Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 8 / 2

We prove the Kahane-Katznelson divergence theorem through an explicit small-norm block construction. Let $\mu$ be normalized Lebesgue measure on the circle.

First establish the compact-set Fourier amplification lemma. If a compact set $K$ satisfieswe can make a trigonometric polynomial $q$ with $\Vert q{\Vert }_{\infty }\le \epsilon$, supported in any sufficiently high interval of positive frequencies, whose partial prefix has magnitude greater than $M$ on $K$.

To construct it, choose a smooth nonnegative function $u$ equal to one near $K$, with values at most one and meanOuter regularity and a smooth cutoff give this choice; the stipulated bound on $\mu (K)$ leaves room between the two exponentials. The Schwarz integral on the unit diskhas positive real part in the disk, $H(0)=\delta$, and boundary real part $u$. Its holomorphic logarithmsatisfies $W(0)=0$ and $|\mathrm{Im}W|<\pi /2$. On $K$, the boundary value has $\mathrm{Re}W\ge \log (1/\delta )>4\pi M/\epsilon$. Smoothness of $u$ makes $H$ continuous at the boundary, and its positive boundary real part near $K$ makes $W$ continuous there.

Choose a radius just below one, then truncate the Taylor series of $W$ at that radius. This gives an analytic polynomial $R(e^{it})$ with zero constant term, degree $d$, andThe radial function is analytic beyond the closed unit disk, so the Taylor truncation is uniform on the whole circle. For $L>d$, putIts frequencies lie between $L-d$ and $L+d$, all positive. Its prefix through frequency $L$ includes exactly the negative-frequency half of $\mathrm{Im}R$ shifted into this interval:Consequently $|S_{L}q(t)|>\epsilon (4\pi M/\epsilon -1)/(2\pi )>M$ on $K$, whereas $\Vert q{\Vert }_{\infty }<\epsilon$. Increasing $L$ places the entire block above any previously used frequency.

We next use compact batching of a small open set to handle an arbitrary null set, without assuming that it is compact or a countable union of compact null sets. SetFor each $j$, choose an open $U_j\supseteq E$ with $\mu (U_j)<\exp (-8\pi j/{\epsilon }_{j,0})$. Decompose $U_j$ into countably many closed subarcs with pairwise disjoint interiors: subdivide each open component into closed pieces accumulating only at its excluded endpoints. Group these subarcs into finite successive batches $K_{j,k}$. After batch $k$, include enough pieces that the remaining total length is less than $\exp (-8\pi j/{\epsilon }_{j,k+1})$. Require each batch endpoint in the enumeration to increase. ThenEach batch is compact; endpoints shared by pieces have zero measure and do not affect the estimates.

Enumerate the pairs $(j,k)$ in diagonal order. Apply the block lemma with target $M=j$ to each $K_{j,k}$, and shift its spectrum above all preceding blocks. Denote the resulting polynomial by $q_{j,k}$ and setSince ${\sum }_{j,k}{\epsilon }_{j,k}=1/2$, this series is uniformly convergent and defines a continuous complex-valued function.

Fix $t\in E$. For every $j$ there is a $k$ with $t\in K_{j,k}$. The difference between the Fourier partial sum just before that block and the sum at its midpoint has magnitude greater than $j$: previous blocks cancel in the difference, and future blocks have not yet entered. As $j\to \infty$, these cutoffs tend to infinity. At least one of the two partial sums therefore has magnitude greater than $j/2$. The Fourier partial sums are unbounded, hence not Cauchy, at $t$. We have provedThis establishes divergence on every prescribed null set, including dense nonclosed null sets; it does not assert that the divergence set is exactly $E$.

Use the uniform bound for harmonic sine polynomialsFor completeness, reduce to $0<t\le \pi$ and split at $K=\min (m,\lfloor 1/t\rfloor )$. The first part is bounded by $Kt\le 1$. Geometric-series summation bounds every interval sum of $\sin kt$ by $1/\sin (t/2)\le \pi /t$. Summation by parts bounds the remaining harmonic-weighted tail by $\pi /[t(K+1)]\le \pi$. Negative $t$ follows by oddness and $t=0$ is immediate.

Let $H_m={\sum }_{k=1}^{m}1/k$. Choose $m_j$ so large that $H_{m_j}\ge C_0{2}^{j+2}$, and positive integers $L_j>m_j$ such that the intervals $[L_j-m_j,L_j+m_j]$ are strictly separated and increase. DefineThe bound $\Vert Q_j{\Vert }_{\infty }\le 2^{-j-2}$ makes this a continuous function with $g(0)=0$.

Each Fourier coefficient of $P_m$ has modulus $1/(2k)$, so the sum of the absolute coefficients is $H_m$. Therefore every prefix of a normalized block $Q_j$ has norm at most one. At any Fourier cutoff, all earlier blocks are complete, at most one block is partial and all later blocks are absent. HenceAt zero, a completed block contributes zero, but its prefix through frequency $L_j$ consists of the negative-frequency sine coefficients and equals $i/2$. ThusBoth index sequences tend to infinity. The uniformly bounded partial sums fail to converge at the origin, even though the function is continuous and zero there.