Regularity, Singularity & Analyticity
1. The Regular Point
Definition
Let \( f: D \longrightarrow \mathbb{C} \). A point \( z_0 \in D \) is said to be a regular point of \( f \) (or \( f \) is regular at \( z_0 \)) if:
There exists an open disc around \( z_0 \) on which \( f \) is differentiable everywhere.
Properties of Set R
Let \( D_f \) be the set of points where \( f \) is differentiable.
Let \( R \) be the set of regular points.
$$ R = \{ z \in D : f \text{ is regular at } z \} $$Key Observations:
- \( R \subset D_f \) (Regularity implies differentiability).
- \( R \) is an Open Set.
- If \( z_0 \in R \), then there exists \( \delta > 0 \) such that every point in the disc \( |z – z_0| < \delta \) is also a regular point.
2. Singular Points & Examples
Example: Differentiable but not Regular
Consider \( f(z) = x^2 + y^2 + 2xy i \) with \( D = \mathbb{C} \).
- \( D_f \) (Differentiable Set): The X-axis (from Cauchy-Riemann equations).
- \( R \) (Regular Set): \( \phi \) (Empty set).
Reason: The X-axis contains no open discs. To be regular, you need a full disc of differentiability.
$$ R’ = \phi \quad \text{and} \quad R’ – R = \phi $$Definition: Singular Point
A point \( \alpha \in \mathbb{C} \) is a singularity of \( f \) if it is a limit point of the set of regular points (\( R \)) but is not regular itself.
$$ \alpha \text{ is a singularity} \iff \alpha \in R’ – R $$Where \( R’ \) is the derived set (limit points) of \( R \).
Note: The set of singularities \( S \) is a closed set.
Definition: Analytic Function
\( f \) is said to be Analytic on \( D \) if it is regular at every point of \( D \).
$$ f \text{ is Analytic on } D \iff D = D_f = R $$3. The Hierarchy of Sets
Alternative Definition of Analyticity
\( f: D \longrightarrow \mathbb{C} \) is analytic on \( D \) if for any \( z_0 \in D \), there exists \( \delta > 0 \) such that \( f'(z) \) exists for all \( z \) in the neighborhood.
i.e., \( f \) is everywhere regular in \( D \).
Relationship: Interior Points
Let \( D_f^\circ \) denote the interior of the set of differentiable points.
Theorem: The set of regular points is exactly the interior of the differentiable set.
$$ D_f^\circ = R $$Proof logic: If \( z_0 \) is an interior point of \( D_f \), there is a disc of differentiable points around it, making it regular.
Double Star: The Golden Hierarchy
For any function \( f: D \longrightarrow \mathbb{C} \), the following inclusion always holds:
$$ R \subseteq D_f^\circ \subseteq D_f \subseteq D $$Since \( R = D_f^\circ \), we usually write:
$$ R = D_f^\circ \subset D_f \subset D $$Conclusion: If \( D \) is open and \( f \) is differentiable on \( D \) (\( D_f = D \)), then \( D_f^\circ = D \), which implies \( R = D \). Thus, \( f \) is analytic.
1. The Regular Point & Detailed Proofs
Definition
Let \( f: D \longrightarrow \mathbb{C} \). A point \( z_0 \in D \) is said to be a regular point of \( f \) if there exists an open disc around \( z_0 \) on which \( f \) is differentiable everywhere.
Property 1: Openness of Regularity
Statement: If \( z_0 \) is a regular point, then every point in the neighborhood \( |z – z_0| < \delta \) is also a regular point.
1. Let \( z_0 \in R \). By definition, \( \exists \delta > 0 \) such that \( f \) is differentiable on the disc \( S = \{ z : |z – z_0| < \delta \} \).
2. Let \( z_1 \) be any arbitrary point inside \( S \) (i.e., \( |z_1 – z_0| < \delta \)).
3. Since \( S \) is an open set, there exists a smaller radius \( \delta_1 > 0 \) such that the disc \( S_1 = \{ z : |z – z_1| < \delta_1 \} \) is completely contained within \( S \).
4. Since \( f \) is differentiable on the entire set \( S \), it is differentiable on the subset \( S_1 \).
5. Therefore, \( f \) is differentiable on a neighborhood of \( z_1 \).
6. Conclusion: \( z_1 \) is a regular point. Thus, \( S \subset R \).
Property 2: Relationship with Interior of \( D_f \)
Statement: \( R \) is the largest open subset of \( D_f \). Equivalently, \( R = D_f^\circ \) (The interior of the differentiable set).
Part A: Prove \( R \subseteq D_f^\circ \)
Since \( R \) is an open set (proved in Property 1) and by definition regular points must be differentiable (\( R \subseteq D_f \)), \( R \) is an open subset contained in \( D_f \).
Since \( D_f^\circ \) is the union of all open subsets of \( D_f \), it must contain \( R \).
\( \implies R \subseteq D_f^\circ \).
Part B: Prove \( D_f^\circ \subseteq R \)
Let \( z \in D_f^\circ \).
By definition of an interior point, there exists a neighborhood \( N \) of \( z \) such that \( N \subset D_f \).
This implies \( f \) is differentiable on the entire neighborhood \( N \).
By definition of regularity, \( z \) is a regular point.
\( \implies z \in R \).
Conclusion: From A and B, \( R = D_f^\circ \).