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

Question

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 !

Answer 1

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 $[0,2\pi]$, $f\equiv 0$.