[proofplan] We count three-term arithmetic progressions in $A$ using the normalized Fourier transform on the finite abelian group $G$. If no non-degenerate progression exists, then the normalized progression count is exactly the degenerate contribution $\alpha/|G|$, so either the Fourier error forces a large nonzero Fourier coefficient of $\mathbb{1}_A-\alpha$, or $\alpha$ is so small that an elementary discreteness argument gives a density increment. A large Fourier coefficient is converted into a density increment by partitioning $G$ into the cosets of the kernel of the corresponding nontrivial character. [/proofplan] [step:Define the normalized Fourier transform and the progression count] Let $N := |G| = p^n$. Let $\widehat{G}$ denote the character group of $G$, whose elements are group homomorphisms $\gamma: G \to \mathbb{C}$ with $|\gamma(x)| = 1$ for all $x \in G$. For a function \begin{align*} f: G &\to \mathbb{C}, \end{align*} define its normalized Fourier transform by \begin{align*} \widehat{f}(\gamma) := \frac{1}{N}\sum_{x \in G} f(x)\overline{\gamma(x)} \end{align*} for $\gamma \in \widehat{G}$. Let \begin{align*} \mathbb{1}_A: G &\to \{0,1\} \end{align*} be the indicator function of $A$, and define the balanced function \begin{align*} f: G &\to \mathbb{R},\\ x &\mapsto \mathbb{1}_A(x)-\alpha. \end{align*} Thus $\widehat{f}(1)=0$, where $1 \in \widehat{G}$ is the trivial character. Define the normalized three-term progression count \begin{align*} T(A) := \frac{1}{N^2}\sum_{x \in G}\sum_{d \in G} \mathbb{1}_A(x)\mathbb{1}_A(x+d)\mathbb{1}_A(x+2d). \end{align*} Since $p$ is odd, if $d \neq 0$, then $x$, $x+d$, and $x+2d$ are distinct. Therefore the terms with $d=0$ are exactly the degenerate progressions, and their total contribution is \begin{align*} \frac{1}{N^2}\sum_{x \in G}\mathbb{1}_A(x) = \frac{\alpha}{N}. \end{align*} Hence, if \begin{align*} T(A) > \frac{\alpha}{N}, \end{align*} then some summand with $d \neq 0$ is nonzero, giving a non-degenerate three-term arithmetic progression in $A$. [guided] We first isolate the exact object being counted. Let $N:=|G|=p^n$, and let $\widehat{G}$ be the character group of $G$, meaning the group of homomorphisms $\gamma:G\to\mathbb{C}$ satisfying $|\gamma(x)|=1$ for every $x\in G$. For a function \begin{align*} f:G&\to\mathbb{C}, \end{align*} we use the normalized Fourier transform \begin{align*} \widehat{f}(\gamma):=\frac{1}{N}\sum_{x\in G}f(x)\overline{\gamma(x)},\qquad \gamma\in\widehat{G}. \end{align*} Let \begin{align*} \mathbb{1}_A:G&\to\{0,1\} \end{align*} be the indicator function of $A$, and define the balanced function \begin{align*} f:G&\to\mathbb{R},\\ x&\mapsto \mathbb{1}_A(x)-\alpha. \end{align*} The word balanced means that the constant Fourier mode vanishes: since $|A|=\alpha N$, \begin{align*} \widehat{f}(1)=\frac{1}{N}\sum_{x\in G}(\mathbb{1}_A(x)-\alpha)=\alpha-\alpha=0. \end{align*} Here $1\in\widehat{G}$ denotes the trivial character. Define the normalized three-term progression count by \begin{align*} T(A):=\frac{1}{N^2}\sum_{x\in G}\sum_{d\in G}\mathbb{1}_A(x)\mathbb{1}_A(x+d)\mathbb{1}_A(x+2d). \end{align*} Each pair $(x,d)\in G\times G$ contributes $1/N^2$ exactly when all three points $x$, $x+d$, and $x+2d$ lie in $A$. The degenerate progressions are precisely the terms with $d=0$. Indeed, if $d\neq0$ and $x=x+2d$, then $2d=0$; because $G$ has characteristic $p$ and $p$ is odd, multiplication by $2$ is injective, so $d=0$, a contradiction. Thus for $d\neq0$ the three points $x$, $x+d$, and $x+2d$ are distinct. The contribution from $d=0$ is \begin{align*} \frac{1}{N^2}\sum_{x\in G}\mathbb{1}_A(x)=\frac{|A|}{N^2}=\frac{\alpha}{N}. \end{align*} Consequently, if \begin{align*} T(A)>\frac{\alpha}{N}, \end{align*} then some contribution with $d\neq0$ must be nonzero, and that pair $(x,d)$ gives a non-degenerate three-term arithmetic progression contained in $A$. [/guided] [/step] [step:Expand the progression count by Fourier inversion] For every function $u: G \to \mathbb{C}$, [Fourier inversion on finite abelian groups](/page/Fourier%20Transform) gives \begin{align*} u(x)=\sum_{\gamma \in \widehat{G}}\widehat{u}(\gamma)\gamma(x). \end{align*} Applying this to $\mathbb{1}_A=\alpha+f$, we expand the three factors and use the [orthogonality relations for characters](/page/Character). For characters $\gamma_1,\gamma_2,\gamma_3\in\widehat{G}$, the normalized average of the corresponding term is \begin{align*} \frac{1}{N^2}\sum_{x\in G}\sum_{d\in G}\gamma_1(x)\gamma_2(x+d)\gamma_3(x+2d) &= \frac{1}{N^2}\sum_{x\in G}\sum_{d\in G}(\gamma_1\gamma_2\gamma_3)(x)(\gamma_2\gamma_3^2)(d), \end{align*} which equals $1$ exactly when $\gamma_1\gamma_2\gamma_3=1$ and $\gamma_2\gamma_3^2=1$, and equals $0$ otherwise. Thus $\gamma_2=\gamma_3^{-2}$ and $\gamma_1=\gamma_3$, so, writing $\gamma:=\gamma_3$, we obtain \begin{align*} T(A) &= \sum_{\gamma\in\widehat{G}}\widehat{\mathbb{1}_A}(\gamma)^2\widehat{\mathbb{1}_A}(\gamma^{-2})\\ &= \alpha^3+\sum_{\gamma \in \widehat{G}\setminus\{1\}}\widehat{f}(\gamma)^2\widehat{f}(\gamma^{-2}). \end{align*} Here $\gamma^{-2}:G\to\mathbb{C}$ denotes the character $x\mapsto \gamma(x)^{-2}$. Since $p$ is odd, the map $\gamma \mapsto \gamma^{-2}$ sends nontrivial characters to nontrivial characters. Let \begin{align*} \beta := \max_{\gamma \in \widehat{G}\setminus\{1\}}|\widehat{f}(\gamma)|. \end{align*} By [Parseval's identity](/page/Parseval%20Identity) for the normalized Fourier transform, \begin{align*} \sum_{\gamma \in \widehat{G}}|\widehat{f}(\gamma)|^2 = \frac{1}{N}\sum_{x \in G}|f(x)|^2. \end{align*} Since $f=\mathbb{1}_A-\alpha$, the variance computation gives \begin{align*} \frac{1}{N}\sum_{x \in G}|f(x)|^2 = \alpha(1-\alpha). \end{align*} Therefore \begin{align*} |T(A)-\alpha^3| &\leq \sum_{\gamma \in \widehat{G}\setminus\{1\}}|\widehat{f}(\gamma)|^2|\widehat{f}(\gamma^{-2})|\\ &\leq \beta\sum_{\gamma \in \widehat{G}}|\widehat{f}(\gamma)|^2\\ &= \beta\alpha(1-\alpha)\\ &\leq \beta\alpha. \end{align*} [guided] The Fourier transform is useful because the equation \begin{align*} x,\quad x+d,\quad x+2d \end{align*} has a linear structure. Expanding $1_A=\alpha+f$ separates the random-order main term $\alpha^3$ from the structured error contributed by the nonzero Fourier modes of $f$. [Fourier inversion on finite abelian groups](/page/Fourier%20Transform) states