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If a subset $A$ of a cyclic group has normalized additive energy at most ${\alpha }^4+ϵ$, every nonconstant Fourier coefficient on a finite abelian group of its indicator has modulus at most ${ϵ}^{1/4}$. Expanding the count of $a+c=2b$ gives the frequency product $\widehat{1_A}(r)\widehat{1_C}(r)\widehat{1_B}(-2r)$. The Cauchy-Schwarz inequality and Parseval identity on a finite group bound its nonconstant part by the displayed quantity. Character doubling has multiplicity $\gcd (2,N)$, so no assumption that the modulus is odd is needed. The error is a normalized density error multiplied by $N^2$.

Suppose $|f|\le 1$ on a finite cyclic group, $H$ has density $\beta$, and for each $h\in H$ the multiplicative derivative has a Fourier coefficient on a finite abelian group of magnitude at least $\eta$ at $\theta (h)$. Align the correlation phases, apply Cauchy-Schwarz, group pairs by their shift and frequency differences, and apply Parseval identity. This proves the displayed lower bound on additive energy of a frequency graph without replacing exact frequency equalities by approximate ones.

There is a normalization omission in the printed conclusion: the discrepancy of a count of pairs must be bounded by a small function of the energy excess times $N^2$. We prove this normalized result, then give a counterexample to the literal bound without $N^2$. Write $ϵ$ for the energy excess parameter to avoid confusing it with an element of the third set.

For Fourier analysis on a finite abelian group, use $e_N(t)=e^{2\pi it/N}$ andCharacter orthogonality, ${\mathbb{E}}_x{e}_N(rx)=1$ for $r=0$ and zero otherwise, gives Fourier inversion and the Parseval identity on a finite group:Put $u=1_A$, $v=1_B$, $w=1_C$. Expanding the constraint $a_1+a_2=a_3+a_4$ by the same character orthogonality gives the additive energy identityThus ${\sum }_{r\ne 0}|\hat{u}(r){|}^4\le ϵ$, and every nonconstant Fourier coefficient on a finite abelian group of $u$ has modulus at most ${ϵ}^{1/4}$. Expanding the constraint in the triple count $T$ givesThe Cauchy-Schwarz inequality yieldsThe first squared norm is $\gamma$ by the Parseval identity on a finite group. For the second, multiplication by $2$ on ${\mathbb{Z}}_N$ has kernel size $h=\gcd (2,N)$ and every element of its image has exactly $h$ preimages. HenceThis proves that additive energy controls three-term progression mixing, including for even moduli:Thus the intended density-error function can be $F(ϵ)=\sqrt{2}{ϵ}^{1/4}$, which tends to zero. For odd moduli the factor $\sqrt{2}$ can be replaced by one.

To see why the literal unnormalized conclusion is false, take $N=q^2$ and let all three sets be the subgroup of multiples of $q$. Their densities are $1/q$. Every choice of three entries of that subgroup determines the fourth entry of an additive quadruple, so its energy is $q^3$. The excess parameter isEvery pair $(a,b)$ in the subgroup determines $c=2b-a$ there, giving $T=q^2$, whereas $\alpha \beta \gamma N^2=q$. Their difference is $q^2-q\to \infty$, which cannot be bounded by any function of $ϵ$ tending to zero. The missing $N^2$ is therefore a genuine source defect, not an alternative Fourier transform normalization.

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.

Write $\beta =|H|/M\gg {\delta }^8$ and $\eta =(\kappa /2{)}^{1/2}\gg {\delta }^4$. Choose unit complex numbers $b_h$ so thatis real and nonnegative. Cauchy-Schwarz, using $|F|\le 1$, bounds $T^2$ byAfter setting $y=x+k$, each summand is a unit phase times the Fourier coefficient on a finite abelian group of ${\partial }_{h-k}F$ at frequency $\theta (h)-\theta (k)$.

