Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 16 / 3 / i / Solution

The expectations use normalized counting measure on a finite abelian group $G$. This is the finite-group convention implicit in the averaging notation; an arbitrary infinite group has no normalized counting probability measure on all its points. For $h\in G$, define ${\Delta }_{h}f(x)=f(x)\overline{f(x+h)}$, and let $\mathcal{C}$ denote complex conjugation. The Gowers U3 norm isAveraging first over $h_3$ gives the Derivative identity for the Gowers U3 normIndeed, if $u(x)={\Delta }_{h_1}{\Delta }_{h_2}f(x)$, the average over $(x,h_3)$ is ${\mathbb{E}}_{x,h_3}u(x)\overline{u(x+h_3)}=|{\mathbb{E}}_{x}u(x){|}^2$. The eighth root is consequently well-defined. Homogeneity follows because the cube has four unconjugated and four conjugated factors, so $\Vert af{\Vert }_{U^3}=|a|\Vert f{\Vert }_{U^3}$. If the norm is zero, every squared term in the finite average vanishes. Taking $h_1=h_2=0$ gives ${\mathbb{E}}_x|f(x){|}^4=0$, so $f=0$ pointwise. This proves positive definiteness.

For the triangle inequality, we prove the needed Gowers-Cauchy-Schwarz inequality. For eight functions $F_{\omega }$ on a product of finite sets $X_1\times X_2\times X_3$, form the box norm multilinear averageFix a coordinate $j$. After averaging its two variables separately, this is $\mathbb{E}A\overline{B}$, where $A$ uses all vertices with ${\omega }_j=0$ and $B$ those with ${\omega }_j=1$, with the remaining parity conjugations. The Cauchy-Schwarz inequality giveswhere $S_j^b$ replaces the function at each vertex by the original function with its $j$th index set to $b$. Each right-hand average is a squared absolute value averaged over the other coordinates, hence nonnegative. Apply this operation successively for $j=1,2,3$, to every resulting factor. The eight final families each use one original $F_{\omega }$ at all vertices. Taking roots gives the box Cauchy-Schwarz inequalityNow take $X_j=G$ and $F_{\omega }(a,b,c)=f_{\omega }(a+b+c)$. With $x=a_0+b_0+c_0$ and $h_j=x_j^1-x_j^0$, these independent variables induce uniform $(x,h_1,h_2,h_3)$: each such tuple has exactly $|G{|}^2$ preimages. Therefore each box norm is the corresponding $U^3$ norm, and the mixed cube obeysExpand the eight factors in the cube expression for $f+g$. Its $2^8$ mixed terms are bounded by this inequality, and their bounds sum to $(\Vert f{\Vert }_{U^3}+\Vert g{\Vert }_{U^3}{)}^8$. Thus $\boxed{\Vert f+g{\Vert }_{U^3}\le \Vert f{\Vert }_{U^3}+\Vert g{\Vert }_{U^3}}$, completing the norm proof.

Two consequences will be useful below. Multiplication by a character of a finite abelian group leaves the norm unchanged, because the alternating character product around a three-dimensional cube equals one. Alsowhere the last inequality uses the $h=0$ term and $\Vert u{\Vert }_{U^2}^4={\sum }_{\chi }|\hat{u}(\chi ){|}^4\ge |\mathbb{E}u{|}^4$. Hence $\Vert f{\Vert }_{U^3}\ge |G{|}^{-1/8}\Vert f{\Vert }_2$.