Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 8 (Introduction to Fourier Analysis)

Use normalized Fourier coefficients and the inner productThe functions $e_n(t)=e^{int}$ are orthonormal. Hence the Fourier partial sum $S_{N}f={\sum }_{|n|\le N}\hat{f}(n)e_n$ is the orthogonal projection onto the trigonometric polynomials of degree at most $N$. For any such polynomial $P$, orthogonality givesso $\Vert f-S_{N}f{\Vert }_2\le \Vert f-P{\Vert }_2$.

Given $\epsilon >0$, the permitted density result supplies a trigonometric polynomial $P$ with $\Vert f-P{\Vert }_{\infty }<\epsilon$. Once $N$ includes its degree,Therefore the Fourier partial sums converge to $f$ in the normalized $L^2$ norm, giving exactly the stated mean-square limit.

Apply part (i) to the continuous difference $h=f-g$. Every Fourier coefficient of $h$ vanishes, so every Fourier partial sum is zero. The orthogonal projection convergence from part (i) therefore gives $\Vert h{\Vert }_2=0$.

If $h(t_0)\ne 0$, continuity gives an interval on which $|h|$ is bounded below by a positive number. That interval would contribute positively to $\Vert h{\Vert }_2^2$, a contradiction. HenceThe continuity hypothesis upgrades equality almost everywhere to pointwise equality.

By the Weierstrass M-test, absolute summability of the Fourier coefficients makes ${\sum }_n\hat{f}(n)e^{int}$ uniformly convergent to a continuous periodic function $H$. Termwise integration is justified by that uniform convergence, and orthogonality givesPart (ii) then implies $H=f$. Thus the Fourier series converges uniformly to the original function, not merely to some continuous limit. In particular,

The absolute Fourier convergence from a square-integrable derivative uses more than the pointwise estimate $|\hat{f}(n)|=O(1/|n|)$. Periodic integration by parts givesBy Bessel's inequality,Now apply the Cauchy-Schwarz inequality:Adding the finite constant coefficient proves absolute summability. Since a continuously differentiable periodic function has $f^{\prime }\in L^2$, all hypotheses of part (iii) hold and its Fourier series converges uniformly.

Part (i) and the Cauchy-Schwarz inequality showDirect integration of this finite Fourier partial sum givesMoreover Bessel's inequality puts both coefficient sequences in ${\ell }^2$, so their product series is absolutely convergent by the Cauchy-Schwarz inequality. Consequently the cross form of Parseval's identity isTaking $g=f$ also gives equality of the squared function norm and the squared coefficient norm.

For the Hurwitz proof of the planar isoperimetric inequality, take a positively oriented regular simple closed curve of length $L$ and enclosed area $A$. Write its complex position as $z(t)=x(t)+iy(t)$, with $0\le t\le 2\pi$ proportional to arc length. Then $|z^{\prime }(t)|=L/(2\pi )$. Translate the curve to make its mean position zero, and write its Fourier coefficients as $c_n$, with $c_0=0$.

Green's theorem gives the signed area, and Parseval's identity computes it:Periodic integration by parts and Parseval's identity applied to $z^{\prime }$ giveSince $n\le n^2$ for every integer $n$,The sums converge absolutely: $\sum |n||c_n{|}^2\le (\sum |c_n{|}^2{)}^{1/2}(\sum n^2|c_n{|}^2{)}^{1/2}$. Equality forces $c_n=0$ unless $n=0$ or $1$; after the mean translation, $z(t)=c_1{e}^{it}$ is a circle. Conversely a circle attains equality. Thus circles uniquely attain equality, up to translation and orientation.

The same proof applies to a rectifiable simple closed curve using its Lipschitz arc length parametrization. Its derivative exists almost everywhere and belongs to $L^2$; periodic mollification converges to the curve in the function and derivative $L^2$ norms. This justifies the derivative coefficient identity, Parseval's identity and area integral by approximation. Reversing orientation, if necessary, makes the enclosed area positive. The printed name “Hurewitz” is read as Hurwitz.

Use the transform convention $F(\lambda )={\int }_{\mathbb{R}}f(t)e^{-i\lambda t}dt$. The given Fourier inversion theorem givesHere $F$ is continuous, vanishes at both endpoints and belongs to $L^2[-\pi ,\pi ]$. It therefore defines a continuous periodic function. Its Fourier coefficient at index $-n$ is $f(n)$. By part (i),Pair this convergence with $e^{it\lambda }$. Since that function has normalized $L^2$ norm one, the Cauchy-Schwarz inequality yields, uniformly in $t$,The elementary integral is the sinc function,Thus the sampling expansion by periodic Fourier projection isThere is no pointwise interchange with an unproved Fourier series: the calculation first uses finite sums and then an $L^2$ limit.

