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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 ?

Atmos
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  • What are your assumptions on $f?$ – Dedalus May 19 '18 at 15:19
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    The reason I am asking is because as the question stands now, your statement does not seem to be true. For example, if $f(t) = e^{it},$ then $\int_{-\pi}^\pi f(t)e^{i n t} dt =0$ for $n \geq 0.$ – Dedalus May 19 '18 at 15:24
  • Yes you are right, i've edited – Atmos May 19 '18 at 15:28
  • Hint: what is $ \overline{\int_{-\pi}^\pi f(t) e^{int} dt}$ if $f(t)$ is real? What does this say about the Fourier coefficients of $f?$ – Dedalus May 19 '18 at 15:34
  • (If you need more assistance, tell me and I will give a further hint) – Dedalus May 19 '18 at 15:37
  • If you know that a continuous function can be uniformly approximated by trigonometric polynomials and that you can interchange limit and integral in case of uniform convergence of integrands then you are done. – Nyfiken May 19 '18 at 16:07
  • Have you had a course in Complex Analysis? – Disintegrating By Parts May 21 '18 at 06:42
  • Only with Laurent series, holomorphism, residue etc – Atmos May 21 '18 at 09:14
  • Do you know Liouville's Theorem where you show that every bounded entire function is constant? A modification of that theorem can give you what you want, where you show that if an entire function is uniformly bounded on a sequence of circles (or squares) centered at the origin with the sequence of radii tending to $\infty$, then the function is constant. Interested? – Disintegrating By Parts May 22 '18 at 01:29
  • Yes, but i dont know about the theorem – Atmos May 22 '18 at 08:30

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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.

Disintegrating By Parts
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