Moment theorem for Fourier's coefficient I would like to show that for $f : \mathbb{R} \rightarrow \mathbb{R}$ a continuous $2 \pi$ - periodic function  if
$$
\forall n \in \mathbb{N}, \ \frac{1}{\pi}\int_{-\pi}^{\pi}f\left(t\right)e^{int}\text{d}t=0
$$
Then $f=0$
I know a close result that is
$$
\forall n \in \mathbb{N}, \int_{a}^{b}t^nf\left(t\right)\text{d}t=0 \Rightarrow f=0
$$
Is there a simple way to show this ?
 A: You can integrate over $[0,2\pi]$ because $f$ is periodic. Consider the function
$$
         F(\lambda)=\frac{1}{1-e^{2\pi i\lambda}}\int_{0}^{2\pi}f(x)e^{i\lambda x}dx.
$$
This is a holomorphic function on $\mathbb{C}\setminus\mathbb{Z}$. Suppose that $\int_{0}^{2\pi}f(x)e^{inx}dx=0$ for all $n\in\mathbb{Z}$. Then $F(\lambda)$ extends to an entire function of $\lambda$ because $1-e^{2\pi i\lambda}$ has first order zeros at $\lambda=0,\pm 1,\pm 2,\cdots$. You can show that $F(\lambda)$ is uniformly bounded on all squares centered at the origin of width $n+1/2$. That is enough to be able use the Cauchy integral representation of $F(\lambda)$ inside such squares in order to conclude that $F$ is uniformly bounded on $\mathbb{C}$ and, hence, must be a constant $K$. The constant $K$ must be $0$, which can be seen by examining $\lim_{r\uparrow\infty}F(ir)$. Therefore
$$
      \int_{0}^{2\pi} f(x)e^{i\lambda x}dx = 0,\;\;\; \lambda\in\mathbb{C}.
$$
All derivatives of the above are $0$ at $\lambda=0$, which gives
$$
        \int_{0}^{2\pi}x^n f(x)dx =0,\;\;\; n=0,1,2,3,\cdots.
$$
So $f=0$ a.e. by what you know.
