Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 8 (Topics in Fourier Analysis and Complex Variable)

We first prove a compact-square form of the Kolmogorov-Arnold representation theorem, including the construction needed for exact representation. We then remove the compactness restriction; a mere limiting sequence of two-variable approximations would not establish a finite expression using one-variable functions.

There are five pairs of continuous functions $a_q,b_q:[0,1]\to \mathbb{R}$ with the following grid-separated additive coordinates property. At every arbitrarily small mesh, choose for each $q$ a finite family of disjoint closed intervals in each coordinate. Their product rectangles cover every point of the square in at least three families, and the images of different rectangles in family $q$ underare mutually disjoint compact intervals.

Here is a proof of that existence assertion. Work in the complete metric space $X=C([0,1]{)}^{10}$ with its maximum uniform norm. For an integer $N\ge 1$, let $U_N$ comprise tuples having the preceding property with every interval shorter than $1/N$. Each $U_N$ is open: fix a witnessing finite collection of rectangles. Their compact image intervals have positive mutual separation within each family, so sufficiently small uniform perturbations of the ten functions preserve separation, while the domain cover is unchanged.

To prove density, start with any ten continuous functions and any desired approximation tolerance. Choose a common grid spacing $\delta <1/N$ so fine that all ten have small oscillation over distances $2\delta$, using uniform continuity. For the five families, use grid cuts shifted successively by $\delta /5$. Remove tiny open gaps about these cuts, of radius less than $\delta /20$. Gaps from different families are disjoint. The complementary components in $[0,1]$ are closed intervals shorter than $\delta$. Every coordinate belongs to an interval in at least four families, so a pair of coordinates belongs to product rectangles in at least three common families.

In each family, approximate $a_q$ and $b_q$ by constants on those intervals and interpolate linearly across the gaps. The constants can be chosen arbitrarily close to sampled original values, so the resulting functions approximate the originals uniformly. Perturb the finitely many constants slightly so that all sums $a_{q,I}+b_{q,J}$ for distinct pairs $(I,J)$ are different. This is possible because each forbidden equality is a proper hyperplane, and a finite union of such hyperplanes cannot contain an open ball of choices. The rectangle images are now distinct singleton sets. Thus $U_N$ is dense. The Baire category theorem gives a tuple in ${\bigcap }_{N\ge 1}{U}_N$. Fix that tuple once and for all.

Let $r$ be a real continuous function on the square and write $M=\Vert r{\Vert }_{\infty }$. If $M=0$ there is nothing to do. Otherwise choose a sufficiently fine witnessing cover that the oscillation of $r$ in every rectangle is at most $M/10$. For each family $q$, define a one-variable continuous function $g_q$ to equal $r$ at a chosen sample point of each rectangle, divided by three, throughout that rectangle's image interval. The image intervals are disjoint, so interpolate linearly between them and extend constantly beyond the extreme intervals. ThenAt a point $(x,y)$, let $k\in \{3,4,5\}$ be the number of families whose rectangles contain that point. In those $k$ families the corresponding summand differs from $r(x,y)/3$ by at most $M/30$. In the other families its absolute value is at most $M/3$. ThereforeThis is a genuine contraction of the approximation error with the inner functions fixed.

Starting with $r_0=F\in C([0,1{]}^2)$, repeat the construction and put $r_{j+1}=r_j-{\sum }_q{g}_{q,j}\circ s_q$. Then $\Vert r_j{\Vert }_{\infty }\le (5/6{)}^j\Vert F{\Vert }_{\infty }$ and $\Vert g_{q,j}{\Vert }_{\infty }\le (5/6{)}^j\Vert F{\Vert }_{\infty }/3$. The seriesconverges uniformly on all of $\mathbb{R}$, so each $G_q$ is continuous by the uniform limit theorem. Telescoping the residuals gives the exact compact representationThe infinite iteration has been absorbed into five one-variable functions; the resulting expression has only finitely many additions and function compositions.

