Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 7 (Topics in Fourier Analysis)

Identify the circle with $\mathbb{T}=\mathbb{R}/(2\pi \mathbb{Z})$ and use normalized Fourier coefficients and convolution:For $0\le r<1$, summing two geometric series gives the Poisson kernel on the circle:This Fourier series converges absolutely and uniformly for each fixed $r<1$. Consequently we can integrate it term by term in the convolution; the substitution $u=t-s$ givesThe resulting series is itself absolutely and uniformly convergent, since $|\hat{f}(n)|\le \Vert f{\Vert }_{\infty }$ and ${\sum }_n{r}^{|n|}<\infty$.

The Poisson kernel on the circle is nonnegative and has normalized integral one, as its constant Fourier coefficient is one. Its mass concentrates near zero. Indeed, for $r\ge 1/2$ and $\delta \le |s|\le \pi$,Hence its normalized integral outside $(-\delta ,\delta )$ tends to zero as $r\uparrow 1$. Given $\eta >0$, uniform continuity of $f$ gives a $\delta >0$ such that $|f(t-s)-f(t)|<\eta$ whenever the circular distance of $s$ from zero is less than $\delta$. Splitting the convolution error into this arc and its complement yieldsThe second term tends to zero, and $\eta$ is arbitrary. Thus the Abel sums converge uniformly to $f$, or $\boxed{\Vert f\ast P_r-f{\Vert }_{\infty }\to 0}$. This is uniform Poisson summability of continuous circle functions: positivity, unit mass and concentration supply the required approximate identity argument.

Use the given integer $u$ satisfying $u^2\equiv -1(\mathrm{mod}p)$. The Euclidean lattice under consideration has basis $(1,u),(0,p)$, because every vector in it is uniquely $(a,ua+bp)$ with $a,b\in \mathbb{Z}$. Its fundamental parallelogram therefore has areaWe use the following two-dimensional form of the Minkowski convex body theorem: a convex, centrally symmetric measurable subset of ${\mathbb{R}}^2$ of area strictly greater than four times the covolume of a full-rank Euclidean lattice contains a nonzero vector of that lattice. Apply it to the closed disk of squared radius $3p/2$. Its area is $3\pi p/2>4p$, so it contains a nonzero lattice vector $(a,b)$ satisfyingOn the other hand the defining modular congruence and $u^2\equiv -1(\mathrm{mod}p)$ implyThere is exactly one positive multiple of $p$ strictly below $2p$, so $\boxed{p=a^2+b^2}$. Neither coordinate can be zero, since a prime is not the square of an integer. This congruence lattice proof of the prime sum of two squares proves the required case of the sum of two squares theorem without assuming any conclusion at the critical area $4p$.

Write $S_{N}f(t)={\sum }_{|n|\le N}\hat{f}(n)e^{int}$ for the symmetric Fourier partial sum, with normalized Fourier coefficients on $\mathbb{T}=\mathbb{R}/(2\pi \mathbb{Z})$. We will prove the stronger conclusionIf $E$ is empty, take $f=0$. Otherwise the proof of the Kahane-Katznelson divergence theorem has three steps: a bounded trigonometric polynomial whose partial sum is large on short arcs, a finite-arc batching of an arbitrary null set, and a uniformly convergent sequence of separated frequency blocks. In particular, no assumption that $E$ is closed or compact is needed.

