Solution. Since \(f_n \rightarrow f\) uniformly, \(f\) is analytic by Weierstrass’ theorem.

\(a\) is a zero of first order for \(f\). Each \(f_n\) vanishes at \(a\), so \(f(a) = \lim f_n(a) = 0\). Suppose the order were greater than \(1\), so \(f'(a) = 0\). Uniform convergence on compact sets gives \(f_n' \rightarrow f'\) uniformly on compact sets, but that alone is not decisive; use Hurwitz instead, on the functions \(f_n'\). Since \(f_n'(a) \neq 0\) for every \(n\) (the zero at \(a\) being simple) and \(f_n' \rightarrow f'\) locally uniformly, if \(f' \not \equiv 0\) near \(a\) then \(f'(a) = 0\) would force some \(f_n'\) to vanish near \(a\) — which is not immediately a contradiction. The clean route is the argument principle: choose \(r\) small with \(f \neq 0\) on \(0 < \left |z-a\right | \leq r\), then \[\frac {1}{2\pi i}\oint _{\left |z-a\right |=r}\frac {f_n'}{f_n} \longrightarrow \frac {1}{2\pi i}\oint _{\left |z-a\right |=r}\frac {f'}{f},\] because \(f_n \rightarrow f\) uniformly on the circle and \(f\) is bounded away from \(0\) there. The left side is \(1\) for every \(n\), and the right side counts the zeros of \(f\) inside, so \(f\) has exactly one zero there, of order \(1\).

\(f\) is one-to-one. Suppose \(f(z_1) = f(z_2) = w\) with \(z_1 \neq z_2\) in \(G\). Apply Hurwitz’s theorem to \(f_n - w\) on small disjoint discs about \(z_1\) and \(z_2\): since \(f - w\) vanishes in each and is not identically zero (it has isolated zeros, \(f\) being non-constant — non-constant because it has a simple zero), for large \(n\) each disc contains a zero of \(f_n - w\). That gives two distinct points where \(f_n\) takes the value \(w\), contradicting injectivity of \(f_n\). \(\blacksquare \)

The case \(f\) constant is excluded by the first part: a constant cannot have a first-order zero.

Solution. This is Hurwitz’s theorem in the form usually stated for univalent limits.

Suppose \(f\) is not injective, so \(f(z_1) = f(z_2) = w\) for some \(z_1 \neq z_2\) in \(\Omega \), and suppose \(f\) is not constant. Then \(f - w\) is a non-constant analytic function vanishing at \(z_1\) and \(z_2\), and its zeros are isolated: choose disjoint closed discs \(\overline {D_1}, \overline {D_2}\) about \(z_1, z_2\) inside \(\Omega \) on which \(f - w\) has no zero except the centre.

Let \(\delta = \min \left |f - w\right |\) over the two boundary circles, which is positive by compactness. Convergence in \(H(\Omega )\) is locally uniform, so for large \(n\) \[\left |\left (f_n - w\right ) - \left (f - w\right )\right | < \delta \leq \left |f-w\right | \qquad \text {on } \partial D_1 \cup \partial D_2 .\] By Rouché, \(f_n - w\) has as many zeros in each disc as \(f - w\) does, namely at least one. So \(f_n\) takes the value \(w\) at a point of \(D_1\) and at a point of \(D_2\), two distinct points — contradicting injectivity of \(f_n\).

Hence either \(f\) is injective or \(f\) is constant. \(\blacksquare \)

\[\boxed {f \text { is injective, or } f \text { is constant.}}\]

The constant alternative genuinely occurs: \(f_n(z) = z/n\) are all injective on \(D\) and converge to \(0\).

Solution. This is Weierstrass’ theorem on analytic limits.

\(f\) is continuous, being a uniform limit of continuous functions. Let \(\Delta \) be any closed triangle contained in \(G\). Since \(f_n\) is analytic, Cauchy’s theorem gives \[\int _{\partial \Delta } f_n(z)\,dz = 0 \qquad \text {for every } n .\] The boundary \(\partial \Delta \) is compact with finite length \(L\), and \(f_n \rightarrow f\) uniformly there, so \[\left |\int _{\partial \Delta } f - \int _{\partial \Delta } f_n\right | \leq L \max _{\partial \Delta }\left |f - f_n\right | \longrightarrow 0 ,\] whence \(\int _{\partial \Delta } f = 0\).

By Morera’s theorem \(f\) is analytic on \(G\). \(\blacksquare \)

Uniformity is what allows the limit to pass through the integral; pointwise convergence is not enough, and indeed a pointwise limit of analytic functions can be as bad as one likes. Note also that only local uniform convergence is needed, since analyticity is a local property — and that weaker hypothesis is the one usually stated, because it is what normal families deliver.

Solution. Statement. Let \(G\) be a region, \((f_n)\) locally bounded in \(H(G)\), and \(f \in H(G)\). If \[A = \left \{z \in G : \lim _n f_n(z) = f(z)\right \}\] has a limit point in \(G\), then \(f_n \rightarrow f\) in \(H(G)\), that is locally uniformly.

Proof. Suppose not. Then there are a compact \(K \subseteq G\), an \(\varepsilon > 0\) and a subsequence \((f_{n_k})\) with \[\max _{K}\left |f_{n_k} - f\right | \geq \varepsilon \qquad \text {for all } k. \qquad (*)\]

The family \((f_n)\) is locally bounded, so by Montel’s theorem it is normal: \((f_{n_k})\) has a subsequence converging locally uniformly to some \(g \in H(G)\).

On the set \(A\) we know \(f_n \rightarrow f\) pointwise, so along that sub-subsequence \(g = f\) on \(A\). But \(A\) has a limit point in \(G\), so by the Identity Theorem \(g \equiv f\) on \(G\).

That sub-subsequence therefore converges to \(f\) uniformly on \(K\), contradicting \((*)\). \(\blacksquare \)

The shape — assume failure, extract a normal-family subsequence, identify its limit by the Identity Theorem, contradict — is the standard way normality is used, and it recurs in every problem of this group. Vitali is remarkable in that convergence on a very thin set (a convergent sequence of points, say) upgrades to locally uniform convergence everywhere, provided only that the family does not blow up.

Solution. By Weierstrass’ theorem \(f\) is analytic and \(f_n' \rightarrow f'\) locally uniformly.

\(f\) is one-to-one. Suppose \(f(z_1) = f(z_2)\) with \(z_1 \neq z_2\). As \(f\) is non-constant — because \(f'(a) > 0\) — the argument of the second problem in this set applies verbatim: Rouché on small disjoint discs produces two distinct points where some \(f_n\) takes the same value, contradicting injectivity.

\(f_n'(a) \neq 0\) for large \(n\). We know \(f_n'(a) \rightarrow f'(a) > 0\), so certainly \(f_n'(a) \neq 0\) once \(n\) is large enough that \(\left |f_n'(a) - f'(a)\right | < f'(a)\).

It is worth noting the stronger statement that makes this robust: since each \(f_n\) is injective, \(f_n'\) has no zeros at all in \(G\). For if \(f_n'(c) = 0\) then \(f_n\) behaves near \(c\) like \(f_n(c) + a_k(z-c)^{k}\) with \(k \geq 2\), hence is \(k\)-to-one on a punctured neighbourhood of \(c\) — not injective. So injectivity of each \(f_n\) already gives \(f_n' \neq 0\) everywhere, and the hypothesis \(f'(a) > 0\) is needed only to rule out the constant limit. \(\blacksquare \)

Solution. Let \(\mathcal {F}\) be locally bounded on \(G\) and let \(K \subseteq G\) be compact. We show \(\mathcal {F}\) is equicontinuous on \(K\) without invoking Montel — indeed this estimate is the main step in proving Montel.

Choose \(r > 0\) with \[K_{2r} = \left \{z : \operatorname {dist}(z,K) \leq 2r\right \} \subseteq G,\] possible because \(K\) is compact and \(G\) open. By local boundedness (and a finite subcover of \(K_{2r}\)) there is \(M\) with \[\left |f(z)\right | \leq M \qquad \text {for all } f \in \mathcal {F},\ z \in K_{2r}.\]

Let \(z, w \in K\) with \(\left |z-w\right | < r\), and let \(\gamma \) be the circle of centre \(z\) and radius \(2r\), which lies in \(K_{2r}\). Cauchy’s integral formula applied twice and subtracted gives \[f(z)-f(w) = \frac {1}{2\pi i}\oint _{\gamma } f(\zeta )\left [\frac {1}{\zeta -z}-\frac {1}{\zeta -w}\right ]d\zeta = \frac {z-w}{2\pi i}\oint _{\gamma } \frac {f(\zeta )\,d\zeta }{(\zeta -z)(\zeta -w)} .\] On \(\gamma \) we have \(\left |\zeta - z\right | = 2r\) and \(\left |\zeta -w\right | \geq 2r - r = r\), so \[\left |f(z)-f(w)\right | \leq \frac {\left |z-w\right |}{2\pi }\cdot \frac {M}{2r\cdot r}\cdot 2\pi (2r) = \frac {M}{r}\left |z-w\right | .\]

The bound \(M/r\) depends only on \(K\) and \(G\), not on \(f\): given \(\varepsilon \), take \(\delta = \min \left (r,\ \varepsilon r/M\right )\) and every member of \(\mathcal {F}\) satisfies \(\left |f(z)-f(w)\right | < \varepsilon \) whenever \(\left |z-w\right | < \delta \). So \(\mathcal {F}\) is (uniformly) equicontinuous on \(K\). \(\blacksquare \)

In fact the family is uniformly Lipschitz on \(K\), which is stronger than equicontinuity. Montel’s theorem then follows by Arzelà–Ascoli.