The convergence can also be made absolute. Parseval's identity gives ${\sum }_n|f(n){|}^2=\Vert F{\Vert }_2^2$, and Bessel's inequality applied to $e^{it\lambda }$ gives ${\sum }_n|D(t-n){|}^2\le 1$. Henceuniformly in $t$. At an integer argument, $D$ is one at zero and zero at the other integers, so the expansion interpolates the samples exactly.

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 $E$.

Use the uniform bound for harmonic sine polynomialsFor completeness, reduce to $0<t\le \pi$ and split at $K=\min (m,\lfloor 1/t\rfloor )$. The first part is bounded by $Kt\le 1$. Geometric-series summation bounds every interval sum of $\sin kt$ by $1/\sin (t/2)\le \pi /t$. Summation by parts bounds the remaining harmonic-weighted tail by $\pi /[t(K+1)]\le \pi$. Negative $t$ follows by oddness and $t=0$ is immediate.

Let $H_m={\sum }_{k=1}^{m}1/k$. Choose $m_j$ so large that $H_{m_j}\ge C_0{2}^{j+2}$, and positive integers $L_j>m_j$ such that the intervals $[L_j-m_j,L_j+m_j]$ are strictly separated and increase. DefineThe bound $\Vert Q_j{\Vert }_{\infty }\le 2^{-j-2}$ makes this a continuous function with $g(0)=0$.

Each Fourier coefficient of $P_m$ has modulus $1/(2k)$, so the sum of the absolute coefficients is $H_m$. Therefore every prefix of a normalized block $Q_j$ has norm at most one. At any Fourier cutoff, all earlier blocks are complete, at most one block is partial and all later blocks are absent. HenceAt zero, a completed block contributes zero, but its prefix through frequency $L_j$ consists of the negative-frequency sine coefficients and equals $i/2$. ThusBoth index sequences tend to infinity. The uniformly bounded partial sums fail to converge at the origin, even though the function is continuous and zero there.

For infinite products, a nonzero limiting product requires that its factors tend to one. A useful sufficient condition isAfter finitely many factors, $|u_j|<1/2$ and the principal holomorphic logarithm satisfies $|\log (1+u_j)|\le 2|u_j|$. Therefore the sum of logarithms converges, and exponentiating it gives a finite nonzero product. The same argument on compact sets proves infinite product convergence from logarithmic tails: a locally uniformly absolutely convergent tail of holomorphic logarithms gives a holomorphic nonvanishing tail product. Finite factors then determine all zeros and their orders.

The Weierstrass elementary factors areFor $|w|<1$,Their only zero is a simple zero at $w=1$.

Work with entire functions on $\mathbb{C}$. As usual for prescribed exact zero orders, the distinct zero locations must have consistent multiplicities. The literal statement allows repeated locations with conflicting orders; that cannot be true, for example if the same point is prescribed order one and order two. Remove consistent repetitions rather than adding their orders, and separate a possible zero at the origin.

List the distinct nonzero locations as $a_j$, with prescribed orders $n_j$, and write $m$ for the prescribed order at zero, or zero if the origin is not prescribed. Choose $p_j$ large enough thatThen the Weierstrass factorization theorem construction isOn every compact set, $|z/a_j|\le 1/2$ for all sufficiently large $j$ because $|a_j|\to \infty$. Its logarithmic tail is bounded byThus the product converges locally uniformly, is entire, and has exactly the specified zeros with exactly their orders. At a prescribed zero, only its own finite factor vanishes; all other finite factors and the tail product are nonzero.

Without escape to infinity the conclusion fails in general. Distinct proposed zeros $1/j$ accumulate at zero. The identity theorem would force an entire function with those zeros to vanish identically, which does not have precisely the prescribed isolated zeros. A finite zero set can of course be realized by a polynomial.

For entire functions with the same zero divisor, the quotient $q=f/g$ extends through every common zero by cancelling equal local powers. It is entire and nowhere zero. Hence $q^{\prime }/q$ is entire and has a primitive on the simply connected plane. Choose $h(0)$ with $e^{h(0)}=q(0)$ and putThe derivative of $q{e}^{-h}$ vanishes, and its value at zero is one. ThereforeIf also $f=e^{k}g$, then $e^{h-k}=1$ everywhere. The continuous difference $h-k$ takes values in the discrete set $2\pi i\mathbb{Z}$ and so is constant on the connected plane:The whole-plane hypothesis matters: a zero-free holomorphic quotient on a multiply connected domain need not possess a global holomorphic logarithm.

Write the finite abelian group additively. A character of a finite abelian group is a group homomorphism $\chi :G\to \{z:|z|=1\}$. We first prove that there are exactly $|G|$ such characters of a finite abelian group, without assuming a structure theorem.

