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

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Hey there,

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

Thanks in advance
 
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thank you very much
 
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].
 
Bacle2 said:
Maybe you could also use the following:

Polynomials are dense in C[a,b] (Weirstrass)

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.
 
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O.K, good point, I was (implicitly) assuming that result for compact intervals.
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.

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