# Proof of Hausdorff's moment theorem using Fourier's series

Let $$f \in \mathcal{C}^0([0,2\pi])$$ such that : $$\forall n \in \mathbb{N}\$$, $$\displaystyle \int_{0}^{2\pi}t^nf(t)\mathrm{d}t=0$$.

I have to show that : $$\forall n \in \mathbb{Z}, \ \displaystyle \int_{0}^{2\pi}f(t)\exp(-int)\mathrm{d}t=0$$ and $$f\equiv 0$$.

Here is the beginning :

Let define $$\widetilde{f}$$ the $$2\pi$$-periodic, even, continuous function on $$\mathbb{R}$$ which fits with $$f$$ on $$]0,2\pi]$$.

Moreover the Fourier's coefficients give : $$\forall n\in \mathbb{Z}$$ : $$c_n(\widetilde{f})=\displaystyle \frac{1}{2\pi}\int_{0}^{2\pi}f(t)\exp(-int)\mathrm{d}t$$.

Then we can notice that : $$\exp(-int)=\displaystyle \sum \limits_{k\ge 0}\frac{(-int)^k}{k!}$$.

So $$c_n(\widetilde{f})=\displaystyle\frac{1}{2\pi}\int_{0}^{2\pi}\left(\sum \limits_{k\ge 0}f(t)\frac{(-int)^k}{k!}\right)\mathrm{d}t$$.

Let's try to exchange $$\displaystyle \sum$$ and $$\displaystyle \int$$ :

Fix $$n\in \mathbb{N}$$.

-For all $$k\ge 0$$, the functions $$t\mapsto f(t)\dfrac{(-int)^k}{k!}$$ are continuous on $$[0,2\pi]$$.

-For all $$t\in [0,2\pi]$$ : $$\Big\vert f(t)\dfrac{(-int)^k}{k!} \Big \vert \le \vert f(t)\vert \Big\vert\dfrac{(-int)^k}{k!}\Big\vert \le M\dfrac{\vert-int\vert^k}{k!}=M\dfrac{(2\pi n)^k}{k!}$$.

That last term is the general term of a convergent series (of sum $$M\exp(2\pi n)$$) so we can deduce that the series converges uniformly on $$[0,2\pi]$$.

Hence : $$c_n(\widetilde{f})=\displaystyle\frac{1}{2\pi}\int_{0}^{2\pi}\left(\sum \limits_{k\ge 0}f(t)\frac{(-int)^k}{k!}\right)\mathrm{d}t = \displaystyle \frac{1}{2\pi} \sum \limits_{k\ge 0}\int_{0}^{2\pi}f(t)\frac{(-int)^k}{k!}\mathrm{d}t = \displaystyle \sum \limits_{k\ge 0} \frac{(-in)^k}{2\pi (k!)} \int_{0}^{2\pi}f(t)t^k\mathrm{d}t =0.$$

We can conclude using Parseval's formula (for $$f \in \mathcal{C}^0([0,2\pi])$$) :

$$\displaystyle \int_{0}^{2\pi} \vert f(t) \vert^2\mathrm{d}t = \sum \limits_{n\in\mathbb{Z}} \vert c_n(\widetilde{f})\vert^2 = 0$$ hence $$f \equiv 0$$.

At the end we want $$[a,b]$$ instead of $$[0,2\pi]$$ but the variable change $$u=a+t\left(\dfrac{b-a}{2\pi}\right)$$ seems to work no ?

Is the general reasoning correct ?

Thanks in advance !

• Is $\mathbb N = \{1,2,3,\dots\}?$
– zhw.
Commented Apr 18, 2018 at 18:01

## 1 Answer

You do not need Fourier series to prove that if $f\in C^0([0,2\pi])$ and $\int_{0}^{2\pi}x^n f(x)\,dx = 0$, then $f\equiv 0$ on $[0,2\pi]$ (so $\int_{0}^{2\pi}f(x) e^{\pm in x}\,dx$ is obviously zero). By the Weierstrass approximation theorem, for any $\varepsilon >0$ there is a polynomial $p(x)$ such that $\left|f(x)-p(x)\right|\leq \varepsilon$ for any $x\in[0,2\pi]$. By the assumptions $$\int_{0}^{2\pi} p(x)\,f(x)\,dx = 0$$ hence $\int_{0}^{2\pi} f(x)^2\,dx$ is arbitrarily close to zero. Since $f$ is continuous on $[0,2\pi]$, $f\equiv 0$.

• I was thinking about Parseval but I need to justify a normal convergence before ... Commented Apr 18, 2018 at 17:33