We prove the Roth theorem on three-term arithmetic progressions by a density increment on an ordinary integer arithmetic progression. The required increment will be small, but depend only on the current subset density.

First establish the increment step. Suppose $A\subseteq [1,L]$ has subset density $\rho =|A|/L>0$ and no nonconstant three-term arithmetic progression. Put $M=4L+1$, regard $[1,L]$ as a subset $I$ of the cyclic group ${\mathbb{Z}}_M$, and set $f=1_A-\rho 1_I$. Then $\mathbb{E}f=0$ and $|f|\le 1$. Use the normalized Fourier coefficient on a finite abelian groupThe character orthogonality geometric-series identity ${\mathbb{E}}_x{e}^{2\pi irx/M}=1$ for $r=0$ and zero otherwise gives Fourier inversion on a finite group, Parseval identity on a finite group, andSince $M$ is odd, multiplication by two permutes its frequencies. If one argument is $f$ and the other two have absolute value at most one, Cauchy-Schwarz and the Parseval identity on a finite group give $|T|\le {\max }_r|\hat{f}(r)|$.

A modular arithmetic progression whose three entries lie in $I$ satisfies the corresponding integer equation: the absolute value of $x+z-2y$ is less than $2L<M$, so congruence to zero forces equality. Thus $T(1_A,1_A,1_A)=|A|/M^2\le 1/(16L)$, since only constant arithmetic progressions remain. Meanwhile, the choices $1\le x\le \lfloor L/2\rfloor$ and $0\le h\le \lfloor L/4\rfloor$ give at least $L^2/16$ arithmetic progressions in $I$ for $L\ge 8$. As $M\le 5L$, this gives $T(1_I,1_I,1_I)\ge 1/400$. Expand $1_A=\rho 1_I+f$. The seven terms other than the main term are each bounded by $\max |\hat{f}|$. For $L\ge 50{\rho }^{-3}$,Consequently a nonzero frequency $r$ obeys $|\hat{f}(r)|\ge \gamma$, where $\gamma ={\rho }^3/6000$. It is nonzero because $\hat{f}(0)=0$.

We turn this coefficient into a local increase. Let $Q=\lfloor \sqrt{L}\rfloor$. Partition the unit circle into $Q$ equal intervals and consider the $Q+1$ points $jr/M$, $0\le j\le Q$, modulo one. Two occupy the same interval, yielding an integer $1\le q\le Q$ with $\Vert qr/M{\Vert }_{\mathbb{R}/\mathbb{Z}}\le 1/Q$. Set $\ell =\lfloor \gamma Q/200\rfloor$. If $L$ exceeds a sufficiently large absolute multiple of ${\rho }^{-6}$, then $\ell \ge 1$ and each residue-class arithmetic progression of step $q$ in $[1,L]$ has at least $\ell$ terms. Partition each into pieces of lengths between $\ell$ and $2\ell$, distributing the final remainder among the pieces.

On each such arithmetic progression $P$, the character of a finite abelian group $\chi (x)=e^{-2\pi irx/M}$ differs from its value at the first point by at most $4\pi \ell /Q<\gamma /4$. Write $s_P={\sum }_{x\in P}f(x)$. Thereforeso ${\sum }_P|s_P|\ge 3\gamma L/4$. But ${\sum }_P{s}_P={\sum }_{I}f=0$, and hence the sum of the positive $s_P$ is at least $3\gamma L/8$. Some piece must consequently satisfy $s_P/|P|\ge \gamma /4$. On that piece,Also, once $\gamma Q\ge 400$, $|P|\ge \ell \ge \gamma \sqrt{L}/800$. This proves the one-frequency density increment on an integer interval: for absolute constants $c,C>0$, if $L\ge C{\rho }^{-6}$, there is a arithmetic progression $P\subseteq [1,L]$ of length at least $c{\rho }^3\sqrt{L}$ on which the subset density rises by at least $c{\rho }^3$. For example one may decrease $c$ to $10^{-7}$ and choose $C$ large enough for the displayed elementary inequalities.

Now fix $0<\delta \le 1$ and suppose the required conclusion fails. On each new arithmetic progression, identify its points in order with $[1,L^{\prime }]$. This affine identification preserves all integer arithmetic progressions, so its copy of $A\cap P$ is still progression-free. All densities remain at least $\delta$. Put $\epsilon =c{\delta }^3$, $H=C{\delta }^{-6}$, and choose an integer $t>1/\epsilon$. Every increment raises the subset density by at least $\epsilon$ and changes the length by $L_{j+1}\ge \epsilon \sqrt{L_j}$. Iterating the latter inequality givesChoose $N>(H/{\epsilon }^2{)}^{2^t}$. Then every length up to step $t$ exceeds $H$, so every increment is available. After $t$ increments the subset density exceeds one, a contradiction. ThusFor $\delta >1$ there are no sets satisfying the subset density assumption. The argument above proves the substantive case without invoking the theorem being asked for.