Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 7 (Topics in Fourier Analysis)

Use $\mathbb{T}=\mathbb{R}/(2\pi \mathbb{Z})$ and normalized Haar measure $dm=dsdt/(2\pi {)}^2$ on the torus. Here $|s|$ means distance to zero modulo $2\pi$, represented in $[-\pi ,\pi ]$; an ordinary coordinate in $[0,2\pi )$ would not express concentration at the origin correctly.

Put $B(s,t)=2+\cos s+\cos t$, $I_N=\int B^{N}dm$, and $A_N=I_N^{-1}$. Since $0\le B\le 4$ and $B(0,0)=4$, $I_N>0$, the kernel $K_N=A_N{B}^N$ is nonnegative, and $\int K_{N}dm=1$ exactly. Fix $0<\delta \le \pi$. On the set where either $|s|\ge \delta$ or $|t|\ge \delta$, $B\le q=3+\cos \delta <4$. Choose $q^{\prime }$ with $q<q^{\prime }<4$. continuity provides a neighborhood $U$ of the origin of positive measure $v$ where $B\ge q^{\prime }$. ConsequentlyFor $0<\epsilon \le \pi$, take $\delta =\epsilon$ and then $N$ large enough that this bound is at most $\epsilon$. For $\epsilon >\pi$ the exterior conditions are empty. This proves all the requested kernel properties, with $\boxed{A={\left[\int (2+\cos s+\cos t{)}^{N}dm\right]}^{-1}}$.

Now let $M=\Vert f{\Vert }_{\infty }$ and define $P_N=K_N\ast f$. The powered-cosine approximate identity on a torus is a trigonometric polynomial, since its base is a finite sum of complex exponentials and its exponent is an integer. Expanding $K_N$ and integrating its finitely many Fourier terms shows that $P_N$ is also a trigonometric polynomial; it is real-valued because both factors are real. Translation invariance and normalization giveGiven $\eta >0$, uniform continuity of $f$ supplies $\delta >0$ making the difference in brackets smaller than $\eta /2$ on the square $|s|,|t|<\delta$, uniformly in $(x,y)$. The contribution from that square is at most $\eta /2$. On its complement it is at most $2M{\sup }_{\max (|s|,|t|)\ge \delta }{K}_N$, which tends to zero by the preceding estimate. Therefore $\boxed{\Vert P_N-f{\Vert }_{\infty }\to 0}$, proving uniform approximation without assuming any Fourier-series convergence of $f$.

The necessary and sufficient condition isEquivalently, $1,u/(2\pi ),v/(2\pi )$ are linearly independent over $\mathbb{Q}$. This is the equidistribution criterion for a torus translation.

For sufficiency, set $w=e^{i(ku+\ell v)}$. For a nonconstant character of the torus the condition gives $w\ne 1$, and the geometric series formula yieldsThe constant torus character has average one. Thus orbit averages converge to integrals against normalized Haar measure for every trigonometric polynomial. Part (i) and a uniform approximation bound extend this to every continuous function $g$: the difference between the orbit average and the integral for $g$ differs from that for an approximating polynomial by at most twice their uniform distance.

For a nonempty half-open rectangle $R$, choose continuous functions $g_{-}\le 1_R\le g_{+}$ with values in $[0,1]$ and $\int (g_{+}-g_{-})dm$ arbitrarily small. One can take distance cutoffs: $g_{-}$ vanishes at the boundary and rises to one inside, while $g_{+}$ is one on the closure and falls to zero outside a thin neighborhood. Their difference is supported in shrinking neighborhoods of the rectangle's finitely many boundary segments, which have Haar measure zero. Squeezing the orbit averages between the two continuous averages gives the normalized area $m(R)=(b-a)(d-c)/(2\pi {)}^2$. Empty rectangles are immediate, and full-circle coordinate intervals can be treated with the constant cutoff in that coordinate.

For necessity, if a nonzero pair is resonant, every orbit point lies in the closed set $H=\{(s,t):e^{i(ks+\ell t)}=1\}$. This is a proper subset because the torus character is nonconstant. Its open complement contains a rectangle of positive area in a coordinate chart, which can be chosen inside the standard coordinate square. That rectangle is never visited, contradicting the required positive limiting frequency. Hence the boxed condition is necessary as well as sufficient.

