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

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 $E$.