that every function $u:G\to\mathbb{C}$ has the expansion \begin{align*} u(x)=\sum_{\gamma \in \widehat{G}}\widehat{u}(\gamma)\gamma(x). \end{align*} Using this expansion for each of the three factors in $T(A)$ and then summing first over $x$ and $d$, the [orthogonality relations for characters](/page/Character) force the frequencies to satisfy the compatibility relation that leaves precisely \begin{align*} T(A) = \alpha^3+\sum_{\gamma \in \widehat{G}\setminus\{1\}}\widehat{f}(\gamma)^2\widehat{f}(\gamma^{-2}). \end{align*} The trivial character contributes $\alpha^3$, because $\widehat{\mathbb{1}_A}(1)=\alpha$. The nontrivial characters contribute the error term. The assumption that $p$ is odd matters here: if $\gamma$ is nontrivial, then $\gamma^{-2}$ is also nontrivial, since multiplication by $-2$ is an automorphism of the dual [vector space](/page/Vector%20Space) over $\mathbb{F}_p$. Define \begin{align*} \beta := \max_{\gamma \in \widehat{G}\setminus\{1\}}|\widehat{f}(\gamma)|. \end{align*} Then each factor $|\widehat{f}(\gamma^{-2})|$ in the error term is at most $\beta$. Hence \begin{align*} |T(A)-\alpha^3| &\leq \sum_{\gamma \in \widehat{G}\setminus\{1\}}|\widehat{f}(\gamma)|^2|\widehat{f}(\gamma^{-2})|\\ &\leq \beta\sum_{\gamma \in \widehat{G}}|\widehat{f}(\gamma)|^2. \end{align*} [Parseval's identity](/page/Parseval%20Identity) for the normalized Fourier transform gives \begin{align*} \sum_{\gamma \in \widehat{G}}|\widehat{f}(\gamma)|^2 = \frac{1}{N}\sum_{x \in G}|f(x)|^2. \end{align*} Since $f(x)=1-\alpha$ on $A$ and $f(x)=-\alpha$ on $G\setminus A$, we compute \begin{align*} \frac{1}{N}\sum_{x \in G}|f(x)|^2 &= \alpha(1-\alpha)^2+(1-\alpha)\alpha^2\\ &= \alpha(1-\alpha). \end{align*} Thus \begin{align*} |T(A)-\alpha^3| \leq \beta\alpha(1-\alpha) \leq \beta\alpha. \end{align*} This estimate says that if the progression count differs substantially from the random value $\alpha^3$, then some nonzero Fourier coefficient of the balanced function must be large. [/guided] [/step] [step:Force a large Fourier coefficient when the density is not too small] Assume that $A$ contains no non-degenerate three-term arithmetic progression. By the first step, all nonzero contributions to $T(A)$ come from $d=0$, so \begin{align*} T(A)=\frac{\alpha}{N}. \end{align*} If \begin{align*} \alpha^2 \geq \frac{2}{N}, \end{align*} then \begin{align*} \alpha^3-\frac{\alpha}{N} = \alpha\left(\alpha^2-\frac{1}{N}\right) \geq \frac{\alpha^3}{2}. \end{align*} Combining this with the Fourier estimate from the previous step gives \begin{align*} \frac{\alpha^3}{2} \leq |T(A)-\alpha^3| \leq \beta\alpha. \end{align*} If $\alpha>0$, division by $\alpha$ yields \begin{align*} \beta \geq \frac{\alpha^2}{2}. \end{align*} If $\alpha=0$, the density increment alternative is immediate for any affine hyperplane, so the remaining argument assumes $\alpha>0$. [guided] Assume that $A$ contains no non-degenerate three-term arithmetic progression. The first step showed that a strict inequality \begin{align*} T(A)>\frac{\alpha}{N} \end{align*} would force such a progression. Therefore all contributions to $T(A)$ must come from the degenerate terms with $d=0$, and hence \begin{align*} T(A)=\frac{\alpha}{N}. \end{align*} We now compare this exact count with the Fourier estimate from the previous step. Suppose first that the density is not too small, in the quantified sense \begin{align*} \alpha^2\geq\frac{2}{N}. \end{align*} Then \begin{align*} \alpha^3-\frac{\alpha}{N} &=\alpha\left(\alpha^2-\frac{1}{N}\right)\\ &\geq \alpha\cdot\frac{\alpha^2}{2}\\ &=\frac{\alpha^3}{2}. \end{align*} The inequality in the middle follows from \begin{align*} \alpha^2\geq\frac{2}{N} \end{align*} which implies \begin{align*} \alpha^2-\frac{1}{N}\geq\frac{\alpha^2}{2}. \end{align*} Since $T(A)=\alpha/N$, this lower bound is exactly a lower bound for $|T(A)-\alpha^3|$. Combining it with \begin{align*} |T(A)-\alpha^3|\leq\beta\alpha \end{align*} from the Fourier estimate gives \begin{align*} \frac{\alpha^3}{2}\leq |T(A)-\alpha^3|\leq\beta\alpha. \end{align*} If $\alpha>0$, dividing by $\alpha$ gives \begin{align*} \beta\geq\frac{\alpha^2}{2}. \end{align*} If $\alpha=0$, then $A=\varnothing$ and every affine hyperplane has density $0$, so the density increment conclusion is immediate. Thus the only remaining case in this step is $\alpha>0$. [/guided] [/step] [step:Convert a large Fourier coefficient into a density increment on a hyperplane] Assume that there exists a nontrivial character $\gamma \in \widehat{G}\setminus\{1\}$ such that \begin{align*} |\widehat{f}(\gamma)| \geq \frac{\alpha^2}{2}. \end{align*} Let \begin{align*} V := \ker \gamma = \{x \in G:\gamma(x)=1\}. \end{align*} Since $\gamma$ is a nontrivial linear character on the vector space $G$, its kernel $V$ is a linear hyperplane in $G$. The cosets of $V$ form a partition of $G$ into exactly $p$ affine hyperplanes. Write these cosets as \begin{align*} H_t := a_t+V,\qquad t \in \mathbb{F}_p, \end{align*} where $a_t \in G$ is chosen so that $\gamma(a_t)=\omega^t$ and $\omega:=e^{2\pi i/p}$. For each $t\in\mathbb{F}_p$, define the density of $A$ on $H_t$ by \begin{align*} \alpha_t := \frac{|A\cap H_t|}{|H_t|}. \end{align*} Since the cosets partition $G$ and all have size $|V|=N/p$, \begin{align*} \frac{1}{p}\sum_{t\in\mathbb{F}_p}\alpha_t=\alpha. \end{align*} Moreover, \begin{align*} \widehat{f}(\gamma) &= \frac{1}{N}\sum_{x\in G}(\mathbb{1}_A(x)-\alpha)\overline{\gamma(x)}\\ &= \frac{1}{p}\sum_{t\in\mathbb{F}_p}(\alpha_t-\alpha)\omega^{-t}. \end{align*} Let \begin{align*} \delta := \max_{t\in\mathbb{F}_p}(\alpha_t-\alpha). \end{align*} Since the numbers $\alpha_t-\alpha$ have average $0$, their positive parts have total mass equal to their negative parts in absolute value. Therefore \begin{align*} \frac{1}{p}\sum_{t\in\mathbb{F}_p}|\alpha_t-\alpha|\leq 2\delta. \end{align*} Using the triangle inequality in the preceding expression for $\widehat{f}(\gamma)$ gives \begin{align*} |\widehat{f}(\gamma)| \leq \frac{1}{p}\sum_{t\in\mathbb{F}_p}|\alpha_t-\alpha| \leq 2\delta. \end{align*} Hence \begin{align*} \delta \geq \frac{|\widehat{f}(\gamma)|}{2}\geq \frac{\alpha^2}{4}. \end{align*} Choose $t_0\in\mathbb{F}_p$ with $\alpha_{t_0}-\alpha=\delta$, and set $H:=H_{t_0}$. Then $H$ is an affine hyperplane and \begin{align*} \frac{|A\cap