First consider a nonempty finite family of arcs with centers ${\theta }_{\nu }$, half-lengths $0<d_{\nu }\le 1$, and $D={\sum }_{\nu }{d}_{\nu }$. On the closed unit disk defineEach summand has positive real part there, and $\Phi (0)=1$. Thus the principal complex logarithm $A=\log \Phi$ is holomorphic on a neighborhood of the closed disk, $A(0)=0$, and $|\mathrm{Im}A|<\pi /2$. To bound its real part on an arc, put $h=t-{\theta }_{\nu }$ with $|h|\le d_{\nu }$ and $q=1+d_{\nu }-\cos h$. ThenAfter multiplication by the weight $d_{\nu }/D$, this contribution is at least $1/(4D)$; all the other contributions have positive real part. Throughout the arc union we therefore haveThis is the positive-real logarithmic amplifier for short arcs. Since $A$ is holomorphic beyond the closed unit disk, its Taylor series converges uniformly there. Choose a polynomial $Q(z)={\sum }_{\nu =1}^d{a}_{\nu }{z}^{\nu }$ with ${\sup }_{|z|\le 1}|Q(z)-A(z)|<1$. There is no constant term because $A(0)=0$. Set $C=\pi /2+1$ andwhere the integer $M>d$ may be chosen arbitrarily large. We have $\Vert P{\Vert }_{\infty }\le 1$, and its frequencies lie in $[M-d,M+d]$. At the cut $N=M$, the positive-frequency half of $Q$ is excluded and the negative-frequency half is included. Henceon the arc union. In particular, for any $K>0$, a total half-length $D\le \frac{1}{4}{e}^{-1-2CK}$ gives $|S_{M}P|\ge K$ there. This proves the needed compact-set Fourier amplification lemma, with the additional freedom to place its entire frequency block above any prescribed frequency.

Next we construct the null-set limsup cover by finite arc families. SetSince $E$ has Lebesgue measure zero, for every row $\ell \ge 1$ choose a countable arc cover $\{I_{\ell ,m}:m\ge 1\}$ of $E$ with positive half-lengths $d_{\ell ,m}$ and ${\sum }_m{d}_{\ell ,m}<2^{-\ell }{D}_1$. We can take closed arcs by slightly enlarging an open cover; their half-lengths are less than one. The whole two-index array has total half-length less than $D_1$. Absolute convergence of this sum lets us choose strictly increasing integers $N_j$ such that the sum outside the finite square $1\le \ell ,m\le N_j$ is less than $D_{j+1}$. Put $N_0=0$ and let ${\mathcal{F}}_j$ contain those arcs for whichEach family is finite. Its total half-length is less than $D_j$: for $j=1$ use the total sum, and for $j>1$ use the tail outside the preceding square. Moreover, every $t\in E$ belongs to arcs with unbounded row index, so belongs to the unions of infinitely many of the families ${\mathcal{F}}_j$. This batching is essential: a dense null set need not have a finite cover of small total length, whereas the infinitely-often finite covers above suffice.

Apply the amplification construction to ${\mathcal{F}}_j$ with height $K_j$, obtaining $P_j$ with $\Vert P_j{\Vert }_{\infty }\le 1$ and $|S_{M_j}{P}_j(t)|\ge K_j$ on that family's union. Choose the modulation integers successively so that the frequency intervals $[M_j-d_j,M_j+d_j]$ are positive and each lies strictly above the preceding one. Empty families may be represented by $P_j=0$ and an unused frequency interval. Define the frequency-separated Fourier block seriesThe bound ${\sum }_j{a}_j\Vert P_j{\Vert }_{\infty }\le 1$ gives absolute uniform convergence, so $f$ is continuous. Uniform convergence also justifies computing every Fourier coefficient term by term. At frequency $M_j$, all earlier blocks have been included in full and all later blocks are absent:For $t$ in the union of ${\mathcal{F}}_j$, the reverse triangle inequality givesEvery $t\in E$ lies in infinitely many such unions, and $M_j\to \infty$. Therefore the boxed unboundedness holds at every point of $E$, completing the proof for arbitrary null sets.

