Basic question on fourier coefficients

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Main Question or Discussion Point

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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  • #2
jasonRF
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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.
[tex]
\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).
[/tex]
Now if [itex] |f(x_m^{-})|[/itex] is an upper bound of [itex]|f|[/itex] on [itex] [(m-1)2\pi/n, (m-1/2)2\pi/n][/itex], and [itex]|f(x_m^{+})|[/itex] is an upper bound of [itex]|f|[/itex] on [itex] [(m-1/2)2\pi/n, m2\pi/n][/itex], then
[tex]
\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^{-})|,
[/tex]
and similarly for the other integral. Hence I get
[tex]
|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.
[/tex]
Now, from the way I defined [itex]x_m^{-}[/itex] and [itex]x_m^{-}[/itex], it must be true that [itex]|x_m^{-}-x_m^{+}| \leq 2\pi/n[/itex]. So I get
[tex]
|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.
[/tex]

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

cheers,

jason
 
Last edited:

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