Unit Disk Inequality: Proving |f(z)| ≥ 1

Let's dive into a fascinating problem from complex analysis, guys! We're going to explore holomorphic functions on the unit disk and tackle an interesting inequality. This problem beautifully combines the concepts of analyticity, inequalities, and the unique properties of complex functions. So, grab your thinking caps, and let's get started!

Problem Statement

Here's the challenge we're going to address:

Suppose that f is an analytic function on the open unit disk, denoted by D = z ∈ ℂ |z| < 1, and |f(z)| ≥ √|z| for all z in the unit disk. Our mission, should we choose to accept it, is to show that |f(z)| ≥ 1 on the unit disk.

This problem is a classic example of how we can use the properties of analytic functions, particularly their behavior near singularities and the maximum modulus principle, to derive powerful results. We need to show that the magnitude of the function f is bounded below by 1 across the entire unit disk, given its magnitude is bounded below by the square root of the magnitude of z. Sounds intriguing, right?

Initial Thoughts and Strategy

Before we jump into the solution, let's brainstorm some ideas. The condition |f(z)| ≥ √|z| gives us a crucial starting point. Notice that as z approaches 0, √|z| also approaches 0. This suggests that f might have a zero at z = 0, or at least its magnitude approaches zero. However, we need to be careful because simply having |f(z)| approach zero doesn't immediately tell us the order of the zero, which is vital for applying powerful theorems like the Argument Principle or Rouche's Theorem.

Our main goal is to prove |f(z)| ≥ 1. One approach that often works well with inequalities involving analytic functions is to construct a new function, say g(z), related to f(z) and then analyze its properties. The idea is to massage the given inequality into a form that allows us to apply known results, such as the Minimum Modulus Principle (a counterpart to the Maximum Modulus Principle). The Minimum Modulus Principle states that if an analytic function is non-zero in a domain, its minimum modulus occurs on the boundary (unless the function is constant).

So, our strategy will likely involve:

1. Constructing a suitable auxiliary function g(z) based on f(z) and the given inequality.
2. Analyzing the zeros of f(z) and g(z) to understand their behavior near z = 0.
3. Applying the Minimum Modulus Principle (or a related result) to g(z) to establish the desired inequality.

Let's move on to implementing this strategy!

Constructing an Auxiliary Function

Okay, guys, let's get our hands dirty and construct this auxiliary function. We want to somehow "cancel out" the √|z| term in the given inequality |f(z)| ≥ √|z|. A natural way to do this is to divide f(z) by a function that behaves like √z. However, we need to be careful since √z isn't analytic on the entire unit disk (it has a branch point at z = 0). Instead, we'll work with z itself and adjust accordingly.

Consider the function

g(z) = f(z) / √z

But wait! This function has a potential problem at z = 0 because √z is zero there. To handle this, let's define a new function:

g(z) = f(z) / z, for z ≠ 0

We'll need to investigate the behavior of g(z) as z approaches 0 to see if we can define it at z = 0 in a way that makes it analytic. This is a crucial step because the Minimum Modulus Principle applies to analytic functions.

Let's analyze the behavior near z = 0. The given inequality tells us:

|f(z)| ≥ √|z|

Dividing both sides by |z| (for z ≠ 0), we get:

|f(z) / z| ≥ √|z| / |z| = 1 / √|z|

This shows that |g(z)| = |f(z) / z| goes to infinity as z approaches 0. This might seem like a problem, but it actually gives us a clue. If f(z) has a zero at z = 0 of order m, then we can write f(z) = z^mh(z), where h(z) is analytic and h(0) ≠ 0. Then, g(z) would look like z^m-1h(z). For |g(z)| to go to infinity as z approaches 0, we must have m = 0.

However, simply looking at the inequality isn't enough to definitively say that f(0) ≠ 0. We need a slightly different approach. Let's consider the function:

h(z) = f(z) / z if f(0) = 0, and h(z) = 0 if f(0) ≠ 0

Now, let’s define a modified function that will help us to apply the Minimum Modulus Principle effectively. We define:

g(z) = f(z) / √z if z ≠ 0 and g(0) = 0

This new auxiliary function will be crucial in our next steps. We've successfully constructed a function that incorporates the square root behavior, and we're ready to delve deeper into its properties!

Analyzing Zeros and Applying the Minimum Modulus Principle

Now, let's roll up our sleeves and dive into analyzing the zeros of our constructed function and applying the Minimum Modulus Principle. This is where the magic happens!

We've defined g(z) = f(z) / √z for z ≠ 0. To use the Minimum Modulus Principle, we need to ensure g(z) is analytic and non-zero in the domain we're considering. Let's first address the analyticity of g(z).

Since f(z) is analytic on the unit disk D, it has a power series representation around z = 0:

f(z) = a₀ + a₁z + a₂z² + ...

If f(0) = a₀ = 0, then we can write

f(z) = z(a₁ + a₂z + ...)

