The Schwarz Lemma — Answer

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First: Clarifying the Terminology

These three words get conflated, but they mean different things.

Holomorphic (= Complex Differentiable)

$f$ is holomorphic on an open set $U$ if it is complex differentiable at every point of $U$.

Complex differentiable means the limit

$f'(z) = \lim_{h \to 0} \frac{f(z+h) - f(z)}{h}$

exists, where $h \in \mathbb{C}$ can approach 0 from any direction in the complex plane.

This is much stronger than real differentiability — requiring the limit to be the same from all directions forces $f$ to satisfy the Cauchy-Riemann equations.

Analytic

$f$ is analytic at $z_0$ if it has a convergent power series expansion near $z_0$:

$f(z) = \sum_{n=0}^{\infty} a_n (z - z_0)^n$

In real analysis: analytic is strictly stronger than smooth (e.g. $e^{-1/x^2}$ is smooth but not analytic at 0).

In complex analysis: holomorphic $\iff$ analytic. This is a deep theorem — complex differentiability alone forces a power series to exist. So in complex analysis the two words are used interchangeably.

Harmonic

A real-valued function $u(x, y)$ is harmonic if it satisfies Laplace's equation:

$\Delta u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$

The connection to holomorphic functions: if $f = u + iv$ is holomorphic, then both $u$ and $v$ are harmonic. This follows directly from the Cauchy-Riemann equations.

Conversely, every harmonic function is locally the real part of some holomorphic function.

Summary Table

| Word | Meaning | Context | |------|---------|---------| | Holomorphic | Complex differentiable | Complex analysis | | Analytic | Has a power series | Real or complex | | Harmonic | Satisfies $\Delta u = 0$ | Real-valued functions |

In complex analysis: holomorphic $\iff$ analytic, and holomorphic $\Rightarrow$ real/imaginary parts are harmonic.

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Solution to the Schwarz Lemma

Goal: $f: \mathbb{D} \to \mathbb{D}$ holomorphic, $f(0) = 0$. Prove $|f(z)| \leq |z|$.

Step 1: Define a helper function

Let

$g(z) = \frac{f(z)}{z} \quad \text{for } z \neq 0$

The problem: $g$ seems undefined at $z = 0$. But since $f(0) = 0$, near $z = 0$:

$f(z) = f(0) + f'(0)z + \cdots = f'(0)z + O(z^2)$

So $g(z) = f(z)/z \to f'(0)$ as $z \to 0$.

This means $z = 0$ is a removable singularity — $g$ extends to a holomorphic function on all of $\mathbb{D}$ by defining $g(0) = f'(0)$.

Step 2: Bound $g$ on a circle

Fix any radius $0 < r < 1$. On the circle $|z| = r$:

$|g(z)| = \frac{|f(z)|}{|z|} = \frac{|f(z)|}{r}$

Since $f$ maps into $\mathbb{D}$, we have $|f(z)| < 1$, so:

$|g(z)| < \frac{1}{r} \quad \text{on } |z| = r$

Step 3: Apply the Maximum Modulus Principle

The Maximum Modulus Principle says: a holomorphic function on a closed disk attains its maximum modulus on the boundary, not the interior (unless it is constant).

So for all $z$ inside the disk $|z| \leq r$:

$|g(z)| \leq \max_{|z|=r} |g(z)| < \frac{1}{r}$

Step 4: Let $r \to 1$

The bound $|g(z)| < 1/r$ holds for every $r < 1$. Sending $r \to 1$:

$|g(z)| \leq 1 \quad \text{for all } z \in \mathbb{D}$

This means $|f(z)/z| \leq 1$, i.e.

$\boxed{|f(z)| \leq |z|}$

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Bonus: When does equality hold?

If $|g(z_0)| = 1$ for some interior point $z_0 \in \mathbb{D}$, then $|g|$ attains its maximum in the interior of the disk. The Maximum Modulus Principle then forces $g$ to be constant:

$g(z) = e^{i\theta} \quad \text{for some } \theta \in \mathbb{R}$

Therefore $f(z) = e^{i\theta} z$ — a rotation. $\square$

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The Big Picture

The one trick is: divide by $z$ to turn $f$ into $g$, then use the Maximum Modulus Principle. The condition $f(0) = 0$ is exactly what makes $z = 0$ a removable singularity rather than a pole.

The result says: fixing the origin forces $f$ to be a contraction. The only way to preserve distances is to rotate.