Disclaimer: Of course, any orthonormal system of functions is linearly independent in the sense of linear algebra, but I am interested in infinite linear combinations with potentially "ugly" coefficients.
Specifically, I am considering Karhunen–Loève style expansions of the form $$ f = \sum_{j=0}^\infty \sum_{k \in \mathbb{Z}} c_{j,k} \psi_{j,k} \tag{$\ast$} $$ with respect to an (inhomogeneous) wavelet orthonormal basis. I am mostly interested in compactly supported wavelet systems, and in particular in Daubechies wavelets, but other cases and also homogeneous wavelet systems would be of interest.
My question is as follows:
If $f = 0$ in $(\ast)$, with the series convergent in the sense of tempered distributions, does it follow that all coefficients $c_{j,k}$ vanish?
The main reason why this question is non-trivial (as far as I see) is that one cannot simply "test the equation $(\ast)$ with $\psi_{i,\ell}$", since $\psi_{i,\ell}$ is not a Schwartz function (it is compactly supported, but only has finite smoothness), so that the assumed convergence in the sense of tempered distributions cannot be applied (as far as I can see).
I am interested in this question, since most of the sources regarding the characterization of Besov spaces via wavelets (for instance Theorem 1.20 in Triebel's book "Function spaces and wavelets on domains") claim without further qualification that the wavelet coefficients for representing a function from a Besov space as in $(\ast)$ are uniquely determined. However, as far as I see, the proof only shows this uniqueness among all sequences that belong to the "discrete Besov sequence space", not among all sequences for which the series converges in the sense of tempered distributions.
