
#1
Oct1909, 09:04 AM

P: 276

1. Royden Chapter 4, # 16, P.94
Establish the RiemannLebesgue 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] 3. 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! 



#2
Oct1909, 09:32 AM

HW Helper
P: 3,309

how about this
start with [tex]\mathop{\lim}\limits_{n \to \infty} \int_{\infty}^{\infty}dx (f  \psi)(cosnx) \leq \mathop{\lim}\limits_{n \to \infty} \int_{\infty}^{\infty}dx (f  \psi)(cosnx)[/tex] and try using some of the properties of absolute values & your step equation to try & get to the desired result 



#3
Oct1909, 10:32 AM

P: 276

lanedance, thanks. I've gotten a bit further, but I'm still stuck. I guess I don't see how showing that
[tex] \mathop{\lim}\limits_{n \to \infty} \int_{\infty}^{\infty}dx (f  \psi)(cosnx) \leq \epsilon [/tex] is proving that the limit is 0 for integrable functions, f. I can picture what this theorem is doing and why it works. The frequency of cos nx becomes as frequent as we please so all of the values of the function are "canceled out" (in a sense) ... But, I'm just not seeing the connection here. It's saying the parts of f, that can't be approximated by a step function, are smaller than epsilon... but they need to be zero... not just "very small" ... 



#4
Oct1909, 10:49 AM

HW Helper
P: 3,309

Real Analysis integral issue...
hmm.. can you clarify how the step function is defined in the theorem?
i was thinking it was more saying that over a set E, a step function psi exists such that the integral of (fpsi) is zero, so thinking of psi as a sort of "average value of f" over E, determined by the integral (& maybe weighted by the length of the step) anyway, regardless, I still think we can get close based on the following [tex]\mathop{\lim}\limits_{n \to \infty} \int_{\infty}^{\infty}dx (f  \psi)(cosnx) = \mathop{\lim}\limits_{n \to \infty} (\int_{\infty}^{\infty}dx (f.cosnx)  \int_{\infty}^{\infty}dx (\psi.cosnx)) [/tex] [tex] = \mathop{\lim}\limits_{n \to \infty} \int_{\infty}^{\infty}dx (f.cosnx)  \mathop{\lim}\limits_{n \to \infty} \int_{\infty}^{\infty}dx (\psi .cosnx) [/tex] should be able to evaluate the step function integral without much trouble and take the limit now playing with the absolute value side of the inequality [tex] \mathop{\lim}\limits_{n \to \infty} \int_{\infty}^{\infty}dx (f  \psi)(cosnx) \leq \mathop{\lim}\limits_{n \to \infty} \int_{\infty}^{\infty}dx (f  \psi)(cosnx) \leq \mathop{\lim}\limits_{n \to \infty} \int_{\infty}^{\infty}dx (f  \psi) < \epsilon [/tex] subtituting these back into the original absolute value integral equality should pretty much do it i think 



#5
Oct1909, 11:17 AM

HW Helper
P: 3,309

ok rereading what you wrote, i see you had pretty much done what i suggested
so what's your definition of integrable here? but i think by a step function you mean effctively a partition of E, into intervals [x_k, x_k+1], where psi takes on some f(c_k) such that c_k is in [x_k, x_k+1]? i think the key is you can choose epsilon as close to zero as you like, and difference in the integral of the parts of you speak of will be less than epsilon and for any partition the integral would become something like [tex]\mathop{\lim}\limits_{n \to \infty} \int_{\infty}^{\infty}dx (\psi .cosnx) = \mathop{\lim}\limits_{n \to \infty} \sum_k \int_{x_k}^{x_k+1}dx f(c_k) cosnx = \mathop{\lim}\limits_{n \to \infty} \sum_k f(c_k) \frac{sinnx}{n} \rightarrow 0 [/tex] which cancel as you say as the sinusoid oscillates infintely rapidly on the constant function as this integral is zero, it shows the required though i am a tiny bit worried about the infinite sum here, maybe you can also show f(x) has to tend to zero at +inf, for the function to be integrable on (inf,inf) 



#6
Oct1909, 11:35 AM

HW Helper
P: 3,309

hope i'm not missing the point, just starting to look at lebesgue measure, if you're talking about from that point of view i think the fact that f is integrable shows that the subsets where f is illbehaved are small (measure zero) and the integral of f can be estimated arbitrarily close by the integral of the step functions, making the ideas above salvagable



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