Inverse Function, Maximum Principle, and Schwarz's Lemma

The problem is not the problem. The problem is your attitude about the problem, Captain Jack Sparrow

Introduction

Definition. Complex sequence A sequence of complex numbers is a function $a: \mathbb{N} \to \mathbb{C}$. We usually denote it by $(a_n)_{n \in \mathbb{N}}$ or simply $(a_n)$, where $a_n := a(n)$. The value $a_1$ is called the first term of the sequence, $a_2$ the second term, and in general $a_n$ the n-th term of the sequence.

Definition. Convergent complex sequence. A complex sequence $(a_n)_{n \in \mathbb{N}}$ is said to converge to a complex number $L \in \mathbb{C}$ if for every $\varepsilon > 0$ there exists an integer $N \in \mathbb{N}$ such that for all $n \geq N$ one has $|a_n - L| < \varepsilon$. In this case we write $\lim_{n \to \infty} a_n = L$ or $a_n \to L$ as $n \to \infty$, and L is called the limit of the sequence $(a_n)_{n \in \mathbb{N}}$.

Definition. Cauchy sequence. A complex sequence $(a_n)_{n \in \mathbb{N}}$ is called a Cauchy sequence if for every $\varepsilon > 0$ there exists an integer $N \in \mathbb{N}$ such that for all $n, m \geq N$ one has $|a_n - a_m| < \varepsilon$.

Definition. Series and partial sums.Let $(a_n)_{n \in \mathbb{N}}$ be a complex sequence. For each n $\in \mathbb{N}$, the finite sum $s_n := a_1 + a_2 + \cdots + a_n = \sum_{k=1}^n a_k$ is called the n-th partial sum of the (infinite) series $\sum_{k=1}^\infin a_k$ which we also denote simply by $\sum a_n$ when the index is clear from the context.

Definition. Convergent series. The series $\sum_{n=1}^{\infty} a_n$ is said to converge to the sum $s \in \mathbb{C}$ if the sequence of partial sums $(s_n)_{n \in \mathbb{N}}$ defined by $s_n = a_1 + a_2 + \cdots + a_n = \sum_{k=1}^n a_k$ converges to s, that is, $\lim_{n \to \infty} s_n = s$. In this case we write $s := \sum_{n=1}^\infin a_n$. If the sequence $(s_n)_{n \in \mathbb{N}}$ does not converge, we say that the series $\sum_{n=1}^{\infty} a_n$ diverges (or does not converge).

Definition. A complex power series centered at 0 in the variable z is a series of the form $a_0 + a_1z + a_2z^2 + \cdots = \sum_{n=0}^\infty a_n z^n$ with coefficients $a_i \in \mathbb{C}$

Definition. A complex power series centered at a complex number $a \in \mathbb{C} $ is an infinite series of the form: $\sum_{n=0}^\infty a_n (z - a)^n,$ where each $a_n \in \mathbb{C}$ is a coefficient, z is a complex variable, and $(z - a)^n$ is the nth power about the center.

Theorem. Given a power series $\sum_{n=0}^\infty a_n z^n$, there exists a unique value R, $0 \le R \le \infin$ (called the radius of convergence) such that:

1. For any z with |z| < R (inside the circle), the series $\sum_{n=0}^\infty a_n z^n$ converges absolutely (this is a “green light” zone).
2. For any z with |z| > R, the series diverges (this is a “red light” zone).

   On the Circle (|z| = R), this theorem gives no information. This is the yellow light zone —the series could converge or diverge.

Differentiability of Power Series. If $f(z) = \sum_{n=0}^{\infty} a_nz^n$ for |z| < R (R > 0), then f is analytic on B(0; R) and $f'(z) = \sum_{n=1}^{\infty} na_nz^{n-1}$ for |z| < R.

Weierstrass M-test. Let $\{u_k(z)\}_{k=0}^\infty$ be a sequence of complex-valued functions defined on a set $\gamma^* \subseteq \mathbb{C}$. If there exists a sequence of non-negative real numbers $\{M_k\}_{k=0}^\infty$ such that:

1. Bounding Condition: $|u_k(z)| \leq M_k$ for all $z \in \gamma^*$ and all $k \in \mathbb{N}$
2. Convergence of Bound: The series $\sum_{k=0}^\infty M_k$ converges

Then, the original series $\sum_{k=0}^\infty u_k(z)$ converges uniformly on $\gamma^*$.

