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Math Help - Show that an integral is well defined

  1. #1
    Ase
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    Show that an integral is well defined

    Hi

    I am trying to show that for f belonging to L^2(-pi;pi) the integral that defines the complex Fourier Coefficients is well defined. In other words what I need to show is that

    int_from -pi to pi(|f(x)*exp(-i*k*x)|dx) < infinity (limited)

    I was thinking that since f belongs to L^2(-pi;pi) then the integral of this will be finite. Further more since the inteval is limited (-pi;pi) and k belonging to the set of integers the complex exponential function would also be finite.

    Am I right?
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  2. #2
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    Quote Originally Posted by Ase View Post
    Hi

    I am trying to show that for f belonging to L^2(-pi;pi) the integral that defines the complex Fourier Coefficients is well defined. In other words what I need to show is that

    int_from -pi to pi(|f(x)*exp(-i*k*x)|dx) < infinity (limited)

    I was thinking that since f belongs to L^2(-pi;pi) then the integral of this will be finite. Further more since the inteval is limited (-pi;pi) and k belonging to the set of integers the complex exponential function would also be finite.

    Am I right?
    You didn't prove anything. Basically, since |e^{-ikx}|=1, you have to prove that if \int_{-\pi}^\pi |f(x)|^2 dx<\infty, then \int_{-\pi}^\pi |f(x)| dx<\infty. The advice is to use Cauchy-Schwarz inequality (or the elementary inequality a\leq\frac{a^2+1}{2}).
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  3. #3
    Ase
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    Hi Laurent

    Thanks a lot. I will get right at it.
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  4. #4
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    well-defined integral

    For f \in L^{2}(-\pi,\pi) I would like to show that:

    \int_{-\pi}^{\pi}|f(x)e^{-ikx}|< \infty


    I have done the following:

    \int_{-\pi}^{\pi}|f(x)e^{-ikx}|dx= \int_{-\pi}^{\pi}|f(x)(\cos(kx)-i\sin(kx)|dx\leq \int_{-\pi}^{\pi}|f(x)dx|



    How do I continue (I tried to use Cauchy-Sjcwarz inequality as suggested but it did not lead my anywhere)

    Thanks a bunch.
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  5. #5
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    Quote Originally Posted by surjective View Post
    For f \in L^{2}(-\pi,\pi) I would like to show that:

    \int_{-\pi}^{\pi}|f(x)e^{-ikx}|< \infty


    I have done the following:

    \int_{-\pi}^{\pi}|f(x)e^{-ikx}|dx= \int_{-\pi}^{\pi}|f(x)(\cos(kx)-i\sin(kx)|dx\leq \int_{-\pi}^{\pi}|f(x)dx|



    How do I continue (I tried to use Cauchy-Sjcwarz inequality as suggested but it did not lead my anywhere)

    Thanks a bunch.
    To apply Cauchy-Schwarz, write \int f(x) dx = \int f(x)g(x)  dx where g(x)=1.

    Alternatively, use the simple inequality I gave in the previous post: \int f\leq \frac{1}{2}\Big((\int f^2)+(\int 1)\Big)=\frac{1}{2}\Big((\int f^2) + 2\pi\Big)<\infty.
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  6. #6
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    well-defined intergral

    Here is what I have done:

    <br />
\int_{-\pi}^{\pi}|f(x)|dx=\int_{-\pi}^{\pi}|f(x)g(x)|dx \leq \int_{-\pi}^{\pi}|f(x)||g(x)|dx \leq \int_{-\pi}^{\pi}|f(x)||f(x)|dx < \infty<br />

    Correct?
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  7. #7
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    Quote Originally Posted by surjective View Post
    Here is what I have done:

    <br />
\int_{-\pi}^{\pi}|f(x)|dx=\int_{-\pi}^{\pi}|f(x)g(x)|dx \leq \int_{-\pi}^{\pi}|f(x)||g(x)|dx \leq \int_{-\pi}^{\pi}|f(x)||f(x)|dx < \infty<br />

    Correct?
    You should take a look at what Cauchy-Schwarz inequality for integrals is; because I can't see how you're applying it here.
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  8. #8
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    well-defined integral revisited

    Hello,

    Let me try one more time:

    \int_{-\pi}^{\pi}|f(x)|dx= \int_{-\pi}^{\pi}|f(x)g(x)|dx \leq \left(\int_{-\pi}^{\pi}|f(x)|^{2}dx \right)^{\frac{1}{2}}\left(\int_{-\pi}^{\pi}|g(x)|^{2}dx \right)^{\frac{1}{2}} \leq

    \left(\int_{-\pi}^{\pi}|f(x)|^{2}dx \right)^{\frac{1}{2}}\left(\int_{-\pi}^{\pi}|f(x)|^{2}dx \right)^{\frac{1}{2}} < \infty

    What do you think?
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  9. #9
    MHF Contributor

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    Quote Originally Posted by surjective View Post
    Hello,

    Let me try one more time:

    \int_{-\pi}^{\pi}|f(x)|dx= \int_{-\pi}^{\pi}|f(x)g(x)|dx \leq \left(\int_{-\pi}^{\pi}|f(x)|^{2}dx \right)^{\frac{1}{2}}\left(\int_{-\pi}^{\pi}|g(x)|^{2}dx \right)^{\frac{1}{2}} \leq

    \left(\int_{-\pi}^{\pi}|f(x)|^{2}dx \right)^{\frac{1}{2}}\left(\int_{-\pi}^{\pi}1 dx \right)^{\frac{1}{2}} =\sqrt{2\pi}\left(\int_{-\pi}^{\pi}|f(x)|^{2}dx \right)^{\frac{1}{2}}< \infty

    What do you think?
    I did the corrections above.
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