One moment please

Let f,g:[a,b]\to\Bbb{R} be continuous functions. We call \int_a^b x^kf(x)\,\mathrm{d}x the k-th moment of f. Prove that if f and g have identical k-th moments for all k\geq 0, then f=g.

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One moment please

One thought on “One moment please

  1. We will prove a more general result. Suppose f,g are in L^1[a,b] and they have identical k-th moments for all k\geq 0. Let h(x)=f(x)-g(x). By hypothesis, \int_a^bp(x)h(x)dx=0 for every polynomial p(x). Thus, by the Weierstrass Approximation Theorem, it follows that \int_a^bc(x)h(x)dx=0 for every continuous function c(x). Consider the continuous function \rho_n (x) = Cne^{\frac{-1}{1-|nx|^2}}\chi_{[-1/n,1/n]}(x) for n \in \mathbb{N} and C such that \int_\mathbb{R}\rho(x)dx=1. Then, the convolution \rho_n\star h(y) = \int_a^b\rho(y-x)h(x)dx=0 \to h(y), as n \to \infty, for almost every y \in [a,b]. Thus, f=g a.e. and, if they are continuous, everywhere.

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