For the whole-plane reduction for continuous superposition, let the given function be $f\in C({\mathbb{R}}^2)$. SetThis is finite on every bounded interval by compactness. Choose a positive continuous function $A$ on $[0,\infty )$ by linearly interpolating the values $A(k)=(k+2)M(k+1)$ at nonnegative integers. For $k\le t\le k+1$, monotonicity of the endpoint values gives $A(t)\ge A(k)\ge (1+t)M(t)$. Extend $A$ constantly to negative arguments. ConsequentlyLet $\kappa (x)=\frac{1}{2}+{\pi }^{-1}\arctan x$. Transport $b$ to $(0,1{)}^2$ with ${\kappa }^{-1}(s)=\tan (\pi (s-1/2))$. Extend the transported function by zero on the boundary of the square. The displayed bound proves continuity at every boundary point, including corners, because approaching the boundary forces at least one original coordinate to tend to infinity. Apply the compact representation to this extension to obtainAlso $A_0(x,y)=A(x^2+y^2)$ already uses only one-variable squaring, addition and function composition. Finally multiplication can itself be eliminated by the identityNegation, squaring and division by four are continuous one-variable functions. Hence the entire expression uses only the permitted one-variable continuous functions and addition, even for an unbounded $f$ on ${\mathbb{R}}^2$.

Use ${\mathbb{T}}^n=(\mathbb{R}/2\pi \mathbb{Z}{)}^n$ and the uniform metric $d(f,g)=\Vert f-g{\Vert }_{\infty }$. Let $N(\epsilon ,E)$ be the least number of radius-$\epsilon$ balls in this metric covering $E$. The Kolmogorov entropy, equivalently metric entropy, isIt is infinite if no finite cover exists; set the entropy of the empty set to zero. Changing the logarithm base or requiring centres in $E$ changes only the constants and resolution factors in the estimates below.

For an explicit appropriate norm, takeAn equivalent standard norm gives the same exponents, with different constants. We prove the two bounds for the entropy of a smooth periodic function ball separately.

For the lower bound, choose a nonnegative bump function $\phi \in C_c^{\infty }((-1,1{)}^n)$ with $0\le \phi \le 1$ and $\phi (0)=1$. Let $K$ bound all its derivatives of orders at most $p$, and choose $c>0$ with $cK\le 1$. For sufficiently small $h\le 1$, place $L\ge c_n{h}^{-n}$ translates of its scaled support disjointly on the torus, for instance by taking grid centres separated by $3h$ in a fundamental cube. Define the real functionsusing periodic extension. Disjointness means that at any point only one summand contributes. For every $|\alpha |\le p$, its derivative is bounded by $c{h}^{p-|\alpha |}K\le 1$. Thus every $f_{\sigma }$ lies in $B_{n,p}$. If $\sigma \ne \tau$, evaluate at a centre where the binary coefficients differ to obtainChoose $h=(4\epsilon /c{)}^{1/p}$ for sufficiently small $\epsilon$. The $2^L$ functions are more than $2\epsilon$ apart, so no radius-$\epsilon$ ball covers two of them. The metric covering number consequently satisfiesTo include the remaining $\epsilon <1/2$, observe that the two constant functions $1,-1$ belong to the ball and require distinct covering balls. Reducing the positive lower-bound constant extends the estimate over that fixed remaining interval.

For the upper bound, use this precise version of multivariable Jackson approximation: for every $f\in C^p({\mathbb{T}}^n)$ and integer $N\ge 1$, there exists a real trigonometric polynomial $P_N$ with $|k_j|\le N$ in every coordinate andChoose $N=\lceil (4J_{n,p}/\epsilon {)}^{1/p}\rceil$. Then the approximation error is at most $\epsilon /4$, and $\Vert P_N{\Vert }_{\infty }\le 2$. Write $P_N(x)={\sum }_{k\in \{-N,\dots ,N{\}}^n}{c}_k{e}^{ik\cdot x}$. There are $D=(2N+1{)}^n$ coefficients, and normalized integration gives $|c_k|\le \Vert P_N{\Vert }_{\infty }\le 2$.