Use extension of a character across a cyclic quotient. Given a subgroup $H$ and $a\notin H$, let $k$ be the least positive integer with $ka\in H$. The subgroup $K=H+⟨a⟩$ has $k$ cosets of $H$. If $\varphi$ is a character of a finite abelian group on $H$, choose any of the $k$ roots ${\lambda }^k=\varphi (ka)$ and defineThis is well-defined: two representations differ by an integer multiple of $ka$, and the root equation exactly cancels that difference. It is a character of a finite abelian group, and every extension arises from one of the $k$ choices of $\lambda$. Build a chain from the trivial subgroup to $G$ by adjoining elements. The character of a finite abelian group count multiplies by the same factor as the subgroup order at every step, so $|\hat{G}|=|G|$.

For a nontrivial character of a finite abelian group $\eta$, choose $a$ with $\eta (a)\ne 1$. Translating the group sum showsso the sum is zero. Applied to $\chi \overline{\psi }$, this provesThe $|G|$ orthogonal nonzero characters of a finite abelian group therefore form a basis of all complex functions on $G$, a vector space of dimension $|G|$.

With the unnormalized Fourier coefficients $\hat{f}(\chi )={\sum }_{x\in G}f(x)\overline{\chi (x)}$, expansion in that basis gives the Fourier inversion on a finite groupUnder a normalized forward-transform convention the prefactor would instead be one. Thus the unspecified constant is determined by the convention, and the inversion itself follows directly from character of a finite abelian group counting and orthogonality.

Use the Chebyshev estimate from central binomial coefficients. DefineThe first is the Chebyshev theta function; the second counts prime powers with the same logarithmic prime weight. For a positive integer $m$, every prime $m<p\le 2m$ divides $\left(\genfrac{}{}{0ex}{}{2m}{m}\right)$, henceSumming over dyadic intervals yields $\vartheta (2^j)\le 2^{j+1}\log 2$. By monotonicity and rounding $x$ upward to a power of two,

For the lower bound, the central binomial coefficient is the largest of the $2m+1$ coefficients whose sum is $4^m$, soThe exponent of a prime in the central coefficient isEach summand is zero or one. Therefore $\log \left(\genfrac{}{}{0ex}{}{2m}{m}\right)\le \psi (2m)$. Higher prime powers contribute onlyCombining the lower bound with $2m$ just below $x$ shows $\vartheta (x)\ge cx$ for sufficiently large $x$, with, for example, $c=(\log 2)/2$.

Let $\pi (x)$ denote the number of primes at most $x$. Since every prime weight is at most $\log x$,For the upper bound, separate the primes at $\sqrt{x}$. The small ones number at most $\sqrt{x}$, and every larger one has weight at least $(\log x)/2$, givingSince $\sqrt{x}=o(x/\log x)$, this provesfor all sufficiently large $n$, for instance with $A=(\log 2)/2$ and $B=8\log 2+1$. This elementary Chebyshev estimate does not assume the Prime number theorem.

The printed differential in the transform integral is $dz$; it must be $dt$, since $z$ is the transform parameter. We prove the Newman Tauberian theorem in this corrected interpretation. Setalmost everywhere. The finite-interval transform $F_T$ is entire.

Fix $R>0$. Because the analytic domain contains the whole imaginary axis, compactness supplies a $\delta >0$ such that the thin rectangle $-\delta \le \mathrm{Re}z\le 0$, $|\mathrm{Im}z|\le R$ lies in the domain. Let $C_{+}$ be the right semicircle of radius $R$, oriented from $-iR$ to $iR$. Join its endpoints by a leftward path $L$ along the other three sides of that rectangle. This is a closed positively oriented contour.

Use contour damping for bounded Laplace transforms withThe residue theorem givesOn $C_{+}$, with $x=\mathrm{Re}z>0$,Thus the integrand has modulus at most $2M/R^2$, and this half-circle contributes at most $M/R$ after division by $2\pi$.

On $L$, the $F$ term tends to zero as $T\to \infty$: every interior point of the path has negative real part, $F$ is bounded on this fixed compact path, and the remaining kernel factor is bounded because the path avoids zero. Dominated convergence applies, with the two endpoints irrelevant to the path integral.

For the $F_T$ term, deform $L$ to the left semicircle $C_{-}$ of radius $R$. This deformation uses only the entire function $F_T$; it does not demand that $F$ extend across a large left half-disk. Both paths lie to the left of zero and their enclosed deformation region avoids the kernel pole. On $C_{-}$, for $x<0$,The same circle factor gives another bound $M/R$. ConsequentlyThe radius $R$ is arbitrary, soThis proves convergence of the ordinary improper integral, not merely a damped limit.

For the weakened domain hypothesis, take $f(t)=1$. Its Laplace transform is $F(z)=1/z$, analytic on the open right half-plane, but ${\int }_0^{T}f(t)dt=T$ diverges. Analyticity only in the open right half-plane is insufficient.