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We prove the three-term Roth theorem by a density increment, keeping the interval boundary rather than replacing the interval by a cyclic group with a different density. The case $\delta >1$ is vacuous; assume $0<\delta \le 1$.

We first prove the following alternative with absolute constants $c,C>0$: if $A\subseteq [1,N]$ has density $\eta$ and no nonconstant three-term arithmetic progression, and $N\ge C{\eta }^{-4}$, then some integer progression $P\subseteq [1,N]$ hasFor this purpose take $M=6\lceil 4N/6\rceil +1$, so $4N<M\le 5N$ for large $N$ and $2,3$ are invertible modulo $M$. On $G=\mathbb{Z}/M\mathbb{Z}$, put $I=1_{[1,N]}$, $a=1_A$, $g=\eta I$ and $f=a-g$. In particular $\sum f=0$, and $f$ vanishes outside the interval. Use normalized Fourier coefficients on a finite abelian group, $\hat{u}(r)=M^{-1}{\sum }_{x}u(x)e(-rx/M)$. Character orthogonality, obtained by summing a finite geometric series, gives Fourier inversion on a finite group and Parseval identity on a finite group.

For $\Lambda (u,v,w)={\mathbb{E}}_{x,d}u(x)v(x+d)w(x+2d)$ these identities giveThere is no wrapping ambiguity when all three entries lie in $[1,N]$: the integer $x-2y+z$ has absolute value below $M$, so a modular relation is an integer relation. If no nonconstant progression exists, $\Lambda (a,a,a)=|A|/M^2$. Meanwhile the choices $1\le x\le \lfloor N/4\rfloor$ and $1\le d\le \lfloor N/8\rfloor$ give $\Lambda (I,I,I)\ge 1/10000$ for sufficiently large $N$.

Expand the difference by telescoping,If $\rho ={\max }_r|\hat{f}(r)|$, each term on the right has modulus at most $\rho \eta N/M$. Indeed multiplication of frequencies by $2$ permutes them, and Cauchy-Schwarz with Parseval identity bounds the two other Fourier factors by their $L^2$ norms; each squared norm is at most $\eta N/M$. For $N\ge 10000{\eta }^{-2}$, the diagonal count is less than half the main density term. ThereforeIn particular there is a nonzero frequency $r$ withThe frequency is nonzero because $\hat{f}(0)=0$.

Here is the conversion of this coefficient into a density increment. Put $Q=\lfloor \sqrt{N}\rfloor$. Of the $Q+1$ numbers $0,\theta ,\dots ,Q\theta$ modulo one, two lie in the same one of $Q$ equal arcs. Their difference gives $1\le q\le Q$ with $\Vert q\theta \Vert \le 1/Q$, where $\theta =-r/M$. This is the needed Dirichlet approximation theorem, proved by the pigeonhole principle.

Partition each residue class modulo $q$ in $[1,N]$ into consecutive progressions of length $L=\lfloor \gamma Q/(100\pi )\rfloor$, leaving its final short tail. Their total tail size is at most $qL\le \gamma N/(100\pi )$. On each full progression the character differs from its value at the first point by at most $2\pi L/Q\le \gamma /50$. Hence removing tails and replacing the character by its initial value showsThe sum of $f$ over all full progressions is the negative of its sum over the tails, so its absolute value is at most $\gamma N/(100\pi )$. Consequently the sum of the positive block sums is at least $\gamma N/5$. Since these blocks contain at most $N$ points in total, at least one has average $f$ at least $\gamma /5$. It gives a density increment at least ${\eta }^2/100000$. For $N\ge C{\eta }^{-4}$ with a sufficiently large absolute $C$, $L\ge c{\eta }^2\sqrt{N}$. This proves the alternative, with one small common absolute $c$.

Reidentify $P$ with $[1,|P|]$ by its affine parametrization. Absence of a nonconstant progression is preserved, while density increases. At every step the density is at least the original $\delta$, so the increment is at least $c{\delta }^2$ and the new interval length is at least $c{\delta }^2\sqrt{N_{\mathrm{old}}}$. After at most $T=\lceil 1/(c{\delta }^2)\rceil +1$ steps, the density would exceed one. Choose the original $N$ so large that the length stays above $C{\delta }^{-4}$ throughout these finitely many steps. Such a choice exists explicitly by working backwards: a required next length $L_0$ is ensured by previous length at least $(L_0/(c{\delta }^2){)}^2$. The alternative could then be applied at every step, giving a contradiction. ThusThis is the Roth density increment on an integer progression proof.

For the pattern with offsets $0,1,3$, replace the counting functional by ${\mathbb{E}}_{x,d}u(x)v(x+d)w(x+3d)$. Its Fourier expression iscoming from the relation $2x-3y+z=0$. Both multipliers are invertible modulo the chosen $M$, so exactly the same Parseval estimates apply. This relation has no wrap in $[1,N]$, since its magnitude is at most $3N<M$, and the earlier restricted choices of $x,d$ still give a fixed positive density main term. The same increment and iteration find a nonconstant triple $\{a,a+d,a+3d\}$. Negative $d$ is allowed, as requested, and the affine reparametrization used in the iteration preserves this pattern as well.