Round their real and imaginary parts with mesh $\delta =\epsilon /(8D)$, respecting $c_{-k}=\overline{c_k}$ so the rounded polynomial remains real. Each coefficient changes by at most $2\delta$, and hence the uniform change in the polynomial is at most $2D\delta =\epsilon /4$. At most $(C/\delta {)}^{2D}$ choices are needed; allowing redundant choices only enlarges this bound. These rounded polynomials form an $\epsilon$-net for the ball, andHere $D\le C_{n,p}{\epsilon }^{-n/p}$, and the last logarithm absorbs the constant and $\log D$ for $\epsilon <1/2$. Together the bounds give

For the unit ball of $C({\mathbb{T}}^n)$ alone, take infinitely many disjoint small neighbourhoods and a height-one continuous bump supported in each. Distinct such functions have uniform distance one. Thus no finite $\epsilon$-net exists when $\epsilon <1/2$, andThe derivative bound was essential for total boundedness; mere continuity gives no common small-scale oscillation bound.

For the final composition estimate, interpret outer periodic functions on ${\mathbb{T}}^m$ by their $2\pi$-periodic extensions to ${\mathbb{R}}^m$. Since $p\ge 1$, each $f\in B_{m,p}$ satisfiesby integrating its first partial derivatives along coordinate segments. This uniform Lipschitz continuity controls errors in the inner functions.

Choose an outer net of radius $\eta =\epsilon /2$ for $B_{m,p}$ and an inner net of radius $\delta =\epsilon /(2m)$ for $E$. For each $g=f(u_1,\dots ,u_m)$ choose corresponding approximants $f_0,v_1,\dots ,v_m$. No derivative bound for $f_0$ is needed: use the derivative bound of $f$ before replacing the outer function. Pointwise,All composed centres are continuous. Multiplying the numbers of outer and inner choices gives the entropy of Lipschitz compositions inequalityApply the proved upper bound with dimension $m$, and the assumed estimate for $E$. Fixed rescalings of $\epsilon$ preserve its exponent and change $\log (1/\epsilon )$ by only an additive constant. Therefore

The Riemann mapping theorem asserts that a nonempty simply connected domain $\Omega ⊊\mathbb{C}$ is biholomorphic to the unit disk $\mathbb{D}$. More precisely, for each $z_0\in \Omega$ there is a unique biholomorphism $f:\Omega \to \mathbb{D}$ satisfying $f(z_0)=0$ and $f^{\prime }(z_0)>0$. We prove existence through an extremal derivative, deriving the compactness and injectivity steps rather than assuming special lemmas for the theorem.

First construct a bounded univalent function on $\Omega$. Choose $a\notin \Omega$ and a holomorphic logarithm of $z-a$, which exists by the permitted logarithm result on the translated simply connected domain. PutThen $u(z{)}^2=z-a$, so $u$ is univalent and never zero. Its image $G=u(\Omega )$ is open, and $G\cap (-G)=\varnothing$: equality $u(x)=-u(y)$ would imply $x=y$ after squaring and then $u(x)=0$.

Fix $w_0\in G$ and choose $r>0$ with $B(w_0,r)\subset G$. It follows that $B(-w_0,r)\subset -G$ is disjoint from $G$, so $|u(z)+w_0|\ge r$ throughout $\Omega$. Henceis a univalent function with $|h|\le 1/2$. For $b\in \mathbb{D}$, defineThe identityand the inverse $T_b^{-1}(w)=(w+b)/(1+\overline{b}w)$ show directly that $T_b$ is an automorphism of the unit disk. Apply $T_{h(z_0)}$ to $h$ and then a rotation to make its nonzero derivative at $z_0$ positive. The family $\mathcal{A}$ of univalent maps $\Omega \to \mathbb{D}$ with value zero and positive derivative at $z_0$ is therefore nonempty.

LetIt is positive. Choose $\rho >0$ with $\overline{B(z_0,\rho )}\subset \Omega$. The Cauchy estimate gives $v^{\prime }(z_0)\le 1/\rho$ for every $v\in \mathcal{A}$, so $M$ is finite. Take a sequence $v_j\in \mathcal{A}$ with $v_j^{\prime }(z_0)\to M$.