The Minkowski convex body theorem states that a centrally symmetric convex measurable set in ${\mathbb{R}}^d$ of volume greater than $2^d\mathrm{covol}(\Lambda )$ contains a nonzero point of the full lattice $\Lambda$. For a compact convex body, the weak inequality also suffices: apply the strict version to $(1+\epsilon )C$, let $\epsilon \downarrow 0$, and choose a constant subsequence of the resulting lattice points in the bounded set $2C$. Its nonzero value belongs to $C$ by closedness.

For $Q>1$, apply this borderline version with $\Lambda ={\mathbb{Z}}^2$ toThe linear transformation $(u,v)\mapsto (u,xu-v)$ has determinant of absolute value one, so this parallelogram has area four. A nonzero lattice point $(n,m)$ satisfies $|n|\le Q$ and $|nx-m|\le Q^{-1}$. If $n=0$, then $|m|<1$, giving the forbidden zero point; hence $n\ne 0$. It follows thatIf $x$ is irrational, these pairs cannot remain in a finite set as $Q\to \infty$: the positive errors $|nx-m|$ of any such finite set have a positive minimum, contradicting their bound $Q^{-1}$. If $x=p/q$ is rational in lowest terms with $q>0$, the pairs $(n,m)=(jq,jp)$ for $j\ge 1$ have zero error and are infinitely many distinct pairs. This proves the assertion for every real $x$.

In this part identify $\mathbb{T}$ with the complex unit circle by $t\mapsto e^{it}$. Write the finite abelian group $G$ additively and let $\hat{G}=\mathrm{Hom}(G,\mathbb{T})$, with pointwise multiplication, be the character group of a finite abelian group of $G$. A linear character has modulus one, since its values have finite order. We first prove the extension and counting facts needed for Fourier analysis on a finite abelian group.

If $H\le G$, $\chi \in \hat{H}$, and $x\notin H$, let $d$ be the least positive integer with $dx\in H$. Choose $z\in \mathbb{T}$ with $z^d=\chi (dx)$ and defineThis is well-defined: two representations differ by $j-j^{\prime }=qd$ and $h-h^{\prime }=-qdx$, whose contributions cancel. It is a group homomorphism and extends $\chi$. Conversely every extension has this form, and the equation for $z$ has exactly $d$ distinct roots. Since $[H+⟨x⟩:H]=d$, successive adjunctions prove that each linear character of $H$ has exactly $[G:H]$ extensions. Taking $H=\{0\}$ gives $\boxed{|\hat{G}|=|G|}$. This is the extension of a character across a cyclic quotient, proved here without a structure theorem. Extending the linear character sending a nonzero element $g$ of order $d$ to $e^{2\pi i/d}$ also proves that linear characters separate points of $G$.

Use the normalized inner product $⟨f,g⟩=|G{|}^{-1}{\sum }_{x\in G}f(x)\overline{g(x)}$. For a nontrivial linear character $\chi$, choose $a$ with $\chi (a)\ne 1$. Translation of the sum gives ${\sum }_x\chi (x)=\chi (a){\sum }_x\chi (x)$, so that sum is zero. Applying this to $\chi \overline{\psi }$ givesThere are $|G|$ such orthonormal vectors in the $|G|$-dimensional space ${\mathbb{C}}^G$, so they form an orthonormal basis. Define the Fourier coefficients by $\hat{f}(\chi )=|G{|}^{-1}{\sum }_{x}f(x)\overline{\chi (x)}$. Expanding in that basis proves Fourier inversion on a finite group and the Parseval identity on a finite group:For normalized convolution $(f\ast g)(x)=|G{|}^{-1}{\sum }_{y}f(y)g(x-y)$, changing the finite summation variables yields $\widehat{f\ast g}(\chi )=\hat{f}(\chi )\hat{g}(\chi )$. Translation $f(x)\mapsto f(x+a)$ multiplies its Fourier coefficient by $\chi (a)$. These formulas fix all normalization and conjugation conventions.

The PDF's double-hat identification is canonical Pontryagin duality: define $e:G\to \hat{\hat{G}}$ by $e_x(\chi )=\chi (x)$. This is a group homomorphism and is injective because linear characters separate points. Applying the already proved character-counting result to $\hat{G}$ shows that both groups have size $|G|$, soThe TeX loses one hat. For completeness, the single-dual isomorphism $\hat{G}\cong G$ also holds, but is generally noncanonical. Here is a proof including the needed finite cyclic decomposition. Choose an element $a$ of maximal order $d$. Every element order divides $d$: for two elements, extract their prime-power components by integer multiples: if an element has order $d$ and $p^{\alpha }$ is the exact power of $p$ in $d$, multiplying it by $d/p^{\alpha }$ gives order $p^{\alpha }$. Then combine the components with the larger order for each prime. Components of coprime orders have a sum of product order, because their cyclic subgroups have trivial intersection. This constructs an element of order the least common multiple of the two original orders, which cannot exceed maximal $d$; hence each order divides $d$.

