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The open mapping theorem
Theorem 1 (On the local structure of a regular map) Let \(f\) be regular in \(D\), \(z_0 \in D\), \(f(z_0) = w_0\). Let \[
f'(z_0) = \ldots = f^{(n-1)}(z_0) = 0, \quad f^{(n)}(z_0) \neq 0
\] Then there exist neighborhoods \(U(z_0)\), \(V(w_0)\) such that for every \(w \in \dot{V}(w_0)\) the equation \(f(z) = w\) has exactly \(n\) distinct solutions.
Indeed, we have \(f(z) - f(z_0) = c_{n}(z - z_0)^n + o((z - z_0)^n) \sim c_{n}(z-z_0)^n\) as \(z \to z_0\). The meaning of the theorem: a holomorphic function behaves, in a neighborhood of each of its points, like raising to a power. Raising to the power \(n\) wraps the circle \(U(z_0)\) around \(n\) times.
Criterion for local univalence (injectivity): the first derivative is nonzero.
Corollary 1 Let \(D\) be a domain, \(f\) regular in \(D\), \(f\) not constant. Then \(f(D)\) is a domain.
Connectedness of \(f(D)\) follows from continuity of \(f\). Let us prove openness. By contradiction: suppose \(z_0 \in D\) is such that \(f(z_0) \not\in \operatorname {int}f(D)\). By the theorem above there is a neighborhood \(V(f(z_0))\) such that every point in it has at least 1 preimage. Contradiction.
An old problem: let \(D\) be a domain, let \(f\) be regular in \(D\), \(\left|f(z)\right| \equiv \operatorname {const}\). Prove that \(f \equiv \operatorname {const}\).
Suppose \(f\) is not constant. By the theorem above, \(f(D)\) is a domain. But \(f(D)\) is a subset of a circle (since \(\left|f\right| \equiv \operatorname {const}\)). No subset of a circle can be open. Contradiction.
The maximum modulus principle
Theorem 2 (The maximum principle, 1st formulation) A function regular in a domain cannot attain the supremum of its modulus. Namely, let \(f\) be regular in the domain \(D\), \(f \not\equiv \operatorname {const}\). Then for every point \(z_0 \in D\)
\[
\left|f(z_0)\right| < \sup _{z \in D} \left|f(z)\right|
\]
Theorem 3 (The maximum principle, 2nd formulation) Let \(f\) be continuous on \(\overline{D}\) and regular in \(D\) (\(D\) a bounded domain). Then
\[
\max _{z \in \overline{D}}\left|f(z)\right| = \max _{z\in \partial D}\left|f(z)\right|
\]
Figure in development. Will be recreated as an interactive figure (Plotly/Three.js).
Plan: TikZ: the maximum modulus principle.
Source: courses/nedosh_courses/courses/01-analysis/CA/contents/specials/310/010_max_modulus_principle_fig.tex
Similarly one can prove the maximum (minimum) principle for \(\operatorname {Re}f\) (\(\operatorname {Im}f\)):
\(f\) regular in a bounded domain \(D\), \(f\) not constant. Then for every \(z_0 \in D\) we have
\[
\operatorname {Re}f(z_0) < \sup _{z \in D}\operatorname {Re}f(z)
\]
similarly for the infimum: \(\operatorname {Re}f(z_0) > \inf_{z \in D}\operatorname {Re}f(z)\) (unlike for the modulus, here no conditions such as \(f \neq 0\) are needed: it suffices to apply the maximum principle to \(-f\));
the same two statements for \(\operatorname {Im}f\) (this is \(\operatorname {Re}(-if)\));
in the 2nd formulation: if \(f\) is continuous on \(\overline{D}\), then \(\max_{\overline{D}}\operatorname {Re}f = \max_{\partial D}\operatorname {Re}f\), and similarly for the minimum and for \(\operatorname {Im}f\).
What about a minimum modulus principle? For example, \(f(z) = z\), \(\left|z\right| < 1\). The minimum is attained: at zero.
Theorem 4 (The minimum modulus principle) \(f\) regular in a bounded domain \(D\), \(f \not\equiv \operatorname {const}\), \(f(z) \neq 0\) on \(D\). Then for every \(z_0 \in D\) \[
\inf _{z \in D}\left|f(z)\right| < \left|f(z_0)\right|
\]
Figure in development. Will be recreated as an interactive figure (Plotly/Three.js).
Plan: TikZ: the minimum modulus principle.
Source: courses/nedosh_courses/courses/01-analysis/CA/contents/specials/310/020_min_modulus_principle_fig.tex
Let \[
P(z) = z^{n} + c_{n-1}z^{n-1} + \ldots + c_1z + c_0, \quad P(z) \not\equiv z^n
\] Prove that there exists a point \(z_0 \in \mathbb {C}\) such that \(\left|z_0\right| = 1\), \(\left|P(z_0)\right| > 1\).