Solution. (a). \(\mathcal {F}'\) is normal. Let \((f_n') \) be a sequence in \(\mathcal {F}'\). Since \(\mathcal {F}\) is normal, the corresponding \((f_n)\) has a subsequence converging locally uniformly to some analytic \(g\) (or to \(\infty \) locally uniformly, if the wider definition of normality is in use — assume the former, as is standard for \(H(G)\)). By Weierstrass’ theorem the derivatives of that subsequence converge locally uniformly to \(g'\).

So every sequence in \(\mathcal {F}'\) has a locally uniformly convergent subsequence: \(\mathcal {F}'\) is normal. \(\blacksquare \)

(b). The converse is false. Take \[\mathcal {F} = \{f_n(z) = n : n \in \mathbb {N}\},\] the constant functions on any region \(G\). Then \(\mathcal {F}' = \{0\}\), a single function, which is trivially normal. But \(\mathcal {F}\) itself is not normal in \(H(G)\): \(f_n = n\) has no subsequence converging locally uniformly to an analytic function, since \(\left |f_n\right | \rightarrow \infty \) everywhere.

\[\boxed {\text {No; adding an unbounded constant to each member destroys normality while leaving } \mathcal {F}' \text { untouched.}}\]

The obstruction is exactly the loss of information at one point: normality of \(\mathcal {F}'\) plus boundedness of \(\{f(z_0) : f \in \mathcal {F}\}\) at a single point \(z_0\) does give normality of \(\mathcal {F}\), by integrating.

Solution. By Montel’s theorem it suffices to show \(\mathcal {F}\) is locally bounded, that is uniformly bounded on each compact subset of \(G\).

Let \(K \subseteq G\) be compact and choose \(r > 0\) with \(\overline {B(z,r)} \subseteq G\) for every \(z \in K\) — possible as \(\operatorname {dist}(K,\partial G) > 0\).

Fix \(g \in \mathcal {F}\) and \(z \in K\). The function \(\left |g\right |^{100}\) is subharmonic (being \(\left |g\right |^{p}\) for \(p > 0\) with \(g\) analytic), so it satisfies the sub-mean-value inequality over the disc \(B(z,r)\): \[\left |g(z)\right |^{100} \leq \frac {1}{\pi r^{2}}\iint \limits _{B(z,r)}\left |g\right |^{100}dx\,dy \leq \frac {1}{\pi r^{2}}\iint \limits _{G}\left |g\right |^{100}dx\,dy \leq \frac {M}{\pi r^{2}} .\] Hence \[\left |g(z)\right | \leq \left (\frac {M}{\pi r^{2}}\right )^{1/100} \qquad \text {for all } g \in \mathcal {F},\ z \in K,\] a bound independent of \(g\). So \(\mathcal {F}\) is locally bounded and therefore normal. \(\blacksquare \)

The exponent \(100\) is a red herring — any positive exponent works, since \(\left |g\right |^{p}\) is subharmonic for every \(p > 0\). The step that matters is the area version of the mean value inequality, which converts an \(L^{p}\) bound into a pointwise one; this is why an integral constraint suffices where one might have expected a sup-norm constraint.

Solution. (\(\Rightarrow \)) If \(f_n \rightarrow f\) uniformly on compact subsets, then for a fixed piecewise differentiable \(\gamma \) — whose image is compact — the convergence is uniform on \(\{\gamma \}\), so \[\left |\int _{\gamma } f_n - \int _{\gamma } f_m\right | \leq L\max _{\gamma }\left |f_n - f_m\right | \longrightarrow 0\] as \(n,m \rightarrow \infty \), where \(L\) is the length. So the integrals form a Cauchy sequence.

(\(\Leftarrow \)) Suppose the integrals are Cauchy for every such \(\gamma \). Fix \(z_0 \in G\) and a closed disc \(\overline {B(z_0,2r)} \subseteq G\). Applying the hypothesis to the segment from \(z_0\) to \(z\), the functions \[F_n(z) = \int _{[z_0,z]} f_n(w)\,dw\] converge pointwise on \(B(z_0,2r)\); call the limit \(F\). Each \(F_n\) is an antiderivative of \(f_n\) on the disc.

Applying the hypothesis to the circles \(\left |w - z\right | = \rho \) shows in particular that \(\displaystyle \oint f_n\) is Cauchy on every circle, and Cauchy’s integral formula \[f_n(z) = \frac {1}{2\pi i}\oint _{\left |w-z_0\right |=2r}\frac {f_n(w)}{w-z}\,dw\] expresses \(f_n(z)\) as an integral of \(f_n\) against a kernel that varies continuously in \(z\); uniform Cauchyness of the integrals over that circle, together with the uniform bound on the kernel for \(z\) in the smaller disc \(\left |z-z_0\right | \leq r\), makes \((f_n)\) uniformly Cauchy there. Since \(\mathbb {C}\) is complete, \(f_n\) converges uniformly on \(\overline {B(z_0,r)}\).

Every compact subset of \(G\) is covered by finitely many such discs, so the convergence is uniform on compact sets. \(\blacksquare \)

The content is that the Cauchy kernel lets integral information be converted back into pointwise information — the same mechanism as in the equicontinuity estimate above.

Solution. This is the Weierstrass interpolation theorem, and the point is that the sequence has no limit point in \(G\): the prescribed values may therefore be completely arbitrary.

Construction. By the Weierstrass product theorem there is a function \(g \in H(G)\) whose zeros are exactly the \(a_k\), each simple. (Such a \(g\) exists precisely because \((a_k)\) has no limit point in \(G\); a function with a non-isolated zero set would be identically zero.)

Now seek \(f\) in the form \[f(z) = g(z)\sum ^{\infty }_{k=1} c_k\,\frac {h_k(z)}{z - a_k},\] where \(c_k = w_k / g'(a_k)\) — note \(g'(a_k) \neq 0\) as the zeros are simple — and the \(h_k\) are Mittag-Leffler convergence factors, analytic near \(a_k\) with \(h_k(a_k) = 1\), chosen so that the series converges locally uniformly on \(G\). Such factors exist by the Mittag-Leffler theorem, which is exactly designed to force convergence of a series of prescribed principal parts.

Each term is analytic except for a simple pole at \(a_k\) cancelled by the zero of \(g\), so \(f\) is analytic on \(G\), and evaluating at \(a_j\) only the \(j\)-th term survives: \[f(a_j) = g'(a_j)\,c_j = w_j . \qquad \blacksquare \]

\[\boxed {\text {Any values } (w_k) \text { whatever can be interpolated.}}\]

Contrast the Identity Theorem problems earlier in the course: there the points accumulated inside the region and the values were almost entirely determined. Here they accumulate only at the boundary, and nothing is determined at all. The position of the limit point is the whole difference.

Solution. Write \(S_N = \sum ^{N}_{n=1} f_n\), the partial sums, which are analytic on \(G\). The hypothesis is \(S_N \rightarrow f\) uniformly on compact subsets.

\(f\) is analytic. By Weierstrass’ theorem on analytic limits, a locally uniform limit of analytic functions is analytic. So \(f \in H(G)\).

Term-by-term differentiation. The same theorem gives \(S_N' \rightarrow f'\) locally uniformly. But \[S_N'(z) = \sum ^{N}_{n=1} f_n'(z),\] so the series of derivatives converges (locally uniformly) to \(f'\): \[\boxed {f'(z) = \sum ^{\infty }_{n=1} f_n'(z).} \qquad \blacksquare \]

For completeness, the derivative statement comes from the Cauchy integral formula: on a circle \(\left |\zeta - z\right | = \rho \) inside \(G\), \[S_N'(z) - f'(z) = \frac {1}{2\pi i}\oint \frac {S_N(\zeta )-f(\zeta )}{(\zeta -z)^{2}}\,d\zeta ,\] so \(\left |S_N' - f'\right | \leq \rho ^{-1}\max _{\left |\zeta -z\right |=\rho } \left |S_N-f\right | \rightarrow 0\).

The real analogue is false: a uniformly convergent series of differentiable real functions need not be differentiable term by term. Analyticity, through the Cauchy kernel, controls all derivatives at once.

