Fourier coefficient problem

  • Thread starter Zaare
  • Start date
  • #1
54
0
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].

(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.
 
Last edited:

Answers and Replies

  • #2
54
0
Ok, here's the way I reason:
In (a) I showed that the fourier coefficient has an upper limit. Then in (b) I showed that a certain function satisfying the desired condition has a Fourier coefficient which is of the same size as the upper limit in (a). Then obviously the upper limit cannot be improved.
Is it this obvious, or am I forgetting something?
 

Related Threads on Fourier coefficient problem

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
765
  • Last Post
Replies
13
Views
2K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
5
Views
3K
Top