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.

Problem 23

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

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

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

Problem 24
  1. State a form of the Cauchy theorem.

  2. State a converse of the Cauchy theorem.

Problem 25

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

  1. Prove that

\[\sum_{n=1}^\infty n |a_n|^2\leq 1.\]
  1. Is the constant 1 the best possible?

Problem 26

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

Problem 27

Find an explicit conformal mapping of the semidisk

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

onto the unit disk.

Problem 28

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

\[|f(z)|<1 \; \text{ for all } \; |z|<1.\]
  1. State the Schwarz lemma, as applied to \(f\).

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


Solution to Problem 23

  1. By Cauchy’s formula, we have

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

for every \(R>0\). Therefore,

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

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

  1. 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 \(w\), there is a sequence \(\{z_n\}\) with \(z_n\rightarrow \infty\) and \(f(z_n)\rightarrow w\) as \(n\rightarrow \infty\). Since this contradicts the given hypotheses, \(f(z)\) cannot be a transcendental function. That is, \(f(z)\) must be a polynomial.

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

\[g(z) = f(1/z) = \sum_{n=-\infty}^\infty b_n z^n\]

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

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

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

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

That is, \(0 = a_{m+1} = a_{m+2} = \cdots\), so

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

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

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

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

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

\[\min_{z\in D_R} |g(z)|\geq \min_{|z|=R}|g(z)| = \epsilon > 0,\]

for some \(\epsilon>0\). Therefore, \(1/g\) 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 \(f(z)\) is a polynomial.

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

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

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

Solution to Problem 25

  1. This is a special case of the following area theorem:

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

    \[A = \pi \sum_{n=1}^\infty n |a_n|^2.\]

    The area of the image of \(D\) under \(f\) is the integral over \(D\) of the Jacobian of \(f\). That is,

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

    Compute \(|f'(z)|\) by differentiating the power series of \(f(z)\) term by term,

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

    Next, take the squared modulus,

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

    This gives,

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

    Letting \(z = r e^{i\theta}\),

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

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

    (9)\[A = 2\pi \sum_{n=1}^\infty n^2 |a_n|^2 \int_0^1r^{2n-1} \, dr= \pi \sum_{n=1}^\infty n^2 |a_n|^2.\]

    To apply this theorem to the problem at hand, note that the hypotheses of the problem imply that \(f\) maps the unit disk bijectively onto its range \(f(D)\), which is contained inside \(D\) and, therefore, has area less or equal to \(\pi\). This and (9) together imply

    \[\pi \geq \pi \sum_{n=1}^\infty n^2 |a_n|^2,\]

    which gives the desired inequality.

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

Solution to Problem 26

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

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

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

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

Solution to Problem 27

(See solution to this problem of Apr 1989, and this problem of Nov 2006.)

Solution to Problem 28

  1. See Schwarz’s lemma.

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

    \[\varphi(z) = \frac{\frac{1}{2} - z}{1-\frac{z}{2}}.\]

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

    (10)\[1 \geq |g'(0)| = |\varphi'(1/2)| |f'(0)|.\]


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

    Therefore, \(\varphi'(1/2) = -4/3\), and it follows from (10) that

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



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


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

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Real Analysis Exams

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