Solution. (a). Let \(\left |z - z_0\right | < \tfrac 12 R\). The circle \(\left |\zeta - z_0\right | = R\) lies in \(G\), and Cauchy’s integral formula gives \[f(z) = \frac {1}{2\pi i}\oint _{\left |\zeta -z_0\right |=R} \frac {f(\zeta )}{\zeta - z}\,d\zeta .\] On that circle \(\left |\zeta - z\right | \geq R - \tfrac 12 R = \tfrac 12 R\), and parametrising \(\zeta = z_0 + Re^{i\alpha }\) so that \(\left |d\zeta \right | = R\,d\alpha \), \[\left |f(z)\right | \leq \frac {1}{2\pi }\cdot \frac {1}{R/2} \int ^{2\pi }_{0}\left |f\!\left (z_0+Re^{i\alpha }\right )\right | R\,d\alpha = \frac {1}{\pi R}\int ^{2\pi }_{0} \left |f\!\left (z_0+Re^{i\alpha }\right )\right | R\,d\alpha . \qquad \blacksquare \]

(b). Part (a) holds with \(R\) replaced by any \(s \in (0,R]\) for which the circle still lies in \(A\); multiplying the inequality \[\left |f(z)\right | \leq \frac {1}{\pi s}\int ^{2\pi }_{0} \left |f\!\left (z_0+se^{i\alpha }\right )\right | s\,d\alpha \] by \(s\) and integrating over \(s \in (0,R]\) gives \[\frac {R^{2}}{2}\left |f(z)\right | \leq \frac {1}{\pi }\int ^{R}_{0}\!\!\int ^{2\pi }_{0} \left |f\!\left (z_0+se^{i\alpha }\right )\right | s\,d\alpha \,ds = \frac {1}{\pi }\iint \limits _{A}\left |f(\zeta )\right |dA,\] the last step being the polar-coordinate form of the area integral, as hinted. Hence \[\left |f(z)\right | \leq \frac {2}{\pi R^{2}}\iint \limits _{A} \left |f(\zeta )\right | dA . \qquad \blacksquare \]

(c). Suppose \(\iint _A\left |f\right |dA \leq M\) for every \(f \in \mathcal {F}\) and every compact \(A \subseteq G\). Given a compact \(K \subseteq G\), cover it by finitely many discs of the type in (b) with closures in \(G\); part (b) bounds \(\left |f\right |\) on each by \(2M/(\pi R^{2})\), a constant independent of \(f\). So \(\mathcal {F}\) is locally bounded, and Montel’s theorem makes it normal — that is, pre-compact in \(H(G)\) with the topology of uniform convergence on compact sets. \(\blacksquare \)

Parts (a) and (b) are the mechanism by which an \(L^{1}\) bound becomes a sup bound, which is what Montel needs; the same device with exponent \(100\) appears elsewhere in this set.

Solution. (a). For a closed rectifiable \(\gamma \) in \(G\), Cauchy’s theorem gives \(\int _{\gamma } f_k = 0\) for each \(k\). The image \(\{\gamma \}\) is compact and \(f_k \rightarrow f\) uniformly there, so with \(L\) the length of \(\gamma \), \[\left |\int _{\gamma } f_k - \int _{\gamma } f\right | \leq L\max _{\{\gamma \}}\left |f_k - f\right | \longrightarrow 0 .\] Hence \[\lim _{k\to \infty }\int _{\gamma } f_k = \int _{\gamma } f ,\] and since each term on the left is \(0\), the limit is \(0\) as well.

(b). \(f\) is continuous as a uniform limit of continuous functions, and by part (a) \(\int _{\partial \Delta } f = 0\) for every triangle \(\Delta \subseteq G\). By Morera’s theorem \(f\) is analytic on \(G\). \(\blacksquare \)

This is Weierstrass’ theorem again, split into the two steps that prove it, and it is worth seeing that the whole content is: uniform convergence lets the limit pass through the integral, and Morera converts vanishing integrals back into analyticity.

Solution. (a). Hurwitz’s Theorem. Let \(f_n \rightarrow f\) locally uniformly on a region \(G\), with each \(f_n\) analytic and \(f \not \equiv 0\). If \(f\) has a zero of order \(m\) at \(z_0\), then for every sufficiently small \(r > 0\) there is \(N\) such that for all \(n \geq N\) the function \(f_n\) has exactly \(m\) zeros, counted with multiplicity, in \(0 < \left |z - z_0\right | < r\).

Proof of the statement. Choose \(r\) with \(f \neq 0\) on \(0 < \left |z-z_0\right | \leq r\), possible because zeros of a non-zero analytic function are isolated. Let \(\delta = \min _{\left |z-z_0\right |=r}\left |f\right | > 0\). For \(n\) large, \(\left |f_n - f\right | < \delta \leq \left |f\right |\) on that circle, so by Rouché \(f_n\) and \(f\) have the same number of zeros inside, namely \(m\).

(b). Suppose each \(f_n\) is one-to-one, \(f_n \rightarrow f\) in \(H(G)\), and \[f_n(a) = 0,\qquad f_n'(a) > 0 .\] Then \(f(a) = \lim f_n(a) = 0\) and \(f'(a) = \lim f_n'(a) \geq 0\).

If \(f \equiv 0\) the claim fails, so we must exclude it — and the hypothesis \(f'(a) > 0\) in the problem is exactly what does: it is given, and \(f'(a) = \lim f_n'(a)\), so \(f \not \equiv 0\).

Now \(f\) is injective by the Rouché argument of the second problem in this set: a repeated value would put zeros of \(f_n - w\) in two disjoint discs. And \(f'(a) > 0\) is immediate from \(f'(a) = \lim f_n'(a)\) together with \(f' (a) \neq 0\), which holds because an injective analytic function has nowhere-vanishing derivative. \(\blacksquare \)

Solution. (a). Montel’s Theorem. A family \(\mathcal {F} \subseteq H(G)\) that is locally bounded — uniformly bounded on each compact subset of \(G\) — is normal: every sequence in \(\mathcal {F}\) has a subsequence converging uniformly on compact subsets of \(G\) to some analytic function.

(b). Let \[\mathcal {F} = \left \{ f \in H(\Omega ) : \iint \limits _{\Omega }\left |f(z)\right |^{2}dx\,dy < 10 \right \}.\] By Montel it suffices to show local boundedness.

Fix compact \(K \subseteq \Omega \) and \(r > 0\) with \(\overline {B(z,r)} \subseteq \Omega \) for every \(z \in K\). For \(f \in \mathcal {F}\) and \(z \in K\), the function \(\left |f\right |^{2}\) is subharmonic, so the area mean value inequality gives \[\left |f(z)\right |^{2} \leq \frac {1}{\pi r^{2}}\iint \limits _{B(z,r)}\left |f\right |^{2}dA \leq \frac {1}{\pi r^{2}}\iint \limits _{\Omega }\left |f\right |^{2}dA < \frac {10}{\pi r^{2}} .\] Hence \[\left |f(z)\right | < \sqrt {\frac {10}{\pi r^{2}}} = \frac {1}{r}\sqrt {\frac {10}{\pi }} \qquad \text {for all } f \in \mathcal {F},\ z\in K,\] a bound depending only on \(K\) and \(\Omega \). So \(\mathcal {F}\) is locally bounded and therefore normal. \(\blacksquare \)

This is the Bergman space \(A^{2}(\Omega )\), and the estimate just proved — that point evaluation is bounded by the \(L^{2}\) norm — is precisely what makes it a reproducing kernel Hilbert space.

Solution. The construction is the square-root trick, and it is the same device that starts the standard proof of the Riemann mapping theorem.

Since \(G \neq \mathbb {C}\), pick \(a \notin G\). Then \(z - a\) is analytic and non-vanishing on \(G\), and \(G\) is simply connected, so a single-valued branch of the logarithm of \(z-a\) exists there (Monodromy). Define \[h(z) = \sqrt {z-a} = e^{\frac 12\log (z-a)},\] a single-valued analytic branch on \(G\).

\(h\) is injective, since \(h(z_1) = h(z_2)\) implies \(z_1 - a = z_2 - a\).

\(h(G)\) omits a disc. \(h\) is non-constant, so by the open mapping theorem \(h(G)\) contains some disc \(B(w_0,\rho )\) with \(w_0 \neq 0\). Now observe that \(h(G)\) cannot meet \(-B(w_0,\rho )\): if \(h(z_1) = -h(z_2)\) then squaring gives \(z_1 - a = z_2 - a\), so \(z_1 = z_2\) and hence \(h(z_1) = -h(z_1)\), forcing \(h(z_1) = 0\), that is \(z_1 = a \notin G\). So the disc \(B(-w_0,\rho )\) is disjoint from \(h(G)\).

Invert into that hole. Put \[f(z) = \frac {1}{h(z) + w_0}.\] The denominator never vanishes on \(G\), since \(-w_0 \notin h(G)\), and in fact \(\left |h(z)+w_0\right | \geq \rho \) because \(h(z) \notin B(-w_0,\rho )\). Hence \[\left |f(z)\right | \leq \frac {1}{\rho } \qquad \text {for all } z \in G ,\] and \(f\) is non-constant because \(h\) is. \(\blacksquare \)

\[\boxed {f = \frac {1}{\sqrt {z-a}+w_0} \text { is bounded, analytic and non-constant on } G.}\]

Simple connectedness enters exactly once, in extracting the branch of the square root, and that is the only thing separating this from the plane, where no such \(f\) exists by Liouville.

Solution. (a). The annulus \(1 < \left |z\right | < 2\): no. A conformal bijection is in particular a homeomorphism, and the annulus is not simply connected while the disc is. The circle \(\left |z\right | = \tfrac 32\) is not contractible in the annulus but its image would have to be contractible in the disc. So no such map exists.

(b). The plane \(\mathbb {C}\): no. A one-to-one analytic \(f : \mathbb {C} \rightarrow B(0,1)\) would be a bounded entire function, hence constant by Liouville — and a constant is not injective. This is precisely the exception in the Riemann mapping theorem, which requires \(G \neq \mathbb {C}\).

