If I take the discrete wavelet transform of a function $f: \mathbb{R} \to \mathbb{R}$ I get a countable set of coefficients, $\omega_{k,l}$, $k,l \in \mathbb{Z}$ where
$$ f(x) = \sum_{k=-\infty}^{\infty} \sum_{l=-\infty}^{\infty}2^{k/2}\omega_{k,l}\psi(2^kx-l) $$
where $\psi$ is the mother wavelet.
Intuitively to me it seems that not every function $f$ should be definable by a countable number of coefficients, since $f$ itself is defined over the uncountable reals. I'd like to build an intuition for what the space of functions is that can be represented by a countable number of wavelet coefficients in this way (or alternatively what functions can not be defined this way)
Analogy to Fourier transform
In the case of a Fourier transform I know that the periodic functions can be represented by a countable number of Fourier coefficients: the sketch of a proof is as follows:
- A periodic function $f(x)$ with period $T$ is the convolution of one period of that function with the infinite set of delta-functions spaced by $T$
- Convolution becomes multiplication under a Fourier transform
- The Fourier transform of a uniformly-spaced train of delta-functions is itself a uniformly-spaced train of delta-functions
- So, the Fourier transform of $f$ is the product of the Fourier transform of one period of the signal with a train of delta functions
- So it only takes non-zero values at locations of the delta
- So it can be uniquely defined by a function over the (countable) locations of the deltas
This argument convinces me that a countable set of Fourier coefficients can only define periodic functions. An aperiodic function will have an uncountable number of Fourier coefficients (i.e. its Fourier Transform is a function over $\mathbb{R}$ not $\mathbb{Z}$.)
Is there a similar argument to show what the limitations of a discrete wavelet transform are? Or am I missing something and is this whole argument wrong-headed?