I'm reading through some material on wavelet and I'm finding it difficult to see why the following is true.

Let $\Psi(\omega)$ is the Fourier transform of the mother wavelet $\psi(t)$


then many paper on wavelet would immediate jump to, is obvious that:


But wouldn't the above yield a 0/0 condition?

I've also tried relaxing the limits and take a derivative on both side, but this would only imply that the mother wavelet is zero at some frequency $\omega$ not $\omega$ = 0. How do you rigorously prove the above relation?


1 Answer 1


It's obvious that Fourier transform of the mother wavelet is zero at the origin, otherwise the integral, whose argument is nonnegative, would be unbounded. Just because both the numerator and the denominator are zero, it does not mean it does not have a limit. So we know that $\Psi(0)=0$, but this is merely necessary, not sufficient for the integral to be bounded.

Consider the function $f: \omega \to |\sin(\omega)|^2/|\omega|$. How does this function's integral from $(-T,T)$ behave as $T \to \infty$. I hope that gives you some ideas.


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