Moment theorem for Fourier's coefficient

Question

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 ?

Answer 1

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.