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Homework Help: Real Analysis integral issue

  1. Oct 19, 2009 #1
    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]

    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]


    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! :tongue:
  2. jcsd
  3. Oct 19, 2009 #2


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    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
  4. Oct 19, 2009 #3
    lanedance, thanks. I've gotten a bit further, but I'm still stuck. I guess I don't see how showing that

    \mathop{\lim}\limits_{n \to \infty} \int_{-\infty}^{\infty}dx (f - \psi)(cosnx) \leq \epsilon

    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" ...
  5. Oct 19, 2009 #4


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    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 (f-psi) 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))
    = \mathop{\lim}\limits_{n \to \infty} \int_{-\infty}^{\infty}dx (f.cosnx) - \mathop{\lim}\limits_{n \to \infty} \int_{-\infty}^{\infty}dx (\psi .cosnx)
    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


    subtituting these back into the original absolute value integral equality should pretty much do it i think
    Last edited: Oct 19, 2009
  6. Oct 19, 2009 #5


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    ok re-reading what you wrote, i see you had pretty much done what i suggested :redface:

    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

    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)
    Last edited: Oct 19, 2009
  7. Oct 19, 2009 #6


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    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 ill-behaved 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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