# Maximum Modulus

## Spring 2020 HW 3.8 #complex/exercise/work

Prove the fundamental theorem of Algebra using the maximum modulus principle.

## Spring 2020.7 #complex/qual/completed

Let $$f$$ be analytic on a bounded domain $$D$$, and assume also that $$f$$ that is continuous and nowhere zero on the closure $$\mkern 1.5mu\overline{\mkern-1.5muD\mkern-1.5mu}\mkern 1.5mu$$.

Show that if $$|f(z)|=M$$ (a constant) for $$z$$ on the boundary of $$D$$, then $$f(z)=e^{i \theta} M$$ for $$z$$ in $$D$$, where $$\theta$$ is a real constant.

By the maximum modulus principle, $${\left\lvert {f} \right\rvert} \leq M$$ in $$\mkern 1.5mu\overline{\mkern-1.5muD\mkern-1.5mu}\mkern 1.5mu$$. Since $$f$$ has no zeros in $$\mkern 1.5mu\overline{\mkern-1.5muD\mkern-1.5mu}\mkern 1.5mu$$, $$g\coloneqq 1/f$$ is holomorphic on $$D$$ and continuous on $$\mkern 1.5mu\overline{\mkern-1.5muD\mkern-1.5mu}\mkern 1.5mu$$. So the maximum modulus principle applies to $$g$$, and $$M^{-1}\geq {\left\lvert {g} \right\rvert} = 1/{\left\lvert {f} \right\rvert}$$, so $${\left\lvert {f} \right\rvert} \leq M$$. Combining these, $${\left\lvert {f(z)} \right\rvert} = M$$, so $$f(z) = \lambda M$$ where $$\lambda$$ is some constant with $${\left\lvert {\lambda} \right\rvert}=1$$. This is on the unit circle, so $$\lambda = e^{i\theta}$$ for some fixed angle $$\theta$$.

## Fall 2020.6 #complex/qual/completed

Suppose that $$U$$ is a bounded, open and simply connected domain in $$\mathbb{C}$$ and that $$f(z)$$ is a complex-valued non-constant continuous function on $$\mkern 1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern 1.5mu$$ whose restriction to $$U$$ is holomorphic.

• Prove the maximum modulus principle by showing that if $$z_{0} \in U$$, then

\begin{align*} \left|f\left(z_{0}\right)\right|<\sup \{|f(z)|: z \in \partial U\} . \end{align*}

• Show furthermore that if $$|f(z)|$$ is constant on $$\partial U$$, then $$f(z)$$ has a zero in $$U$$ (i.e., there exists $$z_{0} \in U$$ for which $$f\left(z_{0}\right)=0$$ ).

Let $$M\coloneqq\sup_{z\in {{\partial}}U}{\left\lvert {f(z)} \right\rvert}$$. If $$M=0$$, then $$f$$ must be the constant zero function, so assume $$M>0$$.

Suppose toward a contradiction that there exists a $$z_0 \in U$$ with $${\left\lvert {f(z_0)} \right\rvert} = M$$. Note that the map $$z\mapsto {\left\lvert {z} \right\rvert}$$ is an open in discs that don’t intersect $$z=0$$. Since $$f$$ is holomorphic, by the open mapping theorem $$f$$ is an open map, so consider $$D_{\varepsilon}(z_0)$$ a small disk not containing $$0$$. Then $$f(D_{\varepsilon}(z_0))$$ is open, and the composition $$z\mapsto f(z) \mapsto {\left\lvert {f(z)} \right\rvert}$$ is an open map $$D_{\varepsilon}(z_0)\to {\mathbb{R}}$$. Now if $$f$$ is nonconstant, $${\left\lvert {f(D_{\varepsilon}(z_0))} \right\rvert} \supseteq(M-{\varepsilon}, M+{\varepsilon})$$ contains some open interval about $$M$$, which contradicts maximality of $$f$$ at $$z_0$$.

See notes for a proof using the mean value theorem.