Use the angular-frequency Fourier transform conventionA precise sufficient interpretation of the regularity assumption is $f\ne 0$, $f\in H^1(\mathbb{R})$ in the first-order Sobolev space and $xf\in L^2(\mathbb{R})$; in particular, nonzero Schwartz functions are admissible. We use the Plancherel theorem, in the normalization $\Vert \hat{f}{\Vert }_2^2=2\pi \Vert f{\Vert }_2^2$, together with its derivative identity $\widehat{f^{\prime }}(\lambda )=i\lambda \hat{f}(\lambda )$. Both identities hold in $L^2$ for $f\in H^1$, by extending their identities for Schwartz functions. Thus the frequency factor isFor a Schwartz function, integration by parts gives $\Vert f{\Vert }_2^2=-2\mathrm{Re}\int x{f}^{\prime }(x)\overline{f(x)}dx$. This remains true under the stated assumptions: insert a smooth cutoff ${\chi }_R(x)=\chi (x/R)$ into the derivative of $x|f{|}^2$, where $\chi$ is one near zero and compactly supported. Integration givesThe cutoff-derivative term tends to zero, since $x{\chi }_R^{\prime }$ is bounded independently of $R$ and supported in a tail. The other terms converge by dominated convergence and the integrability of $|f^{\prime }||xf|$, furnished by the Cauchy-Schwarz inequality. HenceAfter squaring and using the Plancherel theorem, we obtain the uncentred Fourier uncertainty principle:If equality holds, both displayed inequalities must be equalities. The equality condition in the Cauchy-Schwarz inequality gives $f^{\prime }=cxf$ almost everywhere for a constant $c\in \mathbb{C}$; the real-part inequality and the identity for $\Vert f{\Vert }_2^2$ then force $c=-a$ with $a>0$ real. The weak derivative equation has the solution $f(x)=C{e}^{-a{x}^2/2}$: locally multiply by $e^{a{x}^2/2}$ to obtain a function whose weak derivative is zero, hence a constant. Conversely, for this Gaussian function the normalized second moments are $1/(2a)$ in position and $a/2$ in angular frequency, whose product is $1/4$. ThereforeThese are raw second moments, rather than variances about their respective means. Consequently translations and nonzero frequency modulations of the Gaussian function are not equality cases here; the equality case of the Heisenberg uncertainty relation for centred variances has those additional freedoms.

Use $\hat{f}(\lambda )=\int f(t)e^{-i\lambda t}dt$ and interpret the regularity assumption to include $f\in L^2(\mathbb{R})$ with the continuous representative furnished by Fourier inversion. Write $F=\hat{f}$. The bandwidth assumption implies $F\in L^1$ as well as $L^2$, by the Cauchy-Schwarz inequality on its finite support, sois continuous and defined at every $t$. The integer samples are precisely the Fourier coefficients of $F$ with the sign convention appropriate to the basis $e^{-in\lambda }$:The completeness and Parseval identity for the exponential orthonormal basis on $[-\pi ,\pi ]$ give $F={\sum }_{n\in \mathbb{Z}}f(n)e^{-in\lambda }$ in $L^2$ andThe last equality is the Plancherel theorem. Apply the inverse integral to the finite symmetric Fourier partial sums. Sincewe obtain the Nyquist–Shannon sampling theorem in this normalization:where the quotient is one at $t=n$. This is a sampling expansion by periodic Fourier projection, not merely an almost-everywhere inversion. Indeed, the inverse integral of an error $G\in L^2[-\pi ,\pi ]$ is bounded, uniformly in $t$, by $\Vert G{\Vert }_2/\sqrt{2\pi }$. Hence the symmetric finite sampling sums converge uniformly on all of $\mathbb{R}$. Moreover, the Parseval identity applied to $e^{it\lambda }$ givesThe Cauchy-Schwarz inequality therefore makes the sampling series absolutely convergent at every $t$, with its absolute tail bounded uniformly by $({\sum }_{|n|>N}|f(n){|}^2{)}^{1/2}$. Thus the integer samples recover the function everywhere.

For the nonuniqueness beyond this bandwidth, fix $0<{ϵ}_0<\min (ϵ,\pi )$ and a nonzero smooth function $\Phi$ compactly supported in $(-{ϵ}_0,{ϵ}_0)$. Let $h$ be its inverse Fourier transform and putThe function $h$, and hence $g$, is a Schwartz function. The two shifted spectral supports are disjoint, so $\hat{g}\ne 0$, but its support is contained strictly inside $(-\pi -ϵ,\pi +ϵ)$. Every integer sample of $g$ is zero. Therefore $g$ and the zero function have identical samples but are distinct, providing a strictly supercritical sampling alias counterexample for every positive bandwidth enlargement.

