Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 82 / Section A

The Fejér kernel for the printed normalization isIts endpoint coefficients at $\pm N$ are zero. It is nonnegative, and integrating its Fourier series gives ${\int }_{\mathbb{T}}{F}_N=1$. Hence ${\tilde{S}}_N(f)=f\ast F_N$ andThus the absolute constant can be $C=1$. Positivity of the Fejér kernel, rather than mere boundedness of its individual Fourier coefficients, is the reason this holds.

We give a quantitative construction for the whole open set, including the case of infinitely many component intervals. Pointwise divergence at a single point is insufficient here. Write $z=e^{2\pi it}$ and use normalized measure on $\mathbb{T}$.

First let $K$ be a finite union of closed arcs with $|K|\le 3\epsilon$, where $\epsilon$ is small. Choose a smooth function nonnegative function $u$, equal to one on $K$, with $u\le 1$ and $\eta :=\int u\le 6\epsilon$. This follows by making the transition regions around the finitely many arc endpoints sufficiently short. Define the analytic Herglotz integralIts real part is $(\eta +P[u](z))/(2\eta )>0$, where $P[u]$ is the disk Poisson integral, and $G(0)=1$. Smoothness of $u$ makes $G$ continuous and smooth function up to the circle; on $K$ its boundary real part is at least $1/(2\eta )$. The branch $H=\log G$ therefore has $H(0)=0$ andThe complex logarithm has no branch problem because $G$ lies in the right half-plane. Its smooth function boundary values give absolutely summable Taylor coefficients. Choose an analytic polynomial $A$, with zero constant coefficient, uniformly within $1/4$ of $H$ on the closed disk. Put $P(t)=\mathrm{Im}A(e^{2\pi it})/\pi$. Then $\Vert P{\Vert }_{\infty }\le 1/2+1/(4\pi )<1$.

If $D=\mathrm{deg}A$ and $m>D$, the modulated trigonometric polynomial $Q(t)=e^{2\pi imt}P(t)$ has spectrum in $[m-D,m+D]$. A sharp Fourier partial sum at $m$ selects exactly the negative-frequency half of $P$, since $A(0)=0$:Consequently, for every $t\in K$,The carrier frequency $m$ can be arbitrarily large. This compact-set Fourier amplification lemma uses a bounded imaginary part to obtain a large one-sided Fourier multiplier; no assertion about pointwise divergence is used.

Now let $E$ be any open set of measure $\delta$. DefineIf this is zero take $f=0$. Otherwise choose a decreasing sequence ${\delta }_j\downarrow 0$, with ${\delta }_1=\delta$, so small for $j\ge 2$ thatFor $j=1$ this inequality also holds: $B(\delta )>4$ and $B(\delta )/2\ge B(\delta )/4+1$. Let $L_j$ be an increasing compact exhaustion of $E$.

Using compact batching of a small open set, we recursively choose finite closed-arc unions $K_j\subset E$, of measure at most $3{\delta }_j$, such that $L_j$ is covered by the interiors of $K_1,\dots ,K_j$ andHere is the measure-theoretic justification. At step $j$, the uncovered portion has measure at most ${\delta }_j$. Its intersection with $L_j$ after removing the previously covered interiors is compact, and can be covered by finitely many closed arcs lying in $E$ with total length at most ${\delta }_j$ plus an arbitrarily small excess. By inner regularity of Lebesgue measure, a further compact subset of the uncovered open portion captures all but ${\delta }_{j+1}$ of its measure; cover it similarly. The combined finite arc union has length at most $3{\delta }_j$. This both covers the outstanding part of $L_j$ and ensures the next remainder estimate. In particular $E\subset {\bigcup }_j{K}_j$.

Apply the finite-arc construction to $K_j$ to get $P_j$ of degree $D_j$. Choose carriers $m_j$ recursively so that the bands $[m_j-D_j,m_j+D_j]$ are positive and strictly ordered with no overlap. SetThe frequency-separated Fourier block series and the Weierstrass M-test give a continuous complex-valued function with $\Vert f{\Vert }_{\infty }\le {\sum }_j{2}^{-j}=1$. At the cutoff $m_j$, all earlier bands have been included in full and every later band is absent. Thus on $K_j$,Every point of $E$ lies in some $K_j$, soValues of $\psi$ outside $0<\delta \le 1$ may be defined arbitrarily. The construction handles an arbitrary open set, not just one point or a finite collection of intervals. It also prevents cancellation explicitly: earlier bands contribute a bounded complete sum, and later bands contribute nothing at the chosen cutoff.

The printed interval has a typo: its right endpoint must be $2/3+\delta$, rather than $1/3+\delta$. Keep $\varphi$ as an unnormalized indicator function; only $\psi$ is normalized to have integral one. Thus useWith the Boolean literal printed interval, the convolution vanishes at $2/3$, so a sequence consisting entirely of $2/3$ is already a counterexample to the proposed majorization. For the corrected interval and every $t\in [1/3,2/3]$, all $t-s$ with $|s|\le \delta$ belong to the support of $\varphi$. Hence $f(t)=1$ there. Everywhere $0\le f\le 1$, giving $S_N\le {\sum }_{n\le N}f(u_n)$, including interval endpoints.

Put $L=1/3+2\delta$. For $k\ne 0$, direct integration and the convolution theorem give the exact Fourier coefficientsIn particular,This also supplies the required justification of the Fourier series. The convolution $f$ is continuous: it is the overlap length of two moving intervals, divided by $2\delta$. Its Fourier coefficients are absolutely summable, so the corresponding Fourier series converges uniformly to a continuous function $g$. Termwise integration gives $\hat{g}(k)=\hat{f}(k)$; the permitted uniqueness assertion applied to $f-g$ yields $g=f$.

Let $A_k={\sum }_{n\le N}{e}^{2\pi ik{u}_n}$. Split the absolutely convergent Fourier series at the arbitrary real cutoff $M>1$:There are fewer than $2M$ low-frequency terms and each coefficient has magnitude at most $L<1$. For the tail, $|A_k|\le N$ and ${\sum }_{|k|\ge M}{k}^{-2}\le C/M$. ThereforeThis proves the requested interval discrepancy estimate without an unjustified sharp truncation of the Fourier series.

For a nonzero integer $b$, let $a$ be the nearest integer to $b\sqrt{2}$. The integer $a^2-2b^2$ is nonzero, and $|a|\le \sqrt{2}|b|+1/2$. The suggested factorization yieldsThis is stronger than the stated lower bound for the fractional part, since the distance to the nearest integer is at most that fractional part. For $u_n=n\sqrt{2}$, the geometric series and $|\sin (\pi t)|\ge 2\Vert t{\Vert }_{\mathbb{R}/\mathbb{Z}}$ giveThus the low-frequency contribution is $O(M^2)$. Choose $M=\lceil 100N^{2/5}\rceil$ and $\delta =(2\sqrt{M}{)}^{-1}$; these choices satisfy $0<\delta <1/10$ for every $N\ge 1$. Both $\delta N$ and $N/(\delta M)$ are $O(N^{4/5})$, and so is $M^2$. Hence