Past exam of the metric spaces and complex analysis course of the University of Oxford / 2014 (4321, A2)

Source paper: Mathematics Part A FOI release, member FOI-20210708-1 - Data/4321.pdf. The original cover identifies Metric Spaces and Complex Analysis, A2, formal code 4321, Friday 20 June 2014, 9.30am to 12.30pm. All four source pages are readable.

A homeomorphism $f:M\to N$ is a bijection for which both $f$ and $f^{-1}$ are continuous maps, using the respective metric topologies.Continuity of a bijection alone is insufficient: continuity of the inverse is a separate requirement.

The closure $\overline{A}$ is the smallest closed subset of $M$ containing $A$, equivalently the intersection of all such closed subsets. In a metric space it is characterized byIf some ball misses $A$, its centre lies outside a closed superset of $A$; conversely the complement of any closed superset is an open neighbourhood disjoint from $A$. This verifies the equivalence of the two descriptions.

False. Removing a point from the real line leaves two disjoint nonempty open rays, so the remainder is disconnected. Removing a point from the Euclidean plane leaves a path-connected space: join any two remaining points by a line segment if it avoids the removed point, or by two segments through a point not on either of the two lines through the removed point and the chosen endpoints.

A proposed homeomorphism would restrict to a homeomorphism between these punctured spaces and preserve connectedness, which is impossible. Hence

False: the target need not equal the image. Take the one-point compact space $M=\{0\}$, $N=\mathbb{R}$, and the continuous map $f(0)=0$. The target is not compact, for example because it is unbounded.

The valid theorem is that a continuous image of a compact space is compact. Thus $f(M)$ is compact; compactness of $N$ follows if surjectivity is added.

True. For $u,v\in f(M)$ choose preimages $x,y\in M$. A path $\gamma :[0,1]\to M$ joining $x$ to $y$ exists because $M$ is path-connected. The composition of continuous functions $f\circ \gamma$ is a path in $f(M)$ from $u$ to $v$.Surjectivity onto a larger target is unnecessary because the conclusion concerns the image itself.

True. For any continuous map, $f(\overline{A})\subseteq \overline{f(A)}$: every open neighbourhood of $f(x)$ pulls back to an open neighbourhood of $x\in \overline{A}$, and therefore meets $A$ in its preimage.

Apply the same inclusion to the continuous inverse of the homeomorphism:Applying $f$ gives the reverse inclusion, provingThe closures are taken in their respective ambient spaces.

True. For every finite partial sum,The nondecreasing partial sums on the left are bounded, so the square series converges. Hence ${\ell }^1\subseteq {\ell }^2$ andApplying this norm inequality to $x-y$ proves $\Vert f(x)-f(y){\Vert }_2\le \Vert x-y{\Vert }_1$. Thus the inclusion is a Lipschitz continuous map, in particular continuous, and it is well-defined on every element of l1 sequence space.

Nonnegativity and symmetry follow from the corresponding metric axioms in the two factors. The maximum is zero exactly when both coordinate distances are zero, hence exactly when the ordered pairs are equal.

For the triangle inequality, put $p=(m_1,n_1)$, $q=(m_2,n_2)$ and $r=(m_3,n_3)$. ThenTherefore the maximum formula defines a metric, the maximum product metric.

Each coordinate distance is at most the maximum, so convergence of the pairs implies convergence of both coordinates. Conversely, for any $ϵ>0$, coordinate convergence gives indices $N_M,N_N$ after which the respective distances are below $ϵ$. Beyond their maximum, both are below $ϵ$, and so is the maximum product metric.The same distance comparison will also handle Cauchy sequences.

Let $(x_n,y_n)$ be a Cauchy sequence in the product. Since each coordinate distance is bounded by $\delta$, both $(x_n)$ and $(y_n)$ are Cauchy in their respective factors. By completeness they have limits $x\in M$, $y\in N$. Part (ii) then gives convergence of the pairs to $(x,y)$.

Thus the product of two complete metric spaces is complete under the maximum product metric. This checks every Cauchy sequence, rather than assuming beforehand that it has a coordinate limit.

For coordinate differences $a,b$, the elementary inequalitiesgiveThe identity map between the two versions of the square is bijective, and both it and its inverse are Lipschitz continuous maps by these inequalities. It is therefore a homeomorphism. In fact the same comparison works on all of ${\mathbb{R}}^2$; restriction to the square causes no difficulty.

Take $M=\mathbb{R}$ and the constant contraction mapping $f(x)=0$. For $p=(0,0)$, $q=(0,1)$,No Lipschitz constant less than one can work, so $F$ need not be a contraction. More generally, differing only in the second input coordinate already prevents a strict contraction whenever $M$ has two points.

If $f$ has contraction constant $0\le c<1$, two iterations giveand consequentlyThus $F^2$ is a contraction mapping. The swap lift of a contraction shows why an iterate can contract even when one step does not.