Let $r(d,\xi )$ count pairs $(h,k)\in H^2$ with $h-k=d$ and $\theta (h)-\theta (k)=\xi$. Grouping terms, another Cauchy-Schwarz and the Parseval identity yieldwhere $E_{\theta }(H)=\sum r(d,\xi {)}^2$ is the additive energy of a frequency graph. The last inequality uses ${\sum }_{\xi }|\widehat{{\partial }_{d}F}(\xi ){|}^2=\mathbb{E}|{\partial }_{d}F{|}^2\le 1$ for each $d$.

ThusThe bound proves that derivative correlations force additive frequency energy. This energy counts exactly the ordered quadruples satisfying the two requested additive relations, after relabelling the difference equality as a sum equality. Shift equalities initially hold modulo $M$, but the shifts lie in $[-N,N]$ and $M>8N$, so they are also integer equalities. Frequency equalities hold in $\mathbb{R}/\mathbb{Z}$, as required. Consequently a common exponent $C=48$ works for both conclusions, after adjusting absolute implicit constants and using $0<\delta \le 1$.

For the quadratic phase, the multiplicative derivative on the overlap is ${\partial }_h{f}_0(x)=e(2\alpha hx+\alpha h^2)$. Hence an explicit choice isTake $|h|\le \lfloor N/2\rfloor$. The correlation magnitude is $(N-|h|)/M\ge 1/32$, and every additive quadruple of shifts satisfies the frequency relation. The interval of shifts has $\gg N^3$ such quadruples by Cauchy-Schwarz. These particular real frequencies need not belong to the grid used to prove existence for general $f$; evaluating them at the interval's integer coordinates gives the stated exact formula.

A genuinely nonlinear example is the bracket-linear frequency functionIt has $\gg N^3$ exact additive quadruples, yet agrees with any affine function $ah+b(\mathrm{mod}1)$ at only $o(N)$ shifts, uniformly in the affine function. The question permits giving this example without proof, but the mechanism is useful. For a pair with sum $s$, the value $\lfloor \sqrt{2}{h}_1\rfloor +\lfloor \sqrt{2}{h}_2\rfloor$ has only two possibilities, $\lfloor \sqrt{2}s\rfloor$ or $\lfloor \sqrt{2}s\rfloor -1$. There are $O(N)$ pair classes, so Cauchy-Schwarz produces $\gg N^3$ collisions with both additive relations.

To see why no long affine agreement occurs, three agreement points force the corresponding lattice points $(h,\lfloor \sqrt{2}h\rfloor )$ to be collinear: eliminating the affine slope gives $\sqrt{3}$ times an integer determinant equal to an integer, and irrationality makes that determinant zero. A rational line of slope $p/q$ contains agreement shifts in one residue class modulo $q$, and closeness of $\lfloor \sqrt{2}h\rfloor$ to $\sqrt{2}h$ bounds their span by $1/|\sqrt{2}-p/q|$. Lines with $q$ large have arbitrarily small possible density; for bounded $q$, irrationality of $\sqrt{2}$ bounds the number uniformly. Thus the maximum agreement is $o(N)$.

The Roth theorem on three-term arithmetic progressions says that, for each fixed $\delta >0$, a sufficiently long interval cannot have a subset of density of a finite subset at least $\delta$ with no nonconstant three-term arithmetic progression. Equivalently, the maximum size of a three-term-progression-free subset of $[N]=\{1,\dots ,N\}$ is $o(N)$. Here is a density increment proof, including its analytic ingredients.

Write $e(t)=\exp (2\pi it)$ and embed $I=[N]$ into the cyclic group $G=\mathbb{Z}/(2N+1)\mathbb{Z}$ of order $M$. For normalized Fourier analysis on a finite abelian group take $\hat{h}(r)=M^{-1}{\sum }_{x}h(x)e(-rx/M)$. The orthogonality of complex exponentials follows from a finite geometric series: its average is zero for a nonzero frequency and one for frequency zero. Expanding in these linear phases consequently proves both Fourier inversion on a finite group and the Parseval identity on a finite group. In particular,The second identity follows by summing the frequency expansion over $x,d$: the two constraints leave precisely the frequencies $(r,-2r,r)$. Since $M$ is odd, multiplication by two permutes the frequencies. Also a congruence $x+z=2y$ with $x,y,z\in I$ is an equality in the integers, since $|x+z-2y|<M$.

