.. ===========
.. 1995 Apr 10
.. ===========

Examination 4
==============

1995 Apr 
---------

**Instructions.** *Work as many of the problems as you can. Each solution should be clearly written on a separate sheet of paper.*

.. -------------------------------------------------------------
   
.. index:: Cauchy's formula, Casorati-Weierstrass theorem, transcendental function, Liouville's theorem, entire function

.. _1995Apr_1:

.. proof:prob::

   Let :math:`f(z) = \sum a_n z^n` be an entire function.

   (a) Suppose that :math:`|f(z)| \leq A |z|^N + B` for all :math:`z\in \mathbb{C}` where :math:`A, B` are finite constants. Show that :math:`f` is a polynomial of degree :math:`N` or less.

   (b) Suppose that :math:`f` satisfies the condition: :math:`|f(z_n)|\rightarrow \infty` whenever :math:`|z_n| \rightarrow \infty`. Show that :math:`f` is a polynomial.

.. -------------------------------------------------------------
   
.. index:: Cauchy's theorem, Cauchy's theorem (partial converse of)

.. _1995Apr_2:

.. proof:prob::

   (a) State a form of the Cauchy theorem.

   (b) State a converse of the Cauchy theorem.

.. -------------------------------------------------------------
   
.. index:: Jacobian

.. _1995Apr_3:

.. proof:prob::

   Let :math:`f(z) = \sum_{n=0}^\infty a_n z^n` be analytic and one-to-one on :math:`|z|<1`. Suppose that :math:`|f(z)|<1` for all :math:`|z|<1`. [1]_

   (a) Prove that

   .. math:: \sum_{n=1}^\infty n |a_n|^2\leq 1.

   (b) Is the constant 1 the best possible?

.. -------------------------------------------------------------
   
.. index:: harmonic function

.. _1995Apr_4:

.. proof:prob::

   Let :math:`u(z)` be a nonconstant, real valued, harmonic function on :math:`\mathbb{C}`. Prove there exists a sequence :math:`\{z_n\}` with :math:`|z_n| \rightarrow \infty` for which :math:`u(z_n)\rightarrow 0`.

.. -------------------------------------------------------------
   
.. index:: conformal mapping

.. .. _1995April10_5:

.. _1995Apr_5:

.. proof:prob::

   Find an explicit conformal mapping of the semidisk

   .. math:: H = \{z : |z| < 1, \mathfrak{Re}(z) > 0\}

   onto the unit disk.

.. -------------------------------------------------------------

.. index:: Schwarz lemma

.. _1995Apr_6:

.. proof:prob::

   Suppose :math:`f(z)` is a holomorphic function on the unit disk that satisfies:

   .. math:: |f(z)|<1 \; \text{ for all } \; |z|<1.

   (a) State the Schwarz lemma, as applied to :math:`f`.

   (b) If :math:`f(0)=\frac{1}{2}`, how large can :math:`|f'(0)|` be?

--------------------------------------

Solutions
---------

.. container:: toggle

  .. container:: header

     Solution to :numref:`Problem {number} <1995Apr_1>`

  (a) By :ref:`Cauchy's formula <cauchy-formula>`, we have

  .. math:: a_n = \frac{f^{(n)}(0)}{n!} = \frac{1}{2\pi i} \int_{|\zeta| = R} \frac{f(\zeta)}{\zeta^{n+1}}\, d\zeta,

  for every :math:`R>0`. Therefore,

  .. math::

      \begin{aligned}
          |a_n| &\leq  \frac{1}{2\pi} \int_{|\zeta| = R} \frac{|f(\zeta)|}{|\zeta|^{n+1}}\,|d\zeta|\\[4pt]
                & \leq  \frac{1}{2\pi} \, \frac{A\, R^N + B}{R^{n+1}}\, 2\pi R\\[4pt]
                &= A \,R^{N-n} + B \, R^{-n}.
      \end{aligned}

  Again, this holds for every :math:`R>0`. Thus, for any :math:`n>N` and :math:`\epsilon >0`, taking :math:`R` large enough forces :math:`|a_n| < \epsilon` (:math:`n=N+1, N+2, \dots`). Since :math:`\epsilon` was arbitrary, we have :math:`a_n = 0` for all :math:`n=N+1, N+2,\dots`. Therefore, :math:`f(z) = \sum_{n=0}^N a_n z^n`.  

  (b) We give three different proofs. The first is the shortest, but relies on the heaviest machinery.

  **Proof 1.** If we take for granted that any transcendental (i.e. non-polynomial) entire function has an essential singularity at infinity, then the Casorati-Weierstrass theorem (see 3 of Nov. ’01) implies that, for any complex number :math:`w`, there is a sequence :math:`\{z_n\}` with :math:`z_n\rightarrow \infty` and :math:`f(z_n)\rightarrow w` as :math:`n\rightarrow \infty`. Since this contradicts the given hypotheses, :math:`f(z)` cannot be a transcendental function. That is, :math:`f(z)` must be a polynomial.

