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Thread: Convolution problem

  1. #1
    FLT
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    Convolution problem

    I'd like to prove that the only possible solution to f = f*f with f \in L^1 is f = 0. So, I have the following;

    1. Take the fourier transform of both sides

    \hat{f} = \widehat{f*f} = \sqrt{2\pi}\hat{f}\hat{f}

    2. Cancelling we get

    \hat{f} = \frac{1}{\sqrt{2\pi}}

    3. Writing the convolution outright and cancelling appropriately, we obtain

    \int e^{-ix\xi} f(x) dx = 1

    In L^1, this must mean that f = 0, given that the indefinite integral is constant.

    I'm unsure about this - any pointers would be much appreciated!
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  2. #2
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    Quote Originally Posted by FLT View Post
    I'd like to prove that the only possible solution to f = f*f with f \in L^1 is f = 0. So, I have the following;

    1. Take the fourier transform of both sides

    \hat{f} = \widehat{f*f} = \sqrt{2\pi}\hat{f}\hat{f}

    2. Cancelling we get

    \hat{f} = \frac{1}{\sqrt{2\pi}}

    3. Writing the convolution outright and cancelling appropriately, we obtain

    \int e^{-ix\xi} f(x) dx = 1

    In L^1, this must mean that f = 0, given that the indefinite integral is constant.

    I'm unsure about this - any pointers would be much appreciated!
    Steps 1 and 2 are fine, but 3 looks unconvincing to me. Having got as far as step 2, I would use the Riemann–Lebesgue lemma, which says that \hat{f}(\xi)\to0 as |\xi|\to\infty, contradicting the assertion that \hat{f} = \frac{1}{\sqrt{2\pi}}. The contradiction comes from the cancellation in step 2, which assumed that \hat{f}\ne0. (So maybe step 2 wasn't quite as fine as I first said. )
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  3. #3
    FLT
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    Quote Originally Posted by Opalg View Post
    Steps 1 and 2 are fine, but 3 looks unconvincing to me. Having got as far as step 2, I would use the Riemann–Lebesgue lemma, which says that \hat{f}(\xi)\to0 as |\xi|\to\infty, contradicting the assertion that \hat{f} = \frac{1}{\sqrt{2\pi}}. The contradiction comes from the cancellation in step 2, which assumed that \hat{f}\ne0. (So maybe step 2 wasn't quite as fine as I first said. )
    Of course! I've overlooked that result;

    If f \in L^1 then \lim_{\xi\to\pm\infty}\hat{f}(\xi) = 0

    Ok, so would it suffice to let the RHS of step 1 tend to 0 and conclude the result from there? Or have I missed a trick completely?
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