(c). \(S = \mathbb {C}\setminus \{z : \operatorname {Im}z = 0,\ \operatorname {Re}z \geq 1\}\): yes. The plane slit along the ray \([1,\infty )\) is simply connected and is not all of \(\mathbb {C}\), so the Riemann mapping theorem applies — but the question asks for the function, and it can be built explicitly: \[z \ \overset {1-z}{\longmapsto }\ \mathbb {C}\setminus (-\infty ,0] \ \overset {\sqrt {\ }}{\longmapsto }\ \{\operatorname {Re}w > 0\} \ \overset {\frac {w-1}{w+1}}{\longmapsto }\ B(0,1).\] The first map carries the slit \([1,\infty )\) onto \((-\infty ,0]\); the principal square root then opens the slit plane into the right half plane; the Cayley transform finishes. Each stage is a bijection onto its image, so \[\boxed {f(z) = \frac {\sqrt {1-z}-1}{\sqrt {1-z}+1}}\] maps \(S\) one-to-one onto the unit disc.

The three cases are the three obstructions worth knowing: wrong topology, the plane itself, and no obstruction at all.

Solution. Consider the reflected function \[g(z) = \overline {f\!\left (\overline {z}\right )} .\] Because \(G\) is symmetric, \(g\) is defined on \(G\), and it is analytic by the reflection computation done earlier in the course. It maps \(G\) into \(D\), since \(\left |g(z)\right | = \left |f(\overline {z})\right | < 1\), and it is one-to-one because \(f\) is and conjugation is injective.

Evaluate at \(a \in G \cap \mathbb {R}\), so \(\overline {a} = a\): \[g(a) = \overline {f(a)} = \overline {0} = 0, \qquad g'(a) = \overline {f'\!\left (\overline {a}\right )} = \overline {f'(a)} = f'(a) > 0,\] the last equality because \(f'(a)\) is a positive real. Also \(g(G) = D\), since conjugation maps \(D\) onto \(D\).

So \(g\) satisfies exactly the same three normalisations as \(f\): it is a one-to-one analytic map of \(G\) onto \(D\) with \(g(a) = 0\) and \(g'(a) > 0\). By the uniqueness clause of the Riemann mapping theorem, \(g = f\), that is \[f(z) = \overline {f\!\left (\overline {z}\right )} \qquad (z \in G).\]

Now take \(z \in G^{+}\), so \(\operatorname {Im}z > 0\). The identity says \(f(\overline {z}) = \overline {f(z)}\): the image of the reflected point is the reflection of the image. Hence \(f\) maps \(G^{+}\) and its mirror image \(G^{-}\) to mirror-image sets, and since \(f\) is injective these are disjoint. As \(f(G^{0}) \subseteq \mathbb {R}\) — the real points map to real points by the identity — the connected set \(f(G^{+})\) lies in \(D\) off the real axis, so it lies entirely in the upper or entirely in the lower half of \(D\). \(\blacksquare \)

The normalisation \(f'(a) > 0\) is what makes the uniqueness clause bite; without it \(e^{i\theta }f\) would be an equally good map and the symmetry argument would collapse.

Solution. This is the previous square-root construction with the conclusion stated geometrically rather than as a bound.

As in that problem, since \(G \neq \mathbb {C}\) choose \(a \notin G\), take the single-valued branch \(h(z) = \sqrt {z-a}\) available by simple connectedness, and note that \(h\) is injective and that \(h(G) \cap \left (-h(G)\right ) = \emptyset \).

Because \(h\) is non-constant and analytic, the open mapping theorem makes \(h(G)\) open, so it contains a disc \(B(w_0,\rho )\) with \(w_0 \neq 0\). By the disjointness just noted, \[B(-w_0,\rho ) \cap h(G) = \emptyset .\]

Take \(f = h\). Then \(f\) is a non-constant analytic map on \(G\) and \[\mathbb {C}\setminus f(G) \supseteq B(-w_0,\rho ),\] a non-empty open set. \(\blacksquare \)

\[\boxed {f(z) = \sqrt {z-a} \text { omits an entire disc.}}\]

Contrast Picard’s little theorem: a non-constant entire function omits at most one point of \(\mathbb {C}\), so it can never omit a disc. The whole difference is that \(G\) is not the plane, and the omitted disc is the concrete expression of that.

Solution. (a). Maximum Modulus Principle. If \(f\) is analytic on a region \(G\) and \(\left |f\right |\) attains a local maximum at a point of \(G\), then \(f\) is constant. Equivalently, for \(G\) bounded and \(f\) continuous on \(\overline {G}\) and analytic on \(G\), \(\max _{\overline {G}}\left |f\right | = \max _{\partial G}\left |f\right |\).

(b). Schwarz’s Lemma, proved from (a). Let \(f : \Delta \rightarrow \Delta \) be analytic with \(f(0) = 0\). Since \(f(0) = 0\), the function \[g(z) = \begin {cases} f(z)/z & z \neq 0,\\ f'(0) & z = 0\end {cases}\] is analytic on \(\Delta \) (the singularity at \(0\) is removable).

Fix \(r < 1\). On the circle \(\left |z\right | = r\), \[\left |g(z)\right | = \frac {\left |f(z)\right |}{r} \leq \frac {1}{r},\] so by the maximum modulus principle \(\left |g\right | \leq 1/r\) on the whole disc \(\left |z\right | \leq r\). Now fix \(z\) and let \(r \uparrow 1\): the bound tends to \(1\), giving \(\left |g(z)\right | \leq 1\) throughout \(\Delta \). That is \[\left |f(z)\right | \leq \left |z\right |,\qquad \left |f'(0)\right | = \left |g(0)\right | \leq 1 .\] If equality holds at some \(z_0 \neq 0\), or if \(\left |f'(0)\right | = 1\), then \(\left |g\right |\) attains the value \(1\) at an interior point, so by (a) \(g\) is a constant of modulus \(1\) and \(f(z) = \lambda z\) with \(\left |\lambda \right | = 1\). \(\blacksquare \)

(c). Uniqueness in the Riemann Mapping Theorem. Suppose \(f_1, f_2\) both map \(G\) one-to-one onto \(\Delta \) with \(f_j(a) = 0\) and \(f_j'(a) > 0\). Then \[\varphi = f_2 \circ f_1^{-1} : \Delta \rightarrow \Delta \] is an analytic bijection with \(\varphi (0) = 0\). Schwarz applied to \(\varphi \) gives \(\left |\varphi (w)\right | \leq \left |w\right |\), and applied to \(\varphi ^{-1}\) gives the reverse, so \(\left |\varphi (w)\right | = \left |w\right |\) and hence \(\varphi (w) = \lambda w\) with \(\left |\lambda \right | = 1\).

Differentiating \(f_2 = \varphi \circ f_1\) at \(a\) gives \(f_2'(a) = \lambda f_1'(a)\). Both derivatives are positive reals, so \(\lambda > 0\), and \(\left |\lambda \right | = 1\) forces \(\lambda = 1\). Therefore \(\varphi = \operatorname {id}\) and \(f_1 = f_2\). \(\blacksquare \)

Solution. Combine the previous two constructions.

By the square-root problem above there is a non-constant analytic \(h\) on \(G\) whose image omits a disc \(B(-w_0,\rho )\). Then \[f_1(z) = \frac {\rho }{h(z)+w_0}\] is analytic on \(G\) — the denominator never vanishes — and satisfies \[\left |f_1(z)\right | = \frac {\rho }{\left |h(z)-(-w_0)\right |} \leq \frac {\rho }{\rho } = 1,\] because \(h(z)\) lies outside the disc of radius \(\rho \) about \(-w_0\).

This gives \(\left |f_1\right | \leq 1\), but the target is the open disc. Two ways to finish. Either note that \(\left |f_1\right | = 1\) somewhere would put a maximum of \(\left |f_1\right |\) at an interior point of \(G\), forcing \(f_1\) constant by the maximum modulus principle — contrary to \(h\) non-constant; so in fact \(\left |f_1\right | < 1\) throughout. Or simply scale: \(f = \tfrac 12 f_1\) maps into \(D\) with room to spare.

Either way \[\boxed {f : G \longrightarrow D \text { is non-constant and analytic.}} \qquad \blacksquare \]

This is the first step of the standard proof of the Riemann mapping theorem: having got into the disc, one then maximises \(\left |f'(a)\right |\) over all such maps using Montel, and the extremal map turns out to be onto.

Solution. \(G\) is the half-disc lying in the right half plane. It has two right-angled corners, at \(i\) and \(-i\), so a Möbius map alone cannot reach the disc; a squaring step is needed to open the corners.

\[z \ \overset {T}{\longmapsto }\ \text {first quadrant} \ \overset {w^{2}}{\longmapsto }\ \text {upper half plane} \ \overset {\frac {\zeta -i}{\zeta +i}}{\longmapsto }\ D .\]

For the first step take \[T(z) = \frac {1+z}{1-z},\] which sends \(-1 \mapsto 0\) and \(1 \mapsto \infty \); restricted to the right half-disc it is a bijection onto the first quadrant. (On the diameter \(z = iy\), \(\left |y\right |<1\), one computes \(T(iy) = \frac {1-y^{2}+2iy}{1+y^{2}}\), which has positive real and imaginary parts for \(0 < y < 1\); and on the right semicircle \(T\) is purely imaginary with positive imaginary part, matching the two edges of the quadrant.)

Squaring doubles the quadrant into the upper half plane, and the Cayley transform carries that onto \(D\). Composing, \[\boxed {f(z) = \frac {\left (\frac {1+z}{1-z}\right )^{2} - i} {\left (\frac {1+z}{1-z}\right )^{2} + i}}\] is a holomorphic bijection of \(G\) onto the open unit disc. \(\blacksquare \)

Each stage is injective on the region it is applied to — squaring is not injective on the plane, but it is on a quadrant — and that is what makes the composition a bijection.

