Krdytkyu

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

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

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 $C^0$.

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
$$ int_0^{2pi}f(x)e^{-inx}dx=-int_0^{2pi}f(x+frac{pi}{n})e^{-inx}dx$$ would give the desired result.

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).