Understanding Smooth Extension to Boundary of D in C (or R^2)

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Discussion Overview

The discussion revolves around the concept of a function "extending smoothly" to the boundary of a domain in complex analysis (C) or two-dimensional real analysis (R²). Participants explore the implications of this smooth extension in the context of Cauchy's integral formula and the definitions involved.

Discussion Character

  • Conceptual clarification
  • Technical explanation

Main Points Raised

  • One participant asks for clarification on what it means for a function to extend smoothly to the boundary of a domain, particularly in relation to Cauchy's integral formula.
  • Another participant defines the smooth extension of a function to the closure of a domain, indicating that the extension is smooth everywhere.
  • A subsequent post questions whether "smooth" implies "complex differentiable" and seeks clarification on how differentiability is understood at boundary points.
  • In response, a participant asserts that smoothness at the boundary only implies continuity, as higher levels of smoothness cannot be defined at boundary points.

Areas of Agreement / Disagreement

Participants appear to agree on the definition of smoothness as continuity at the boundary, but there is uncertainty regarding the implications of differentiability at boundary points and the interpretation of smoothness in this context.

Contextual Notes

The discussion does not resolve the nuances of differentiability at boundary points or the specific implications of smoothness in relation to complex differentiability.

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What does it mean to say something "extends smoothly" to a boundary in C (or R^2)?

I'm studying Cauchy's integral formula, and one of the assumptions of the theorem is that a function be analytic on a domain D and extend smoothly to the boundary of D. What does that mean, exactly?
 
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Let D be an (open) set on which a function f(z) is defined, and denote by D' the closure of D (i.e. D union its boundary).

Then the smooth extension of f to D' is the function g(z) defined by
g(z) = f(z) for all z in D
g(z) is smooth everywhere
 


CompuChip said:
Let D be an (open) set on which a function f(z) is defined, and denote by D' the closure of D (i.e. D union its boundary).

Then the smooth extension of f to D' is the function g(z) defined by
g(z) = f(z) for all z in D
g(z) is smooth everywhere

Thanks. And does "smooth" in this context mean "complex differentiable?" And if so, how do we make sense of differentiability at a point on the boundary?
 


Exactly, you can't. Smooth simply means it's continuous there (as of course you cannot define higher classes of smoothness on a boundary point)
 

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