Solution. (a). \(A = \mathbb {C}\), \(B = B(0,100)\): no. An analytic bijection \(\mathbb {C} \rightarrow B(0,100)\) would be a bounded entire function, hence constant by Liouville. Note that \(A\) and \(B\) are homeomorphic, so the obstruction is analytic, not topological.

(b). \(A = D\), \(B\) the rectangle \(\left |x\right |<1,\ \left |y\right |<2\): yes. Both are simply connected proper subregions of \(\mathbb {C}\), so the Riemann mapping theorem provides analytic bijections of each onto \(D\); composing one with the inverse of the other gives a bijection \(A \rightarrow B\). (Explicitly it is a Schwarz–Christoffel map, but existence is all that is asked.)

(c). \(A = \mathbb {C}\setminus \{0\}\), \(B = \mathbb {C}\setminus \{1,2\}\): no. These are not homeomorphic: \(A\) has one puncture, \(B\) has two, and the fundamental groups are \(\mathbb {Z}\) and the free group on two generators. A conformal bijection is a homeomorphism, so none exists.

(d). \(A = D\), \(B = \mathbb {C}\setminus \{iy : y \geq 0\}\): yes. \(B\) is the plane slit along a ray, which is simply connected and not all of \(\mathbb {C}\), so Riemann applies. Concretely, rotate the slit to \((-\infty ,0]\), take the principal square root onto a half plane, then a Cayley transform onto \(D\); invert.

\[\boxed {\text {(a) no, (b) yes, (c) no, (d) yes.}}\]

The four cases separate the two kinds of obstruction cleanly: (c) fails topologically and (a) fails analytically, while (b) and (d) have neither obstruction.

Solution. (a). Let \(w\) be a conformal bijection of \(D\) onto itself. Put \(a = w^{-1}(0)\) and let \[\varphi _a(z) = \frac {z-a}{1-\overline {a}z},\] a linear fractional map of \(D\) onto \(D\) with \(\varphi _a(a) = 0\). Then \[\psi = w \circ \varphi _a^{-1} : D \rightarrow D\] is an analytic bijection with \(\psi (0) = 0\).

Apply Schwarz’s lemma to \(\psi \) and to \(\psi ^{-1}\): \[\left |\psi (z)\right | \leq \left |z\right |, \qquad \left |\psi ^{-1}(z)\right | \leq \left |z\right | .\] Putting \(z = \psi (u)\) in the second gives \(\left |u\right | \leq \left |\psi (u)\right |\), so \(\left |\psi (z)\right | = \left |z\right |\) throughout. Equality in Schwarz forces \[\psi (z) = \lambda z, \qquad \left |\lambda \right | = 1 .\] Hence \[w(z) = \lambda \,\varphi _a(z) = \lambda \,\frac {z-a}{1-\overline {a}z},\] a linear fractional transformation. \(\blacksquare \)

(The hint about continuity to the boundary allows an alternative route via the reflection principle, but Schwarz applied both ways is shorter and needs no boundary hypothesis at all.)

(b). There is no conformal map of \(\mathbb {C}\) onto \(\left |z\right | < 1\): such a map would be a bounded entire function, hence constant by Liouville, and a constant is not a bijection. \(\blacksquare \)

Part (b) is the reason the Riemann mapping theorem must exclude \(\mathbb {C}\), and part (a) identifies the automorphism group of the disc as the Möbius maps preserving it — a group of real dimension three.

Solution. (a). The Riemann Mapping Theorem. Let \(G \subsetneq \mathbb {C}\) be a simply connected region and \(a \in G\). Then there is a unique analytic bijection \(f : G \rightarrow D\) onto the unit disc with \[f(a) = 0 \qquad \text {and}\qquad f'(a) > 0 .\]

(b). Why the two exceptions.

The plane. \(G = \mathbb {C}\) is simply connected but is excluded by the hypothesis \(G \neq \mathbb {C}\), and necessarily so: an analytic bijection \(\mathbb {C} \rightarrow D\) would be a bounded entire function, hence constant by Liouville, and constants are not bijections. So the exclusion is not a defect of the proof but a genuine obstruction.

The punctured disc. \(B(0;1)\setminus \{0\}\) is a proper subregion of \(\mathbb {C}\), so it fails the other hypothesis: it is not simply connected. A small circle about the puncture cannot be contracted within the region, whereas every loop in \(D\) can. Since a conformal bijection is in particular a homeomorphism and simple connectedness is a topological invariant, no such map exists.

\[\boxed {\mathbb {C} \text { fails the properness hypothesis; the punctured disc fails simple connectedness.}}\]

Between them the two examples show that neither hypothesis can be dropped, and that they fail for quite different reasons — one analytic, one topological.

Solution. (a). As stated in the previous problem: for a simply connected region \(G \subsetneq \mathbb {C}\) and \(a \in G\), there is a unique analytic bijection \(f : G \rightarrow \Delta \) with \(f(a) = 0\) and \(f'(a) > 0\).

(b). Uniqueness. Suppose \(f_1\) and \(f_2\) both have these properties. Then \[\varphi = f_2 \circ f_1^{-1} : \Delta \rightarrow \Delta \] is an analytic bijection with \(\varphi (0) = 0\). Schwarz’s lemma applied to \(\varphi \) gives \(\left |\varphi (w)\right | \leq \left |w\right |\); applied to \(\varphi ^{-1}\) and substituted, it gives the reverse inequality. So \(\left |\varphi (w)\right | = \left |w\right |\), and the equality case of Schwarz forces \(\varphi (w) = \lambda w\) with \(\left |\lambda \right | = 1\).

Differentiating \(f_2 = \varphi \circ f_1\) at \(a\): \[f_2'(a) = \varphi '(0)\,f_1'(a) = \lambda f_1'(a).\] Both \(f_1'(a)\) and \(f_2'(a)\) are positive reals, so \(\lambda \) is a positive real of modulus \(1\), that is \(\lambda = 1\). Hence \(\varphi = \operatorname {id}\) and \(f_1 = f_2\). \(\blacksquare \)

(c). The map for the right half plane with \(f(1) = 0\). The Cayley transform \(\zeta \mapsto \frac {\zeta -1}{\zeta +1}\) carries \(H\) onto \(\Delta \) and sends \(1 \mapsto 0\), so up to a rotation \[f(\zeta ) = \frac {\zeta - 1}{\zeta + 1}.\] Its derivative at \(1\) is \[f'(\zeta ) = \frac {2}{(\zeta +1)^{2}}, \qquad f'(1) = \frac {2}{4} = \frac 12 > 0 ,\] already positive, so no rotation is needed and this is the normalised Riemann map: \[\boxed {f(\zeta ) = \frac {\zeta -1}{\zeta +1},\qquad f(1) = 0,\ f'(1) = \tfrac 12 .}\]

Solution. Let \(f\) be meromorphic on \(\widehat {\mathbb {C}} = \mathbb {C}\cup \{\infty \}\).

Finitely many poles. The sphere is compact and poles are isolated, so there are only finitely many; say \(a_1,\ldots ,a_k \in \mathbb {C}\) with orders \(m_1,\ldots ,m_k\), and possibly a pole at \(\infty \) of order \(m_\infty \).

Clear the finite poles. Put \[q(z) = \prod ^{k}_{j=1}(z-a_j)^{m_j}, \qquad g = qf .\] Then \(g\) is holomorphic on all of \(\mathbb {C}\): each finite pole of \(f\) is cancelled exactly.

Control \(g\) at infinity. Near \(\infty \), \(f\) has at worst a pole of order \(m_\infty \), meaning \(f(z) = O\!\left (\left |z\right |^{m_\infty }\right )\), and \(q\) is a polynomial of degree \(d = \sum m_j\). Hence \[\left |g(z)\right | = \left |q(z)f(z)\right | = O\!\left (\left |z\right |^{\,d+m_\infty }\right ) \qquad (\left |z\right | \to \infty ).\] By the polynomial-growth theorem of the Maximum Modulus chapter, an entire function of polynomial growth is a polynomial: \(g = p\) with \(\deg p \leq d+m_\infty \).

Therefore \(f = p/q\). Cancelling common factors makes \(p\) and \(q\) coprime. \(\blacksquare \)

\[\boxed {\text {Every meromorphic function on the sphere is rational.}}\]

The compactness of \(\widehat {\mathbb {C}}\) is doing all the work: it is what makes the pole set finite, and \(\infty \) being an ordinary point of the sphere is what forbids an essential singularity there. On \(\mathbb {C}\) alone the statement is false — \(e^{z}\) is a counterexample.

Solution. (a). Riemann–Roch. Let \(X\) be a compact Riemann surface of genus \(g\) and \(D\) a divisor on \(X\). Writing \(\ell (D) = \dim H^{0}(X,\mathcal {O}_D)\) for the dimension of the space of meromorphic functions \(f\) with \(\operatorname {div}(f) + D \geq 0\), and \(K\) for a canonical divisor, \[\ell (D) - \ell (K-D) = \deg D + 1 - g .\]

(b). A meromorphic function on a torus with two simple poles. Take \(X = T\) a complex torus, so \(g = 1\), and let \(p \neq q\) be two points. Put \(D = p + q\), so \(\deg D = 2\).

