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Let $\psi$ be a wavelet. Can its Fourier transform $\hat{\psi}$ be also wavelet? produce an example or prove that it is not possible. A wavelet is a function $\psi:\mathbb R\to\mathbb R$ such that (i) $\psi \in L^1(R) \cap L^2(R)$, (ii) $\int_{-\infty}^{\infty} \psi(t) dt = 0$, fourier transform is $\hat f(\omega) =\int_{-\infty}^{\infty} f(t)e^{-i\omega t} dt$.

J. M. ain't a mathematician
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  • Please, since you are proficient already in LaTeX, enclose the mathematics portions of your questions in dollar signs $. This way it will invoke the automatic MathJax Renderer to make your post much more readable. – Willie Wong Oct 26 '10 at 09:36
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    Not sure, if s/he'll listen to you, Willie; if you'll have a look at this user's history on this site, s/he's not very responsive to feedback. :P – J. M. ain't a mathematician Oct 26 '10 at 09:43
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    Yes it can...but to gain the good will of the community it is helpful to avoid the imperative mood when phrasing your questions. – Robin Chapman Oct 26 '10 at 09:44
  • How about taking a rectangular step? So from $-1$ to $0$ it is $-1$, from $0$ to $1$ it is $1$ and $0$ otherwise. This has compact support so it is in $L^1 \cap L^2$, but the Fourier transform doesn't have compact support. Stronger: The Fourier transform is some thing that looks like a $\text{sinc}$-function, and this one certainly is not in $L^1$! Oh, wait, nevermind, you're asking for a case where it is possible. – JT_NL Oct 26 '10 at 10:06
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    As stated $\phi = 0$ satisfies. – JT_NL Oct 26 '10 at 10:16
  • @Jonas: There is not $f\in L^2(\mathbb{R})$ such that $f$ and $\hat{f}$ has compact support. (If $f\ne 0$ has compact support then $\hat{f}$ is entire - Paley-Wiener). – AD - Stop Putin - Oct 26 '10 at 11:20
  • Guessing..What about $\sin(x)\cdot\exp(-x^2)$? – AD - Stop Putin - Oct 26 '10 at 11:21
  • @AD: I know, the function I give has compact support so the Fourier transform does not. But I misinterpreted the question. By the way, your guess works. Can you tell me how you how you thought up that example? – JT_NL Oct 26 '10 at 11:23
  • @Jonas: I wrote down the details (I thought that $exp(-x^2)$ is even with fast decay and that $\sin$ is bounded and odd hence the same should be true for $\hat{f}$). – AD - Stop Putin - Oct 26 '10 at 12:03

1 Answers1

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$f(x)=\sin(x)\cdot\exp(-x^2)$ should do, because:

  • The decay of $f$ ensure $f,\hat{f}\in L^1\cap L^\infty$.
  • $f$ is odd $f(-x)=-f(x)$.
  • $\hat{f}(-\xi)=\int_{-\infty}^\infty e^{-ix(-\xi)}f(x)dx=\int_{-\infty}^\infty e^{-i(-x)\xi}f(x)dx=\int_{\infty}^{-\infty} e^{-it\xi}f(-t)(-dt)=-\hat{f}(\xi)$ where in the last step we used that $f$ is odd.

EDIT: If $g$ is integrable and odd then $$\int_{-\infty}^0g(x)dx =\int_{-\infty}^0-g(-x)dx=\int_{+\infty}^0g(t)dt=-\int_0^{+\infty}g(t)dt$$ hence $$\int_{-\infty}^\infty g(t)dt = 0.$$ This imply that $$\int_{-\infty}^\infty f dx= \int_{-\infty}^\infty\hat{f}d\xi=0$$

AD - Stop Putin -
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  • Is this meant to be an example of a wavelet whose Fourier transform ψ is also wavelet? If yes why you consider f is odd unless it implies that $\hat{f}$ is wavelet too? –  Nov 22 '10 at 19:22
  • @alvoutila: Do you see it now? (I am honestly very happy that you asked, for your own sake). – AD - Stop Putin - Nov 22 '10 at 21:44
  • In fact this is part of a family of wavelets, the Gabor wavelets, which is closed under Fourier transform. – Peter Taylor Jan 30 '12 at 18:42