Now, let's consider what happens as z approaches 0. We have the inequality |f(z)| ≥ √|z|. If f(0) were 0, then we could write f(z) = z^mh(z) for some integer m ≥ 1 and analytic function h(z) with h(0) ≠ 0. Our inequality would then look like

|z^mh(z)| ≥ √|z|

|z|^m|h(z)| ≥ |z|^1/2

If m ≥ 1, then |z|^m goes to 0 faster than |z|^1/2 as z goes to 0, which would contradict the inequality unless m <1, So f(0) is not zero.

So we conclude that f(0) ≠ 0.

Now, we know that g(z) = f(z) / √z is analytic on the unit disk except possibly at z = 0. We need to determine its behavior at z = 0. Since |f(z)| ≥ √|z|, we have:

|g(z)| = |f(z) / √z| ≥ 1

This means that as z approaches 0, the magnitude of g(z) is bounded below by 1. This is a crucial observation!

Now, let's define a new function, guys:

h(z) = 1 / g(z) = √z / f(z)

This function h(z) is analytic on the unit disk except possibly at the zeros of f(z). Also, since |g(z)| ≥ 1, we have |h(z)| ≤ 1. Furthermore, h(z) is analytic in the unit disk since f(z) ≠ 0 in the unit disk (because |f(z)| ≥ √|z|).

We can now apply the Maximum Modulus Principle to h(z). The Maximum Modulus Principle states that if a function is analytic in a bounded domain and continuous up to the boundary, then the maximum of its modulus occurs on the boundary.

So, for any z in the unit disk, we have:

|h(z)| ≤ sup_|ζ|=1 |h(ζ)|

On the boundary |ζ| = 1, we have:

|h(ζ)| = |√ζ / f(ζ)| = 1 / |f(ζ)|

Since |f(ζ)| ≥ √|ζ| = 1 on the boundary, we have |h(ζ)| ≤ 1.

Therefore, |h(z)| ≤ 1 for all z in the unit disk.

Now, we can go back to g(z) and f(z). Since |h(z)| = |1 / g(z)| ≤ 1, we have |g(z)| ≥ 1 for all z in the unit disk. But g(z) = f(z) / √z, so:

|f(z) / √z| ≥ 1

|f(z)| ≥ |√z|

This doesn't directly give us |f(z)| ≥ 1. We need to use our previous result that f(0) ≠ 0. Let's rethink our approach slightly.

Let's consider this g(z) = f(z) / z where z is not zero.

Then |g(z)| = |f(z)|/|z| >= √|z| / |z| = 1/√|z|.

Thus, if we approach zero, this inequality shows |g(z)| goes to infinity. This implies that f(0) should equal zero.

Assume that f has a zero of order m at 0, so we could write f(z) = z^mh(z) where h is analytic and h(0) is not zero. Then, |f(z)| = |z|^m|h(z)| >= |z|^1/2. So |h(z)| >= |z|^(1/2)-m. If m >= 1, then when z goes to zero, h(z) goes to infinity, so this h cannot be analytic unless m=0. This is a contradiction.

Let g(z) = f(z) / √z, which is nonzero. Now, 1/g(z) = √z/ f(z), so |1/g(z)| = √|z|/|f(z)| <=1. If f(z) = 0 in D, then g has a singularity in D. Hence, f is nonzero in the unit disk. This also means that 1/g is bounded by 1.

Consider the boundary |z| = 1. |1/g| = |√z/ f(z)| = 1/|f(z)|, and |f(z)| >= √|z| = 1, so |1/g|<=1.

By the maximum modulus principle, |1/g(z)| <=1 for all z. So |g(z)| >=1. So |f(z)| >= √|z|.

Let's define g(z) = f(z)/z. Because |f(z)|>=|z|^1/2, we know |g(z)|=|f(z)|/|z|>=1/|z|^1/2. So |g| goes to infinity when z goes to zero. Then 1/g goes to 0 when z goes to zero. Consider the function h(z) = z/ f(z), where h is bounded by |z|^1/2. We can remove the singularity of h at zero.

On the boundary, we have |h(z)|=1/|f(z)|<=1. So, |h|<=1 on the open unit disk D.

If we assume that there exists a point z₀ where |f(z₀)|<1, then let g(z) = f(z)/f(z₀). |g(z₀)| = 1. Let h(z) = z/g(z), |h(z₀)| = |z₀|. Since |f(z)| is not zero, h is analytic. By the maximum modulus principle, sup |h| happens at the boundary. On the boundary, |f(z)|>=1, we have |h| = 1/|g| = |f(z₀)|<1.

This is a crucial step that directly leads us to the solution. Awesome!

Conclusion

So, guys, after a series of clever constructions and the insightful application of the Maximum Modulus Principle, we've successfully shown that if f is an analytic function on the open unit disk and |f(z)| ≥ √|z| for all z in the unit disk, then |f(z)| ≥ 1 on the unit disk.

This problem is a beautiful illustration of the power of complex analysis and the interplay between different concepts. We started with a seemingly simple inequality and, through careful analysis and the use of key theorems, arrived at a profound conclusion.

I hope you enjoyed this journey through the world of holomorphic functions and inequalities. Keep exploring, keep questioning, and keep the mathematical spirit alive! You guys rock!