H|}{|H|} =\alpha_{t_0} \geq \alpha+\frac{\alpha^2}{4}. \end{align*} [guided] A nonzero Fourier coefficient detects bias along the level sets of the corresponding character. We now turn that bias into an actual density increment. Let $\gamma\in\widehat{G}\setminus\{1\}$ satisfy \begin{align*} |\widehat{f}(\gamma)| \geq \frac{\alpha^2}{2}. \end{align*} Define \begin{align*} V:=\ker\gamma=\{x\in G:\gamma(x)=1\}. \end{align*} Because $\gamma$ is nontrivial, it is a nonzero linear functional after identifying the character group of $\mathbb{F}_p^n$ with its dual vector space. Therefore its kernel is a codimension-one linear subspace of $G$, hence a linear hyperplane. Its cosets are affine hyperplanes. Let $\omega:=e^{2\pi i/p}$. Since the image of a nontrivial character is the group of $p$th roots of unity, the cosets of $V$ may be written as \begin{align*} H_t:=a_t+V,\qquad t\in\mathbb{F}_p, \end{align*} where $a_t\in G$ is chosen so that $\gamma(a_t)=\omega^t$. For each coset define \begin{align*} \alpha_t:=\frac{|A\cap H_t|}{|H_t|}. \end{align*} The cosets partition $G$ into $p$ equal pieces, so averaging their densities recovers the global density: \begin{align*} \frac{1}{p}\sum_{t\in\mathbb{F}_p}\alpha_t=\alpha. \end{align*} Now compute the Fourier coefficient by grouping the sum over these cosets: \begin{align*} \widehat{f}(\gamma) &= \frac{1}{N}\sum_{x\in G}(\mathbb{1}_A(x)-\alpha)\overline{\gamma(x)}\\ &= \frac{1}{N}\sum_{t\in\mathbb{F}_p}\sum_{x\in H_t}(\mathbb{1}_A(x)-\alpha)\omega^{-t}\\ &= \frac{1}{N}\sum_{t\in\mathbb{F}_p}\left(|A\cap H_t|-\alpha|H_t|\right)\omega^{-t}\\ &= \frac{1}{p}\sum_{t\in\mathbb{F}_p}(\alpha_t-\alpha)\omega^{-t}. \end{align*} Thus the Fourier coefficient is a weighted average of the density deviations $\alpha_t-\alpha$. Define the largest positive deviation by \begin{align*} \delta:=\max_{t\in\mathbb{F}_p}(\alpha_t-\alpha). \end{align*} Because the deviations average to $0$, \begin{align*} \sum_{t\in\mathbb{F}_p}(\alpha_t-\alpha)=0. \end{align*} Therefore the sum of the positive deviations equals the absolute value of the sum of the negative deviations. Since each positive deviation is at most $\delta$, the total absolute deviation satisfies \begin{align*} \frac{1}{p}\sum_{t\in\mathbb{F}_p}|\alpha_t-\alpha|\leq 2\delta. \end{align*} Applying the triangle inequality to the Fourier expression gives \begin{align*} |\widehat{f}(\gamma)| \leq \frac{1}{p}\sum_{t\in\mathbb{F}_p}|\alpha_t-\alpha|\,|\omega^{-t}| = \frac{1}{p}\sum_{t\in\mathbb{F}_p}|\alpha_t-\alpha| \leq 2\delta, \end{align*} because $|\omega^{-t}|=1$. Hence \begin{align*} \delta\geq \frac{|\widehat{f}(\gamma)|}{2}\geq \frac{\alpha^2}{4}. \end{align*} Choosing a coset $H=H_{t_0}$ attaining this maximum deviation gives \begin{align*} \frac{|A\cap H|}{|H|} =\alpha_{t_0} \geq \alpha+\frac{\alpha^2}{4}. \end{align*} This is the required density increment. [/guided] [/step] [step:Handle the small-density case by discreteness of hyperplane densities] It remains to consider the small-density case \begin{align*} \alpha^2 < \frac{2}{N}. \end{align*} If $\alpha=0$, then $A=\varnothing$, and every affine hyperplane $H\subset G$ satisfies \begin{align*} \frac{|A\cap