At $z=x+iy$, compute the complex derivative first along real increments and then along imaginary increments. Holomorphy givesEquating real and imaginary parts proves the Cauchy-Riemann equations:The partial derivatives exist because these directional difference quotients are limits of the holomorphic derivative. This argument proves the necessary equations; it does not claim that their pointwise existence alone is sufficient for holomorphy.

Use the principal complex logarithm:The excluded nonpositive real ray allows a continuous choice of argument, and $L(1)=0$. For $z=x+iy$, its real part is $u=\frac{1}{2}\log (x^2+y^2)$. Differentiating the chosen local argument givesThese continuous partial derivatives satisfy the Cauchy-Riemann equations, proving holomorphy on the slit domain. Its derivative is

A holomorphic primitive on the slit domain is $P(z)=zL(z)-z$, since the product rule and $L^{\prime }=1/z$ give $P^{\prime }=L$. The line segment stays in that domain. The fundamental theorem for line integrals therefore givesHenceThe value $L(i)=\pi i/2$ is fixed by the chosen branch.

In the upper half-plane, the argument lies strictly between zero and $\pi$, while $\log |z|$ can be any real number. Conversely, if $w=s+it$ with $0<t<\pi$, then $z=e^w$ is in the upper half-plane and $L(z)=w$.Thus the principal complex logarithm is a conformal bijection from the upper half-plane onto that horizontal strip, with inverse the exponential function restricted to the strip.

Write $\alpha =a+ib$ with $b>0$. ThenIt preserves the upper half-plane, has inverse $(z-a)/b$, and sends $i$ to $\alpha$. Its nonzero derivative $b$ verifies that this bijection is conformal.

The map $h(z)=e^{\pi z/2}$ is a conformal bijection from the width-two strip to the upper half-plane, and $h(i)=i$. Its inverse there is $h^{-1}(w)=2L(w)/\pi$.

Write $e^{\pi \beta /2}=A+iB$, where $B>0$. Conjugate the preceding affine half-plane map by $h$:Every argument of $L$ lies in the upper half-plane. The formula is therefore a conformal bijection of the strip onto itself, andThe final equality uses $0<\mathrm{Im}(\pi \beta /2)<\pi$, so no multiple of $2\pi i$ is lost. This is a strip automorphism with a prescribed image of its midpoint.

Morera's theorem says that a continuous complex-valued function $f$ on an open set $U\subseteq \mathbb{C}$ is holomorphic iffor every triangle $T$ whose closed interior is contained in $U$. Triangle boundaries are traversed with positive orientation. The continuity hypothesis matters: the zero-integral condition is combined with continuity, not used in isolation.

For the usual piecewise smooth complex path $\gamma$, let $\ell (\gamma )$ denote its finite length. The estimation lemma givesBy uniform convergence, the supremum tends to zero, soThe uniform limit is continuous on the path, so its integral exists. This is uniform convergence and contour integration; the path length is fixed while the functions vary.

Let the common domain be an open set $U$. The uniform limit theorem makes the limit $f$ continuous. For every triangle whose closed interior lies in $U$, Cauchy's integral theorem gives ${\int }_{\partial T}{f}_n=0$. Part (ii) passes this identity to the limit, giving ${\int }_{\partial T}f=0$. Morera's theorem now proves that $f$ is holomorphic.

The same reasoning works for locally uniform convergence, since each triangle boundary is compact. No passage of derivatives through a limit has been assumed.

Fix any bounded disk avoiding the integers, and choose $R$ with $|z|\le R$ on it. For sufficiently large $|n|$, say $|n|\ge 2R$ and $|n|\ge 1$,The convergent numerical series ${\sum }_{n\ne 0}{n}^{-2}$ and the Weierstrass M-test give absolute and uniform convergence of the tails. The finitely many remaining summands are holomorphic away from their integer poles. Thus the partial sums converge locally uniformly on $\mathbb{C}\setminus \mathbb{Z}$.

By part (a)(iii), the sum is a holomorphic function on $\mathbb{C}\setminus \mathbb{Z}$. The term $n=0$ is $-1/z^2$ and is harmless on this domain; it must not be omitted.

For fixed $z\notin \mathbb{Z}$, integratecounterclockwise around the specified square of half-side $R_N=N+1/2$, choosing $N>|z|$. The integer poles inside it are $-N,\dots ,N$, and their residues are $1/(n^2-z^2)$ because $\pi \cot (\pi w)$ has residue one at each integer.

The two other possible poles are $w=z$ and $w=-z$. Their residues are each $\pi \cot (\pi z)/(2z)$, hence sum to $\pi \cot (\pi z)/z$. This formula still applies when the numerator vanishes and a possible pole is removable. The residue theorem givesOn the square, $|w|\ge R_N$, its perimeter is $8R_N$, and the supplied cotangent bound givesThe series converges absolutely by part (b), so its symmetric partial sums have the same limit as the full sum. ThereforeAt half-integers the cotangent form makes the value zero explicit; the reciprocal-tangent notation is understood by its continuous extension there. The cotangent contour summation rests on the contour estimate, not only on formally matching poles.

