Certainly! To prove the Schwarz Lemma using the Maximum Modulus Principle in complex analysis, let’s proceed step by step.

Schwarz Lemma Statement:

Let $f$ be a holomorphic function on the unit disk $\mathbb{D} = \{ z \in \mathbb{C} : |z| < 1 \}$, satisfying:

1. $f(0) = 0$,
2. $|f(z)| \leq |z|$ for all $z \in \mathbb{D}$.

Then, the following hold:

1. $|f(z)| \leq |z|$ for all $z \in \mathbb{D}$,
2. $|f'(0)| \leq 1$,
3. If equality holds for any non-zero $z$, then $f(z)$ is a rotation, i.e., $f(z) = \lambda z$ for some constant $\lambda$ with $|\lambda| = 1$.

Proof using the Maximum Modulus Principle:

Let’s prove the first part and then deduce the remaining results.

Step 1: Define a new function

Consider the function $g(z) = \frac{f(z)}{z}$ for $z \neq 0$ in the unit disk, and define $g(0) = f'(0)$. Notice that $g(z)$ is holomorphic on the unit disk, since $f$ is holomorphic and $f(0) = 0$, ensuring that $\frac{f(z)}{z}$ is well-defined and analytic at $z = 0$.

Step 2: Apply the Maximum Modulus Principle

Since $|f(z)| \leq |z|$ for all $z \in \mathbb{D}$, it follows that for all $z \in \mathbb{D}$, we have:

$|g(z)| = \left|\frac{f(z)}{z}\right| \leq 1 \quad \text{for} \quad z \neq 0.$

Thus, the function $g(z)$ is bounded in the unit disk and holomorphic in $\mathbb{D}$. By the Maximum Modulus Principle, the maximum value of $|g(z)|$ occurs on the boundary of $\mathbb{D}$. But since $|g(z)| \leq 1$ for all $z \in \mathbb{D}$, this maximum value is at most 1. Therefore:

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

Step 3: Conclude the form of $f(z)$

From the above, we conclude that $|f(z)| \leq |z|$ for all $z \in \mathbb{D}$, and this establishes the first part of the Schwarz Lemma.

Step 4: Derivative at the origin

Now, we consider the behavior of $f'(0)$. Since $g(z)$ is holomorphic and $g(0) = f'(0)$, we know that:

$|f'(0)| = |g(0)| \leq 1.$

This proves the second part of the lemma.

Step 5: Equality condition

Finally, we look at the condition where equality holds. If $|f(z)| = |z|$ for all $z \in \mathbb{D}$, then from the Maximum Modulus Principle, we conclude that $f(z)$ must be a constant multiple of $z$, i.e., $f(z) = \lambda z$ where $|\lambda| = 1$.

Conclusion:

This proof utilizes the Maximum Modulus Principle and establishes that $f(z)$ cannot grow faster than $z$, and that the derivative at the origin is bounded by 1. The equality case reveals that $f(z)$ is a rotation, $f(z) = \lambda z$ with $|\lambda| = 1$.

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Would you like more details on any part of this proof or other questions on complex analysis?

Here are some related questions that may interest you:

1. What is the Maximum Modulus Principle and how does it apply to analytic functions?
2. Can you prove the Rouché's Theorem using the Maximum Modulus Principle?
3. What is the Conformal Mapping and how does it relate to distance-preserving properties of analytic functions?
4. How does the Schwarz-Pick Theorem generalize the Schwarz Lemma?
5. Can you explain the Cauchy Integral Formula and how it’s related to analytic functions and their properties?

Tip: When dealing with complex analysis, always check whether a function is bounded, holomorphic, or meets the conditions for using powerful results like the Maximum Modulus Principle or Schwarz Lemma—it can save a lot of time and effort!