  **Proof 2.** Since :math:`f\in H(\mathbb{C})`, the series :math:`f(z) = \sum a_n z^n` converges locally uniformly in :math:`\mathbb{C}`. The hypotheses imply that the function :math:`f(1/z)` has a pole at :math:`z=0`. Let
  
  .. math:: g(z) = f(1/z) = \sum_{n=-\infty}^\infty b_n z^n

  be the Laurent series expansion of the function :math:`g` about :math:`z=0`. Suppose the pole at :math:`z=0` is of order :math:`m`. Clearly :math:`m` is finite, by the criterion for a pole (i.e.,\ :math:`\lim_{z\rightarrow 0} f(1/z) = \infty`). Therefore, we can write

  .. math::

        \label{eq:200} g(z) = f(1/z) = \sum_{n=-m}^\infty b_n z^n = b_{-m}z^{-m} + b_{-m+1}z^{-m+1} + \cdots b_{-1}z^{-1} + b_0 + b_1 z + \cdots

  Now :math:`f` is entire, so it has the form :math:`f(z) = \sum_{n=0}^\infty a_n z^n`, which implies that :math:`f(1/z) =  a_0 + a_1z^{-1} + a_2 z^{-2} + \cdots`. Compared with ([eq:200]),

  .. math::

        a_0 + a_1z^{-1} + a_2 z^{-2} + \cdots  = f(1/z) =  b_{-m}z^{-m} + b_{-m+1}z^{-m+1} + \cdots b_{-1}z^{-1} + b_0 + b_1 z + \cdots

  That is, :math:`0 = a_{m+1} = a_{m+2} = \cdots`, so

  .. math:: f(z) = \sum_{n=0}^m a_n z^n.

  **Proof 3.** By the hypotheses, there is an :math:`R>0` such that :math:`|f(z)|>0` for all :math:`|z|>R`. Therefore, the zeros of :math:`f` are confined to a closed disk :math:`\overline{D}_R = \{|z|\leq R\}`. Since the zeros of :math:`f` are isolated, there are at most finitely many of them in any compact subset of :math:`\mathbb{C}`. In particular, :math:`\overline{D}_R` contains only finitely many zeros of :math:`f`. This proves that :math:`f` has only finitely many zeros in :math:`\mathbb{C}`.

  Let :math:`\{\alpha_1,\dots, \alpha_N\}` be the collections of all zeros of :math:`f` (counting multiplicities). Consider the function

  .. math::

      \label{eq:400} g(z) = \frac{f(z)}{(z-\alpha_1)\cdots (z-\alpha_N)}.

  This is defined and holomorphic in :math:`\mathbb{C}\setminus \{\alpha_1, \dots, \alpha_N\}`, but the :math:`\alpha_i`\ ’s are removable singularities, so :math:`g(z)` is a nonzero entire function. In particular, for any :math:`R>0`,

  .. math:: \min_{z\in D_R} |g(z)|\geq \min_{|z|=R}|g(z)| = \epsilon > 0,

  for some :math:`\epsilon>0`. Therefore, :math:`1/g` is a bounded entire function, hence constant, by Liouville's theorem. What we have shown is that the left hand side of ([eq:400]) is constant, and this proves that :math:`f(z)` is a polynomial.

  **Remark.** A nice corollary to part (b) is the following:

  If :math:`f` is an injective entire function, then :math:`f(z) = a z + b` for some constants :math:`a` and :math:`b`.

  The proof appears below in section [sec:limiting-behavior].

.. -----------------------------------------------------------------

.. container:: toggle

  .. container:: header

     Solution to :numref:`Problem {number} <1995Apr_2>`

  (a) See :ref:`Cauchy's theorem <cauchy-h1>`.

  (b) See the :ref:`partial converse of Cauchy's theorem <conv-cauchy>`.

.. ----------------------------------------------------------------------

.. container:: toggle

  .. container:: header

     Solution to :numref:`Problem {number} <1995Apr_3>`

  (a) This is a special case of the following area theorem:

      Suppose :math:`f(z) = \sum_{n=0}^\infty a_n z^n` is a holomorphic function which maps the unit disk :math:`D= \{|z|<1\}` bijectively onto a domain :math:`f(D) = G` having area :math:`A`. Then

      .. math:: A = \pi \sum_{n=1}^\infty n |a_n|^2.

      The area of the image of :math:`D` under :math:`f` is the integral over :math:`D` of the Jacobian of :math:`f`. That is,

      .. math:: A = \iint_D |f'(z)|^2 \, dx \,dy.

      Compute :math:`|f'(z)|` by differentiating the power series of :math:`f(z)` term by term,

      .. math:: f'(z) = \sum_{n=1}^\infty n a_n z^{n-1}.

      Next, take the squared modulus,

      .. math:: |f'(z)|^2 = \sum_{m,n=1}^\infty m \, n \, a_m \overline{a}_n z^{m-1} \overline{z}^{n-1}.

