# Does the set (z^n ; n\in N) span L^2[0,1]?

1. Mar 1, 2013

### LikeMath

Hey there,

Does the set (z^n ; n\in N) span L^2[0,1]?

2. Mar 1, 2013

3. Mar 1, 2013

### LikeMath

thank you very much

4. Mar 4, 2013

### Bacle2

Maybe you could also use the following:

Polynomials are dense in C[a,b] (Weirstrass) ; Continuous functions ( in [a,b] , i.e., with compact support), are dense in simple functions, which are themselves dense in L2[a,b].

5. Mar 5, 2013

### micromass

Staff Emeritus
If you're going to use Weierstrass, then you have to know that this is for the $\| ~\|_\infty$ - norm. The theorem itself doesn't say anything for the $\|~\|_2$-norm.
Of course, on a compact interval, we have $\|~\|_2 \leq C \|~\|_\infty$ for some C that I'm too lazy to calculate. So density in $\|~\|_\infty$ would imply density in the $\|~\|_2$ norm.

Last edited: Mar 5, 2013
6. Mar 5, 2013

### Bacle2

O.K, good point, I was (implicitly) assuming that result for compact intervals.