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It is known that the function defined as

$$ f (\theta) = \frac{1}{2}(\pi - \theta )$$

for $0< \theta < 2 \pi $, and $f(0)=0$, and extended periodically to the whole axis has the Fourier series

$$ \sum_{n=1}^\infty \frac{\sin n\theta }{n } .$$

Here the Fourier coefficients $F_n$ decay as $O(1/n)$.

My problem is, what theorem tells us this behavior without calculation? The Riemann-Lebesgue theorem only gives $F_n \rightarrow 0$. We also have the theorem that if $f$ is in $C^k$, then $F_n = O(1/n^k)$. But here the specific $f$ is not even in $C^0$.

pie
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2 Answers2

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For functions of bounded variation, we can show that its Fourier coefficient is $\mathcal{O}(1/n)$. For a proof, you can start with a monotone function. Finally, noting that $$ \int_0^{2\pi}f(x)e^{-inx}dx=-\int_0^{2\pi}f(x+\frac{\pi}{n})e^{-inx}dx$$ would give the desired result.

Myunghyun Song
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Another argument is that by differentiating the jump becomes a delta-function, whose Fourier series is constant, but differentiation in Fourier space causes the $n$th Fourier coefficient to be multiplied by $n$. So $1/n$ behaviour indicates a jump of finite size (and can sometimes be useful in checking Fourier calculations).

Richard Martin
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