Suppose toward a contradiction that $$f$$ has no zeros in $$U$$. Then $$g(z) \coloneqq 1/f(z)$$ is holomorphic in $$U$$. Now if $${\left\lvert {f(z)} \right\rvert} = C$$ on $${{\partial}}U$$, we have $${\left\lvert {g(z)} \right\rvert} = {\left\lvert {1\over f(z)} \right\rvert} = {1\over C}$$ on $${{\partial}}U$$, so $$\max_{z\in U} {\left\lvert {f(z)} \right\rvert} = C$$ and $$\min_{{\left\lvert {f(z)} \right\rvert}} = {1\over C}$$. Since $${\left\lvert {f(z)} \right\rvert}$$ is constant on the boundary, we must have $$\max {\left\lvert {f(z)} \right\rvert} = \min {\left\lvert {f(z)} \right\rvert} = C$$, so $$f$$ is constant on $${{\partial}}U$$. By the identity principle, $$f$$ is constant on $$U$$, a contradiction.

## Spring 2020 HW 3, SS 3.8.15 #complex/exercise/completed

Use the Cauchy inequalities or the maximum modulus principle to solve the following problems:

• Prove that if $$f$$ is an entire function that satisfies \begin{align*} \sup _{|z|=R}|f(z)| \leq A R^{k}+B \end{align*} for all $$R>0$$, some integer $$k\geq 0$$, and some constants $$A, B > 0$$, then $$f$$ is a polynomial of degree $$\leq k$$.

• Show that if $$f$$ is holomorphic in the unit disc, is bounded, and converges uniformly to zero in the sector $$\theta < \arg(z) < \phi$$ as $${\left\lvert {z} \right\rvert} \to 1$$, then $$f \equiv 0$$.

• Let $$w_1, \cdots w_n$$ be points on $$S^1 \subset {\mathbb{C}}$$. Prove that there exists a point $$z\in S^1$$ such that the product of the distances from $$z$$ to the points $$w_j$$ is at least 1. Conclude that there exists a point $$w\in S^1$$ such that the product of the above distances is exactly 1.

• Show that if the real part of an entire function is bounded, then $$f$$ is constant.

\begin{align*} {\left\lvert { f(z_0) } \right\rvert} &= {\left\lvert { {1\over 2\pi i} \oint_{{\left\lvert {z-z_0} \right\rvert} = R } {f(z) \over (z-z_0)^{n+1} } \,dz} \right\rvert} \\ &\leq {1\over 2\pi } \oint_{{\left\lvert {z-z_0} \right\rvert} = R } {\left\lvert {f(z)} \right\rvert} R^{-(n+1)} \,dz\\ &\leq {1\over 2\pi }\sup_{{\left\lvert {z-z_0} \right\rvert} = R} {\left\lvert {f(z)} \right\rvert} R^{-(n+1)} \cdot 2\pi R \\ &= \sup_{{\left\lvert {z-z_0} \right\rvert} = R} {\left\lvert {f(z)} \right\rvert} R^{-n} \\ &\leq (AR^k + B)R^{-n} \qquad \text{ if } z_0 = 0 \\ &= AR^{k-n} + BR^{-n} \\ &\to 0 ,\end{align*} provided $$k-n< 0$$, so $$n>k$$. Since $$f$$ is entire, write \begin{align*} f(z) = \sum_{n\geq 0} f^{(n)}(0) {z^n\over n!} = \sum_{0\leq n\leq k} f^{(n)}(0) {z^n\over n!} ,\end{align*} making $$f$$ a polynomial of degree at most $$k$$.

Write $$S_\phi \coloneqq\left\{{0<\operatorname{Arg}(z) < \phi}\right\}$$ and choose $$n$$ large enough so that \begin{align*} {\mathbb{D}}\subseteq S \cup\zeta_n S \cup\zeta_n^2 S \cup\cdots\cup\zeta_{n}^{n-1}S ,\end{align*} i.e. so that the rotated sectors cover the disc. By uniform convergence of $$f$$ to $$0$$ on $$S$$, choose $$r<1$$ small enough so that $${\left\lvert {f(z)} \right\rvert} < {\varepsilon}$$ for $${\left\lvert {z} \right\rvert} < r$$ in $$S$$. Note that $${\mathbb{D}}_r \subseteq \displaystyle\bigcup_{k=0}^{n-1} \zeta_n^k S_r$$, where $$S_r \coloneqq\left\{{z\in S {~\mathrel{\Big\vert}~}{\left\lvert {z} \right\rvert} \leq r}\right\}$$ is a subsector of radius $$r$$.

