# When is a continuous wavelet in the Schwartz space?

I was thinking if the continuous wavelet transform can be in the Schwartz space or not? If someone knows can help me and tell me what are the conditions on the mother wavelet or the signal so that the wavelet transform is in the Schwartz space if it is possible of course?

Thanks

Not an expert on Schwartz space, but here are some pointers following definitions on Wiki. It reads, a function's in Schwartz space if

• A. all of its derivatives exist everywhere in $$\mathbb R$$ (are continuously differentiable)
• B. go to zero as $$x \rightarrow \pm \infty$$ faster than $$1/x^p$$, where $$p$$ is real

and gives an example satisfying these:

$$x^\alpha e^{-a|x|^2} \tag{1}$$

where $$a > 0$$ and is real.

### Practical

All CWT wavelets I've seen decay as fast or faster than $$(1)$$ (satisfy B), and some satisfy A:

• A, B: Morlet, Complex Mexican Hat
• B: Bump, Generalized Morse Wavelets

B is a natural criterion since we desire good time-frequency localization, which requires strong decay in both time and frequency domains. Despite their usefulness, strictly analytic wavelets like Generalized Morse, while satisfying B, don't satisfy A per forced discontinuity at $$\omega = 0$$.

### General

Following Wavelet Tour, ch4, a wavelet is a function with zero mean. A function that satisfies the admissibility criterion,

$$C_\psi = \int_0^{+\infty}\frac{|\hat\psi(\omega)|^2}{\omega} d\omega < +\infty \tag{2}$$

is necessarily zero-mean (and is hence a wavelet), and additionally, is continuously differentiable, which follows from it having sufficient time decay:

$$\int_{-\infty}^{+\infty}(1 + |t|)|\psi(t)|dt < \infty \tag{3}$$

As far as I can tell, this satisfies A partialy, because "continuously differentiable != all derivatives are continuously differentiable", also B at least partially, because I don't know if $$(3)$$ implies B. So not all wavelets, nor all admissible wavelets, are in Schwartz space, but latter are at least somewhat close to.