.. 2006 Nov 13 .. =========== Examination 7 ============== 2006 Nov ---------- **Notation.** :math:`\mathbb{C}` is the set of complex numbers, :math:`D= \{z\in \mathbb{C}: |z|<1\}` is the open unit disk, :math:`\Pi^+` and :math:`\Pi^-` are the upper and lower half-planes, respectively, and, given an open set :math:`G\subset \mathbb{C}`, :math:`H(G)` is the set of holomorphic functions on :math:`G`. .. index:: removable singularity, isolated singularity, Laurent expansion .. _2006Nov_1: .. proof:prob:: (a) Suppose that :math:`f \in H(D\setminus \{0\})` and that :math:`|f(z)| < 1` for all :math:`0<|z|<1`. Prove that there is :math:`F\in H(D)` with :math:`F(z) = f(z)` for all :math:`z\in D\setminus \{0\}`. (b) State a general theorem about isolated singularities for holomorphic functions. .. _2006Nov_2: .. proof:prob:: (a) Explicitly construct, through a sequence of mappings, a one-to-one holomorphic function mapping the disk :math:`D` onto the half disk :math:`D\cap \Pi^+`. (b) State a general theorem concerning one-to-one mappings of :math:`D` onto domains :math:`\Omega\subset \mathbb{C}`. .. index:: Schwarz's lemma, extremal function .. _2006Nov_3: .. proof:prob:: (a) State the Schwarz lemma. (b) Suppose that :math:`f\in H(\Pi^+)` and that :math:`|f(z)|<1` for all :math:`z\in \Pi^+`. If :math:`f(i)=0` how large can :math:`|f'(i)|` be? Find the extremal functions. .. index:: Cauchy's theorem, Cauchy's theorem (partial converse of). .. _2006Nov_4: .. proof:prob:: (a) State Cauchy's theorem and its converse. (b) Suppose that :math:`f` is a continuous function defined on the entire complex plane. Assume that (i) :math:`f\in H(\Pi^+\cup \Pi^-)` (ii) :math:`f(\bar{z}) = \overline{f(z)}` all :math:`z\in \mathbb{C}`. Prove that :math:`f` is an entire function. .. index:: normal family, Arzela-Ascoli theorem, Montel's theorem, locally bounded .. _2006Nov_5: .. proof:prob:: (a) Define what it means for a family :math:`\mathcal{F}\subset H(\Omega)` to be a *normal family*. State the fundamental theorem for normal families. (b) Suppose :math:`f\in H(\Pi^+)` and :math:`|f(z)|<1` all :math:`z\in \Pi^+`. Suppose further that .. math:: \lim{t\rightarrow 0+} f(it) = 0. Prove that :math:`f(z_n) \rightarrow 0` whenever the sequence :math:`z_n \rightarrow 0` and :math:`z_n \in \Gamma` where .. math:: \Gamma = \{ z\in \Pi^+ : |\mathfrak{Re}(z)| \leq \mathfrak{Im}(z)\}. *Hint.* Consider the functions :math:`f_t(z) = f(tz)` where :math:`t>0`. ------------------------------------------ Solutions ---------- .. container:: toggle .. container:: header Solution to :numref:`Problem {number} <2006Nov_1>` **Lemma.** Suppose :math:`G\subset \mathbb{C}` is an open set and :math:`f` is holomorphic in :math:`G` except for an isolated singularity at :math:`z_0 \in G`. If .. math:: \limsup_{z\to z_0} \{|f(z)| < \infty : z\in G\}, then :math:`z_0` is a removable singularity and :math:`f` may be extended holomorphically to all of :math:`G`. *Proof.