= Solution
Let $I=[N]$ and let $f=1_A-\delta1_I$ be the balanced <indicator function>, extended by zero to the <integers>. For finitely supported functions define the <linear configuration count>
$$
\Lambda(g_1,g_2,g_3)=\sum_{x+z=2y}g_1(x)g_2(y)g_3(z).
$$
Since $A$ has no nonconstant three-term <arithmetic progression>, $\Lambda(1_A,1_A,1_A)=|A|=\delta N$. On the other hand, $\Lambda(1_I,1_I,1_I)\geq N^2/4$. Hence, once $N$ is sufficiently large in terms of $\delta$,
$$
\left|\Lambda(1_A,1_A,1_A)-\delta^3\Lambda(1_I,1_I,1_I)\right|\geq c_0\delta^3N^2
$$
for an absolute $c_0>0$.
Use the unnormalized <Fourier transform> $\widehat h(\theta)=\sum_xh(x)e(-\theta x)$ on $\mathbb R/\mathbb Z$. Telescoping $1_A=\delta1_I+f$ writes the preceding difference as
$$
\Lambda(f,1_A,1_A)+\delta\Lambda(1_I,f,1_A)+\delta^2\Lambda(1_I,1_I,f).
$$
Let $M=\|\widehat f\|_\infty$. The Fourier-integral formula for $\Lambda$, followed by the <Cauchy-Schwarz inequality> and the <Parseval identity>, bounds these three terms respectively by
$$
M\delta N,qquad M\delta^{3/2}N,qquad M\delta^2N.
$$
They are all at most $M\delta N$, so
$$
M\geq c_1\delta^2N
$$
for an absolute $c_1>0$. Choose $\theta$ with $|\widehat f(\theta)|=M$.
By the <Dirichlet approximation theorem>, some integer $1\leq d\leq\sqrt N$ satisfies $\|d\theta\|_{\mathbb R/\mathbb Z}\leq N^{-1/2}$. 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 yields
$$
\sum_j\left|\sum_{x\in P_j}f(x)\right|
\geq |\widehat f(\theta)|-4\pi\eta N
\geq\frac{c_1}{2}\delta^2N.
$$
The sums over all cells add to $\sum_If=0$, so their positive parts total half their absolute values. At least one $P_j$ consequently satisfies
$$
\sum_{x\in P_j}f(x)\geq\frac{c_1}{4}\delta^2|P_j|.
$$
This is the <Roth density-increment step>, and it gives
$$
\boxed{|P_j|\geq\eta\sqrt N,qquad |A\cap P_j|\geq(\delta+c\delta^2)|P_j|}
$$
with $c=c_1/4$ and $\eta>0$ depending only on $\delta$.
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