Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 76 / 5

Use ${\Delta }_{h}f(x)=f(x+h)\overline{f(x)}$ and the unnormalized Fourier transformWe state the eighth-power convention explicitly. A bounded function is $\alpha$-quadratically uniform ifwhere $\mathcal{C}$ means conjugation and every expectation is uniform. The expression is real and nonnegative. Applying Parseval identity on a finite group to the multiplicative derivatives gives the equivalent conditionThis is the Derivative identity for the Gowers U3 norm. A set $A$ of density $\delta$ is quadratically uniform when its balanced subset indicator $1_A-\delta$ satisfies that condition, not when its unbalanced indicator does. Some definitions bound $\Vert f{\Vert }_{U^3}$ rather than its eighth power; that norm parameter is ${\alpha }^{1/8}$ in the convention here. Quadratic uniformity controls three-dimensional additive cubes and four-term progression counts; small ordinary linear Fourier coefficients alone are not the same condition.

Define the normalized progression averageWe first prove the needed four-term progression bound with torsion factors, valid for the arbitrary cyclic modulus in the question:For the factor at slope zero, write $y=x+3d$, $t=d$, and apply Cauchy-Schwarz to discard $f_3(y)$:For concise bookkeeping put $D_{h}f(x)=f(x)\overline{f(x+h)}$, the conjugate of the preceding derivative convention, and $F_{i,h}=D_{-(3-i)h}{f}_i$. Expanding the square with $t,t+h$ and setting $z=y-t$ gives ${\mathbb{E}}_{z,t,h}{F}_{0,h}(z-2t)F_{1,h}(z-t)F_{2,h}(z)$. Another Cauchy-Schwarz, now in $(z,h)$, discards $F_{2,h}$ and yieldsExpand with $t,t+k$, set $w=z-t$, and apply Cauchy-Schwarz a third time to discard $D_{-k}{F}_{1,h}(w)$. The remaining $t$-average is a full average over $x=w-t$, soNegating $h,k$ does not change the average. The nonnegative integrand depends only on $2k$ and $3h$. Averaging over the subgroups $2{\mathbb{Z}}_N$ and $3{\mathbb{Z}}_N$ costs at most their indices $t_2=\gcd (2,N)$ and $t_3=\gcd (3,N)$ compared with unrestricted increments. The unrestricted expression is exactly $\Vert f_0{\Vert }_{U^3}^8$, obtained by expanding the last squared average. Thus the bound is $(t_2{t}_3{)}^{1/8}\Vert f_0{\Vert }_{U^3}\le 6^{1/8}\Vert f_0{\Vert }_{U^3}$.

Reversing the progression handles slope three. To isolate slope one, eliminate slopes three, zero and two successively in the same calculation; the surviving increments have coefficients $-2,1,-1$, so only the index $t_2$ is needed. Reversal handles slope two. This proves the minimum-over-factors bound, without assuming that $2$ or $3$ is invertible modulo $N$.

If $\delta =0$, the requested lower bound is zero and is immediate. Otherwise let $g=1_A-\delta$. Telescoping the product of four copies of $1_A=\delta +g$ against the constant product ${\delta }^4$ gives four error terms, each with one factor $g$ and three factors bounded by one. HenceIn particular, choosing $\alpha \le {\delta }^{32}/(6\cdot 8^8)$ givesThese are ordered modular progression parameters $(x,d)$, as requested; small-order differences and $d=0$ are included. A claim of distinct nontrivial progressions would need a separate subtraction and size condition.

Normalize the selected derivative correlations byThey satisfy $|c_k|\le 1$, and the hypothesis gives ${\sum }_{k\in B}|c_k{|}^2\ge \alpha N$. Thus $T={\sum }_{k\in B}|c_k|\ge \alpha N$. Choose phases ${\omega }_k$ of modulus one so that ${\omega }_k{c}_k=|c_k|$, with any choice if $c_k=0$. ThenOne Cauchy-Schwarz givesExpand into pairs $a,b\in B$ and group them by $t=a-b$, $r=\varphi (a)-\varphi (b)$. After translating $y=x+b$, the derivative correlation isThe grouped coefficient isConsequently $T^2\le {\sum }_{t,r}\gamma (t,r)C(t,r)$, whose right-hand side is the real nonnegative expanded square. A second Cauchy-Schwarz yieldsIf $m(t,r)$ counts the pairs in the displayed coefficient, then $|\gamma (t,r)|\le m(t,r)$. Thus $\sum |\gamma {|}^2\le \sum m^2=E(\Gamma )$, the additive energy of a frequency graph $\Gamma =\{(k,\varphi (k)):k\in B\}$. Equality of two pair differences is equivalent, by permuting the four labels, to equality of the two requested sums in each coordinate.

For every $t$, Parseval identity on a finite group in normalized Fourier convention givesThere are $N$ shifts, so $T^4\le NE(\Gamma )$. Since $T\ge \alpha N$, we concludeThis is precisely the number of ordered $(a,b,c,d)\in B^4$ satisfying both $a+b=c+d$ and $\varphi (a)+\varphi (b)=\varphi (c)+\varphi (d)$, all modulo $N$. It proves squared derivative correlations force frequency-graph energy with the constant and exact equalities stated, rather than approximate frequency matching.