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

  1. Jan 23, 2010 #1
    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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  3. Jan 23, 2010 #2

    CompuChip

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

    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
     
  4. Jan 23, 2010 #3
    Re: What does it mean to say something "extends smoothly" to a boundary in C (or R^2)

    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?
     
  5. Jan 24, 2010 #4
    Re: What does it mean to say something "extends smoothly" to a boundary in C (or R^2)

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