      This gives,

      .. math:: A = \iint_D \sum_{m,n=1}^\infty m\, n \,a_m \overline{a}_n z^{m-1} \overline{z}^{n-1} \, dx \,dy.

      Letting :math:`z = r e^{i\theta}`,

      .. math:: A = \sum_{m,n=1}^\infty m \, n \, a_m \overline{a}_n \int_0^1\int_0^{2\pi} r^{m+n-1} e^{i (m-n)\theta}\, d\theta \,dr.

      Now, for all :math:`k\neq 0`, the integral of :math:`e^{i k \theta}` over :math:`0 \leq \theta < 2\pi` vanishes, so the only non-vanishing terms of the series are those for which :math:`m=n`. That is,

      .. math:: A = 2\pi \sum_{n=1}^\infty n^2 |a_n|^2 \int_0^1r^{2n-1}  \, dr= \pi \sum_{n=1}^\infty n^2 |a_n|^2.
         :label: 500

      To apply this theorem to the problem at hand, note that the hypotheses of the problem imply that :math:`f` maps the unit disk bijectively onto its range :math:`f(D)`, which is contained inside :math:`D` and, therefore, has area less or equal to :math:`\pi`. This and :eq:`500` together imply

      .. math:: \pi \geq \pi \sum_{n=1}^\infty n^2 |a_n|^2,

      which gives the desired inequality.  

  (b) The identity function :math:`f(z) = z` satisfies the given hypotheses and its power series expansion has coefficients :math:`a_1 = 1` and :math:`0=a_0 = a_2 = a_3 = \cdots`. This shows that the upper bound of 1 is obtained and is therefore the best possible.

.. ----------------------------------------------------------------------

.. container:: toggle

  .. container:: header

     Solution to :numref:`Problem {number} <1995Apr_4>`

  Suppose, by way of contradiction, that there is no such sequence. Then :math:`u(z)` is bounded away from zero for all :math:`z` in some neighborhood of infinity, say, :math:`\{|z|>R\}`, for some :math:`R>0`.

  Since :math:`u` is continuous, either :math:`u(z)>0` for all :math:`|z|>R`, or :math:`u(z) < 0` for all :math:`|z|>R`. Assume without loss of generality that :math:`u(z)>0` for all :math:`|z|>R`.

  Since :math:`u` is continuous on the compact set :math:`\{|z|\leq R\}`, it attains its minimum on that set. Thus, there is an :math:`M>0` such that :math:`-M\leq u(z)` for all :math:`|z|\leq R`.

  Consider the function :math:`U(z) = u(z) + M`. By construction, :math:`U(z)\geq 0` for all :math:`z\in \mathbb{C}`, and :math:`U` is harmonic in :math:`\mathbb{C}`. But this implies :math:`U(z)`, hence :math:`u(z)`, must be constant. [2]_ This contradicts the hypothesis that :math:`u(z)` be nonconstant and completes the proof.

.. --------------------------------

Solution to :numref:`Problem {number} <1995Apr_5>`

(See solution to :ref:`this problem <1989Apr_3>` of Apr 1989, and :ref:`this problem <2006Nov_2>` of Nov 2006.)

.. -------------------------------------

.. container:: toggle

  .. container:: header

     Solution to :numref:`Problem {number} <1995Apr_6>`

  (a) See :ref:`Schwarz's lemma <schwarz>`.

  (b) Assume :math:`f` satisfies the given hypotheses. In particular, :math:`f(0)=\frac{1}{2}`. Consider the map

      .. math:: \varphi(z) = \frac{\frac{1}{2} - z}{1-\frac{z}{2}}.

      This is a holomorphic bijection of the unit disk, with :math:`\phi(1/2) = 0`. Therefore, :math:`g = \varphi \circ f` satisfies the hypotheses of Schwarz’s lemma. In particular, :math:`|g'(0)| \leq 1`. Since :math:`g'(z) = \varphi'(f(z))f'(z)`, we have

      .. math:: 1 \geq |g'(0)| = |\varphi'(1/2)| |f'(0)|.
         :label: 300

      Now,

      .. math:: \varphi'(z) = \frac{-\left(1-\frac{z}{2}\right)+\left(\frac{1}{2}\right)\left(\frac{1}{2}-z\right)} {\left(1-\frac{z}{2}\right)^2}.

      Therefore, :math:`\varphi'(1/2) = -4/3`, and it follows from :eq:`300` that

      .. math:: |f'(0)| \leq \frac{1}{|\varphi'(1/2)|} = 3/4.

----------------------------

.. rubric:: Footnotes

.. [1]
   On the original exam, the power series representation was given as :math:`f(z) = \sum_{n=1}^\infty a_n z^n`. However, the problem can be solved without assuming :math:`a_0=0` *a priori*.

.. [2]
   Recall Problem 1(b), April 1989, where we proved that a real valued harmonic function :math:`u(z)` satisfying :math:`u(z)\geq 0` for all :math:`z\in \mathbb{C}` must be constant.

-----------------------------------------

.. blank