Can someone explain why the admissability of wavelets allows us to conclude the limit of the Fourier transform of a wavelet approaches 0 when $\omega $ approaches 0. Then if the Fourier transform of the wavelet is continuous it equals 0 when $\omega$ is 0. This then concludes the integral of the wavelet over the reals is 0?
Thanks
So, to repeat the definition, a function $f \in L^2(\Bbb{R})$ is called admissible if
$$ \int |\widehat{f}|^2/x \, dx <\infty. $$
This alone does not imply $\hat{f}(\xi) \to 0$ as $\xi \to 0$.
To see this, let (e.g.)
$$ \hat{f} := \sum_{n=1}^\infty \chi_{(2^{-n}, 2^{-n}+4^{-n})}. $$
Then $2^{-n}+4^{-n} \leq 2^{-n}+ 2^{-n}=2^{-(n-1)}$, which implies that the intervals above are pairwise disjoint. Hence, $$ \int |f|^2/\xi \, d\xi = \sum_n \int_{2^{-n}}^{2^{-n}+4^{-n}} 1/\xi \, d\xi \leq \sum_n \int_{2^{-n}}^{2^{-n}+4^{-n}} 2^n \, d\xi = \sum 2^n / 4^n <\infty. $$
Also, $\hat{f}$ is clearly $L^2$, so by Plancherel is the Fourier transform of some $f \in L^2$.
But $f\notin L^1$, because otherwise $\hat{f}$ would be continuous.
So let us now assume that $\hat{f}$ is continuous. This implies $\hat{f}(0)=0$, because otherwise $|\hat{f}(\xi)|\geq \epsilon$ on $(-\delta, \delta)$ for suitable $\epsilon, \delta >0$. But using $\int_0^1 1/x \, dx =\infty$, this yields that $f$ cannot be admissible.
Now, let us finally assume $f \in L^1$. It is well known that this implies continuity of $\hat{f}$ and hence (by the above)
$$ 0=\hat{f}(0) = \int f(x) \cdot e^{-2 \pi i x \cdot 0} \, dx =\int f\, dx, $$ If $f$ is admissible.
