Oops...I just realized that I didn't need to prove that the functional itself was continuous. Oh well...it's late.
If we call a moment of .
Problem: Prove that if have the same moments (for ) then the must be the same function.
Proof: Define the functional by . We first prove that is continuous.
Lemma: is continuous.
Proof: Note that if we're done since the integral of a continuous non-negative function is zero only when the function is identically zero. Thus, we may assume that
Let be given. Then choosing such that we see that
The conclusion follows. .
But notice by our assumption that for every or equivalently .
But, we in fact see that given any polynomial that
And since by Weirstrass's theorem we have that the polynomials are dense in we have that agrees with the zero function on a dense subset of it's domain. It follows that for every . The conclusion follows from this.
Does that look right? I feel like I made this way harder than it need be.
If anyone is interested, now that I'm awake there is a much more clean way to say this.
Proof: Define, the functional by .
Using the same rational as above we may conclude that is continuous.
Also, it is apparent that is functional and by our conditions we have that .
Thus, let be a polynomial, then .
It follows that if is the set of all polynomials in that . But, it follows from Weierstrass's approximation theorem that is dense. And since is closed we have that . It follows that . In particular and thus by prior remarks the conclusion follows.