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A sufficient absolute constant is $\boxed{C=16}$. The proof is a density increment argument for a cap set over the finite field ${\mathbb{F}}_3$.

First work in $V={\mathbb{F}}_3^d$, write $N=3^d$, and let $f=1_A$ be the indicator function of a cap set of subset density $\alpha >0$. All expectations below are uniform. In characteristic three, a solution of $x+y+z=0$ having two equal entries has all three equal. Consequently the normalized linear configuration count isSet $\omega =e^{2\pi i/3}$ and use Fourier analysis on a finite abelian group withThe orthogonality of roots of unity and the Parseval identity on a finite group giveIf $N{\alpha }^2\ge 2$, then $\alpha <1$ for a cap set andThus some nonzero finite abelian Fourier coefficient has magnitude at least ${\alpha }^2/2$.

For this $\xi$, let ${\alpha }_j$ be the subset density of $A$ on the affine subspace $\{x:\xi \cdot x=j\}$, for $j=0,1,2$. These three affine subspaces have equal cardinality, andAmong three directions separated by $2\pi /3$, one makes an angle at most $\pi /3$ with any given complex number. Hence ${\max }_j({\alpha }_j-\alpha )\ge |\hat{f}(\xi )|$. Restricting to that hyperplane gives the hyperplane density increment for cap setsTranslate the affine subspace to its underlying vector space. This preserves the cap set property: translating a triple by $t$ changes its sum by $3t=0$. The same argument can therefore be iterated.

For completeness, the density increment iteration gives an explicit uniform bound. If $n<16$, the hypothesis $\alpha \ge 16/n$ is impossible. Suppose $n\ge 16$ and a cap set has ${\alpha }_0\ge 16/n$. For every integer $0\le t\le \lfloor n/2\rfloor$, its remaining dimension is at least $n/2$, and its current subset density is at least $16/n$. ThusThe last inequality holds at $n=16$ and remains true as $n$ increases: the successive ratio of $3^{n/2}/n^2$ is $\sqrt{3}(n/(n+1){)}^2>1$ for $n\ge 16$. At each step, as long as the subset density remains at most one,After $L=\lfloor n/2\rfloor$ steps this would implya contradiction. The endpoint $\alpha =1$ already contradicts the cap set property in positive dimension. This proves the claimed existence of three distinct points and the Meshulam bound for cap sets.

Let $I=[N]$ and let $f=1_A-\delta 1_I$ be the balanced indicator function, extended by zero to the integers. For finitely supported functions define the linear configuration countSince $A$ has no nonconstant three-term arithmetic progression, $\Lambda (1_A,1_A,1_A)=|A|=\delta N$. On the other hand, $\Lambda (1_I,1_I,1_I)\ge N^2/4$. Hence, once $N$ is sufficiently large in terms of $\delta$,for an absolute $c_0>0$.

Use the unnormalized Fourier transform $\hat{h}(\theta )={\sum }_{x}h(x)e(-\theta x)$ on $\mathbb{R}/\mathbb{Z}$. Telescoping $1_A=\delta 1_I+f$ writes the preceding difference asLet $M=\Vert \hat{f}{\Vert }_{\infty }$. The Fourier-integral formula for $\Lambda$, followed by the Cauchy-Schwarz inequality and the Parseval identity, bounds these three terms respectively byThey are all at most $M\delta N$, sofor an absolute $c_1>0$. Choose $\theta$ with $|\hat{f}(\theta )|=M$.

By the Dirichlet approximation theorem, some integer $1\le d\le \sqrt{N}$ satisfies $\Vert d\theta {\Vert }_{\mathbb{R}/\mathbb{Z}}\le N^{-1/2}$. Choose a sufficiently small absolute multiple $\eta$ of ${\delta }^2$. Partition each residue-class progression of common difference $d$ into progressions $P_j$ whose lengths lie between $\eta \sqrt{N}$ and $2\eta \sqrt{N}$; for sufficiently large $N$, short final pieces can be joined to the preceding piece. On each $P_j$, the linear phase $e(-\theta x)$ varies by at most $4\pi \eta$. Choosing $\eta$ small enough compared with $c_1{\delta }^2$ therefore yieldsThe sums over all cells add to ${\sum }_{I}f=0$, so their positive parts total half their absolute values. At least one $P_j$ consequently satisfiesThis is the Roth density-increment step, and it giveswith $c=c_1/4$ and $\eta >0$ depending only on $\delta$.

For $g\in {\mathcal{H}}_N$, writeThe orthogonality of complex exponentials converts the linear configuration count intoExpand the difference between the products for $g_1$ and $g_2$ by changing one factor at a time. A typical term iswhere each $h_j$ is either $g_1$ or $g_2$. The assumed uniform norm bound controls the first factor by $\epsilon N$. The substitution $\alpha \mapsto k\alpha$ preserves an integral over the circle group, so Hölder's inequality and the three supplied $L^3$ bounds giveEach of the four terms is therefore $O(\epsilon N^3)$, and henceThis is Fourier stability of a linear configuration count.