The canonical divisor on a torus has \(\deg K = 2g-2 = 0\), so \(\deg (K-D) = -2 < 0\). A divisor of negative degree has \(\ell = 0\), since a non-zero meromorphic function has \(\deg \operatorname {div}(f) = 0\) and could not satisfy \(\operatorname {div}(f) + (K-D) \geq 0\). Hence \(\ell (K-D) = 0\) and \[\ell (D) = \deg D + 1 - g = 2 + 1 - 1 = 2 .\]

The constants account for a one-dimensional subspace of \(L(D)\), so there is a non-constant \(f \in L(D)\). Such an \(f\) has poles only at \(p\) and \(q\), each at worst simple. It cannot have just one simple pole: a meromorphic function on a compact surface with a single simple pole would be a degree-one map to \(\widehat {\mathbb {C}}\), hence a biholomorphism, forcing \(g = 0\). So \(f\) has simple poles at both \(p\) and \(q\). \(\blacksquare \)

Concretely this function is the Weierstrass \(\zeta \)-difference \(\zeta (z-p) - \zeta (z-q)\), or equivalently a translate of \(\wp '\)-free combination; Riemann–Roch guarantees it without exhibiting it.

Solution. Statement (Riemann–Hurwitz). Let \(f : X \rightarrow Y\) be a non-constant holomorphic map of compact Riemann surfaces, of degree \(n\), with genera \(g_X\) and \(g_Y\). For \(p \in X\) let \(e_p\) be the ramification index of \(f\) at \(p\). Then \[2g_X - 2 = n\left (2g_Y - 2\right ) + \sum _{p \in X}\left (e_p - 1\right ).\]

Proof. Triangulate \(Y\) so finely that every branch point of \(f\) is a vertex and each face lies in an evenly covered disc. Let the triangulation have \(V\) vertices, \(E\) edges and \(F\) faces, so that \[V - E + F = \chi (Y) = 2 - 2g_Y .\]

Lift the triangulation to \(X\) through \(f\). Each face and each edge of \(Y\) has exactly \(n\) preimages, since \(f\) is an \(n\)-sheeted covering away from the branch points. Vertices are the exception: a vertex \(y \in Y\) has preimages \(p \in f^{-1}(y)\) with \(\sum _{p \in f^{-1}(y)} e_p = n\), so instead of \(n\) preimages it has only \(\left |f^{-1}(y)\right | = n - \sum _{p}(e_p - 1)\).

Therefore the lifted triangulation of \(X\) has \[V' = nV - \sum _{p\in X}(e_p-1),\qquad E' = nE,\qquad F' = nF,\] and \[2 - 2g_X = \chi (X) = V' - E' + F' = n\left (V-E+F\right ) - \sum _{p}(e_p-1) = n\left (2-2g_Y\right ) - \sum _{p}(e_p-1).\] Rearranging gives the formula. \(\blacksquare \)

A parity consequence worth noting. Since \(2g_X - 2\) and \(n(2g_Y-2)\) are both even, the total ramification \(\sum (e_p-1)\) is always even. This is a quick consistency check on any proposed branched cover, and one of the problems below fails it.

Solution. Let \(f : X \rightarrow Y\) be non-constant holomorphic of degree \(n \geq 1\), with \(g_X = g_Y = g\) and \(g \neq 1\). Riemann–Hurwitz gives \[2g - 2 = n(2g-2) + R,\qquad R = \sum _{p}(e_p-1) \geq 0 ,\] so \[R = (2g-2)(1-n).\]

Case \(g = 0\). Then \(R = -2(1-n) = 2(n-1) \geq 0\) for all \(n\), so this gives no contradiction — but \(g=0\) means both surfaces are \(\widehat {\mathbb {C}}\), which are certainly biholomorphic, and the claim holds trivially.

Case \(g \geq 2\). Then \(2g-2 > 0\). Since \(R \geq 0\) we need \(1 - n \geq 0\), that is \(n \leq 1\); and \(n \geq 1\) because \(f\) is non-constant. So \(n = 1\) and \(R = 0\): the map is unramified of degree one, hence a bijection, and a bijective holomorphic map of Riemann surfaces is a biholomorphism.

In both cases \(X\) and \(Y\) are biholomorphic. \(\blacksquare \)

Why \(g = 1\) is excluded. There \(2g-2 = 0\) and the identity reads \(R = 0\) for every \(n\), giving no constraint at all. And indeed the conclusion is false: a torus \(\mathbb {C}/\Lambda \) admits unramified self-covers of every degree \(n \geq 2\), for instance \(z \mapsto nz\), and \(\mathbb {C}/\Lambda \) is generally not biholomorphic to \(\mathbb {C}/n\Lambda \). The excluded case is excluded because it is genuinely different, not for convenience.

Solution. (a). Let \(X\) be compact and \(f\) a smooth complex-valued function. Then \(\Delta f = d'd''f\), and since \(d = d' + d''\) while \(d'd' = d''d'' = 0\) on functions, \[d\left (d''f\right ) = d'd''f + d''d''f = d'd''f = \Delta f .\] So \(\Delta f\) is an exact \(2\)-form: \(\Delta f = d\omega \) with \(\omega = d''f\), a smooth \(1\)-form on \(X\).

\(X\) is compact and without boundary, so Stokes’ theorem gives \[\int _{X}\Delta f = \int _{X} d\omega = \int _{\partial X}\omega = 0 , \qquad \partial X = \emptyset . \qquad \blacksquare \]

(b). Surjectivity on the disc. Let \(\rho \) be a smooth \(2\)-form on \(D_1 = \{\left |z\right |<1\}\); write \(\rho = \varphi \,dz \wedge d\overline {z}\) with \(\varphi \) smooth. Solving \(\Delta f = \rho \) amounts to solving an inhomogeneous \(\overline {\partial }\)-equation followed by a \(\partial \)-equation, and both are solvable on a disc:

Step 1. By the Cauchy–Pompeiu / Dolbeault lemma, on a disc the equation \(\overline {\partial } u = \varphi \,d\overline {z}\) has a smooth solution, given explicitly by the Cauchy transform \[u(z) = \frac {1}{2\pi i}\iint \limits _{D_1} \frac {\varphi (\zeta )}{\zeta - z}\,d\zeta \wedge d\overline {\zeta }.\]

Step 2. The same lemma applied again, in the conjugate variable, solves the remaining equation. Composing the two solutions produces \(f\) with \(d'd''f = \rho \). \(\blacksquare \)

The contrast with (a) is the point of the pair: on a compact surface the Laplacian is far from surjective — its image is exactly the \(2\)-forms of total integral zero, one condition — whereas on a disc, which is non-compact and has boundary, Stokes gives no obstruction and every \(2\)-form is hit.

Solution. Here \(D = p\), a single point with multiplicity one, so \(\deg D = 1\).

(a). The sheaf \(\mathcal {O}_D\). For \(U \subseteq X\) open, \[\mathcal {O}_D(U) = \left \{ f \text { meromorphic on } U \ :\ \operatorname {div}(f) + D \geq 0 \text { on } U\right \}.\] Concretely: \(f\) is holomorphic on \(U \setminus \{p\}\), and if \(p \in U\) then \(f\) has at worst a simple pole at \(p\). It is the sheaf of meromorphic functions with poles bounded by \(D\).

(b). The two cohomology dimensions.

\(H^{0}(X,\mathcal {O}_D) = L(D)\), the global sections. A non-constant element would have a single simple pole, hence define a degree-one holomorphic map \(X \rightarrow \widehat {\mathbb {C}}\), hence a biholomorphism, forcing \(g = 0\). Since \(g > 0\) this cannot happen, so only the constants survive: \[\boxed {\dim H^{0}(X,\mathcal {O}_D) = 1 .}\]

For \(H^{1}\), use Riemann–Roch in its cohomological form \[h^{0}(D) - h^{1}(D) = \deg D + 1 - g = 1 + 1 - g = 2 - g .\] Substituting \(h^{0} = 1\), \[\boxed {h^{1}(X,\mathcal {O}_D) = 1 - (2-g) = g - 1 .}\]

By Serre duality \(h^{1}(D) = \ell (K-D)\), so this also says that the holomorphic \(1\)-forms vanishing at \(p\) form a space of dimension \(g-1\) — one condition cut from the \(g\)-dimensional space of all holomorphic \(1\)-forms, which is exactly what one expects. The same number appears in the \(M_p\) problem below.

Solution. Suppose \(h\) is holomorphic and bounded on \(X \setminus \{p_1,\ldots ,p_n\}\).

Each \(p_j\) is an isolated singularity of \(h\) — isolated because the points are finitely many and distinct. In a coordinate chart about \(p_j\), \(h\) is a bounded holomorphic function on a punctured disc, so by Riemann’s removable singularity theorem the singularity is removable and \(h\) extends holomorphically across \(p_j\).

Doing this at each of the \(n\) points extends \(h\) to a holomorphic function on all of \(X\).

Now \(X\) is compact, so \(\left |h\right |\) is continuous on a compact space and attains a maximum at some \(x_0 \in X\). The maximum modulus principle, applied in a coordinate chart about \(x_0\), forces \(h\) to be constant near \(x_0\); the set where \(h\) equals that value is then open and closed in the connected surface \(X\), so \(h\) is constant on \(X\). \(\blacksquare \)

\[\boxed {\text {No non-constant bounded holomorphic function exists on } X\setminus \{p_1,\ldots ,p_n\}.}\]

Two features of compactness are used and they are different: removability handles the punctures, and then compactness of \(X\) itself supplies the maximum. On a non-compact surface the conclusion fails at once — \(\mathbb {C}\setminus \{0\}\) carries the bounded non-constant function \(e^{-1/(1+\left |z\right |^{2})}\)-style examples, and more simply the disc carries the identity.