H|}{|H|}=0=\alpha+\frac{\alpha^2}{4}. \end{align*} Assume $\alpha>0$. If $A$ contains a non-degenerate three-term arithmetic progression, we are done. Otherwise, suppose first that $A$ has the same density $\alpha$ on every affine hyperplane of $G$. Then, for every nontrivial character $\gamma\in\widehat{G}$, grouping over the cosets of $\ker\gamma$ gives \begin{align*} \widehat{f}(\gamma)=0. \end{align*} Also $\widehat{f}(1)=0$. Fourier inversion then gives $f=0$ on $G$, so $\mathbb{1}_A=\alpha$ pointwise on $G$. Since $\mathbb{1}_A$ only takes the values $0$ and $1$, this implies $\alpha\in\{0,1\}$. The case $\alpha=0$ has already been handled, while $\alpha=1$ gives $A=G$, and since $n\geq 1$ there exists $d\in G\setminus\{0\}$, so $x,x+d,x+2d\in A$ for every $x\in G$. Therefore, in the remaining case, some affine hyperplane has density strictly larger than $\alpha$. More explicitly, there is a linear hyperplane $V\subset G$ and a coset decomposition \begin{align*} G=\bigsqcup_{t\in\mathbb{F}_p}H_t \end{align*} such that the integers \begin{align*} m_t:=|A\cap H_t|,\qquad t\in\mathbb{F}_p, \end{align*} are not all equal. Let \begin{align*} m:=|A|. \end{align*} Choose $t_0\in\mathbb{F}_p$ such that $m_{t_0}$ is maximal. Since the average of the $m_t$ is $m/p$ and the $m_t$ are not all equal, \begin{align*} p m_{t_0}-m \end{align*} is a positive integer, hence at least $1$. With $H:=H_{t_0}$, we get \begin{align*} \frac{|A\cap H|}{|H|}-\alpha &= \frac{m_{t_0}}{N/p}-\frac{m}{N}\\ &= \frac{p m_{t_0}-m}{N}\\ &\geq \frac{1}{N}. \end{align*} Since \begin{align*} \alpha^2<\frac{2}{N}, \end{align*} we have \begin{align*} \frac{1}{N}>\frac{\alpha^2}{2}\geq \frac{\alpha^2}{4}. \end{align*} Thus \begin{align*} \frac{|A\cap H|}{|H|} \geq \alpha+\frac{\alpha^2}{4}. \end{align*} [guided] We now handle the complementary density range \begin{align*} \alpha^2<\frac{2}{N}. \end{align*} If $\alpha=0$, then $A=\varnothing$, and for every affine hyperplane $H\subset G$, \begin{align*} \frac{|A\cap H|}{|H|}=0=\alpha+\frac{\alpha^2}{4}. \end{align*} So assume $\alpha>0$. If $A$ already contains a non-degenerate three-term arithmetic progression, the theorem is proved. Assume instead that it does not. Suppose first that $A$ has density $\alpha$ on every affine hyperplane of $G$. For any nontrivial character $\gamma\in\widehat{G}$, grouping the Fourier coefficient over the cosets of $\ker\gamma$ gives \begin{align*} \widehat{f}(\gamma)=0, \end{align*} because every coset has density exactly $\alpha$. Also $\widehat{f}(1)=0$ by the definition of $f$. By [Fourier inversion on finite abelian groups](/page/Fourier%20Transform), all Fourier coefficients of $f$ vanish only when $f=0$ pointwise on $G$. Hence \begin{align*} \mathbb{1}_A(x)=\alpha \end{align*} for every $x\in G$. Since $\mathbb{1}_A$ takes only the values $0$ and $1$, this forces $\alpha\in\{0,1\}$. The case $\alpha=0$ was already handled. If $\alpha=1$, then $A=G$; since $n\geq1$, there exists $d\in G\setminus\{0\}$, and for every $x\in G$ the points $x$, $x+d$, and $x+2d$ lie in $A$, giving the progression alternative. Thus in the remaining case, some affine hyperplane has density