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 \(G\).

Problem 42

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.

Problem 43

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 \(\Omega\subset \mathbb{C}\).

Problem 44

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.

Problem 45

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.

Problem 46

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¶

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

\[\limsup_{z\to z_0} \{|f(z)| < \infty : z\in G\},\]

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

\[f(z) = \sum_{n=-\infty}^{\infty} a_n (z-z_0)^n\]

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

\[a_n = \frac{1}{2\pi i}\int_C \frac{f(\zeta)}{(\zeta - z_0)^{n+1}} d\zeta.\]

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

\[\zeta = z_0 + \epsilon e^{i\theta} \quad \Rightarrow \quad d\zeta = i\,\epsilon e^{i\theta} d\theta\]

the coefficients are

\[a_n = \frac{1}{2\pi\, i} \int_0^{2\pi} \frac{f(z_0 + \epsilon e^{i\theta})}{(z_0 + \epsilon e^{i\theta} - z_0)^{n+1}} \; i \epsilon e^{i\theta} d\theta.\]

Therefore,

\[|a_n| \leq \frac{1}{2\pi} \int_0^{2\pi} \frac{M}{\epsilon^{n+1}} \, \epsilon d|\theta| = \frac{M}{\epsilon^n},\]

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

\[f(z) = \sum_{n=0}^{\infty} a_n (z-z_0)^n.\]

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 \(m\).

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.

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)

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 the fundamental theorem for normal families. However, the question asks specifically about the special case when \(\mathcal{F}\) is a family of holomorphic functions, 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 bounded on \(\Omega\) 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.