We now justify a locally uniform convergence subsequence. On any compact subset of $\Omega$, the bound $|v_j|\le 1$ and the Cauchy estimate on slightly larger interior disks give a uniform derivative bound. They therefore give equicontinuity. Apply the Arzelà-Ascoli theorem on a compact exhaustion of $\Omega$, then take a diagonal subsequence for locally uniform convergence. Its limit $v$ is holomorphic: pass to the limit in the Cauchy integral formula on every small interior circle. The same formula for derivatives givesThus $v$ is nonconstant. Initially $|v|\le 1$, and the maximum modulus principle improves this to $|v|<1$ everywhere.

For completeness, the needed locally uniform limit of univalent functions argument is short. If $x\ne y$ and $v(x)=v(y)$, the nonconstant holomorphic function $v-v(y)$ has an isolated zero at $x$. Choose a small closed disk around $x$ excluding $y$, with no zero on its boundary. Uniform convergence there gives $v_j-v_j(y)\to v-v(y)$. The Rouche theorem then makes $v_j-v_j(y)$ have a zero inside that disk for all sufficiently large $j$. Its only possible zero, by injectivity of $v_j$, is $y$, outside the disk. This contradiction proves injectivity of $v$. Therefore $v\in \mathcal{A}$ and attains the supremum $M$.

It remains to prove surjectivity. Suppose $c\in \mathbb{D}$ is omitted by $v$; since $v(z_0)=0$, necessarily $c\ne 0$. The domain $T_c(v(\Omega ))$ is simply connected, because both constituent maps are homeomorphisms onto their images, and it excludes zero. The permitted holomorphic logarithm gives a holomorphic square rootThis maps $\Omega$ into $\mathbb{D}$ and is univalent, since equality of its values implies equality after squaring and hence equality of the original arguments. Write $\gamma =g(z_0)$. Then ${\gamma }^2=T_c(0)=-c$ and $|\gamma |=\sqrt{|c|}$. Normalize again with $V=T_{\gamma }\circ g$ and a rotation. Its derivative magnitude isThe strict inequality follows from $(1-\sqrt{|c|}{)}^2>0$. This square-root improvement of a normalized conformal map contradicts extremality. Thus $v(\Omega )=\mathbb{D}$. An injective holomorphic map has nonzero derivative and a holomorphic local inverse; these inverses agree, so the resulting bijection is biholomorphic. We have proved existence.

For uniqueness, two normalized maps give an automorphism of the unit disk $v_2\circ v_1^{-1}$ fixing zero. Apply the Schwarz lemma, proved in the next part, to this automorphism and its inverse. Both inequalities force equality, so the equality case makes it a rotation. Its derivative at zero is the positive real number $v_2^{\prime }(z_0)/v_1^{\prime }(z_0)$, forcing that rotation to be the identity. ThereforeThe proper-domain hypothesis is necessary: a bounded entire holomorphic function has derivative at most $1/R$ on disks of arbitrarily large radius by the Cauchy estimate, and hence is constant. Conversely, a domain biholomorphic to the disk is simply connected because the map is a homeomorphism.

The Schwarz lemma states that a holomorphic function $f:\mathbb{D}\to \mathbb{D}$ with $f(0)=0$ satisfiesIf equality holds in the first inequality at any nonzero point, or in the derivative inequality, then $f(z)=\lambda z$ for a constant $|\lambda |=1$. Conversely those rotations give equality.

To prove it, the function $g(z)=f(z)/z$ extends holomorphically at zero with value $f^{\prime }(0)$, by the power series of $f$. On $|z|=r<1$, $|g(z)|\le 1/r$. The maximum modulus principle gives this same bound inside that circle. For each fixed $z$, let $r\uparrow 1$ to conclude $|g(z)|\le 1$. This proves both inequalities. Equality in either makes $g$ attain its maximum modulus at an interior point, so the maximum modulus principle makes $g$ a constant of modulus one. This proves the equality cases as well.

