# Basic question on fourier coefficients

If f(x) has a period of 2*pi and |f(x)-f(y)| <= c*|x-y|^a where a and c are positive constants, why are are n-th Fourier coefficients <= c*(pi/n)^a ?

Help or hints would be appreciated.

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jasonRF
Gold Member
I made a stab at this, but ended up with a slightly looser bound. But perhaps it will still be of some use. Here I just do the "sine" coeffs; the argument for cos is essentially the same.
$$\left| b_n \right| \leq \frac{1}{\pi} \int_0^{2 \pi}\,dx\,|f(x)| |sin(nx)|=\frac{1}{\pi}\sum_{m=1}^{n} \int_{(m-1)2\pi/n}^{m2\pi/n}\,dx\,|f(x)|\,|sin(nx)| =\frac{1}{\pi}\sum_{m=1}^{n} \int_{(m-1)2\pi/n}^{(m-1/2)2\pi/n}\,dx\,|f(x)|\,sin(nx) - \int_{(m-1/2)2\pi/n}^{m2\pi/n}\,dx\,|f(x)|\,sin(nx).$$
Now if $|f(x_m^{-})|$ is an upper bound of $|f|$ on $[(m-1)2\pi/n, (m-1/2)2\pi/n]$, and $|f(x_m^{+})|$ is an upper bound of $|f|$ on $[(m-1/2)2\pi/n, m2\pi/n]$, then
$$\int_{(m-1)2\pi/n}^{(m-1/2)2\pi/n}\,dx\,|f(x)|\,sin(nx) \leq |f(x_m^{-})| \int_{(m-1)2\pi/n}^{(m-1/2)2\pi/n}\,dx \, sin(nx) = \frac{2}{n} |f(x_m^{-})|,$$
and similarly for the other integral. Hence I get
$$|b_n| \leq \frac{2}{n\pi} \sum_{m=1}^{n} |f(x_m^{-})|-|f(x_m^{+})| \leq \frac{2}{n\pi} \sum_{m=1}^{n} |f(x_m^{-})-f(x_m^{+})| \leq \frac{2}{n\pi} \sum_{m=1}^{n} c |x_m^{-}-x_m^{+}|^a.$$
Now, from the way I defined $x_m^{-}$ and $x_m^{-}$, it must be true that $|x_m^{-}-x_m^{+}| \leq 2\pi/n$. So I get
$$|b_n| \leq \frac{2c}{n\pi}\sum_{m=1}^{n} \left( \frac{2\pi}{n}\right)^a = \frac{2c}{\pi}\left( \frac{2\pi}{n}\right)^a \leq c \left( \frac{2\pi}{n}\right)^a.$$

So I have an extra $2^a$. Perhaps you can see a better way!

cheers,

jason

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