Let $A\subseteq I$ have subset density $\alpha =|A|/N\ge \delta$, and suppose it contains no nonconstant three-term arithmetic progression. Its balanced indicator function of a finite subset is $f=1_A-\alpha 1_I$, extended by zero outside $I$. Then $\hat{f}(0)=0$. Counting endpoints with the same parity givesIf $N\ge 4{\alpha }^{-2}$, their difference in absolute value is at least ${\alpha }^3{N}^2/(4M^2)$. Telescope the difference of the products into three terms, each containing $f$ and two factors chosen from $1_A,\alpha 1_I$. Put $s={\max }_r|\hat{f}(r)|$. The Cauchy-Schwarz inequality and the Parseval identity on a finite group bound each term by $s\alpha N/M$, since the squared normalized $L^2$ norm of either other factor is at most $\alpha N/M$. Therefore some nonzero frequency $r$ satisfiesThis is the Fourier detection of a progression-free subset of an interval step; it has just been proved rather than assumed.

We now turn this Fourier coefficient on a finite abelian group into a density increment. Set $Q=\lfloor \sqrt{N}\rfloor$. Partition the unit interval into $Q$ equal pieces and place the $Q+1$ fractional parts of $0,r/M,\dots ,Qr/M$ in them. The pigeonhole principle gives $1\le q\le Q$ with $\Vert qr/M\Vert \le 1/Q$. This proves the particular Dirichlet approximation theorem needed here. Put $\kappa ={\alpha }^2/(1000\pi )$ and $L=\lfloor \kappa Q\rfloor$. For sufficiently large $N$ depending only on $\delta$, $L\ge \kappa Q/2$ and $2L\le Q$. Each residue chain modulo $q$ in $I$ has at least $Q$ elements. Split each such chain into arithmetic progressions with between $L$ and $2L-1$ elements, absorbing its final short remainder into the previous block.

The linear phase changes by at most $4\pi L/Q\le {\alpha }^2/250$ within any block. Freezing the linear phase at the first element of each block therefore changes the preceding correlation by at most ${\alpha }^{2}N/250$, because $|f|\le 1$. The triangle inequality givesThe block sums of the balanced indicator function of a finite subset add to zero, so their positive total is half their absolute total. Some block consequently satisfiesHere $c>0$ is absolute, and shrinking it absorbs the floors. An affine parametrization of $P$ identifies $A\cap P$ with a subset of $[|P|]$ and preserves arithmetic progressions.

Repeat this Roth density-increment step. Every step raises the subset density by at least ${\delta }^2/144$, so fewer than $T=\lceil 144/{\delta }^2\rceil +1$ steps are possible. All the lengths satisfy $N_{j+1}\ge c{\delta }^2\sqrt{N_j}$ while above a threshold depending only on $\delta$. Choose the initial $N$ so large that this lower recursion stays above that threshold for $T$ steps; this is possible by working backwards through finitely many squarings. We would then force a subset density greater than one. Thus every fixed positive density eventually forces a nonconstant three-term arithmetic progression. For an infinite subset of the positive integers with positive upper asymptotic density, apply the finite conclusion along initial intervals whose subset density is bounded away from zero; this gives the usual infinite formulation as well.

Use normalized Fourier analysis on a finite abelian group, identifying $A$ with its indicator function $a=1_A$ on the cyclic group ${\mathbb{Z}}_n$. The conventions areThe last operation is normalized convolution on a finite group. With these conventions the Fourier coefficients on a finite abelian group use a normalized average, while sums over frequencies use counting measure.

Expanding the normalized convolution on a finite group and putting $z=x-y$ givesFor the translation of a function $g(x)=f(x-u)$, putting $z=x-u$ yieldsThe two transforms are thereforeThe negative sign in the translation factor follows from the negative sign in our Fourier coefficient on a finite abelian group convention.