Fourier coefficients

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[SOLVED] Fourier coefficients

Homework Statement


For [itex] f \in C^{2\pi}\cap C^1[-\pi,\pi] [/itex], I have to show that

[tex] \sum_{n\in\mathbb{Z}}|c_n(f)| < \infty [/tex]

where c_n(f) is the Fourier coefficient of f;

[tex] c_n(f) = (f, e_n) = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(t)e^{-int}\,dt [/tex]

[itex] f \in C^{2\pi} [/itex] means f continuous and that [itex] f(-\pi) = f(\pi)[/itex].
Hint: Use Cauchy-Schwartz (CS) inequality.

([itex] e_n = e^{int} [/itex])

The Attempt at a Solution



I just can't seem to use CS in a useful way, I keep running into dead ends:

It is easily shown that [itex] c_n(f') = inc_n(f) [/itex]. So by using this and splitting the sum up in 2 parts and using that [itex] e_{-n} = \bar{e}_{n} [/itex], I get

[tex] \sum_{n\in\mathbb{Z}}|c_n(f)| = c_0(f) + \sum_{1}^{\infty}\frac{|(f',e_n)| - |(f',\bar{e}_n)|}{n} [/tex]

Using triangle inequality I can get

[tex] |(f',e_n)| - |(f',\bar{e}_n)| \leq |(f', e_n-\bar{e}_n)| =2|(f', \sin(nt))|[/tex].

Even here CS won't be useful. Is there a different an easier approach?
 
Last edited:

Answers and Replies

  • #2
morphism
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What theorems do you know? Try looking at Parseval and its proof.
 
  • #3
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I kept looking at the wrong inner product to use CS on, that is that inner product of [itex]L_2[/itex], but I should look at the inner product of [itex]\ell_2[/itex] space. So by using that (replacing the minus sign with + in the afore mentioned equation for c_n(f))

[tex] \sum_{n\in\mathbb{Z}}|c_n(f)| \leq c_0(f) + \sum_{1}^{\infty}\frac{|(f',e_n)| + |(f',\bar{e}_n)|}{n} [/tex]

and then using that for [itex]|(f',\bar{e}_n)|/n[/itex] (and the same for [itex]|(f',e_n)|/n[/itex]) we get the summation by CS on l_2 innerproduct to be

[tex] \sum_{1}^{\infty}\frac{1}{n}|(f',e_{-n})| \leq \left(\sum_{1}^{\infty}\frac{1}{n}\right)^{1/2}\left(\sum_{1}^{\infty} |(f',e_{-n})|^2 \right)^{1/2} [/tex]

Were the [itex]C^2 = \sum \frac{1}{n^2}[/itex] is convergent. Before I can use Parseval's theorem, I need to extend the last sum to go from -infinity to +infinity so (1 more inequality), so I get (f', e_n) instead of (f',e_{-n}),

[tex] \sum_{1}^{\infty}\frac{1}{n}|(f',e_{-n})| \leq C\left(\sum_{-\infty}^{\infty} |(f',e_{n})|^2 \right)^{1/2} = C\|f'\| < \infty [/tex]

Is this correct, or have I done some 'illegal' steps? If I haven't then the problem is solved.
 
Last edited:

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