MHB Proving Uniqueness of Fourier Coefficients for Continuous Periodic Functions

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For a continuous periodic function \( f:\mathbb R\to\mathbb R \) with period \( 2\pi \), if both integrals \( \int_0^{2\pi}f(x)\cos(nx)\,dx=0 \) for \( n=0,1,\ldots \) and \( \int_0^{2\pi}f(x)\sin(nx)\,dx=0 \) for \( n=1,2,\ldots \) hold, it can be proven that \( f(x)=0 \) for all \( x \in \mathbb R \). This result is linked to the uniqueness of Fourier coefficients, indicating that if all coefficients are zero, the function itself must be identically zero. The discussion references Fejér's theorem, which supports this conclusion by stating that \( f \) is the uniform limit of its Fourier series' Cesàro sums. Therefore, the condition of zero coefficients leads to the conclusion that the function must vanish everywhere. This establishes a strong connection between Fourier analysis and the properties of continuous periodic functions.
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Let $f:\mathbb R\to\mathbb R$ be a continuous function of period $2\pi.$ Prove that if $\displaystyle\int_0^{2\pi}f(x)\cos(nx)\,dx=0$ for $n=0,1,\ldots$ and $\displaystyle\int_0^{2\pi}f(x)\sin(nx)\,dx=0$ for $n=1,2,\ldots,$ then $f(x)=0$ for all $x\in\mathbb R.$

I know this has to do with the uniqueness of the Fourier coefficients, but I don't know how to solve it.
Thanks!
 
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Markov said:
Let $f:\mathbb R\to\mathbb R$ be a continuous function of period $2\pi.$ Prove that if $\displaystyle\int_0^{2\pi}f(x)\cos(nx)\,dx=0$ for $n=0,1,\ldots$ and $\displaystyle\int_0^{2\pi}f(x)\sin(nx)\,dx=0$ for $n=1,2,\ldots,$ then $f(x)=0$ for all $x\in\mathbb R.$

I know this has to do with the uniqueness of the Fourier coefficients, but I don't know how to solve it.
Thanks!

Sounds an awful lot like the Riemann-Lebesgue Lemma. Are the tools of Lebesgue integration available to you?
 
Markov said:
Let $f:\mathbb R\to\mathbb R$ be a continuous function of period $2\pi.$ Prove that if $\displaystyle\int_0^{2\pi}f(x)\cos(nx)\,dx=0$ for $n=0,1,\ldots$ and $\displaystyle\int_0^{2\pi}f(x)\sin(nx)\,dx=0$ for $n=1,2,\ldots,$ then $f(x)=0$ for all $x\in\mathbb R.$

I know this has to do with the uniqueness of the Fourier coefficients, but I don't know how to solve it.
Thanks!
This requires some fairly heavy machinery. One method is to use Fejér's theorem, which says that $f$ is the uniform limit of the Cesàro sums $s_n(f)$ of its Fourier series. If all the Fourier coefficients of $f$ are zero then $s_n(f)=0$ for all $n$, and hence $f=0.$
 
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