Can somebody help me understand this quote from Wikipedia:
In mathematics, a dual wavelet is the dual to a wavelet. In general, the wavelet series generated by a square-integrable function will have a dual series, in the sense of the Riesz representation theorem. However, the dual series is not itself in general representable by a square-integrable function.
What do they mean by "dual series, in the sense of the RRT"? Is this related to adjoint maps? As far as I understand, adjoint maps are the representatives of dual maps (between the dual spaces)? But why are those dual series not always representable? Isn't RRT more about ensuring a representation, not about ensuring a dual (which might or might not be representable) like the quote suggests?
It reminds me about distributions, which are not always representable (like the Dirac delta function), but I forgot the reason why they evade the RRT, is this related?
Or maybe they don't mean the dual of the mother wavelet itself, but the dual of the map from $L^2$ to the open half-plane (from integrating a given function against all scales and translations of the mother wavelet), but such a dual would be (by RRT) an element of the open half-plane, not of $L^2$?
That article is strange. I would call $\psi$ a wavelet only when the affine frame of it is actually a frame of $L^2$.
The distinction of $\tilde\psi^{jk}$ and $\tilde\psi_{jk}$ makes no sense. It would be a little better if the first were the coordinate functional and the second the point of the Hilbert space obtained via Riesz duality.
The contradiction would then lie in the inability to fit this dual frame, or any other collection of coordinate functions, into an affine frame spanned by a single function $\psi$.
If you have actual wavelets, related to a compactly supported scaling function, then the factorization of the biorthogonality condition can lead to situations where the dual scaling function does not exist (the dual refinement equation has no solution) as element of $L^2$ in the worst case, or is not (piecewise) continuous in a (mildly) bad case.