# can somebody help me to prove that C2pi is complete

Discussion in 'General Math' started by Kal, Nov 21, 2004.

1. ### KalGuest

Hi
Can somebody help me to prove that the space of 2pi-periodic continuous
functions is complete and separable?

Kal, Nov 21, 2004

2. ### Lynn KurtzGuest

Hint 1: You need to specify a metric.

Hint2: Do you have the fact that C[0, 2pi] is complete and separable?

Hint3: If a sequence {f_n} is cauchy in the periodics, what about it
in C[0, 2pi]? Would it have a limit? Would it be periodic? Etc.

Hope that helps.

--Lynn

Lynn Kurtz, Nov 21, 2004

3. ### KalGuest

can I use ||f(R)||=sup{abs(f(x)) where x is in R}?
it's not given but I can assume it.
it is not necesarily a cauchy but I can find a subsequence in C[0,2pi] that
is cauchy

Kal, Nov 21, 2004
4. ### The World Wide WadeGuest

can I use ||f(R)||=sup{abs(f(x)) where x is in R}?[/QUOTE]

What do you mean "can I"? It's your question.

The World Wide Wade, Nov 22, 2004
5. ### Lynn KurtzGuest

Hey, it's your problem, use whatever metric you want to state the
problem. It's nice, though, to use a metric for which your statement
is true. Let's call P the set of periodic functions on R. You might
note that for f in P, ||f|| also can be computed by taking the sup
over [0, 2pi] as well as over R. In fact you might wish to think about
the restriction of the functions in P to [0, 2pi]. After all, if a
sequence of periodic functions converges there......

The functions {f_n} are in S. Are you sure about that?

--Lynn

Lynn Kurtz, Nov 22, 2004