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.

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