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For fixed $c>0$ and prime $p$, if $A\subseteq {\mathbb{F}}_p^N$ has at least $c|A{|}^3$ additive quadruples, then $2A-2A$ contains a vector subspace of size at least $c^{\prime }(c,p)|A|$. Combine the Balog-Szemerédi-Gowers theorem, the Freiman-Ruzsa theorem over a finite field, and the Finite-field Bogolyubov lemma.

The printed statement needs $A\ne \varnothing$: for the empty set every linear subspace has cardinality at least one, so the requested bound is impossible. We prove the intended nonempty version of the Freiman-Ruzsa theorem over a finite field. Necessarily $C\ge 1$, because any one translate of $A$ inside $A+A$ has size $|A|$.

We first prove all sumset estimates needed below. For finite nonempty $X$ and finite $U,V$ in an abelian group, choose one representation $z=u_z-v_z$ for each $z\in U-V$. The mapis injective: subtracting the outputs recovers $z$, hence its chosen representation and then $x$. This proves the Ruzsa triangle inequality

Choose nonempty $X\subseteq B$ minimizing $c^{\prime }=|X+B|/|X|\le C$, where $B$ is any set with $|B+B|\le C|B|$. We prove the Petridis minimal-growth lemma $|X+B+D|\le c^{\prime }|X+D|$ for every finite $D$. List $D=\{d_1,\dots ,d_s\}$, and let $X_i$ consist of those $x\in X$ for which $x+d_i$ was not already in $X+\{d_1,\dots ,d_{i-1}\}$. The disjoint pieces $X_i+d_i$ partition $X+D$.

Every $(X\setminus X_i)+B+d_i$ was already covered in $X+B+\{d_1,\dots ,d_{i-1}\}$. The new part at step $i$ therefore has size at mostHere minimality of $X$ applies to every nonempty subset $X\setminus X_i$, and the empty case has size zero. Summing proves the lemma. Iteration gives $|X+kB|\le C^k|X|$. Apply the Ruzsa triangle inequality with $U=kB$, $V=\ell B$ to obtainThis derives the required Plünnecke-Ruzsa inequality. In characteristic two, subtraction is addition, so $|rB|\le C^r|B|$ for every positive integer $r$.

Fix $a\in A$ and translate to $B=A+a$, which contains zero, has size $|A|$, and has the same doubling constant. Choose $m=\lceil {\log }_2(2|12B|)\rceil$. A uniformly random linear map $L:{\mathbb{F}}_2^N\to G^{\prime }={\mathbb{F}}_2^m$ kills each specified nonzero vector with probability $2^{-m}$. The union bound shows that some $L$ satisfiesBecause $0\in B$, all lower iterated sumsets lie in $12B$. In particular $L$ is injective on $B$, since the difference of two elements belongs to $2B$. For $B^{\prime }=L(B)$, the density $\alpha =|B^{\prime }|/|G^{\prime }|$ is at least $1/(4C^{12})$. This proves injective linear modelling of a small sumset.

We next prove the Finite-field Bogolyubov lemma in this model. Write $f=1_{B^{\prime }}$ and use normalized real Fourier coefficients on a finite abelian group $\hat{f}(\xi )={\mathbb{E}}_{x}f(x)(-1{)}^{\xi \cdot x}$. LetBy Parseval identity, $|S|\le 2{\alpha }^{-2}$. Thus $W$ is a vector subspace of codimension at most $D=\lceil 32C^{24}\rceil$.

For $w\in W$, Fourier inversion of the fourfold normalized convolution givesAll terms with $\xi \in S$ are nonnegative, and the zero-frequency term is ${\alpha }^4$. The absolute contribution from the remaining frequencies is at mostHence the convolution is positive and $W\subseteq 4B^{\prime }$. Also $|W|\ge 2^{-D}|G^{\prime }|\ge 2^{-D}|B|$.

For each $w\in W$, choose its lift $\sigma (w)\in 4B$. It is unique because two lifts differ by an element of $8B\cap \ker L$, which is zero. For $w_1,w_2\in W$, the discrepancybelongs to $12B\cap \ker L$, so it vanishes. Also $\sigma (0)=0$. Therefore $\sigma$ is a linear map and its image $\tilde{W}\subseteq 4B$ is a vector subspace with $|\tilde{W}|=|W|\ge 2^{-D}|B|$. This establishes lifting a subspace through a Freiman model without assuming that arbitrary choices of preimages preserve addition.

