= 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 $K$ satisfies
$$
\mu(K)\leq\exp(-8\pi M/\varepsilon),\qquad0<\varepsilon\leq1,\quad M\geq1,
$$
we can make a <trigonometric polynomial> $q$ with $\|q\|_\infty\leq\varepsilon$, 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 mean
$$
0<\delta=\int u\,d\mu<\exp(-4\pi M/\varepsilon).
$$
<Outer 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 disk>
$$
H(z)=\int_{\mathbb T}\frac{e^{it}+z}{e^{it}-z}\,u(t)\,d\mu(t)
$$
has positive real part in the disk, $H(0)=\delta$, and boundary real part $u$. Its <holomorphic logarithm>
$$
W(z)=\log H(z)-\log\delta
$$
satisfies $W(0)=0$ and $|\operatorname{Im}W|<\pi/2$. On $K$, the boundary value has $\operatorname{Re}W\geq\log(1/\delta)>4\pi M/\varepsilon$. 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 $d$, and
$$
|\operatorname{Im}R(t)|<\pi,\qquad
\operatorname{Re}R(t)>4\pi M/\varepsilon-1\quad(t\in K).
$$
The radial function is analytic beyond the closed unit disk, so the Taylor truncation is uniform on the whole circle. For $L>d$, put
$$
q(t)=\frac{\varepsilon}{\pi}e^{iLt}\operatorname{Im}R(t).
$$
Its frequencies lie between $L-d$ and $L+d$, all positive. Its prefix through frequency $L$ includes exactly the negative-frequency half of $\operatorname{Im}R$ shifted into this interval:
$$
S_Lq(t)=-\frac{\varepsilon}{2\pi i}e^{iLt}\overline{R(t)}.
$$
Consequently $|S_Lq(t)|>\varepsilon(4\pi M/\varepsilon-1)/(2\pi)>M$ on $K$, whereas $\|q\|_\infty<\varepsilon$. 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. Set
$$
\varepsilon_{j,k}=2^{-j-k-2},\qquad j\geq1,\ k\geq0.
$$
For each $j$, choose an open $U_j\supseteq E$ with $\mu(U_j)<\exp(-8\pi j/\varepsilon_{j,0})$. 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 $\exp(-8\pi j/\varepsilon_{j,k+1})$. Require each batch endpoint in the enumeration to increase. Then
$$
U_j=\bigcup_{k\geq0}K_{j,k},\qquad
\mu(K_{j,k})\leq\exp(-8\pi j/\varepsilon_{j,k}).
$$
Each 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 set
$$
f=\sum_{j\geq1,\ k\geq0}q_{j,k}.
$$
Since $\sum_{j,k}\varepsilon_{j,k}=1/2$, 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 proved
$$
\boxed{E\subseteq\{t:\limsup_N|S_Nf(t)|=\infty\}.}
$$
This establishes divergence on every prescribed null set, including dense nonclosed null sets; it does not assert that the divergence set is exactly $E$.
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