Let
\[
f(\zeta ) := \zeta ^{n}P(1/\zeta ) = 1 + c_{n-1}\zeta + \ldots + c_{1}\zeta ^{n-1} + c_0\zeta ^{n}
\]
\(f(0) = 1\), \(f\) is holomorphic in \(\left|\zeta \right| < 1\), continuous in \(\left|\zeta \right| \leq 1\). The maximum is not attained in the interior, hence
\[
1 < \max _{\left|\zeta \right| \leq 1}\left|f(\zeta )\right| = \max _{\left|\zeta \right| = 1}\left|f(\zeta )\right| = \left|f(\zeta _0)\right|
\]
for some point \(\zeta_0 \in \left\{ \left|\zeta \right| = 1\right\}\). Let \(z_0 = \frac{1}{\zeta_0}\). Then
\[
\left|P(z_0)\right| = \left|z_0^n\right| \cdot \left|f\left(\frac{1}{z_0}\right)\right| > 1
\]
Where did we use \(P(z) \not\equiv z^n\)? To be able to say that \(f(\zeta ) \not\equiv \operatorname {const}\).
Let \(D\) be a bounded domain, \(f\) regular in \(D\), continuous on \(\overline{D}\), \(\left|f(z)\right| \equiv \operatorname {const}\) on \(\partial D\), \(f(z) \not\equiv \operatorname {const}\) on \(D\). Prove that there exists \(z_0 \in D\) with \(f(z_0) = 0\).
By contradiction: if \(f(z) \neq 0\) in \(D\), then for every \(z_0 \in D\) we have, by the maximum principle and the minimum principle (since the function does not vanish), \[
\min _{z \in \partial D}\left|f(z)\right| =\min _{z \in \overline{D}}\left|f(z)\right| \leq \left|f(z_0)\right| \leq \max _{z \in \overline{D}}\left|f(z)\right| = \max _{z \in \partial D}\left|f(z)\right|
\] The outer quantities are equal, so all the inequalities become equalities. Hence \(\left|f(z)\right| \equiv \operatorname {const}\) in \(D\), and hence also \(f \equiv \operatorname {const}\) on \(D\). Contradiction.
Given a square \(ABCD\) with side \(l\). Find the point \(P\) (in the closed square) at which the product of the distances to the vertices of the square attains its maximum. Find the value of this maximum.
Let \(z_A = 0\), \(z_{B} = il\), \(z_{C} = l + il\), \(z_{D} = l\). Let \(D = \left\{ 0 < \operatorname {Re}z, \operatorname {Im}z < l\right\}\) (the interior of the square). Consider the function \(f(z) = (z-z_A) \cdot \ldots \cdot (z-z_D)\). This is a polynomial, so \(f\) is holomorphic in the domain \(D\) and continuous up to the boundary. By the maximum principle, \[
\max _{z \in \overline{D}}\left|f(z)\right| = \max _{z \in \partial D}\left|f(z)\right|
\] \(\partial D\) consists of 4 segments. By the symmetry of the square, it does not matter which segment we consider. Let us consider the one from \(z_A\) to \(z_D\): \(z = z(t) = (1-t)z_A + tz_D = tl\), \(t \in [0,1]\). Then \[
\begin{align} \left|f(z(t))\right| & = \left|tl - 0\right|\cdot \left|tl - il\right|\cdot \left|tl - l - il\right| \cdot \left|tl - l\right| = \left|l\right|^4 \cdot \left|t\right| \cdot \sqrt{t^2 + 1} \cdot \sqrt{(t-1)^2 + 1} \cdot \sqrt{(t-1)^2} = \\ & = \left|l\right|^4 \cdot \sqrt{t^2(1-t)^2(t^2 + 1)((t-1)^2 + 1)} \end{align}
\] It is not hard to compute (an exercise in real analysis) that the function under the square root attains its maximum on \([0,1]\) at \(t = \frac{1}{2}\), with value there equal to \(\frac{1}{4} \cdot \frac{1}{4} \cdot \frac{5}{4} \cdot \frac{5}{4} = \frac{5^2}{4^4}\). In total the maximum of the product equals \[
l^4\sqrt{\frac{5^2}{4^4}} = \frac{5}{16}\, l^4,
\] attained at the midpoint of the side (\(t = \frac{1}{2}\)).
\(P\) is the midpoint of an arbitrary side, and the maximum equals \(\frac{5}{16}\, l^4\).
Schwarzβs lemma
Lemma 1 Let \(f\) be regular in \(D = \left\{ \left|z\right| < 1\right\}\), \(f(0) = 0\), \(\left|f(z)\right| \leq 1\), \(z \in D\). Then \[
\left|f(z)\right| \leq \left|z\right|, \; \forall z \in D, \qquad \left|f'(0)\right| \leq 1
\] moreover, if \(\left|f(z_0)\right| = \left|z_0\right|\) at even one point \(z_0 \neq 0\), or \(\left|f'(0)\right| = 1\), then \(f(z) = e^{i\alpha }z\), \(\alpha \in \mathbb {R}\).