Coefficients of power series. Let f(z) = $\sum_{k=0}^\infty c_kz^k$ where this power series has radius of convergence R > 0. Then,the n-th coefficient of a power series $c_n$ can be extracted using the integral formula, $c_n = \frac{1}{2\pi i} \int_{C_r} \frac{f(z)}{z^{n+1}}dz, 0 \le r \lt R, n \ge 0$ where $C_r$ is a circle of radius r centered at 0 and oriented positively.

Taylor’s Theorem. If f is analytic on an open disk B(a; R) (a disk of radius R centered at a), then f(z) can be represented exactly by a unique power series within that disk: $f(z) = \sum_{n=0}^{\infty}a_n (z - a)^n, \forall z \in B(a; R)$

This theorem bridges the gap between differentiability and power series. It guarantees that if a function behaves well (it is analytic) in a disk, it must also be infinitely differentiable and expressed or representable by a power series (an infinite polynomial) within that disk.

Furthermore, there exist unique constants $a_n = \frac{f^{(n)}(a)}{n!} = \frac{1}{2\pi i}\int_{C_r} \frac{f(w)}{(w-a)^{n+1}}dw$ where $C_r$ is a circle of radius r < R centered at a and oriented in the counterclockwise direction (positively oriented).

Mapping Theorems, the Maximum Principle, and Schwarz’s Lemma

Analytic functions map regions of the complex plane in highly structured, predictable ways. The foundational result describing this behavior is the Local Mapping Theorem.

The Local Mapping Theorem. Suppose $f$ is analytic at $z_0$ with $f(z_0) = w_0$, and that the shifted function $f(z)-w_0$ has a zero of order $n \ge 1$ at $z_0$. Then, for all sufficiently small $\varepsilon > 0$, there exists a corresponding $\delta > 0$ such that for every target value $a$ strictly inside the punctured disk $0 < |a - w_0| < \delta$, the equation $f(z) = a$ has exactly $n$ distinct roots inside the domain disk $|z - z_0| < \varepsilon$.

Every point $a$ near the center (except the center itself $w_0$) comes from exactly $n$ points in the domain. Essentially, it shows that an analytic function behaves like a polynomial $w = z^n$ near a point, mapping a small neighborhood to another small neighborhood in an n-to-1 fashion. 

A direct and profound consequence of this local behavior is a global topological property:

Corollary. Open mapping theorem. A non-constant analytic function maps open sets to open sets.

Proof.

Let $f$ be a non-constant analytic function on an open set $G$. We must prove the image set $f(G)$ is strictly open. To prove a set is open, we must show that every point in the set is surrounded by a “safety bubble” (a small disk) that is entirely contained within the set.

Pick any $z_0 \in G$, and let $w_0 = f(z_0)$. Because $f$ is non-constant, the function $f(z) - w_0$ does not vanish entirely; it must have a zero of some finite order $n \ge 1$ at $z_0$.

By the Local Mapping Theorem, there exists an $\varepsilon$-disk around $z_0$ and a corresponding $\delta$-disk around $w_0$ such that every single value $a$ in the target disk $B(w_0; \delta)$ is “hit” by the function exactly $n$ times by points inside $B(z_0; \varepsilon)$.

Because every point $a \in B(w_0; \delta)$ has a pre-image in $G$, the entire disk $B(w_0; \delta)$ is part of the image set $f(G)$.

$$ B(w_0; \delta) \subseteq f(B(z_0; \varepsilon)) \subseteq f(G) $$

Since every point $w_0 \in f(G)$ has such a safety bubble, $f(G)$ is an open set. $\blacksquare$

Remark. If the zero of $f(z)-w_0$ is simple at $z_0$ (i.e., n = 1), then then near $z_0$, every nearby value is attained exactly once, so f is a local bijection. There is a one-to-one and onto correspondence between an open neighborhood U of $z_0$ and an open neighborhood V of $w_0$. This means an inverse function $f^{-1}$ exists on that neighborhood.

Definition. A conformal map is a function $f: U\subseteq \mathbb{C}\rightarrow \mathbb{C}$ that preserves angles between smooth curves at each point of its domain, including orientation. More precisely, if two smooth curves intersect at a point $z_0$, then their images under f intersect at $f(z_0)$with the same angle and the same orientation.

Injective analytic functions are conformal

Proposition. If f is analytic (holomorphic) on a domain U, then at any point $z_0$ where $f'(z_0)\neq 0$, the function is conformal at $z_0$. So nonzero derivative means: (i) local linear behavior; (ii) local invertibility; (iii) and angle preservation.

Proposition. Let G be an open set and let f be analytic and one-to-one in G. Then, f is conformal in G.

One-to-One (Injective): $f(z_1) = f(z_2) \implies z_1 = z_2$. The function never maps two different points to the same spot. In other words, it never “overlaps” itself.

Proof.

