# Removable Singularity of L_a^2(G): Proof of Limit at z=0

• Nusc
In summary: This is because the function f(z) is bounded by 2/sqrt(pi) ||f||_2, which goes to 0 as z approaches 0.
Nusc

## Homework Statement

If G = { z in C: 0<|Z|<1} show that every f in L_{a}^{2}(G) has a removable singularity at z = 0

Proof:

We must show that lim z->0 z*f(z) = 0 for all f in L_{a}^{2}(G)

By a corollary 1.12, if f in L_{a}^{2}(G), a in G and 0<r<dist(a,bdr G), thne
|f(a)| <= 1/(r\sqrt(\pi))||f||_2,

for |z| < 1/2 we have that

|f(z)| <= 1/(|z|\sqrt(pi)/sqrt(2)) ||f||_2 = 2/(|Z|sqrt(pi) ||f||_2

so that |zf(z)| = |z||f(z)| <= 2/sqrt(pi) ||f||_2.

how do I proceed from there

## The Attempt at a Solution

We must show that lim z->0 z*f(z) = 0 for all f in L_{a}^{2}(G)By a corollary 1.12, if f in L_{a}^{2}(G), a in G and 0<r<dist(a,bdr G), thne|f(a)| <= 1/(r\sqrt(\pi))||f||_2,for |z| < 1/2 we have that|f(z)| <= 1/(|z|\sqrt(pi)/sqrt(2)) ||f||_2 = 2/(|Z|sqrt(pi) ||f||_2so that |zf(z)| = |z||f(z)| <= 2/sqrt(pi) ||f||_2.Now as z approaches 0, zf(z) tends to 0 since the limit of the product of two functions is the product of their limits. Therefore, lim z->0 zf(z) = 0 and hence, f(z) has a removable singularity at z = 0.

## 1. What is the Removable Singularity of L_a^2(G)?

The Removable Singularity of L_a^2(G) is a concept in complex analysis that refers to a singularity at a point z=0 that can be removed by analytic continuation. In other words, the function is undefined at z=0, but by extending the function to a larger domain, the singularity can be removed and the function becomes well-defined at z=0.

## 2. What is the significance of the Removable Singularity of L_a^2(G)?

The Removable Singularity of L_a^2(G) is significant because it allows us to extend the domain of a function to include points that were previously undefined. This allows for a more complete understanding of the behavior of the function and can lead to new insights and applications in mathematics and other fields.

## 3. How is the Removable Singularity of L_a^2(G) proven at z=0?

To prove the Removable Singularity of L_a^2(G) at z=0, we use a combination of techniques from complex analysis, including analytic continuation and Laurent series. By constructing a function that is analytic in a larger domain and has the desired behavior at z=0, we can show that the singularity is removable.

## 4. What is the role of the function L_a^2(G) in the proof of the Removable Singularity at z=0?

The function L_a^2(G) plays a crucial role in the proof of the Removable Singularity at z=0. It is a function that is analytic in a larger domain and has the desired behavior at z=0. By constructing this function, we can show that the singularity is removable and extend the domain of the function.

## 5. Can the Removable Singularity of L_a^2(G) occur at points other than z=0?

Yes, the Removable Singularity of L_a^2(G) can occur at points other than z=0. In fact, it can occur at any point where the function is undefined but can be extended by analytic continuation. The concept of removable singularities is not limited to a specific point, but rather refers to a general property of functions in complex analysis.

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