Extend the primitive linear character of $⟨a⟩$ to $\chi :G\to \mathbb{T}$. All its values are $d$th roots because all element orders divide $d$. Define $\rho (g)=ja$ when $\chi (g)=e^{2\pi ij/d}$. This is a group homomorphism fixing $⟨a⟩$, and $g=\rho (g)+(g-\rho (g))$ shows $G=⟨a⟩\oplus \ker \rho$. Induction on group size decomposes $G$ into finite cyclic groups. A cyclic group of order $d_j$ has precisely the linear characters $j\mapsto e^{2\pi imj/d_j}$ and hence a character group of a finite abelian group isomorphic to itself. The dual of a finite direct product of groups is the product of its factor duals, since a linear character is determined by its restrictions. This proves $\hat{G}\cong G$ after choices of cyclic factors and generators, without using an unproved finite-abelian-group structure theorem.

The improper integral is assumed to have a finite limit. Fix $\lambda >1$ and put $I(U,V)={\int }_U^V(\beta (x)-x)x^{-2}dx$. Convergence implies $I(X,\lambda X)\to 0$ and $I(X/\lambda ,X)\to 0$ as $X\to \infty$. By monotonicity,andThusBoth outer bounds tend to one as $\lambda \downarrow 1$, giving $\boxed{\beta (X)/X\to 1}$. This monotone integral Tauberian lemma uses monotonicity to control the value at a point from integral averages on adjacent multiplicative intervals; convergence of the integral alone would not control arbitrary local spikes.

Let $M=\sup |f|$ and $F_T(z)={\int }_0^{T}f(t)e^{-tz}dt$. Each $F_T$ is an entire function: on any compact set of $z$, differentiation under this finite-interval integral is justified by a bounded integrable majorant. We prove $\boxed{F_T(0)\to F(0)}$, which gives ordinary convergence of the improper integral, not just an Abel limit.

Fix $R>0$. Since $\Omega$ contains the imaginary segment from $-iR$ to $iR$ and is open, choose $0<\delta <R$ so small that its closed left strip of width $\delta$ belongs to $\Omega$. Let $C_{+}$ be the right semicircle $|z|=R$ oriented from $-iR$ to $iR$, and let $\Gamma$ be the left half-ellipse $z=\delta \cos \theta +iR\sin \theta$, $\pi /2\le \theta \le 3\pi /2$, oriented back to $-iR$. The enclosed right half-disk and thin left region lie in $\Omega$. The residue theorem, applied to the analytic difference with its simple pole at zero, givesFor $x=\mathrm{Re}z>0$,On the right circle, $|1+z^2/R^2|=2x/R$ and $|z|=R$. The exponential cancels the displayed damping, so the absolute integrand per unit arc length is at most $2M/R^2$. The right-arc contribution is therefore at most $M/R$ after the factor $1/(2\pi )$.

For the $F_T$ term on $\Gamma$, deform that left path to the left semicircle $C_{-}$ of radius $R$. The Cauchy integral theorem allows this because $F_T$ is an entire function and the region between the paths excludes the pole at zero. On $C_{-}$, $x<0$ andThe same factor $|1+z^2/R^2|=2|x|/R$ bounds this contribution by $M/R$. The arc endpoints can be omitted from the estimates, since they have zero arc length and the integrands are continuous there.

Finally, the $F$ term on the original $\Gamma$ tends to zero as $T\to \infty$. This compact path stays away from zero, so $F(z)(1+z^2/R^2)/z$ is bounded there. Its interior has $\mathrm{Re}z<0$, hence $e^{Tz}\to 0$, and $|e^{Tz}|\le 1$. Dominated convergence on the finite-length path gives the claim. Combining the three pieces yieldsSince $R$ is arbitrary, the difference tends to zero. This proves the Newman Tauberian theorem in the required setting; the proof crucially uses analytic continuation across every finite segment of the imaginary axis.

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.