Claim: f is conformal in G or, alternatively, $f'(z) \neq 0$ for all $z \in G$. We will assume the derivative is zero at some point and show that this forces the function to overlap itself (proof by contradiction).

Suppose there exists a point $a \in G$ where the derivative vanishes: $f'(a) = 0$.

Let $w_0 = f(a)$. Consider the Taylor expansion of $f(z) - w_0$ around a, $f(z) - w_0 = c_1(z-a) + c_2(z-a)^2 + \dots$ where $c_1 = f'(a) =[\text{By our previous assumption}] 0$. Therefore, the first non-zero term must be at least the second power $c_2$ or higher. This means the function $f(z) - w_0$ has a zero of order $n \ge 2$.

We established in the Local Mapping Theorem that if the order is n, then there is a $\varepsilon$-neighborhood around “a” and a corresponding $\delta$-neighborhood around “f(a) = $w_0$” such that for every target value $w \in \mathbb{B}(w_0; \delta), w \ne w_0$, the equation f(z) = w has exactly n distinct roots in $\mathbb{B}(a; \varepsilon)$ (f(w) is “hit” n times).

Since $n \ge 2$, the equation $f(z) = w$ has at least 2 distinct roots, say $z_1$ and $z_2, z_1 \neq z_2$ where $f(z_1) = w \text{ and } f(z_2) = w$. This means f maps two different points to the same location. Therefore, f is not one-to-one (not injective). This contradicts the theorem’s hypothesis that f is one-to-one on G, therefore our initial assumption that f’(a) = 0 must be false, $f'(z) \neq 0, \quad \forall z \in G$

Recall: Proposition. If f is analytic (holomorphic) on a domain U, then at any point $z_0$ where $f'(z_0)\neq 0$, the function is conformal.

Since the derivative is never zero, the mapping is conformal everywhere.

Inverse Function, Maximum Principle, and Schwarz’s Lemma

Inverse Function Theorem. Let f be analytic and one-to-one (injective) on an open set G. Then, the inverse image $f^{-1}$ is analytic on the image set f(G), and its derivative is given by $(f^{-1})'(b) = \frac{1}{f'(f^{-1}(b))}, \forall b \in f(G), b = f(a).$

Proof.

👣 Does $f^{-1}$ exist? Since f is one-to-one on G, it is indeed a bijection from G onto its image f(G). Therefore, the inverse function is well-defined: $f^{-1}: f(G) \to G$.

Recall. Open mapping theorem. A non-constant analytic function maps open sets to open sets. Therefore, by the Open Mapping Theorem, f(G) is open.

👣 is $f^{-1}$ is continuous? We will prove that the inverse image of an arbitrary open set V in G is open in f(G). In other words, the pre-image of V under $f^{-1}$ is open in f(G).

Recall. Topological Definition of Continuity. A function $h: X \to Y$ is continuous if and only if for every open set V in the target space Y, the pre-image $h^{-1}(V)$ is an open set in the source space X.

Continuous functions pull open sets back to open sets.

Let V be an arbitrary open set in G.

We know f is a non-constant analytic function. The Open Mapping Theorem tells us that f maps open sets to open sets. Since V is open, then f(V) is open. Since f is one-to-one correspondence from V to f(V), then the “inverse of the inverse” is just the original function, $f(V) = (f^{-1})^{-1}(V)$. So $(f^{-1})^{-1}(V)$ is open in f(G). Conclusion: the pre-image of V under $f^{-1}$ is open in f(G).

👣 Is $f^{-1}$ differentiable? (Analyticity)

Let $a \in G, f(a) = b$, then $f^{-1}(b) = a$. By previous preposition, since f is one-to-one, then f is conformal, $f'(f^{-1}(b)) \ne 0$

$f^{-1}$ is continuous, then $\lim_{z \to b} f^{-1}(z) = f^{-1}(b)$. We want to compute the derivative of the inverse function $f^{-1}(z)$ at b:

$$ \begin{aligned} \lim_{z \to b} \frac{f^{-1}(z) - f^{-1}(b)}{z-b} &=\lim_{z \to b} \frac{f^{-1}(z) - f^{-1}(b)}{f(f^{-1}(z))-f(f^{-1}(b))} \\[2pt] &\text{f is one-to-one and } f^{-1} \text{ is continuous} \\[2pt] &=\lim_{f^{-1}(z) \to f^{-1}(b)} \frac{1}{\frac{f(f^{-1}(z))-f(f^{-1}(b))}{f^{-1}(z) - f^{-1}(b)}} \\[2pt] &\text{Since f is analytic and conformal, }f'(f^{-1}(b)) \ne 0 \\[2pt] &=\frac{1}{f'(f^{-1}(b))} \end{aligned} $$

The limit (derivative) exists, $f^{-1}$ is analytic at b (i.e. f(a)) and since f is onto f(G), every point in f(G) is $b = f(a)$ for some a, so $f^{-1}$ is analytic in f(G) and $(f^{-1})'(b) = \frac{1}{f'(f^{-1}(b))}$.

