- #1
futurebird
- 272
- 0
1. Royden Chapter 4, # 16, P.94
Establish the Riemann-Lebesgue Theorem:
If f is integrable on [tex]( - \infty, \infty)[/tex] then,
[tex]\mathop{\lim}\limits_{n \to \infty}\int_{\infty}^{\infty}f(x) \cos nx dx =0[/tex]
2. The hint says to use this theorem:
Let f be integrable over E then given [tex]\epsilon > 0[/tex] there is a step function such that
[tex]\int_{E}| f - \psi| < \epsilon[/tex]
If f is analytic then we can just integrate by parts:
[tex]\mathop{\lim}\limits_{n \to \infty}\int_{\infty}^{\infty}f(x) \cos nx dx = \mathop{\lim}\limits_{a \to \infty} \left( f(x) \frac{\sin nx}{n} - \frac{1}{n}\int_{\infty}^{\infty}f'(x) \sin nx dx \right)[/tex]
[tex]=0[/tex]
but otherwise we can find a step function that is very close for f... and then use the theorem above but I don't know how. My first issue is that the theorem is an integral over a set E... but can [tex]E=( - \infty, \infty)[/tex]?
I could really use some help. Please go slowly with me, this stuff makes me deeply confused!
Establish the Riemann-Lebesgue Theorem:
If f is integrable on [tex]( - \infty, \infty)[/tex] then,
[tex]\mathop{\lim}\limits_{n \to \infty}\int_{\infty}^{\infty}f(x) \cos nx dx =0[/tex]
2. The hint says to use this theorem:
Let f be integrable over E then given [tex]\epsilon > 0[/tex] there is a step function such that
[tex]\int_{E}| f - \psi| < \epsilon[/tex]
The Attempt at a Solution
If f is analytic then we can just integrate by parts:
[tex]\mathop{\lim}\limits_{n \to \infty}\int_{\infty}^{\infty}f(x) \cos nx dx = \mathop{\lim}\limits_{a \to \infty} \left( f(x) \frac{\sin nx}{n} - \frac{1}{n}\int_{\infty}^{\infty}f'(x) \sin nx dx \right)[/tex]
[tex]=0[/tex]
but otherwise we can find a step function that is very close for f... and then use the theorem above but I don't know how. My first issue is that the theorem is an integral over a set E... but can [tex]E=( - \infty, \infty)[/tex]?
I could really use some help. Please go slowly with me, this stuff makes me deeply confused!