By the MMP, let $$M$$ be the maximum of $$f$$ on $${\mathbb{D}}$$, which is attained at some point on $$S^1$$. Then $${\left\lvert {f} \right\rvert} < M$$ on every $$\zeta_n^k S_r$$. Now define \begin{align*} g(z) \coloneqq f(z) \prod_{k=1}^{n-1} f(\zeta_n^k z) \coloneqq f(z) \prod_{k=1}^{n-1}f_k(z) .\end{align*} Note that $${\left\lvert {f(z)} \right\rvert}\leq {\varepsilon}$$ and $${\left\lvert {f_k(z)} \right\rvert} \leq M$$, so \begin{align*} {\left\lvert {g(z)} \right\rvert}\leq {\varepsilon}\cdot M^{n-1} \overset{{\varepsilon}\to 0}\longrightarrow 0 .\end{align*} since $$M$$ is a constant. So $$g(z) \equiv 0$$ on $${\mathbb{D}}_r$$, and by the identity principle, on $${\mathbb{D}}$$. Thus some factor $$f_k(z)$$ is identically zero. But if $$f(\zeta_n^k z)\equiv 0$$ on $${\mathbb{D}}$$, then $$f(z) \equiv 0$$ on $${\mathbb{D}}$$, since every $$z\in {\mathbb{D}}$$ can be written as $$\zeta_n^k w$$ for some $$w\in {\mathbb{D}}$$.

Consider \begin{align*} f(z) \coloneqq\prod_{1\leq k \leq n} (w_k - z) .\end{align*} Then $$f$$ is holomorphic and nonconstant on $${\mathbb{D}}$$, so attains a maximum $$M$$ on $$S^1$$. Moreover, $${\left\lvert {f(z)} \right\rvert} = \prod {\left\lvert {w_k-z} \right\rvert}$$ is exactly the product of distances from $$z$$ to the $$w_k$$. Moreover, since $${\left\lvert {f(0)} \right\rvert} = \prod{\left\lvert {w_k} \right\rvert} = 1$$, we must have $$M>1$$.

Now note that $$f(w_k) = 0$$ and $$f$$ is continuous in $${\mathbb{D}}$$. So $${\left\lvert {f(z)} \right\rvert} \in [0, M] \subseteq {\mathbb{R}}$$ where $$M>1$$, so by the intermediate value theorem, $${\left\lvert {f(z)} \right\rvert} = 1$$ for some $$z$$.

Write $$f=u+iv$$ where by assumption $$u$$ is bounded. Both $$u$$ and $$v$$ are harmonic, so if $${\left\lvert {u} \right\rvert} \leq M$$ on $${\mathbb{C}}$$, then there is some disc where $${\left\lvert {u} \right\rvert} = M$$ for some point in the interior. By the MMP for harmonic functions, $$u$$ is constant on $${\mathbb{C}}$$. So $$u_x, u_y = 0$$, and by Cauchy-Riemann, $$v_x, v_y = 0$$, so $$v'=0$$ and $$v$$ is constant, making $$f$$ constant.

Consider $$g(z) \coloneqq e^{f(z)}$$, then $${\left\lvert {g(z)} \right\rvert} = e^{\Re(z)}$$ is entire and bounded and thus constant by Liouville’s theorem. So $$g'(z) = 0$$, but on the other hand $$g'(z) = f'(z) e^{f(z)} = 0$$, so $$f'(z) = 0$$ and $$f$$ must be constant since $$e^f$$ is nonvanishing.

## Spring 2020 HW 3, 3.8.17 #complex/exercise/work

Let $$f$$ be non-constant and holomorphic in an open set containing the closed unit disc.

• Show that if $${\left\lvert {f(z)} \right\rvert} = 1$$ whenever $${\left\lvert {z} \right\rvert} = 1$$, then the image of $$f$$ contains the unit disc.