* Under the stated hypotheses, there is an :math:`\epsilon >0` and an :math:`M > 0` such that the deleted neighborhood :math:`B^o \triangleq \{z\in \mathbb{C}: 0 < |z-z_0| \leq \epsilon\}` is contained in :math:`G` and such that :math:`|f(z)| \leq M` for all :math:`z\in B^o`. Let .. math:: f(z) = \sum_{n=-\infty}^{\infty} a_n (z-z_0)^n be the Laurent expansion of :math:`f` for :math:`z\in B^o`, where .. math:: a_n = \frac{1}{2\pi i}\int_C \frac{f(\zeta)}{(\zeta - z_0)^{n+1}} d\zeta. Here :math:`C` denotes the positively oriented circle :math:`|\zeta -z_0|=\epsilon`. Changing variables, .. math:: \zeta = z_0 + \epsilon e^{i\theta} \quad \Rightarrow \quad d\zeta = i\,\epsilon e^{i\theta} d\theta the coefficients are .. math:: 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, .. math:: |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 :math:`n<0`, then :math:`|a_n|` can be made arbitrarily small, by choosing a sufficiently small :math:`\epsilon`. This proves that :math:`a_n = 0` for negative :math:`n`, and so .. math:: f(z) = \sum_{n=0}^{\infty} a_n (z-z_0)^n. Thus, :math:`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 :math:`G\subset \mathbb{C}` be open. and suppose :math:`f(z)` is holomorphic for all :math:`z\in G` except for an isolated singularity at :math:`z=z_0\in G`. Then (i) :math:`z_0` is a pole of :math:`f` if and only if :math:`\lim_{z\to z_0} |f(z)| = \infty`; (ii) if :math:`m>0` is the smallest integer such that :math:`\limsup_{z\to z_0} |(z-z_0)^m f(z)|` remains bounded, then :math:`z_0` is a pole of order :math:`m`. .. ------------------------------------------------------------------------- .. container:: toggle .. container:: header Solution to :numref:`Problem {number} <2006Nov_2>` (a) [1]_ Let :math:`\phi_0(z) = \frac{1-z}{1+z}`. Our strategy will be to show that :math:`\phi_0` maps the fourth quadrant onto :math:`D\cap \Pi^+`, and then to construct a conformal mapping, :math:`f`, of the unit disk onto the fourth quadrant. Then the composition :math:`\phi_0\circ f` will have the desired properties. Consider the boundary of the first quadrant. Note that :math:`\phi_0` maps the real line onto itself. Furthermore, :math:`\phi_0` takes 0 to 1 and takes :math:`\infty` to -1. Since :math:`\phi_0(1) = 0`, we see that the positive real axis :math:`(0,\infty)` is mapped onto the segment :math:`(-1,1)`. Now, since :math:`\phi_0` maps the right half-plane :math:`P^+` onto the unit disk, it must map the boundary of :math:`P^+` (i.e., the imaginary axis) onto the boundary of the unit disk. Thus, as :math:`0\mapsto 1` and :math:`\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 :math:`\phi_0(i) = -i`, it is clear that the positive imaginary axis is mapped to the lower half-circle :math:`\{e^{i\theta}: -\pi < \theta < 0\}`. Therefore, in mapping the right half-plane onto the unit disk, :math:`\phi_0` maps the first quadrant to the lower half-disk :math:`D\cap \Pi^-`, and must therefore map the fourth quadrant to the upper half-disk. That is, :math:`\phi_0 : Q_4 \rightarrow D\cap \Pi^+`, where .. math:: 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 :math:`\phi_1(z) = iz`, then :math:`\phi_1 \circ \phi_0: D\rightarrow \Pi^+`. Let :math:`\phi_2(z) = z^{1/2}` be a branch of the square root function on :math:`\Pi^+`. Then :math:`\phi_2` maps :math:`\Pi^+` onto the first quadrant, .. math:: Q_1 := \{z\in \mathbb{C}: \mathfrak{Re}(z) > 0, \mathfrak{Im}(z) >0\}. Let :math:`\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, :math:`f = \phi_3 \circ \phi_2 \circ \phi_1 \circ \phi_0` is a conformal bijection of :math:`D` onto :math:`Q_4`. Therefore, :math:`\phi_0 \circ f` is a conformal bijection of :math:`D` onto :math:`D\cap \Pi^+`. (b) (Riemann) Let :math:`\Omega \subset \mathbb{C}` be a simply connected region such that :math:`\Omega \neq \mathbb{C}`. Then :math:`\Omega` is conformally equivalent to :math:`D`. That is, there is a conformal bijection, :math:`\phi`, of :math:`\Omega` onto the unit disk. Moreover, if we specify that a particular :math:`z_0\in \Omega` must be mapped to :math:`0`, and we specify the value of :math:`\arg \phi(z_0)`, then the conformal mapping is unique. .. todo:: look up more precise statement of rmt. .. ------------------------------------------------------------------------- .. container:: toggle .. container:: header Solution to :numref:`Problem {number} <2006Nov_3>` (a) See :ref:`Schwarz's lemma `. (b) In order to apply :ref:`Schwarz's lemma `, map the disk to the upper half-plane with the Möebius map :math:`\phi: D\rightarrow \Pi^+` defined by .. math:: \phi(z) = i\frac{1-z}{1+z}. Then, :math:`\phi(0) = i`. Therefore, the function :math:`g = f\circ \phi: D\xrightarrow{\phi} \Pi^+\xrightarrow{f} D` satisfies :math:`|g(z)|\leq 1` and :math:`g(0) = f(\phi(0)) = f(i) = 0`. By :ref:`Schwarz's lemma `, then, :math:`|g'(0)|\leq 1`. Finally, observe that :math:`g'(z) = f'(\phi(z))\phi'(z)`, and then check that :math:`\phi'(0) = -2i`. Whence, :math:`g'(0) = f'(\phi(0))\phi'(0) = f'(i)(-2i)`, which implies :math:`1 \geq |g'(0)| = 2|f'(i)|`. Therefore :math:`|f'(i)|\leq 1/2`. .. ------------------------------------------------------------------------- .. .. container:: toggle .. .. container:: header Solution to :numref:`Problem {number} <2006Nov_4>` (a) See :ref:`Cauchy's theorems ` and the :ref:`partial converse `. (b) (coming soon) .. ------------------------------------------------------------------------- .. container:: toggle .. container:: header Solution to :numref:`Problem {number} <2006Nov_5>` (a) Let :math:`\Omega` be an open subset of the plane. A family :math:`\mathcal{F}` of functions in :math:`\Omega` is called a *normal family* if every sequence of functions in :math:`\mathcal{F}` has a subsequence which converges locally uniformly in :math:`\Omega`. (The same definition applies when the family :math:`\mathcal{F}` happens to be contained in :math:`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 :math:`\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 :math:`\mathcal{F}\subset C(\Omega, S)` be a family of continuous functions from an open set :math:`\Omega\subseteq \mathbb{C}` into a metric space :math:`(S,d)`. Then :math:`\mathcal{F}` is a normal family if and only if (i) :math:`\mathcal{F}` is equicontinuous on each compact subset of :math:`\Omega`, and (ii) for each :math:`z\in \Omega`, the set :math:`\{f(z):f\in \mathcal{F}\}` is contained in a compact subset of :math:`S`. Recall that a family :math:`\mathcal{F}` of functions is called *locally bounded* on :math:`\Omega` iff for all compact :math:`K\subset \Omega` there is a constant :math:`M_K` such that :math:`|f(z)|\leq M_K` for all :math:`f\in \mathcal{F}` and :math:`z\in K`. Assume the set-up of the Arzela-Ascoli theorem, and suppose :math:`S=\mathbb{C}` and :math:`\mathcal{F}\subset H(\Omega)`. Then :math:`\mathcal{F}` is a normal family if and only if it is locally bounded. (b) Fix a sequence :math:`\{z_n\}\subset \Gamma` with :math:`z_n\rightarrow 0` as :math:`n\rightarrow \infty`. We must prove :math:`f(z_n)\rightarrow 0`. Define :math:`f_n(z) = f(|z_n|z)`. Then, since :math:`z\in \Gamma \Rightarrow |z_n|z\in \Gamma`, we have .. math:: |f_n(z)| = |f(|z_n|z)| < 1, \text{ for all } z\in \Gamma \text{ and } n\in \mathbb{N}. Therefore, :math:`\mathcal{F}` is a normal family in :math:`\Gamma`. Also note that each :math:`f_n` is holomorphic in :math:`\Gamma` since :math:`f(tz) \in H(\Gamma)` for any constant :math:`t>0`. Thus, :math:`\mathcal{F}` is a normal family of holomorphic functions in :math:`\Gamma`. Let :math:`g` be a normal limit of :math:`\{f_n\}`; i.e.,there is some subsequence :math:`n_k` such that, as :math:`k\rightarrow \infty`, :math:`f_{n_k} \rightarrow g` locally uniformly in :math:`\Gamma`. Consider the point :math:`z=i`. Since :math:`f(it)\rightarrow 0` as :math:`t\downarrow 0`, .. math:: 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 :math:`z=iy` with :math:`y>0`, we have :math:`g(z)=0`. Since :math:`g` is holomorphic in :math:`\Gamma`, the identity theorem implies that :math:`g\equiv 0` in :math:`\Gamma`. Next, consider .. math:: f_n\left(\frac{z_n}{|z_n|}\right)= f\left(|z_n|\frac{z_n}{|z_n|}\right)= f(z_n). :label: 06-5-1 The numbers :math:`z_n/|z_n|` lie in the compact set :math:`\gamma = \{z\in \Gamma: |z|=1\}`. Since :math:`f_{n_k}\rightarrow g` uniformly in :math:`\gamma`, for any :math:`\epsilon >0`, there is a :math:`K>0` such that :math:`|f_{n_k}(z) - g(z)| = |f_{n_k}(z)| < \epsilon`, for all :math:`k\geq K` and all :math:`z\in \gamma`. That is, .. math:: \lim_{k\rightarrow \infty} \sup\{|f_{n_k}(z)|: z\in \gamma\} \triangleq \lim_{k\rightarrow \infty} \|f_{n_k}\|_\gamma = 0, and, since :math:`z_{n_k}/|z_{n_k}|\in \gamma`, .. math:: \left|f_{n_k}\left(\frac{z_{n_k}}{|z_{n_k}|}\right)\right| \leq \|f_{n_k}\|_\gamma. .. math:: \therefore \quad \lim_{k\to \infty} f_{n_k}\left(\frac{z_{n_k}}{|z_{n_k}|}\right) = 0. By :eq:`06-5-1`, then, :math:`\lim_{k\rightarrow \infty} f(z_{n_k}) = 0`. Finally, recall that :math:`f(z_n)\rightarrow 0` iff every subsequence :math:`z_{n_j}` has a further subsequence :math:`z_{n_{j_k}}` such that :math:`f(z_{n_{j_k}})\rightarrow 0`, as :math:`k\rightarrow \infty`. Now, if :math:`z_{n_j}` is any subsequence, then :math:`\{f(z_{n_j})\}` is a normal family, and, repeating the argument above, there is, indeed, a further subsequence :math:`z_{n_{j_k}}` such that :math:`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 :math:`\lim_{k\rightarrow \infty} f(z_{n_k}) = 0`, as above, and suppose :math:`f(z_n)` does not converge to 0 as :math:`n\rightarrow \infty`. Then there is a :math:`\delta > 0` and a subsequence :math:`\{z_{n_j}\}` such that :math:`|f(z_{n_j})|>\delta` for all :math:`j\in \mathbb{N}`. Relabel this subsequence :math:`\{z_n\}`. Then :math:`\{f(z_n)\}` is itself a normal family and we can repeat the argument above to get a further subsequence :math:`\{z_{n_k}\}` with :math:`\lim_{k \rightarrow \infty} f(z_{n_k})=0`. This contradicts the assumption that :math:`|f(z_{n})|>\delta` for all :math:`n\in \mathbb{N}`. Therefore, :math:`f(z_n) \rightarrow 0`, as desired. ---------------------- .. rubric:: Footnotes .. [1] See also :ref:`this problem <1989Apr_3>` of April 1989 and :ref:`this problem <1995Apr_5>` of April 1995. .. [2] Despite the wording of the problem, the family need not satisfy :math:`\mathcal{F}\subset H(\Omega)` in order to be normal. .. [3] Part (b) of :ref:`this problem <2001Nov_5>` of Nov 2001 asks for a proof of :ref:`Montel's theorem ` using the :ref:`Arzela-Ascoli theorem `. ----------------------------------- .. blank