proved \begin{align*} V\subset 2A-2A. \end{align*} The large-spectrum estimate gave the codimension bound \begin{align*} \operatorname{codim}_{\mathbb{F}_p}V\leq 2\alpha^{-2}. \end{align*} Thus $2A-2A$ contains a subspace whose codimension is at most $2\alpha^{-2}$. This is the stated conclusion with the admissible choice $C_p=2$. Since $2$ is independent of $n$, $A$, and $\alpha$, it depends only on $p$ in the sense required by the theorem. [/guided] [/step]