The maximum principle. Let f be a function that is analytic and non-constant on a region (an open connected set) G. Then, the absolute value |f(z)| cannot have a maximum at any point inside G.

Imagine the graph of the absolute value $|f(z)|$ as a landscape or a tent sheet. The maximum principle states that if the function is “well behaved” and “smooth” (analytic) and not flat (constant), you will never find a mountain peak strictly inside the region. The true maximum height can only be found at the edges (the boundary) of the region.

Proof by Contradiction.

Assume there exists a point $z_0$ (a maximum) strictly inside the region G such that $|f(z_0)|$ is the maximum value, $|f(z_0)| \ge |f(z)|, \forall z \text{ near } z_0$

Let $w_0 = f(z_0)$ be the value at this peak. Since G is a region, it is an open set, so there is some “elbow room” (safety bubble) around $z_0$. We can find a small radius r > 0 such that the entire ball (disk) fits inside G. Formally, $\exist r \gt 0, \mathbb{B}(z_0; r) \subseteq G$.

Now we look at the image of this small ball under the function f. We invoke the Open Mapping Theorem. Non-constant analytic functions map open sets to open sets. Since $\mathbb{B}(z_0; r)$ is an open set, its image $f(\mathbb{B}(z_0; r))$ must also be an open set.

Because the image is open and contains $w_0$, there must be a “elbow room” or safety bubble (a small disk of radius $\delta$) around $w_0$ that fits entirely inside the image, $\mathbb{B}(w_0; \delta) = \mathbb{B}(f(z_0); \delta) \subseteq f(\mathbb{B}(z_0; r))$.

💡 The function f takes the neighborhood of $z_0$ and spreads it out to cover a neighborhood of $w_0$ in all directions.

In such a ball, $|f(z_0)|$ cannot be the maximum value of |f(z)|, you can always move slightly away from the origin in the direction of $w_0$ to find a point with a larger modulus, e.g., $w_{new} = w_0 + \frac{\delta}{2} \frac{w_0}{|w_0|}$ (moving radially outward).

1. $w_{new} \in \mathbb{B}(w_0; \delta)$. This point is inside the disk $\mathbb{B}(w_0; \delta) \subseteq f(\mathbb{B}(z_0; r))$, hence inside the image set.
2. Therefore, there exists some $z_{new}$ in our original domain such that $f(z_{new}) = w_{new}$.
3. Its magnitude is $|w_{\mathrm{new}}|=\left| w_0+\frac{\delta }{2}\frac{w_0}{|w_0|}\right| =|w_0|+\frac{\delta }{2}>|w_0|.$
4. This contradicts our initial assumption that $|f(z_0)|$ was the maximum. Therefore, no maximum can exist inside the region.

Schwarz’s Lemma. Suppose f is analytic on $\mathbb{B}(0; r)$, f(0) = 0 and $|f(z)| \le M, \forall z \in \overline{\mathbb{B}(0; r)}$, then $|f(z)| \le \frac{M}{r}|z|, \forall |z| \le r.$ If equality holds for some point inside the disk $z_0$ with $0 < |z_0| < r$, then there is a real constant m such that $f(z) = \frac{Mze^{im}}{r}$

If an analytic function f on a disk of radius r is pinned at the origin (f(0) = 0) and bounded in modulus by a ceiling $|f(z)| \le M$, then |f(z)| is squeezed by a straight line. If the function ever touches this upper ceiling boundary inside the disk, its “analytic flexibility” breaks, and it is instantly forced to be a perfectly straight, rigid line everywhere.

Proof

We cannot apply the Maximum Modulus Principle to f(z) directly to get the result we want. We need to “peel off” or “remove” the zero at the origin.

Since f is analytic near 0, it has a Taylor series expansion: $f(z) = a_0 + a_1z + a_2z^2 + a_3z^3 + \dots$ with radius of convergence at least r. Because we are given $f(0) = 0$, the constant term $a_0$ must be 0. $f(z) = 0 + a_1z + a_2z^2 + a_3z^3 + \dots =[\text{We can factor out a z from every term:}] z \cdot \underbrace{(a_1 + a_2z + a_3z^2 + \dots)}_{g(z)}$

Motivated by this, let’s define, $g(z) = \begin{cases} \frac{f(z)}{z} & z \neq 0 \\ f'(0) & z = 0 \end{cases}$. Notice that

1. $z \ne 0, \frac{f(z)}{z} = \sum_{n=1}^{\infty} a_n z^{n-1} = \sum_{n=0}^{\infty} a_{n+1} z^n$. Even though we divided by z, g(z) is defined by a power series in z with the same radius of convergence r.

