Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 12 / 2

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.

Since $a$ is an indicator function of density of a finite subset $\alpha$, the triangle inequality gives $|\hat{a}(r)|\le {\mathbb{E}}_{x}a(x)=\alpha$ for every frequency. Also, character orthogonality and the Parseval identity on a finite group giveFor completeness, the character orthogonality used here iswhich follows by summing a finite geometric series. Expanding the squared Fourier coefficients on a finite abelian group and using this identity proves the displayed Parseval identity on a finite group directly.

Combining the uniform bound with that identity gives the fourth-moment boundIn particular, the Lp norm on the frequency side here is a sum, not a normalized average. This is the fourth Fourier moment bound for an indicator function.

For the physical-space L2 norm use $\Vert h{\Vert }_2^2={\mathbb{E}}_x|h(x){|}^2$. By part (i) and the Parseval identity on a finite group,Split this sum into the large spectrum $K$ and its complement. Because $u$ belongs to the Bohr set in the question, for $r\in K$,Consequently the contribution from $K$ is at most ${\epsilon }^2{\sum }_r|\hat{a}(r){|}^4\le {\epsilon }^2{\alpha }^3$ by part (ii). Outside $K$, $|\hat{a}(r)|<\theta$, and $|1-{\omega }^{-ru}|\le 2$. Thus that contribution is at mostAdding the estimates provesThe argument also covers empty $A$ or empty $K$. As usual the radius and threshold are nonnegative; a negative radius makes the premise empty whenever $K$ is nonempty. The estimate expresses Bohr-set almost periodicity of a convolution: a translation of a function by an element of the Bohr set barely changes the large Fourier coefficients on a finite abelian group, while the small ones have little total energy.