Write the Laurent series about $a$ as ${\sum }_{k\in \mathbb{Z}}{c}_k(z-a{)}^k$. The isolated singularity is:

These cases exhaust the possible principal parts. The residue isFor the first example, $\sin z=z+O(z^3)$ givesThus zero is a double pole, with residue zero. For the other example,so the ratio is $-z/2+O(z^2)$. Its singularity is removable, with extended value zero and residue zero. The numerator's zero has higher order than the denominator's, so there is no pole.

A simple zero means $h(a)=0$ and $h^{\prime }(a)\ne 0$. The local factorization is $h(z)=(z-a)h_1(z)$, where $h_1$ is holomorphic and $h_1(a)=h^{\prime }(a)\ne 0$. ThusThe numerator is holomorphic near $a$, so its constant Taylor coefficient is the coefficient of $(z-a{)}^{-1}$ in the quotient. ThereforeIf $g(a)=0$, the possible pole cancels and the residue is zero; the formula does not require $g/h$ actually to have a pole. This is the residue formula for a quotient at a simple denominator zero.

Consider the counterclockwise unit circle and setThe parametrization $z=e^{i\theta }$ givesThe imaginary part is zero: replacing $\theta$ by $2\pi -\theta$ conjugates the integrand, since $n$ is an integer and $x$ is real. The real part is precisely the defining Bessel integral, so $I=J_n(x)$.

Both exponential series converge absolutely and uniformly on the circle. Their product may therefore be integrated term by term:Uniform absolute convergence can also be checked by summing the absolute coefficient bounds, which gives $e^{|x|}$. The residue is obtained exactly when $n-1+j-k=-1$, or $k=n+j$. The Cauchy coefficient formula consequently yieldsThis proves the Bessel function of the first kind series directly from its integral, including the sign and factorial normalization.

The line through the north pole and $p=(a,b,c)$ isIts intersection with the equatorial complex plane has third coordinate zero, giving $t=1/(1-c)$ when $p$ is not the north pole. The first two coordinates are therefore $a/(1-c)$ and $b/(1-c)$, henceThe north pole maps to the point at infinity. This is stereographic projection onto the Riemann sphere.

For $z=x+iy$ and $r^2=|z{|}^2$, solving the sphere and projection equations givesFor $z\ne 0$, the reciprocal has real and imaginary parts $x/r^2$ and $-y/r^2$, and squared modulus $1/r^2$. Substituting these into the inverse formula producesThusThe reciprocal exchanges zero and infinity, corresponding to exchanging the south and north poles, so the formula holds there too. On the sphere this is a rotation through $\pi$ about the first coordinate axis, not the antipodal map.

When all three points are finite, useThe distinctness assumptions ensure that it is a nonconstant Möbius transformation, and direct substitution gives the three required values. If one of the specified points is infinity, the limiting formulas areThese cover every case because the points are distinct.

For uniqueness, if $g$ and $\tilde{g}$ have the same three images, the Möbius map $\tilde{g}\circ g^{-1}$ fixes $0,1,\infty$. Fixing zero and infinity makes it $z\mapsto cz$ with $c\ne 0$; fixing one gives $c=1$. Therefore the normalized map is unique. This is the sharp three-transitivity of Möbius transformations.

Let $G$ be the subgroup preserving the three-point set. Its action gives a group homomorphism $\Phi :G\to S_3$. If $\Phi (h)$ is the identity permutation, then $h$ fixes all three points, so part (i) forces $h$ to be the identity Möbius map. Thus the kernel is trivial and $\Phi$ is injective.

For any permutation $\sigma$, part (i) gives a normalization $g_{\sigma }$ sending the ordered triple $(z_{\sigma (1)},z_{\sigma (2)},z_{\sigma (3)})$ to $(0,1,\infty )$. If $g$ normalizes the original triple, then $h_{\sigma }=g_{\sigma }^{-1}\circ g$ sends $z_j$ to $z_{\sigma (j)}$. Hence every permutation is attained, so $\Phi$ is surjective.This is the setwise stabilizer; its six maps need not fix each point individually.

Choose a Möbius map $T$ with $T(z_1)=0$ and $T(z_2)=\infty$, for example by normalizing the ordered triple $(z_1,z_3,z_2)$. Conjugation $h\mapsto T\circ h\circ T^{-1}$ takes the requested subgroup to the subgroup fixing zero and infinity individually.

Such a transformation must be $z\mapsto cz$ with $c\in {\mathbb{C}}^{\ast }$: the zero condition removes the constant numerator term and the infinity condition removes the nonconstant denominator term. Conversely every such dilation is Möbius and fixes the two points. Composition multiplies its nonzero parameter.This identifies the Möbius pointwise stabilizer of two points. Allowing an interchange of the two fixed points would give a larger group, but those interchanges are excluded by the stated equalities.