   A power series converges inside its radius of convergence. Since the original series for f(z) converges for |z| < r, this new series for g(z) also converges for $|z| < r$. Any convergent power series defines an analytic function.

2. For $z \ne 0, \frac{f(z)}{z}$ is analytic. Since f is differentiable at 0 and f(0) = 0, $\lim_{z\to 0}\frac{f(z)}{z} = f'(0)$ exists. So z = 0 is a removable singularity of the function $z\mapsto f(z)/z$. Defining g(0) = f’(0) removes the singularity and gives an analytic extension. Thus, g is analytic on $\mathbb{B}(0;r)$.

We apply the Maximum Modulus Principle to g(z) on a slightly smaller disk of radius $r_1$ (where $r_1 < r$). It states that the maximum of $|g(z)|$ occurs on the boundary $|z| = r_1$.

Max Modulus Principle, weaker version. If a function f(z) is analytic and not constant in a connected open set 𝐺, then the modulus (absolute value) of the function, |𝑓(𝑧)|, has no local or global maximum point within 𝐺. Max Modulus Principle, strong version. If a function 𝑓(𝑧) is continuous on a closed and bounded region 𝑅 and is analytic (non-constant) in its interior, then the maximum value of |𝑓(𝑧)| in 𝑅 must occur on the boundary of 𝑅.

For any point z on this boundary circle ($|z| = r_1$): $|g(z)| = \left| \frac{f(z)}{z} \right| = \frac{|f(z)|}{|z|} \le[\text{|f(z)| ≤ M from Schwarz’s Lemma statement}] \le \frac{M}{r_1}$.

Since the maximum on the boundary is $\frac{M}{r_1}$, the Maximum Modulus Principle guarantees that for all z inside this smaller disk: $|g(z)| \le \frac{M}{r_1}$

We have the bound $|g(z)| \le \frac{M}{r_1}, \forall z \in \overline{\mathbb{B}(0; r_1)}$ for any radius $r_1 < r$. We can choose $r_1$ to be as close to $r$ as we want. Therefore, on the limit, as we push $r_1 \to r^-$, the value $\frac{M}{r_1}$ approaches $\frac{M}{r}$. Then, for any fixed z inside the disk: $|g(z)| \le \frac{M}{r}, \forall z \in \overline{\mathbb{B}(0; r)}$. Notice that for ∣z∣ = r (any fixed z on the disk), the inequality follows directly from the definition of g and the hypothesis on ∣f∣: |f(z)| ≤ M.

Recall that $f(z) = z \cdot g(z) \leadsto[\text{Taking the absolute value}] |f(z)| = |z| \cdot |g(z)|$.

Substitute the previous bound for |g(z)|, $|f(z)| \le |z| \cdot \frac{M}{r}, \forall z \in \overline{\mathbb{B}(0; r)}$

What happens if “equality holds”? If $|f(z_0)| = \frac{M}{r}|z_0|, \text{ for some } z_0 \in \mathbb{B}(0; r), z_0 \ne 0$, $|f(z_0)| = \frac{M}{r}|z_0| \implies |z_0| \cdot |g(z_0)| = \frac{M}{r}|z_0|$

Divide by $|z_0|$ (which is non-zero): $|g(z_0)| = \frac{M}{r}$.

We previously established that for all z in the disk, $|g(z)| \le \frac{M}{r}$ for all ∣z∣ ≤ r. So, $\frac{M}{r}$ is the maximum possible value (the ceiling) for $|g(z)|$. However, we just found a point $z_0$ strictly inside the disk where g actually hits this ceiling.

The Maximum Modulus Principle states that a non-constant analytic function cannot reach its maximum modulus inside a domain (connected, open set G). Therefore, g is a constant function, g(z) = c, and we already know the magnitude of this constant, $|c| = \frac{M}{r}$

Any complex number c of modulus $\frac{M}{r}$ can be written in polar form as $c = \frac{M}{r} e^{im}$ for some m real number (the angle or “phase”). Finally, substitute g(z) = c back into our function definition, $f(z) = z \cdot g(z) = z \cdot \left( \frac{M}{r} e^{im} \right) = \frac{M e^{im}}{r} z$

The function is a simple linear rotation. $\blacksquare$