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Math Help - Fourier transform

  1. #1
    Super Member Deadstar's Avatar
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    Fourier transform

    I suck horribly at using limits and absolute value signs to manipulate integrals but here is my attempt at the attached question. Im pretty close to the answer... I imagine I'm doing something wrong in the absolute value part...

    While I was typing this I may have solved it but anyway here goes...

    a) f(x+h) - f(x) = \int_{-\infty}^{\infty} \hat{f(\xi)}(e^{2\pi i \xi (x+h)} - e^{2\pi i \xi x})d\xi.

    b) \int_{-\infty}^{\infty} \hat{f(\xi)}(e^{2\pi i \xi (x+h)} - e^{2\pi i \xi  x})d\xi = \int_{|\xi| > 1/|h|} \hat{f(\xi)}(e^{2\pi i \xi (x+h)} - e^{2\pi i \xi  x})d\xi + \int_{|\xi| \leq 1/|h|} \hat{f(\xi)}(e^{2\pi i \xi (x+h)} - e^{2\pi i \xi  x})d\xi

    Now here's where my 'say it to get the right answer' logic comes in.

    Since f is of moderate decrease (or perhaps due to the inequality on |\hat{f(\xi)}|...) then we can find a h such that the first integral on the RHS is zero yes?

    So assuming the above is true...

    We apply absolute values to get...

    f(x+h) - f(x) = \bigg{|} \int_{|\xi| \leq 1/|h|} \hat{f(\xi)}(e^{2\pi i \xi  (x+h)} - e^{2\pi i \xi x})d\xi \bigg{|}

    \leq \int_{|\xi| \leq 1/|h|} |\hat{f(\xi)}| d\xi (Can I do that? Take the absolute value inside, thus getting rid of the exponentials..?)

    At this point onwards I'm just working things out as I go along...

    \leq \int_{|\xi| \leq 1/|h|} \frac{C}{\xi^{1+\alpha}} d\xi

    \leq \int \frac{C}{(1/|h|)^{1+\alpha}} d\xi (?)

    = \frac{C}{\alpha} |h|^{\alpha} = A|h|^{\alpha}.

    So...

    Where do I go wrong (I say that since I have a feeling something is not quite right despite getting the right answer!)
    Attached Thumbnails Attached Thumbnails Fourier transform-sgfhadadh.jpg  
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Deadstar View Post
    I suck horribly at using limits and absolute value signs to manipulate integrals but here is my attempt at the attached question. Im pretty close to the answer... I imagine I'm doing something wrong in the absolute value part...

    While I was typing this I may have solved it but anyway here goes...

    a) f(x+h) - f(x) = \int_{-\infty}^{\infty} \hat{f(\xi)}(e^{2\pi i \xi (x+h)} - e^{2\pi i \xi x})d\xi.

    b) \int_{-\infty}^{\infty} \hat{f(\xi)}(e^{2\pi i \xi (x+h)} - e^{2\pi i \xi  x})d\xi = \int_{|\xi| > 1/|h|} \hat{f(\xi)}(e^{2\pi i \xi (x+h)} - e^{2\pi i \xi  x})d\xi + \int_{|\xi| \leq 1/|h|} \hat{f(\xi)}(e^{2\pi i \xi (x+h)} - e^{2\pi i \xi  x})d\xi

    Now here's where my 'say it to get the right answer' logic comes in.

    Since f is of moderate decrease (or perhaps due to the inequality on |\hat{f(\xi)}|...) then we can find a h such that the first integral on the RHS is zero yes?

    So assuming the above is true...

    We apply absolute values to get...

    f(x+h) - f(x) = \bigg{|} \int_{|\xi| \leq 1/|h|} \hat{f(\xi)}(e^{2\pi i \xi  (x+h)} - e^{2\pi i \xi x})d\xi \bigg{|}

    \leq \int_{|\xi| \leq 1/|h|} |\hat{f(\xi)}| d\xi (Can I do that? Take the absolute value inside, thus getting rid of the exponentials..?)

    At this point onwards I'm just working things out as I go along...

    \leq \int_{|\xi| \leq 1/|h|} \frac{C}{\xi^{1+\alpha}} d\xi

    \leq \int \frac{C}{(1/|h|)^{1+\alpha}} d\xi (?)

    = \frac{C}{\alpha} |h|^{\alpha} = A|h|^{\alpha}.

    So...

    Where do I go wrong (I say that since I have a feeling something is not quite right despite getting the right answer!)
    I didn't look at it closely at all, but shouldn't you have 2|f(x)| as the integrand \left|e^{i\text{stuff}}-e^{i\text{other stuff}}\right|=\text{the sum of the moduli of two points on the unit circle}=2
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  3. #3
    Super Member Deadstar's Avatar
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    Quote Originally Posted by Drexel28 View Post
    I didn't look at it closely at all, but shouldn't you have 2|f(x)| as the integrand \left|e^{i\text{stuff}}-e^{i\text{other stuff}}\right|=\text{the sum of the moduli of two points on the unit circle}=2
    Probably, that won't change much though, just another constant to get sucked into the A at the end.

    It's the moderate decrease part, the absolute value part and the second last line I'm more worried about.
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  4. #4
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    Quote Originally Posted by Deadstar View Post
    It's the moderate decrease part, the absolute value part and the second last line I'm more worried about.
    So much worry! Let's take it in steps.

    - About the second last line: why are you worrying? certainly because you noticed you used the inequality in the wrong direction... so you could have just said it's wrong. In fact, you could even have noticed that \int_0^{1/h}\frac{d\xi}{\xi^{1+\alpha}} is infinite (this is a usual divergent integral)...

    - moderate decrease just tells (if my google search was successful) that |f(x)|\leq\frac{A}{1+x^2}, so it's just a condition that ensures the existence of the Fourier transform and probably the applicability of the inversion theorem, it is not needed in another way here. Note however that if implies that f is bounded by A. In particular, |f(x+h)-f(x)|\leq 2A, hence the conclusion is obvious for |h|>1 (or more generally |h|>c for some c). We don't actually need this argument in the following, but it is good to have in mind that we are interested in small values of h (close to 0).

    - the asumption on \widehat{f}(\xi) tells that it decreases "quickly" when |\xi| gets large; it's a good hunch for bounding the part when |\xi|>1/|h| (which has no reason to equal 0, btw). It indeeds works.

    - It remains to deal with \int_{|\xi|\leq 1/|h|} \widehat{f}(\xi)(e^{2i\pi(x+h)\xi}-e^{2i\pi x\xi})d\xi. Remember |h| is "small", so the integral ranges over a very large interval. Therefore it is a very bad idea to bound the exponentials by 1 since we then get an integral that is about equal to \int_{\mathbb{R}}|\widehat{f}|, i.e. a constant. And a constant can't be less than |h|^{\alpha} for small h... So we need the other part; we need to say that when h is small, then e^{2i\pi\xi(x+h)}\simeq e^{2i\pi \xi x}. To do that, you can prove e^{ia}-e^{ib}=2ie^{i\frac{a+b}{2}}\sin\frac{a-b}{2}, use it here. And use the fact that |\sin u|\leq |u|. Then bound \widehat{f} using the asumption, justify why the integral converges and compute it. It again gives a multiple of |h|^\alpha.
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  5. #5
    Super Member Deadstar's Avatar
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    Bloody hell I've got a lot to fix...
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