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        riddles >> putnam exam (pure math) >> Prime power moments
        
(Message started by: Aryabhatta on Jun 18^th, 2008, 7:26pm)

Title: Prime power moments
Post by Aryabhatta on Jun 18^th, 2008, 7:26pm

if f:[0,1] -> R is a continuous function such that

Integral {0 to 1} of f(x) dx = 0
and
Integral {0 to 1} of x^p f(x) dx = 0

for all primes p, show that f(x) = 0 for all x in [0,1].

Title: Re: Prime power moments
Post by Aryabhatta on Jun 20^th, 2008, 1:24am

Sorry, this one might be unfairly hard. The only proof I know uses a theorem from functional analysis which I somehow stumbled upon recently.

I do not know of any elementary proof and I am not sure how well known that theorem would be among undergraduates.

Title: Re: Prime power moments
Post by towr on Jun 20^th, 2008, 2:28am

If by any chance any function g(x) can be generated by a series a₀+http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifa_i x^P(i) for some sequence a_i, then I could see how to prove it. Because then we can exclude there is any range [a..b] http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subset.gif [0..1] where f(x) is non-zero. Which excludes that f(x) a continuous non-zero function.
It'd be akin to a Fourier transform; and I seem to recall it holds if you include all polynomials, so perhaps limiting oneself to prime polynomials works as well.

Title: Re: Prime power moments
Post by Aryabhatta on Jun 20^th, 2008, 9:18am

Good thinking! You are right... restricting oneself to use only {1, x^2, x^3, x^5, ..., x^p,...} works.

Perhaps an elementary proof would be to show that t^m for any positive integer m, can be written in that form...

Title: Re: Prime power moments
Post by Obob on Jun 20^th, 2008, 10:59am

But this cannot be done. First of all, if a continuous function has a power series expansion, it is necessarily analytic. Secondly, if an analytic function has a power series expansion, that expansion is unique. So there are lots and lots of continuous functions which cannot be written in the form proposed by towr, and moreover any function t^m with m not prime cannot be written in that form, since the power series expansion of t^m around 0 is t^m=t^m.

What needs to be shown is that the span of {x^p:p prime or 0} forms a dense subspace of C([0,1]) with respect to the L^2 norm. For then the annihilator of this subspace is trivial, and there can be no such f. Conversely, if there were an f orthogonal to x^p for all p, then the span of {x^p:p prime or 0} would not be dense. So this is an equivalent statement.

The main tool we all know of for trying to prove this kind of statement is the Stone-Weierstrass theorem. The set {x^p:p prime or 0} does in fact seperate points and contains a nonzero constant. But it is not a subalgebra, so that theorem cannot be applied.

However, all we need to prove is L^2 convergence, not uniform convergence, which is weaker than what Stone-Weierstrass provides. So perhaps some other argument can be supplied. Certainly it would be enough to show that t^m for any m can be written as an infinite series sum a_i t^(p_i) in the L^2 norm, for then we could apply Stone-Weierstrass. But I have no reason to believe this can be done. Note that it is not possible to write t^m as such a series in the sup-norm, since this violates the uniqueness of Taylor series. So this sum will fail to converge to t^m pointwise.

Title: Re: Prime power moments
Post by Aryabhatta on Jun 20^th, 2008, 1:40pm

Yes, If I recall correctly, in fact, even in the "normal" fourier series, uniform (or even pointwise) convergence need not hold.

If we consider pointwise convergence, I think you are right that t^4 etc cannot be represented as such a series.

Also, even if we are talking about L2 convergence, we might need to ortho-normalize {1} U {x^p} first (though I am not sure about it).

Also, I do need to read up on these concepts, so please pardon me If I am speaking nonsense.

Title: Re: Prime power moments
Post by Obob on Jun 20^th, 2008, 1:47pm

If the function is piecewise C1, then the normal Fourier series converges pointwise to the function anywhere the function is continuous. At discontinuities, the value of the Fourier series is the average of the left and right limits of the function at that point.

I doubt that orthonormalizing {1} U {x^p} will help much.

What is the theorem from functional analysis that you want to apply, Aryabhatta?

Title: Re: Prime power moments
Post by Aryabhatta on Jun 20^th, 2008, 1:50pm

That theorem is called [hide] Muntz's Theorem [/hide]

Title: Re: Prime power moments
Post by Obob on Jun 20^th, 2008, 1:57pm

That is quite an interesting result!

I doubt though that there would be a simpler proof for the case of prime exponents than the general case, though.

Title: Re: Prime power moments
Post by Aryabhatta on Jun 20^th, 2008, 2:21pm

on 06/20/08 at 13:57:51, Obob wrote:

That is quite an interesting result!

Yes, I was surprised when I first came across it!

on 06/20/08 at 13:47:31, Obob wrote:

If the function is piecewise C1, then the normal Fourier series converges pointwise to the function anywhere the function is continuous.

I was talking about C (and not C1), but I guess you had t^m in mind and for that, it is definitely true. In fact if I recall correctly, for C1, the convergence is uniform.

Title: Re: Prime power moments
Post by Obob on Jun 20^th, 2008, 2:38pm

Fourier series of general continuous functions are terribly behaved. If you fix a point x in the unit interval, there is a dense subset of the space of continuous functions whose Fourier series fail to converge to the function at x. See Rudin, Real and Complex Analysis, p. 102.

Piecewise C1 is the hypothesis that gives you a nice theory.