Finally, $B+\tilde{W}\subseteq 5B$ is a union of $\tilde{W}$-cosets. The number of these cosets is at mostChoose representatives $b_1,\dots ,b_t$ from $B$. Then $B\subseteq \tilde{W}+\{b_1,\dots ,b_t\}$, so adjoining these representatives and $a$ to $\tilde{W}$ gives a vector subspace $V$ containing $A$. Over ${\mathbb{F}}_2$, adjoining each vector multiplies the size by at most two. Since $|\tilde{W}|\le |4B|\le C^4|B|$, the Freiman-Ruzsa bound from Bogolyubov and linear modelling yieldsThis bound depends only on $C$, as required, and every sumset estimate used in its derivation has been proved.

The hypothesis says that the additive energy satisfies $E(A)\ge c|A{|}^3$. The Balog-Szemerédi-Gowers theorem supplies $A^{\prime }\subseteq A$ withBy the Freiman-Ruzsa theorem over a finite field, $A^{\prime }$ lies in a subspace $H$ withThus $A^{\prime }$ has density at least $C_2^{-1}$ in $H$. Applying the Finite-field Bogolyubov lemma inside $H$ gives a subspaceof codimension bounded in terms of $c,p$ alone. Thereforewhich is the additive energy produces a large subspace in a fourfold difference set result.

The Freiman-Ruzsa theorem over a finite field states that if $A\subseteq {\mathbb{F}}_p^n$ and $|A+A|\le K|A|$, then $A$ is contained in a vector subspace $H$ withAfter translating $A$, assume $0\in A$. Put $S=A-A$, and choose $X\subseteq A+S$ maximal subject to the translates $x+A$, $x\in X$, being pairwise disjoint. Since $X+A\subseteq 2A+S=3A-A$, the Plünnecke-Ruzsa inequality givesso $|X|\le K^4$.

Maximality gives $A+S\subseteq X+S$: if $y\in A+S$ is not already in $X$, then $(y+A)\cap (x+A)\ne \varnothing$ for some $x\in X$, whence $y\in x+A-A=x+S$. Inductively, $mA+S\subseteq ⟨X⟩+S$ for every positive integer $m$. Because $0\in A$, every element of $⟨A⟩$ belongs to some $mA$ in the finite vector space, and thereforeFinally, $|⟨X⟩|\le p^{|X|}$ and $|S|\le K^2|A|$ by the Plünnecke-Ruzsa inequality, so

Let $H\le {\mathbb{F}}_p^n$ be a subspace with $|H|$ much larger than $K^2$, and choose linearly independent vectors $e_1,\dots ,e_{K-1}$ whose images are independent modulo $H$. SetThen $|A|=|H|+K-1\sim |H|$, while $A+A$ is the union of $H$, the $K-1$ disjoint cosets $e_i+H$, and at most $K^2$ exceptional sums $e_i+e_j$. HenceEvery subspace containing $A$ must contain $H$ and all $e_i$, so it has at least $p^{K-1}|H|\sim p^{K-1}|A|$ elements. Thus the exponential dependence on $K$ in the Freiman-Ruzsa theorem over a finite field cannot in general be replaced by a subexponential one.

The hypothesis says that the normalized additive energy of $A$ is at least $\eta$. By the Balog-Szemerédi-Gowers theorem, for an absolute $C_0$ there is $A^{\prime }\subseteq A$ such thatThe finite-field Bogolyubov-Ruzsa consequence of the Freiman-Ruzsa theorem over a finite field says that a set of doubling at most $K$ has a vector subspacewith $|V|\ge p^{-K^{C_1}}|A^{\prime }|$ for an absolute $C_1$. Taking $K={\eta }^{-C_0}$ and enlarging the absolute exponent givesThis is an energy form of the Bogolyubov lemma: the usual lemma assumes positive density in an ambient group, whereas the Balog-Szemerédi-Gowers theorem first extracts a dense structured model from the many additive quadruples. The resulting bound depends on the energy parameter rather than on the possibly tiny ambient density of $A$.