Let $\psi$ be a wavelet. Can its Fourier transform $\hat{\psi}$ be also wavelet? 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 $\hat f(\omega) =\int_{-\infty}^{\infty} f(t)e^{-i\omega t} dt$.

$f(x)=\sin(x)\cdot\exp(-x^2)$ 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 $$\int_{-\infty}^\infty f dx= \int_{-\infty}^\infty\hat{f}d\xi=0$$