For the classification of automorphisms of the unit disk, choose $a\in \mathbb{D}$ and defineDirect expansion givesThus $T_a$ is a biholomorphism of the disk and sends $a$ to zero. If $F$ is any conformal map of the disk onto itself, take $a=F^{-1}(0)$ and put $h=F\circ T_a^{-1}$. Both $h$ and $h^{-1}$ fix zero. The Schwarz lemma gives $|h(z)|\le |z|$ and, applied to $h^{-1}$ at $h(z)$, $|z|\le |h(z)|$. Equality holds, so $h(z)=e^{i\theta }z$. Consequently all the holomorphic conformal automorphisms are preciselyEvery displayed map is indeed a conformal automorphism by the verified inverse and a rotation. Here conformal maps have the usual holomorphic, orientation-preserving meaning; if orientation reversal is allowed, their complex conjugates give the antiholomorphic alternatives.

The precise conclusion is that finite one-sided limits must agree, so a jump discontinuity is impossible. We prove this using explicit kernels and derive every conjugate-series identity used. The distinction from a removable discontinuity is addressed at the end.

Write the Fourier coefficients asTranslation multiplies each coefficient by a unimodular factor and preserves the hypothesis, so place the proposed discontinuity at zero. Suppose the finite limits are $A=f(0^{-})$ and $B=f(0^{+})$. For $0<r<1$, geometric summation gives the Poisson kernel on the circle and the conjugate Poisson kernel on the circle:These series converge uniformly and absolutely for each $r<1$. Therefore they may be integrated term by term against the Lebesgue integrable function $f$. With normalized convolution, $P_r$ has multiplier $r^{|n|}$, while $Q_r$ has multiplier $-i\mathrm{sgn}(n)r^{|n|}$ for $n\ne 0$ and zero at $n=0$. More explicitly, at zero,The assumed vanishing of negative Fourier coefficients thus proves, rather than presupposes, the conjugate relation

We next show that the right-hand side is bounded as $r\uparrow 1$. The kernel $P_r$ is positive and even, and its normalized integral is one by its series. Its mass outside any fixed $(-\delta ,\delta )$ tends to zero; in fact its supremum there is $O_{\delta }(1-r)$. On $(-\delta ,0)$, $f$ is uniformly close to $A$ for sufficiently small $\delta$, and on $(0,\delta )$ it is uniformly close to $B$. Evenness places asymptotically half the mass on each side. The outside contribution tends to zero because $f$ is integrable and the outside kernel tends uniformly to zero. Bounding the inside errors by the one-sided approximation errors givesHence $(Q_r\ast f)(0)$ is bounded by the proved conjugate relation.

On the other hand, oddness of $Q_r$ givesFor $0<\delta <\pi$, differentiating $1-2r\cos t+r^2$ proves the exact identityAlso $Q_r\ge 0$ on $(0,\pi )$ and is uniformly bounded on $[\delta ,\pi ]$ as $r\uparrow 1$. Write $f(-t)-f(t)=A-B+e(t)$ with $e(t)\to 0$ as $t\downarrow 0$. The integral over $[\delta ,\pi ]$ is $O_{\delta }(1)$ by integrability. Given any $\eta >0$, choose $\delta$ so that $|e(t)|\le \eta$ on $(0,\delta )$. Positivity and the preceding identity show that the near-zero error, divided by $\log (1/(1-r))$, has limiting upper bound at most $\eta /\pi$. Letting $\eta \downarrow 0$ provesBoundedness of the numerator makes the left side tend to zero. Therefore $A=B$. This proves one-sided Fourier spectrum excludes jumps, and supplies the required conjugate-sum argument in full.

There is a point-value qualification to the phrase “first kind.” If it means finite unequal one-sided limits, the proof establishes exactly the stated result. If the definition also includes removable discontinuities, the literal statement is false for an arbitrary integrable representative: the function that equals zero everywhere except $f(0)=1$ has every Fourier coefficient zero but a removable discontinuity at zero. Integral hypotheses cannot detect a change at one point. Thus no jump is possible; equal one-sided limits need not equal an arbitrarily assigned point value.