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The admissibility condition for a wavelet $\psi$ is:

$\int \frac{|\hat\psi(x)|^2}{|x|} dx < \infty$, with $\hat\psi$ the Fourier transform of $\psi$.

A necessary and sufficient condition should be:

$$\int \psi(x) dx = 0\quad \mathrm{(or} \;\hat\psi(0) = 0\mathrm{)}$$

But I can't see how you should get from one to the other...

Does someone have some suggestions?

1 Answers1

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The admissibility condition (at least in the most common versions in 1D) can be stated as follows: function $\phi$ is a synthesis (or reconstruction) wavelet for analysis wavelet $\psi$ if the two constants:

$$c^{\pm}_{\phi,\psi} = 2\pi \int_{0}^{+\infty}\tfrac{\hat{\overline{\psi}}(\pm\omega)\hat\phi(\pm\omega)}{\omega} d\omega $$ are finite, non null and equal (see M. Holschneider, Wavelets, an analysis tool, 1995, p. 65 sq.).

This can be satisfied by a lot of couples of functions. If you now want the synthesis wavelet to be the same as the analysis wavelet, ie $\phi = \psi$, then a necessary condition is that $$\int \psi = 0$$ to avoid the singularity at $0$. This is the root for using discrete orthogonal wavelets (otherwise, you have for instance biorthogonal wavelets, used in JPEG 2000 image compression). In other words, the wavelet has zero-mean, which is equivalent to say (when integrals exists) that its spectrum vanished at zero (on the frequency axis).

But it is not sufficient in theory, and requires additional smoothed/boundedness/localization on $\psi$. So my suggestion is to refer to accurate sources for sufficient conditions.