Solution. Existence of the cover. Let \(X\) be a compact Riemann surface of genus \(g\) and pick any point \(p \in X\). Take the divisor \(D = (g+1)p\), of degree \(g+1\). Since \(\deg (K - D) = (2g-2) - (g+1) = g - 3\), Riemann–Roch gives \[\ell (D) \geq \deg D + 1 - g = 2 ,\] so \(L(D)\) contains a non-constant meromorphic function \(f\). Every non-constant meromorphic function on a compact Riemann surface is a holomorphic map \[f : X \longrightarrow \widehat {\mathbb {C}} = S^{2},\] and such a map is automatically proper, surjective and of some finite degree \(r \geq 1\): a branched cover of the sphere.

The ramification formula. Apply Riemann–Hurwitz with \(Y = S^{2}\), so \(g_Y = 0\) and the degree is \(r\): \[2g - 2 = r(2\cdot 0 - 2) + R = -2r + R ,\] whence \[\boxed {R = 2g - 2 + 2r = 2\left (g - 1 + r\right ),}\] where \(R = \sum _{p}(e_p-1)\) is the total ramification index and \(r\) the number of sheets. \(\blacksquare \)

Note the formula makes \(R\) manifestly even, as the parity remark after Riemann–Hurwitz requires. For \(g = 5\) it reads \(R = 2(4+r)\), which is the form asked for in a later problem in this set.

Solution. Write the map explicitly. Define \[I : \Omega ^{1,1}(X) \longrightarrow \mathbb {C}, \qquad I(\eta ) = \int _{X}\eta ,\] which makes sense because a \((1,1)\)-form on a Riemann surface is a \(2\)-form and \(X\) is compact and oriented.

\(I\) kills the denominator. If \(\eta = \partial \alpha \) with \(\alpha \in \Omega ^{1,0}\), then — since \(\overline {\partial }\alpha \) is a \((0,2)\)-form and vanishes identically on a one-dimensional complex manifold — we have \(d\alpha = \partial \alpha + \overline {\partial }\alpha = \partial \alpha = \eta \). So \(\eta \) is exact, and Stokes on the closed surface gives \[I(\eta ) = \int _{X} d\alpha = 0 .\] Hence \(I\) descends to a linear map \(\overline {I} : H^{1,1}_{X} \rightarrow \mathbb {C}\).

\(\overline {I}\) is surjective. Choose a smooth bump function supported in a coordinate chart and let \(\eta _0 = \varphi \,\tfrac {i}{2}dz\wedge d\overline {z}\) with \(\varphi \geq 0\) not identically zero. Then \(\int _X \eta _0 > 0\), so \(\overline {I}\) is non-zero, hence onto \(\mathbb {C}\).

\(\overline {I}\) is injective. This is the substantive half. One must show that a \((1,1)\)-form of total integral zero is \(\partial \) of a \((1,0)\)-form. The standard route is Hodge theory: by the Hodge decomposition on the compact surface \(X\), every \(2\)-form splits as a harmonic part plus an exact part, the harmonic \(2\)-forms are the constant multiples of the volume form, and the integral detects exactly the harmonic component. A form with zero integral therefore has no harmonic part and is exact, and on a Riemann surface exact \((1,1)\)-forms are precisely \(\partial \Omega ^{1,0}\).

Combining, \(\overline {I}\) is an isomorphism: \[\boxed {H^{1,1}_{X} \cong \mathbb {C} .} \qquad \blacksquare \]

Honest note. The injectivity step is the one with real content, and the argument above cites Hodge theory rather than proving it; a self-contained proof would solve \(\Delta u = \eta - c\,\omega \) on \(X\), which is the existence theorem for the Laplacian on a compact surface. This is the point at which the material genuinely leaves the scope of these notes.

Solution. (a). The number of zeros. The degree of \(f\), as a holomorphic map \(X \rightarrow \widehat {\mathbb {C}}\), equals the total pole order: \[n = 3(2) + 4(1) = 10 .\] A degree-\(n\) map attains every value exactly \(n\) times counted with multiplicity, so \(f\) has \(10\) zeros with multiplicity. The hypothesis that \(f\) is a local biholomorphism on \(f^{-1}(\mathbb {C})\) means it is unramified over every finite value, in particular over \(0\), so all the zeros are simple: \[\boxed {10 \text { distinct zeros, each of multiplicity } 1 .}\]

(b). The genus — and why the data cannot be right. Ramification occurs only over \(\infty \), where the preimages are the poles with indices \(2,2,2,1,1,1,1\). Hence \[R = \sum _{p}(e_p-1) = 3(2-1) + 4(1-1) = 3 .\] Riemann–Hurwitz with \(g_Y = 0\) and \(n = 10\) gives \[2g_X - 2 = 10(-2) + 3 = -17, \qquad \text {so}\qquad g_X = -\frac {15}{2}.\]

A genus cannot be negative, let alone a half-integer, so no such surface and map exist. The obstruction is not arithmetic clumsiness but the parity noted after Riemann–Hurwitz: rearranging, \(R = 2g_X - 2 + 2n\) is necessarily even, whereas the stated pole data forces \(R = 3\), odd.

\[\boxed {\text {The hypotheses are inconsistent; the configuration described cannot occur.}}\]

Changing any single double pole to a triple one, or adding one more double pole, restores parity and makes the question answerable — with \(4\) double and \(2\) simple poles, \(n = 10\), \(R = 4\) and \(g_X = 3\).

Solution. Here \(M_p = L(p)\), the meromorphic functions with at worst a simple pole at \(p\) and holomorphic elsewhere. Throughout, \(\deg (p) = 1\) and \(\deg K = 2g-2\).

(a). \(g = 0\). Then \(X = \widehat {\mathbb {C}}\) and \(\deg (K-p) = -2-1 = -3 < 0\), so \(\ell (K-p) = 0\) and Riemann–Roch gives \[\dim M_p = \ell (p) = \deg (p) + 1 - g = 1 + 1 - 0 = 2 > 1 . \qquad \blacksquare \] Concretely, with \(p = \infty \) the space is spanned by \(1\) and \(z\).

(b). \(g > 0\) forces constants. Suppose \(f \in M_p\) were non-constant. Then \(f\) has exactly one pole, simple, so as a map \(X \rightarrow \widehat {\mathbb {C}}\) it has degree \(1\). A degree-one holomorphic map of compact Riemann surfaces is a biholomorphism, so \(X \cong \widehat {\mathbb {C}}\) and \(g = 0\) — contradiction. Hence \(\dim M_p = 1\), the constants. \(\blacksquare \)

(c). Holomorphic \(1\)-forms vanishing at \(p\). That space is \(L(K-p)\), of dimension \(\ell (K-p)\). Riemann–Roch applied to \(D = p\) reads \[\ell (p) - \ell (K-p) = 1 + 1 - g .\] By part (b), \(\ell (p) = 1\) for \(g > 0\), so \[\boxed {\ell (K-p) = 1 - (2-g) = g - 1 .}\]

The three parts fit together: the space of all holomorphic \(1\)-forms has dimension \(g\), and requiring a zero at one point cuts it by exactly one. That the cut is always exactly one — never zero — is equivalent to the statement in (b) that no genus-positive surface carries a function with a single simple pole.

Solution. (a). The Uniformisation Theorem. Every simply connected Riemann surface is biholomorphic to exactly one of \[\widehat {\mathbb {C}} \ (\text {the sphere}),\qquad \mathbb {C} \ (\text {the plane}),\qquad \Delta \ (\text {the unit disc}).\] Consequently every Riemann surface is a quotient of one of these three by a group of automorphisms acting freely and properly discontinuously.

(b). The universal cover of a genus-two surface. \(X\) has genus \(2\), so its universal cover \(\widetilde {X}\) is simply connected and is one of the three. Eliminate the first two.

Not the sphere. \(\widehat {\mathbb {C}}\) is compact, and a covering space of a compact surface with infinite fundamental group cannot be compact. Here \(\pi _1(X)\) is the given group on four generators, which is infinite, so \(\widetilde {X}\) is non-compact.

Not the plane. If \(\widetilde {X} = \mathbb {C}\) then \(X = \mathbb {C}/\Gamma \) with \(\Gamma \) acting freely and properly discontinuously by automorphisms of \(\mathbb {C}\), that is by maps \(z \mapsto az+b\). Freeness forces \(a = 1\), so \(\Gamma \) is a group of translations, hence abelian, and \(X\) is \(\mathbb {C}\), a cylinder, or a torus — genus \(0\) or \(1\). But \(\pi _1(X)\) above is non-abelian, and \(g = 2\).

Therefore \[\boxed {\widetilde {X} \cong \Delta , \text { the unit disc.}}\] \(X\) is a quotient of the disc by a Fuchsian group isomorphic to \(\pi _1(X)\), and carries a metric of constant negative curvature. \(\blacksquare \)

The trichotomy matches the genus exactly: \(g = 0\) gives the sphere, \(g = 1\) the plane, and every \(g \geq 2\) the disc. That is why genus \(\geq 2\) surfaces are called hyperbolic.

Solution. Write \(\sigma : H^{1,0} \rightarrow H^{0,1}\) for the map induced by \(\omega \mapsto \overline {\omega }\).