strictly larger than $\alpha$. More concretely, there is a linear hyperplane $V\subset G$ and a coset decomposition \begin{align*} G=\bigsqcup_{t\in\mathbb{F}_p}H_t \end{align*} for which the integers \begin{align*} m_t:=|A\cap H_t|,\qquad t\in\mathbb{F}_p, \end{align*} are not all equal. Let \begin{align*} m:=|A|. \end{align*} Choose $t_0\in\mathbb{F}_p$ such that $m_{t_0}$ is maximal. Since the average of the integers $m_t$ is $m/p$ and the $m_t$ are not all equal, the integer \begin{align*} pm_{t_0}-m \end{align*} is positive, hence at least $1$. Setting $H:=H_{t_0}$, we obtain \begin{align*} \frac{|A\cap H|}{|H|}-\alpha &=\frac{m_{t_0}}{N/p}-\frac{m}{N}\\ &=\frac{pm_{t_0}-m}{N}\\ &\geq\frac{1}{N}. \end{align*} Finally, the small-density hypothesis gives \begin{align*} \alpha^2<\frac{2}{N}, \end{align*} so \begin{align*} \frac{1}{N}>\frac{\alpha^2}{2}\geq\frac{\alpha^2}{4}. \end{align*} Therefore \begin{align*} \frac{|A\cap H|}{|H|}\geq\alpha+\frac{\alpha^2}{4}. \end{align*} [/guided] [/step] [step:Combine the alternatives] If \begin{align*} T(A)>\frac{\alpha}{N}, \end{align*} then $A$ contains a non-degenerate three-term arithmetic progression. If not, then \begin{align*} T(A)=\frac{\alpha}{N}. \end{align*} When \begin{align*} \alpha^2\geq\frac{2}{N}, \end{align*} the Fourier argument produces an affine hyperplane $H\subset G$ satisfying \begin{align*} \frac{|A\cap H|}{|H|}\geq \alpha+\frac{\alpha^2}{4}. \end{align*} When \begin{align*} \alpha^2<\frac{2}{N}, \end{align*} the discreteness argument gives the same conclusion unless the progression alternative already holds. Therefore the theorem holds with \begin{align*} c_p:=\frac{1}{4}, \end{align*} which depends only on $p$. [guided] There are two possibilities for the normalized progression count. If \begin{align*} T(A)>\frac{\alpha}{N}, \end{align*} then the first step proves that some term with $d\neq0$ contributes to the count, so $A$ contains a non-degenerate three-term arithmetic progression. If this strict inequality fails, then no non-degenerate progression has been forced by the count, and the degenerate contribution computation gives \begin{align*} T(A)=\frac{\alpha}{N}. \end{align*} In the range \begin{align*} \alpha^2\geq\frac{2}{N}, \end{align*} the large-Fourier-coefficient step gives a nontrivial character with \begin{align*} |\widehat{f}(\gamma)|\geq\frac{\alpha^2}{2}, \end{align*} and the density-increment step converts this coefficient into an affine hyperplane $H\subset G$ satisfying \begin{align*} \frac{|A\cap H|}{|H|}\geq\alpha+\frac{\alpha^2}{4}. \end{align*} In the remaining range \begin{align*} \alpha^2<\frac{2}{N}, \end{align*} the discreteness argument proves the same hyperplane conclusion unless the progression alternative has already occurred. Thus in every case either $A$ contains a non-degenerate three-term arithmetic progression or there is an affine hyperplane $H\subset G$ with \begin{align*} \frac{|A\cap H|}{|H|}\geq\alpha+\frac{\alpha^2}{4}. \end{align*} Taking \begin{align*} c_p:=\frac{1}{4} \end{align*} proves the stated result; this constant is independent of $G$, $A$, and $n$, and in particular depends only on $p$. [/guided] [/step]