# Examination 7¶

## 2006 Nov¶

**Notation.** \(\mathbb{C}\) is the set of complex numbers, \(D= \{z\in \mathbb{C}: |z|<1\}\) is the open unit disk, \(\Pi^+\) and \(\Pi^-\) are the upper and lower half-planes, respectively, and, given an open set \(G\subset \mathbb{C}\), \(H(G)\) is the set of holomorphic functions on \(G\).

Suppose that \(f \in H(D\setminus \{0\})\) and that \(|f(z)| < 1\) for all \(0<|z|<1\). Prove that there is \(F\in H(D)\) with \(F(z) = f(z)\) for all \(z\in D\setminus \{0\}\).

State a general theorem about isolated singularities for holomorphic functions.

Explicitly construct, through a sequence of mappings, a one-to-one holomorphic function mapping the disk \(D\) onto the half disk \(D\cap \Pi^+\).

State a general theorem concerning one-to-one mappings of \(D\) onto domains \(\Omega\subset \mathbb{C}\).

State the Schwarz lemma.

Suppose that \(f\in H(\Pi^+)\) and that \(|f(z)|<1\) for all \(z\in \Pi^+\). If \(f(i)=0\) how large can \(|f'(i)|\) be? Find the extremal functions.

State Cauchy’s theorem and its converse.

Suppose that \(f\) is a continuous function defined on the entire complex plane. Assume that

\(f\in H(\Pi^+\cup \Pi^-)\)

\(f(\bar{z}) = \overline{f(z)}\) all \(z\in \mathbb{C}\).

Prove that \(f\) is an entire function.

Define what it means for a family \(\mathcal{F}\subset H(\Omega)\) to be a

*normal family*. State the fundamental theorem for normal families.Suppose \(f\in H(\Pi^+)\) and \(|f(z)|<1\) all \(z\in \Pi^+\). Suppose further that

\[\lim{t\rightarrow 0+} f(it) = 0.\]Prove that \(f(z_n) \rightarrow 0\) whenever the sequence \(z_n \rightarrow 0\) and \(z_n \in \Gamma\) where

\[\Gamma = \{ z\in \Pi^+ : |\mathfrak{Re}(z)| \leq \mathfrak{Im}(z)\}.\]

Hint.Consider the functions \(f_t(z) = f(tz)\) where \(t>0\).

## Solutions¶

Solution to Problem 42

**Lemma.** Suppose \(G\subset \mathbb{C}\) is an open set and \(f\) is holomorphic in \(G\) except for an isolated singularity at \(z_0 \in G\). If

then \(z_0\) is a removable singularity and \(f\) may be extended holomorphically to all of \(G\).

*Proof.* Under the stated hypotheses, there is an \(\epsilon >0\) and an \(M > 0\) such that the deleted neighborhood \(B^o \triangleq \{z\in \mathbb{C}: 0 < |z-z_0| \leq \epsilon\}\) is contained in \(G\) and such that \(|f(z)| \leq M\) for all \(z\in B^o\).

Let

be the Laurent expansion of \(f\) for \(z\in B^o\), where

Here \(C\) denotes the positively oriented circle \(|\zeta -z_0|=\epsilon\). Changing variables,

the coefficients are

Therefore,

which makes it clear that, if \(n<0\), then \(|a_n|\) can be made arbitrarily small, by choosing a sufficiently small \(\epsilon\). This proves that \(a_n = 0\) for negative \(n\), and so

Thus, \(f\in H(G)\).

The lemma solves part (a) and is also an example of a general theorem about isolated singularities of holomorphic functions, so it answers part (b).

Here is another answer to part (b).

Let \(G\subset \mathbb{C}\) be open. and suppose \(f(z)\) is holomorphic for all \(z\in G\) except for an isolated singularity at \(z=z_0\in G\). Then

\(z_0\) is a pole of \(f\) if and only if \(\lim_{z\to z_0} |f(z)| = \infty\);

if \(m>0\) is the smallest integer such that \(\limsup_{z\to z_0} |(z-z_0)^m f(z)|\) remains bounded, then \(z_0\) is a pole of order \(m\).

Solution to Problem 43

1 Let \(\phi_0(z) = \frac{1-z}{1+z}\). Our strategy will be to show that \(\phi_0\) maps the fourth quadrant onto \(D\cap \Pi^+\), and then to construct a conformal mapping, \(f\), of the unit disk onto the fourth quadrant. Then the composition \(\phi_0\circ f\) will have the desired properties.

Consider the boundary of the first quadrant. Note that \(\phi_0\) maps the real line onto itself. Furthermore, \(\phi_0\) takes 0 to 1 and takes \(\infty\) to -1. Since \(\phi_0(1) = 0\), we see that the positive real axis \((0,\infty)\) is mapped onto the segment \((-1,1)\).

Now, since \(\phi_0\) maps the right half-plane \(P^+\) onto the unit disk, it must map the boundary of \(P^+\) (i.e., the imaginary axis) onto the boundary of the unit disk. Thus, as \(0\mapsto 1\) and \(\infty\mapsto -1\), the positive imaginary axis is mapped to either the upper half-circle or the lower half-circle, and similarly for the negative imaginary axis.