For the Riemann-Lebesgue lemma, a half-period shift of the oscillation gives a direct proof. For a nonzero integer $r$, set $h=\pi /r$. Periodicity and a change of variable show that the Fourier coefficient of $t\mapsto f(t+h)$ at $r$ is $e^{irh}\hat{f}(r)=-\hat{f}(r)$. ThusThe right side tends to zero as $|r|\to \infty$ by uniform continuity on the circle. Consequently $\boxed{\hat{f}(r)\to 0\text{ as }|r|\to \infty }$.

There is nevertheless no prescribed rate shared by all continuous functions. Choose increasing positive integers $n_j$ with $k(n_j)\ge j{2}^j$, which is possible since $k(r)\to \infty$. DefineThe series converges absolutely and uniformly, so $g$ is continuous. Termwise integration is justified by uniform convergence, and the orthogonality of distinct exponential modes gives $\hat{g}(n_j)=2^{-j}$. HenceThis arbitrarily slow Fourier coefficient decay is compatible with the Riemann-Lebesgue lemma: for this particular function all Fourier coefficients still tend to zero, but along the selected subsequence they beat the proposed decay scale by an unbounded factor.

Put $L=2\pi$ and represent the circle by $[0,L)$. The full orthonormal basis of Haar wavelets includes the constant $H_{\ast }=L^{-1/2}$ and, for every dyadic interval $I$ of length $\ell =L{2}^{-j}$, the normalized detailTake half-open intervals and the corresponding values at their boundaries, so every point belongs to exactly one interval at each level. The constant basis function is necessary: if it were omitted, even $f\equiv 1$ could not satisfy the asserted convergence. We use the full conventional Haar wavelet system in the threshold sum.

Let $A_{J}f$ be the function equal to the mean of $f$ on each level-$J$ dyadic interval. Comparing the means on a parent and its two children proves the Haar refinement identity directly: the difference between the child mean and the parent mean is the appropriate value of $c_I{H}_I$. Inductively this gives the Haar projectionWriting ${\omega }_f(s)={\sup }_{\mathrm{dist}(x,y)\le s}|f(x)-f(y)|$ for the modulus of continuity, the cell-average formula yieldsFor a detail on a cell $I$, its integral is zero. Subtracting $f(x_I)$ for any $x_I\in I$ therefore gives the coefficient estimateIn particular $|c_I|\le 2\Vert f{\Vert }_{\infty }\sqrt{\ell }$, so for every $\delta >0$ only finitely many details can have $|c_I|\ge \delta$. The hard-threshold sumis thus a well-defined finite sum. Its chosen details need not form a complete set of levels; this is why convergence of the Haar projections alone does not prove the result.

Fix $\eta >0$. Choose $J\ge 1$ so that ${\omega }_f(L{2}^{-J})\le \eta$. All finer details satisfy $|c_I|\le \eta \sqrt{L}{2}^{-j/2}$ at level $j\ge J$. For sufficiently small $\delta$, every nonzero coefficient at levels below $J$, and the constant coefficient if nonzero, is retained. Also choose the unique integer $K\ge J$ such thatFor $j>K$ we have $|c_I|\le \eta \sqrt{L}{2}^{-j/2}<\delta$, so no such detail is retained. Consequently $T_{\delta }f$ differs from $A_{K+1}f$ only by omitted details at levels $J$ through $K$. At any point there is at most one supported detail per level, and an omitted detail has magnitude at most $\delta /\sqrt{L{2}^{-j}}$. The geometric bound for omitted Haar details givesSince $\Vert A_{K+1}f-f{\Vert }_{\infty }\le \eta$, we concludeThe right side can be made arbitrarily small, proving uniform hard-threshold convergence of normalized Haar expansions:The square-root scaling imposed by the specified $L^2$ normalization is what makes the omitted contributions a controlled geometric series.