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

1. Jan 23, 2010

### 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?

2. Jan 23, 2010

### CompuChip

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

3. Jan 23, 2010

### AxiomOfChoice

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?

4. Jan 24, 2010

### elibj123

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)