Checking that \(\phi_0(i) = -i\), it is clear that the positive imaginary axis is mapped to the lower half-circle \(\{e^{i\theta}: -\pi < \theta < 0\}\). Therefore, in mapping the right half-plane onto the unit disk, \(\phi_0\) maps the first quadrant to the lower half-disk \(D\cap \Pi^-\), and must therefore map the fourth quadrant to the upper half-disk. That is, \(\phi_0 : Q_4 \rightarrow D\cap \Pi^+\), where

\[Q_4 := \{z\in \mathbb{C}: \mathfrak{Re}(z) > 0, \mathfrak{Im}(z) <0\}.\]Next construct a mapping of the unit disk onto the fourth quadrant as follows:

If \(\phi_1(z) = iz\), then \(\phi_1 \circ \phi_0: D\rightarrow \Pi^+\).

Let \(\phi_2(z) = z^{1/2}\) be a branch of the square root function on \(\Pi^+\). Then \(\phi_2\) maps \(\Pi^+\) onto the first quadrant,

\[Q_1 := \{z\in \mathbb{C}: \mathfrak{Re}(z) > 0, \mathfrak{Im}(z) >0\}.\]Let \(\phi_3(z) = e^{-i\pi/2}z = -iz\), which takes the first quadrant to the fourth quadrant.

Finally, since all of the mappings are conformal bijections, \(f = \phi_3 \circ \phi_2 \circ \phi_1 \circ \phi_0\) is a conformal bijection of \(D\) onto \(Q_4\).

Therefore, \(\phi_0 \circ f\) is a conformal bijection of \(D\) onto \(D\cap \Pi^+\).

(Riemann) Let \(\Omega \subset \mathbb{C}\) be a simply connected region such that \(\Omega \neq \mathbb{C}\). Then \(\Omega\) is conformally equivalent to \(D\). That is, there is a conformal bijection, \(\phi\), of \(\Omega\) onto the unit disk. Moreover, if we specify that a particular \(z_0\in \Omega\) must be mapped to \(0\), and we specify the value of \(\arg \phi(z_0)\), then the conformal mapping is unique.

Todo

look up more precise statement of rmt.

Solution to Problem 44

See Schwarz’s lemma.

In order to apply Schwarz’s lemma, map the disk to the upper half-plane with the Möebius map \(\phi: D\rightarrow \Pi^+\) defined by

\[\phi(z) = i\frac{1-z}{1+z}.\]Then, \(\phi(0) = i\). Therefore, the function \(g = f\circ \phi: D\xrightarrow{\phi} \Pi^+\xrightarrow{f} D\) satisfies \(|g(z)|\leq 1\) and \(g(0) = f(\phi(0)) = f(i) = 0\).

By Schwarz’s lemma, then, \(|g'(0)|\leq 1\).

Finally, observe that \(g'(z) = f'(\phi(z))\phi'(z)\), and then check that \(\phi'(0) = -2i\).

Whence, \(g'(0) = f'(\phi(0))\phi'(0) = f'(i)(-2i)\), which implies \(1 \geq |g'(0)| = 2|f'(i)|\). Therefore \(|f'(i)|\leq 1/2\).

Solution to Problem 45

See Cauchy’s theorems and the partial converse.

(coming soon)

Solution to Problem 46

Let \(\Omega\) be an open subset of the plane. A family \(\mathcal{F}\) of functions in \(\Omega\) is called a

*normal family*if every sequence of functions in \(\mathcal{F}\) has a subsequence which converges locally uniformly in \(\Omega\). (The same definition applies when the family \(\mathcal{F}\) happens to be contained in \(H(\Omega)\).) 2

The Arzela-Ascoli theorem is arguably

thefundamental theorem for normal families. However, the question asks specifically about the special case when \(\mathcal{F}\) is a family ofholomorphicfunctions, so the student is probably expected to state the version of Montel’s theorem stated below, which is an easy consequence of the Arzela-Ascoli theorem. 3

- Let \(\mathcal{F}\subset C(\Omega, S)\) be a family of continuous functions from an open set \(\Omega\subseteq \mathbb{C}\) into a metric space \((S,d)\). Then \(\mathcal{F}\) is a normal family if and only if

\(\mathcal{F}\) is equicontinuous on each compact subset of \(\Omega\), and

for each \(z\in \Omega\), the set \(\{f(z):f\in \mathcal{F}\}\) is contained in a compact subset of \(S\).

Recall that a family \(\mathcal{F}\) of functions is called

locally boundedon \(\Omega\) iff for all compact \(K\subset \Omega\) there is a constant \(M_K\) such that \(|f(z)|\leq M_K\) for all \(f\in \mathcal{F}\) and \(z\in K\).Assume the set-up of the Arzela-Ascoli theorem, and suppose \(S=\mathbb{C}\) and \(\mathcal{F}\subset H(\Omega)\). Then \(\mathcal{F}\) is a normal family if and only if it is locally bounded.

