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

[AxiomOfChoice]

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?

 

[CompuChip]

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

 

[AxiomOfChoice]

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?

 

[elibj123]

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)