Hint: Show that $$f(z) = w_0$$ has a root for every $$w_0 \in {\mathbb{D}}$$, for which it suffices to show that $$f(z) = 0$$ has a root. Conclude using the maximum modulus principle.

• If $${\left\lvert {f(z)} \right\rvert} \geq 1$$ whenever $${\left\lvert {z} \right\rvert} = 1$$ and there exists a $$z_0\in {\mathbb{D}}$$ such that $${\left\lvert {f(z_0)} \right\rvert} < 1$$, then the image of $$f$$ contains the unit disc.

## Spring 2020 HW 3, 3.8.19 #complex/exercise/work

Prove that maximum principle for harmonic functions, i.e.

• If $$u$$ is a non-constant real-valued harmonic function in a region $$\Omega$$, then $$u$$ can not attain a maximum or a minimum in $$\Omega$$.

• Suppose $$\Omega$$ is a region with compact closure $$\mkern 1.5mu\overline{\mkern-1.5mu\Omega\mkern-1.5mu}\mkern 1.5mu$$. If $$u$$ is harmonic in $$\Omega$$ and continuous in $$\mkern 1.5mu\overline{\mkern-1.5mu\Omega\mkern-1.5mu}\mkern 1.5mu$$, then \begin{align*} \sup _{z \in \Omega}|u(z)| \leq \sup _{z \in \mkern 1.5mu\overline{\mkern-1.5mu\Omega \mkern-1.5mu}\mkern 1.5mu-\Omega}|u(z)| .\end{align*}

Hint: to prove (a), assume $$u$$ attains a local maximum at $$z_0$$. Let $$f$$ be holomorphic near $$z_0$$ with $$\Re(f) = u$$, and show that $$f$$ is not an open map. Then (a) implies (b).

## Spring 2020 HW 3.9 #complex/exercise/work

Let $$f$$ be analytic in a region $$D$$ and $$\gamma$$ a rectifiable curve in $$D$$ with interior in $$D$$.

Prove that if $$f(z)$$ is real for all $$z\in \gamma$$, then $$f$$ is constant.

## Spring 2020 HW 3.14 #complex/exercise/work

Let $$f$$ be nonzero, analytic on a bounded region $$\Omega$$ and continuous on its closure $$\overline \Omega$$.

Show that if $${\left\lvert {f(z)} \right\rvert} \equiv M$$ is constant for $$z\in \partial \Omega$$, then $$f(z) \equiv Me^{i\theta}$$ for some real constant $$\theta$$.

## Tie’s Extra Questions: Spring 2015 #complex/exercise/work

Let $$\displaystyle{\psi_{\alpha}(z)=\frac{\alpha-z}{1-\mkern 1.5mu\overline{\mkern-1.5mu\alpha\mkern-1.5mu}\mkern 1.5muz}}$$ with $$|\alpha|<1$$ and $${\mathbb D}=\{z:\ |z|<1\}$$. Prove that

• $$\displaystyle{\frac{1}{\pi}\iint_{{\mathbb D}} |\psi'_{\alpha}|^2 dx dy =1}$$.

• $$\displaystyle{\frac{1}{\pi}\iint_{{\mathbb D}} |\psi'_{\alpha}| dx dy =\frac{1-|\alpha|^2}{|\alpha|^2} \log \frac{1}{1-|\alpha|^2}}$$.

## Tie’s Extra Questions: Spring 2015 #complex/exercise/work

Let $$\Omega$$ be a simply connected open set and let $$\gamma$$ be a simple closed contour in $$\Omega$$ and enclosing a bounded region $$U$$ anticlockwise. Let $$f: \ \Omega \to {\mathbb C}$$ be a holomorphic function and $$|f(z)|\leq M$$ for all $$z\in \gamma$$. Prove that $$|f(z)|\leq M$$ for all $$z\in U$$.

## Tie’s Extra Questions: Fall 2015 #complex/exercise/work

Assume $$f(z)$$ is analytic in region $$D$$ and $$\Gamma$$ is a rectifiable curve in $$D$$ with interior in $$D$$. Prove that if $$f(z)$$ is real for all $$z \in \Gamma$$, then $$f(z)$$ is a constant.