Fix a sequence \(\{z_n\}\subset \Gamma\) with \(z_n\rightarrow 0\) as \(n\rightarrow \infty\). We must prove \(f(z_n)\rightarrow 0\). Define \(f_n(z) = f(|z_n|z)\). Then, since \(z\in \Gamma \Rightarrow |z_n|z\in \Gamma\), we have

\[|f_n(z)| = |f(|z_n|z)| < 1, \text{ for all } z\in \Gamma \text{ and } n\in \mathbb{N}.\]Therefore, \(\mathcal{F}\) is a normal family in \(\Gamma\). Also note that each \(f_n\) is holomorphic in \(\Gamma\) since \(f(tz) \in H(\Gamma)\) for any constant \(t>0\). Thus, \(\mathcal{F}\) is a normal family of holomorphic functions in \(\Gamma\).

Let \(g\) be a normal limit of \(\{f_n\}\); i.e.,there is some subsequence \(n_k\) such that, as \(k\rightarrow \infty\), \(f_{n_k} \rightarrow g\) locally uniformly in \(\Gamma\).

Consider the point \(z=i\). Since \(f(it)\rightarrow 0\) as \(t\downarrow 0\),

\[g(i) = \lim_{k\rightarrow \infty} f_{n_k}(i) = \lim_{k\rightarrow \infty} f(|z_{n_k}|i) = 0.\]In fact, for any point \(z=iy\) with \(y>0\), we have \(g(z)=0\). Since \(g\) is holomorphic in \(\Gamma\), the identity theorem implies that \(g\equiv 0\) in \(\Gamma\).

Next, consider

(20)¶\[f_n\left(\frac{z_n}{|z_n|}\right)= f\left(|z_n|\frac{z_n}{|z_n|}\right)= f(z_n).\]The numbers \(z_n/|z_n|\) lie in the compact set \(\gamma = \{z\in \Gamma: |z|=1\}\). Since \(f_{n_k}\rightarrow g\) uniformly in \(\gamma\), for any \(\epsilon >0\), there is a \(K>0\) such that \(|f_{n_k}(z) - g(z)| = |f_{n_k}(z)| < \epsilon\), for all \(k\geq K\) and all \(z\in \gamma\). That is,

\[\lim_{k\rightarrow \infty} \sup\{|f_{n_k}(z)|: z\in \gamma\} \triangleq \lim_{k\rightarrow \infty} \|f_{n_k}\|_\gamma = 0,\]and, since \(z_{n_k}/|z_{n_k}|\in \gamma\),

\[\left|f_{n_k}\left(\frac{z_{n_k}}{|z_{n_k}|}\right)\right| \leq \|f_{n_k}\|_\gamma.\]\[\therefore \quad \lim_{k\to \infty} f_{n_k}\left(\frac{z_{n_k}}{|z_{n_k}|}\right) = 0.\]By (20), then, \(\lim_{k\rightarrow \infty} f(z_{n_k}) = 0\).

Finally, recall that \(f(z_n)\rightarrow 0\) iff every subsequence \(z_{n_j}\) has a further subsequence \(z_{n_{j_k}}\) such that \(f(z_{n_{j_k}})\rightarrow 0\), as \(k\rightarrow \infty\). Now, if \(z_{n_j}\) is any subsequence, then \(\{f(z_{n_j})\}\) is a normal family, and, repeating the argument above, there is, indeed, a further subsequence \(z_{n_{j_k}}\) such that \(f(z_{n_{j_k}})\rightarrow 0\). This completes the proof.

Remarks. In the last paragraph, we made use of the fact that a sequence converges to zero iff any subsequence has, in turn, a further subsequence that converges to zero. An alternative concluding argument that doesn’t rely on this result, but proceeds by way of contradiction, runs as follows: Assume we have already shown \(\lim_{k\rightarrow \infty} f(z_{n_k}) = 0\), as above, and suppose \(f(z_n)\) does not converge to 0 as \(n\rightarrow \infty\). Then there is a \(\delta > 0\) and a subsequence \(\{z_{n_j}\}\) such that \(|f(z_{n_j})|>\delta\) for all \(j\in \mathbb{N}\). Relabel this subsequence \(\{z_n\}\). Then \(\{f(z_n)\}\) is itself a normal family and we can repeat the argument above to get a further subsequence \(\{z_{n_k}\}\) with \(\lim_{k \rightarrow \infty} f(z_{n_k})=0\). This contradicts the assumption that \(|f(z_{n})|>\delta\) for all \(n\in \mathbb{N}\). Therefore, \(f(z_n) \rightarrow 0\), as desired.

Footnotes

- 1
See also this problem of April 1989 and this problem of April 1995.

- 2
Despite the wording of the problem, the family need not satisfy \(\mathcal{F}\subset H(\Omega)\) in order to be normal.

- 3
Part (b) of this problem of Nov 2001 asks for a proof of Montel’s theorem using the Arzela-Ascoli theorem.

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