Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 7 / 4

Put $w(x)=(\sin x/x{)}^2$, with $w(0)=1$, and $M={\sup }_r|A_r|<\infty$. Since $0\le w\le 1$, the finite-prefix bound is $\left|{\sum }_{r=1}^n{A}_{r}w(rk)\right|\le nM$. For $k\ne 0$, $w(rk)\le 1/(r^2{k}^2)$ gives absolute convergence of the tail and the correct estimateThe PDF omits the factor $k^{-2}$ from this bound. Its printed version is false: choose $A_4=1$ and all other $A_r=0$, take $n=3$, and let $k\to 0$. The left side tends to one while the printed right side is $2/3$. Thus the intended bound must be qualified as above. At $k=0$, even absolute convergence need not hold for a general sequence tending to zero.

For each fixed $n$, the displayed finite-prefix limit follows immediately: $|k{\sum }_{r=1}^n{A}_{r}w(rk)|\le |k|nM\to 0$. The stronger statement needed for part (ii) isHere is a proof, so the fixed-prefix formula is not silently substituted for an infinite-series result. Set $N=\lfloor 1/|k|\rfloor$ for small nonzero $k$. The multiplied tail is at most $2{\sup }_{r>N}|A_r|/(N|k|)$, which tends to zero because $N|k|\to 1$. For the prefix, choose $J$ with $|A_r|<\eta$ for $r>J$. ThenIts upper limit is at most $\eta$, and $\eta$ is arbitrary. This proves sinc-squared summation of a sequence tending to zero, including approach from either sign of $k$.

For the unheaded piecewise-affine continuation, continuity gives $Ac+B=A^{\prime }c+B^{\prime }$. Evaluate the asserted symmetric quotient at $t=c$. For sufficiently small positive $h$,Its zero limit forces $\boxed{A=A^{\prime }}$, and the continuity relation then forces $\boxed{B=B^{\prime }}$. Only the condition at the corner is needed; at points away from it such a quotient is automatically zero even for unequal slopes.

Write $S_n(t)={\sum }_{r=-n}^n{a}_r{e}^{irt}$. First we prove decay of coefficients of an almost-everywhere convergent trigonometric series rather than assuming the series can be integrated term by term. For $n\ge 1$, $T_n(t)=S_n(t)-S_{n-1}(t)=a_n{e}^{int}+a_{-n}{e}^{-int}$ tends to zero outside the finite set $E$. Suppose $M_n=|a_n|+|a_{-n}|$ does not tend to zero, and choose a subsequence with $M_n\ge \epsilon >0$. Then $|T_n/M_n|\le 1$ and $T_n/M_n\to 0$ almost everywhere. The dominated convergence theorem givesorthogonality of the two distinct Fourier modes instead makes the integral $(|a_n{|}^2+|a_{-n}{|}^2)/M_n^2\ge 1/2$, a contradiction. Thus $\boxed{a_n,a_{-n}\to 0}$.

Define on the real line the twice-integrated continuous functionThe series has both absolute convergence and uniform convergence because its coefficients are bounded. Its symmetric second quotient is therefore, for nonzero $h$,We next prove that this tends to zero whenever $t\notin E$, lifted periodically to the line. Let $b_0=a_0$ and $b_n=a_n{e}^{int}+a_{-n}{e}^{-int}$ for $n\ge 1$, so the partial sums of the $b_n$ are exactly $S_n(t)\to 0$. Put $k=|h|/2$ and $w_n=w(nk)$. Summation by parts givesThe boundary term vanishes because $S_n(t)$ is bounded and $w_n\to 0$ for fixed nonzero $k$. MoreoverIndeed $w^{\prime }(x)=O(x)$ near zero and $|w^{\prime }(x)|\le 2/x^2+2/x^3$ for $x\ge 1$. Each fixed weight difference tends to zero as $k\to 0$. Splitting the last series at an index beyond which $|S_n(t)|<\eta$ therefore bounds its upper limit by $C\eta$. Letting $\eta \to 0$ proves the claimed zero second quotient. This explicitly proves the needed Riemann summation of a convergent trigonometric series, despite the oscillations in its weights.

We also need the elementary fact that a continuous function with zero symmetric second derivative at every interior point of an interval is affine. For a real-valued function, subtract its endpoint chord on any compact subinterval $[a,b]$. If the difference were positive somewhere, adding $\epsilon (t-a)(t-b)$ for sufficiently small $\epsilon >0$ would retain a positive maximum at an interior point. At that maximum each symmetric second difference is nonpositive, while its quotient tends to $2\epsilon >0$, a contradiction. Thus the function lies below its chord. Apply the argument to its negative to get the reverse inequality. Applying this to real and imaginary parts proves the complex-valued case as well. This establishes that vanishing symmetric second derivative forces an affine function. Hence $F$ is an affine function on each open interval between the finitely many exceptional points in any period.

It remains to exclude corners at those points. Multiplying the formula for the second quotient by $h$, and using $h=2k$ and the evenness of $w$, givesfor every $t$, by the infinite-series version of part (i), since $a_n,a_{-n}\to 0$. continuity extends the adjacent affine pieces to each exceptional point, and the fact that symmetric second differences detect an affine corner makes their slopes and intercepts agree. Thus $F$ is globally an affine function on the real line.

The periodic part of its definition gives $F(t+2\pi )-F(t)=2\pi a_{0}t+2{\pi }^2{a}_0$. An affine function has constant increments, so $a_0=0$. The resulting $F$ is both an affine function and a periodic function, hence constant. Finally, absolute uniform convergence justifies integrating its defining series against $e^{-irt}$, and for $r\ne 0$ the resulting Fourier coefficient is $-a_r/r^2$. A constant has all these coefficients zero. Therefore $\boxed{a_r=0\text{ for every }r\in \mathbb{Z}}$, proving uniqueness of a trigonometric series outside a finite set with all exceptional points accounted for.