- #1

Zaare

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I need help with the last part of the following problem:

Let [tex]f(x)[/tex] be a [tex]2\pi[/tex]-periodic and Riemann integrable on [tex][-\pi,\pi][/tex].

Show that

I have done (a). And I was able to show that the sum satisfies the condition. What I don't understand is:

how can I use it to show that the result in (a) cannot be improved?

I thought that the sum can be iterpreted as the Fourier series of [tex]g(x)[/tex], which would say that the Fourier coefficient of [tex]g(x)[/tex] is [tex]1/{n^\alpha}[/tex], where [tex]n=2^k[/tex]. That is the resemblence I see between (a) and (b). But I don't know what to do with that.

Let [tex]f(x)[/tex] be a [tex]2\pi[/tex]-periodic and Riemann integrable on [tex][-\pi,\pi][/tex].

**(a)**Assuming [tex] f(x) [/tex] satisfies the Hölder condition of order [tex] \alpha [/tex][tex] \left| {f\left( {x + h} \right) - f\left( x \right)} \right| \le C\left| h \right|^\alpha ,[/tex]

for some [tex]0 < \alpha \le 1[/tex], some [tex] C > 0 [/tex] and all [tex] x, h [/tex].Show that

[tex] {\hat f\left( n \right)} \le O ( \frac {1} { \left| n \right| ^ \alpha} ) [/tex]

where [tex] \hat f\left( n \right) [/tex] is the Fourier koefficient.**(b)**Proove that the above result cannot be improved by showing that the function[tex] g\left( x \right) = \sum\limits_{k = 0}^\infty {2^{ - k\alpha } e^{i2^k x} } [/tex]

also with [tex]0 < \alpha \le 1,[/tex] satisfies[tex] \left| {g\left( {x + h} \right) - g\left( x \right)} \right| \le C\left| h \right|^\alpha .[/tex]

I have done (a). And I was able to show that the sum satisfies the condition. What I don't understand is:

how can I use it to show that the result in (a) cannot be improved?

I thought that the sum can be iterpreted as the Fourier series of [tex]g(x)[/tex], which would say that the Fourier coefficient of [tex]g(x)[/tex] is [tex]1/{n^\alpha}[/tex], where [tex]n=2^k[/tex]. That is the resemblence I see between (a) and (b). But I don't know what to do with that.

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