Proving Uniqueness of Fourier Coefficients for Continuous Periodic Functions

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SUMMARY

The discussion centers on proving that a continuous periodic function \( f:\mathbb{R}\to\mathbb{R} \) with period \( 2\pi \) is identically zero if its Fourier coefficients are all zero. Specifically, if \( \int_0^{2\pi} f(x) \cos(nx) \, dx = 0 \) for \( n=0,1,\ldots \) and \( \int_0^{2\pi} f(x) \sin(nx) \, dx = 0 \) for \( n=1,2,\ldots \), then it follows that \( f(x) = 0 \) for all \( x \in \mathbb{R} \). This conclusion is supported by Fejér's theorem, which states that \( f \) is the uniform limit of the Cesàro sums of its Fourier series, leading to the result that if all Fourier coefficients are zero, then \( f \) must be zero.

PREREQUISITES
  • Understanding of Fourier series and Fourier coefficients
  • Knowledge of continuous functions and periodic functions
  • Familiarity with Fejér's theorem and its implications
  • Basic concepts of Lebesgue integration
NEXT STEPS
  • Study the proof of Fejér's theorem in the context of Fourier series
  • Learn about the Riemann-Lebesgue lemma and its applications
  • Explore the properties of Cesàro sums in Fourier analysis
  • Investigate the implications of zero Fourier coefficients on function behavior
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Mathematicians, students of analysis, and anyone studying Fourier analysis and its applications in continuous periodic functions.

Markov2
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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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