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Let $f$ be a continuous, $2\pi$-periodic function and suppose that its Fourier coefficients, $a_n$ and $b_n$, equal $0$ when $n$ is odd, i.e., \begin{equation} \int_{-\pi}^\pi f(x)\cos (nx)\,dx=\int_{-\pi}^\pi f(x)\sin (nx)\,dx=0,\quad n \ne 2k. \end{equation}

Is there a way to prove that $f$ is $\pi$-periodic?

Piyo
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water
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  • Just write down the Fourier series. It will be of the type $ \sum a_{2n}\cos (2nx)+\sum a_{2n} \sin (2nx)$ . – Kavi Rama Murthy Oct 01 '21 at 09:25
  • @KaviRamaMurthy Thanks. I now see that because $S_n(x)=S_n(x+\pi)$ then $\sigma_{n}(x)=\sigma_{n}(x+\pi)$ and because $f(x)=lim_{n \to \infty}\sigma_n(x)$ then $f$ is $\pi$-periodic – water Oct 01 '21 at 11:12

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