Consider \(g(z) = \frac{1}{z}f(z)\) (at zero we set it, by continuity, to \(f'(0)\) β the singularity is removable since \(f(0) = 0\)). It will be regular in the unit disk. Fix \(r < 1\): on the circle \(\left|z\right| = r\) we have \(\left|g(z)\right| \leq \frac{1}{r}\), hence, by the maximum principle (2nd formulation, for the disk \(\left|z\right| \leq r\)), \(\left|g(z)\right| \leq \frac{1}{r}\) throughout the disk \(\left|z\right| \leq r\) as well. Letting \(r \to 1-\), we obtain \(\left|g(z)\right| \leq 1\) everywhere, i.e. \(\left|f(z)\right| \leq \left|z\right|\) and \(\left|f'(0)\right| = \left|g(0)\right| \leq 1\).
Either equality case means that \(\left|g\right|\) attains the value \(1\) (its maximum) at an interior point of the disk. By the maximum principle, \(g \equiv \operatorname {const}\), and the constant has modulus \(1\): \(g \equiv e^{i\alpha }\), i.e. \(f(z) = e^{i\alpha }z\).
Does there exist a regular \(f: B_{1}(0) \to B_{1}(0)\) such that \(f(0) = \frac{1}{2}\), \(f\left(\frac{1}{2}\right) = \frac{7}{8}\)?
First attempt: consider \(g(z) = \frac{f(z) - 1/2}{3/2}\). \(g(0) = 0\), \(\left|g(z)\right| < 1\). We have
\[
g\left(\frac{1}{2}\right) = \frac{f(1/2) - 1/2}{3/2} = \frac{7/8 - 1/2}{3/2} = \frac{1}{4}
\]
No contradiction with Schwarzβs lemma: \(\frac{1}{4} \leq \frac{1}{2}\). Normalizing by a shift is βtoo crudeβ β let us try differently.
The general form of a fractional-linear (MΓΆbius) automorphism \(\varphi : B_{1}(0) \to B_{1}(0)\):
\[
\varphi (z) = \frac{z - z_0}{1 - \overline{z_0}z}e^{i \alpha }
\]
In our case let \(\varphi (z) = \frac{z - \frac{1}{2}}{1 - \frac{1}{2}z} = \frac{2z - 1}{2-z}\). Let \(g(z) = \varphi (f(z))\). Then \(g: B_{1}(0) \to B_{1}(0)\), \(g(0) = 0\),
\[
g\left(\frac{1}{2}\right) = \varphi \left(\frac{7}{8}\right) = \frac{2 \cdot \frac{7}{8} - 1}{2 - \frac{7}{8}} = \frac{3/4}{9/8} = \frac{2}{3}
\]
The modulus is greater than \(\frac{1}{2}\) β but Schwarzβs lemma for \(g\) requires \(\left|g\left(\frac{1}{2}\right)\right| \leq \frac{1}{2}\). Contradiction: no such function \(f\) exists.
No: composing with a fractional-linear automorphism of the disk violates Schwarzβs lemma.
Let \(P(z)\) be a polynomial of degree \(n\) and \(M(r) = \max_{|z| = r}|P(z)|\). Prove that for \(0 < r_{1} < r_{2}\) the inequality \(\frac{M\left(r_{1}\right)}{r_{1}^{n}} \geq \frac{M\left(r_{2}\right)}{r_{2}^{n}}\) holds.
Let the function \(f(z)\) be holomorphic in the unit disk \(\mathbb {D}\) with \(|f(z)| < 1\) for \(z \in \mathbb {D}, f(0) = a \neq 0\). Prove that \(f(z) \neq 0\) for \(|z| < |a|\).
Let \(a_{1}, \ldots , a_{n}\) be the vertices of a regular \(n\)-gon inscribed in the unit disk \(\mathbb {D}\). Prove that \[
\sup _{z \in \mathbb {D}} \prod _{k = 1}^{n}\left|z-a_{k}\right| = 2
\]
Let \(P\) be a polynomial with leading coefficient \(1\). Prove that on the unit circle there is a point \(z_{0}\) at which \(\left|P\left(z_{0}\right)\right| \geq 1\).
Let the functions \(f\) and \(g\) be holomorphic in the unit disk \(\mathbb {D}\) and continuous on the closure \(\overline{\mathbb {D}}\). Prove that \[
\max _{z \in \overline{\mathbb {D}}}(|f(z)|+|g(z)|) = \max _{|z| = 1}(|f(z)|+|g(z)|)
\]