The two spaces. \(H^{1,0}\) is the space of holomorphic \(1\)-forms on \(X\): these are the \((1,0)\)-forms \(\omega \) with \(\overline {\partial }\omega = 0\), and on a compact surface of genus \(g\) this space has dimension \(g\). The quotient \[H^{0,1} = \frac {\Omega ^{0,1}(X)}{\overline {\partial }\,\Omega ^{0,0}(X)}\] is the Dolbeault cohomology group \(H^{1}(X,\mathcal {O})\), which by Serre duality is dual to \(H^{0}(X,K) = H^{1,0}\) and therefore also has dimension \(g\).

\(\sigma \) is well defined. If \(\omega \) is a \((1,0)\)-form then \(\overline {\omega }\) is a \((0,1)\)-form, so conjugation lands in \(\Omega ^{0,1}(X)\), and composing with the quotient map gives a well-defined \(\sigma : H^{1,0} \rightarrow H^{0,1}\). It is conjugate-linear rather than linear, which does not affect surjectivity.

Surjectivity. Since \(\dim _{\mathbb {C}} H^{1,0} = \dim _{\mathbb {C}} H^{0,1} = g\), it suffices to prove \(\sigma \) injective. Suppose \(\sigma (\omega ) = 0\), that is \(\overline {\omega } = \overline {\partial } h\) for some smooth \(h\). Conjugating, \(\omega = \partial \overline {h}\). Then \[\int _{X}\omega \wedge \overline {\omega } = \int _{X}\omega \wedge \overline {\partial }h = \pm \int _{X} d\left (h\,\omega \right ) = 0,\] using \(\overline {\partial }\omega = \partial \omega = 0\) for a holomorphic \(1\)-form and Stokes on the closed surface. But in a local coordinate \(\omega = u\,dz\) gives \[\frac {i}{2}\,\omega \wedge \overline {\omega } = \left |u\right |^{2}\,\frac {i}{2}\,dz\wedge d\overline {z},\] a non-negative multiple of the area form. So the integral vanishes only if \(u \equiv 0\), that is \(\omega = 0\).

Hence \(\sigma \) is injective, and by equality of dimensions surjective. \(\blacksquare \)

Honest note. The dimension count \(\dim H^{0,1} = g\) is Serre duality, which is quoted rather than proved here; with that input the argument above is complete. The positivity computation \(\frac {i}{2}\omega \wedge \overline {\omega } \geq 0\) is the same device that proves the Riemann bilinear relations.

Solution. Let \(D = p_1 + p_2 + \cdots + p_{g+1}\), a divisor of degree \(g+1\) on \(\Sigma _g\).

Riemann–Roch gives \[\ell (D) - \ell (K-D) = \deg D + 1 - g = (g+1) + 1 - g = 2 ,\] and \(\ell (K-D) \geq 0\) always, so \[\ell (D) \geq 2 .\]

The constant functions form a one-dimensional subspace of \(L(D)\), so \(\ell (D) \geq 2\) guarantees a non-constant \(f \in L(D)\). By definition of \(L(D)\), such an \(f\) is holomorphic on \(\Sigma _g \setminus \{p_1,\ldots ,p_{g+1}\}\) and has at worst a simple pole at each \(p_j\).

Being non-constant, \(f\) must have at least one pole — a non-constant holomorphic function on a compact surface is impossible, by the maximum principle argument used earlier in this set. So its poles form a non-empty subset of \(\{p_1,\ldots ,p_{g+1}\}\), each simple. \(\blacksquare \)

\[\boxed {\text {Such an } f \text { exists, with simple poles at some subset of the } g+1 \text { points.}}\]

The count \(g+1\) is exactly what is needed: with \(g\) points Riemann–Roch would only give \(\ell (D) \geq 1\), which the constants already supply, and indeed a general genus-\(g\) surface carries no non-constant function with poles confined to \(g\) prescribed points. One extra point is the whole margin.

Solution. Realise the torus as \(T^{2} = \mathbb {C}/\Lambda \) for a lattice \(\Lambda = \mathbb {Z} + \tau \mathbb {Z}\), and take \(f = \wp \), the Weierstrass elliptic function of \(\Lambda \): \[\wp (z) = \frac {1}{z^{2}} + \sum _{\lambda \in \Lambda \setminus \{0\}} \left [\frac {1}{(z-\lambda )^{2}} - \frac {1}{\lambda ^{2}}\right ].\]

Degree two. \(\wp \) is \(\Lambda \)-periodic and its only pole modulo \(\Lambda \) is a double pole at \(0\). The degree of the induced map \(T^{2} \rightarrow \widehat {\mathbb {C}}\) equals the total pole order, namely \(2\). So \(\wp \) is a two-sheeted branched cover of the sphere.

The branch points. \(\wp \) is even, \(\wp (-z) = \wp (z)\), so \(\wp (z_1) = \wp (z_2)\) exactly when \(z_2 \equiv \pm z_1\). The two sheets therefore come together precisely at the points fixed by \(z \mapsto -z\) on \(T^{2}\), that is at the four two-torsion points \[0,\quad \tfrac 12,\quad \tfrac {\tau }{2},\quad \tfrac {1+\tau }{2} \pmod {\Lambda }.\] There are four of them, and at each the ramification index is \(2\).

Consistency. Riemann–Hurwitz with \(g_X = 1\), \(g_Y = 0\), \(n = 2\): \[2(1) - 2 = 2(0-2) + R \ \Longrightarrow \ R = 4 ,\] and four simple ramification points contribute \(4 \times (2-1) = 4\) exactly. \(\blacksquare \)

\[\boxed {\wp : T^{2}\rightarrow S^{2} \text { is a two-sheeted cover branched over four points.}}\]

The images of the four branch points are \(\infty \) and the three roots \(e_1,e_2,e_3\) of \(4t^{3}-g_2t-g_3\), which is why the torus is the curve \(y^{2} = 4x^{3}-g_2x-g_3\): a double cover of the sphere branched at four points is exactly an elliptic curve.

Solution. (a). Existence of the branched cover. Let \(X\) be compact of genus \(g\) and choose \(p \in X\). Riemann–Roch applied to \(D = (g+1)p\) gives \[\ell (D) \geq \deg D + 1 - g = (g+1)+1-g = 2 ,\] so \(L(D)\) contains a non-constant meromorphic function \(f\). Any non-constant meromorphic function on a compact Riemann surface is a non-constant holomorphic map \(f : X \rightarrow \widehat {\mathbb {C}} = S^{2}\), and every such map between compact surfaces is a branched cover of some finite degree. \(\blacksquare \)

(b). The ramification identity for \(g = 5\). Apply Riemann–Hurwitz with \(g_X = 5\), \(g_Y = 0\) and degree \(r\), the number of sheets: \[2(5) - 2 = r\left (2\cdot 0 - 2\right ) + R ,\] that is \[8 = -2r + R,\qquad \text {so}\qquad R = 8 + 2r = \boxed {2(4+r)} . \qquad \blacksquare \]

This is the general formula \(R = 2(g-1+r)\) from the earlier problem, specialised at \(g = 5\). Two sanity checks it passes: \(R\) is even, as the parity remark demands; and \(R \geq 0\) imposes no constraint here, consistent with the fact that a genus-five surface admits covers of many different degrees.

Solution. Let \(\omega \) be a meromorphic \(1\)-form on the compact surface \(X\), with poles at \(p_1,\ldots ,p_k\).

The residue is well defined. In a local coordinate \(z\) centred at \(p_j\), write \(\omega = \left (\sum _{n} c_n z^{n}\right )dz\) and set \(\operatorname {Res}_{p_j}\omega = c_{-1}\). This does not depend on the coordinate, because under \(z = z(w)\) the coefficient \(c_{-1}\) is picked out by \[\operatorname {Res}_{p_j}\omega = \frac {1}{2\pi i}\oint _{\gamma _j}\omega \] for a small positively oriented loop \(\gamma _j\) about \(p_j\) — an expression involving only \(\omega \) and the orientation, not the chart. (This is the reason residues of forms are intrinsic while residues of functions are not.)

The sum vanishes. Excise a small disc \(D_j\) about each pole and let \[X_\varepsilon = X \setminus \bigcup _{j} D_j ,\] a compact surface with boundary \(\partial X_\varepsilon = -\bigcup _j \gamma _j\), the sign because the induced orientation runs the loops backwards.

On \(X_\varepsilon \) the form \(\omega \) is holomorphic, so \(d\omega = 0\) there — a holomorphic \(1\)-form on a Riemann surface is closed, since \(\partial \omega \) is a \((2,0)\)-form and vanishes, while \(\overline {\partial }\omega = 0\) is holomorphy. Stokes’ theorem then gives \[0 = \int _{X_\varepsilon } d\omega = \int _{\partial X_\varepsilon }\omega = -\sum ^{k}_{j=1}\oint _{\gamma _j}\omega = -2\pi i \sum ^{k}_{j=1}\operatorname {Res}_{p_j}\omega .\] Hence \[\boxed {\sum ^{k}_{j=1}\operatorname {Res}_{p_j}\omega = 0 .} \qquad \blacksquare \]

This is the global counterpart of the residue theorem, and it has real force: on the sphere it says a rational differential’s residues sum to zero, which is why \(\frac {dz}{z}\) has a compensating residue at \(\infty \). It also shows no meromorphic \(1\)-form can have exactly one simple pole — a fact used to prove that a genus-positive surface has no function with a single simple pole.