Basic question on fourier coefficients

[neginf]

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.

 

[jasonRF]

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.
<br /> \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)| <br /> =\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).<br />
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
<br /> \int_{(m-1)2\pi/n}^{(m-1/2)2\pi/n}\,dx\,|f(x)|\,sin(nx)<br /> \leq |f(x_m^{-})| \int_{(m-1)2\pi/n}^{(m-1/2)2\pi/n}\,dx \, sin(nx)<br /> = \frac{2}{n} |f(x_m^{-})|,<br />
and similarly for the other integral. Hence I get
<br /> |b_n| \leq \frac{2}{n\pi} \sum_{m=1}^{n} |f(x_m^{-})|-|f(x_m^{+})|<br /> \leq<br /> \frac{2}{n\pi} \sum_{m=1}^{n} |f(x_m^{-})-f(x_m^{+})|<br /> \leq<br /> \frac{2}{n\pi} \sum_{m=1}^{n} c |x_m^{-}-x_m^{+}|^a.<br />
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
<br /> |b_n| \leq \frac{2c}{n\pi}\sum_{m=1}^{n} \left( \frac{2\pi}{n}\right)^a<br /> = \frac{2c}{\pi}\left( \frac{2\pi}{n}\right)^a<br /> \leq c \left( \frac{2\pi}{n}\right)^a.<br />

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

cheers,

jason