I will start with an averaging filter analogy. The simplest (non trivial) and cheapest way to average a discrete signal $x[n]$ is to compute the mean of two adjacent samples, and remplace the two samples by this average. This amounts to compute the discrete system:
$$\frac{x[2n]+x[2n+1]}{2}\,.$$
This can be recast as the convolution $$y[n] = x[n]\ast \frac{ [1\; 1]}{2}$$
You might remember that $\frac{ [1\; 1]}{2}\ast \frac{ [1\; 1]}{2} = \frac{ [1\;2\; 1]}{4}$. And that $\frac{ [1\;2\; 1]}{4}\ast \frac{ [1\; 1]}{2} = \frac{ [1\;3\;3\; 1]}{8}$
This should remind you of the Pascal or binomial triangle, that evolve to a Gaussian shape:
Here, starting with a crude, and a simple finite-impulse response two-point averaging filters, via iterations, you can end up with an IIR, not recursive, infinite-support Gaussian filter. This example is more a metaphor of how discrete filters generate a continuous function.
To be able to add downsampling within the filters, without losing information, contraints the data smoothing picture. In the discrete context, this requires to have a downsampled smoothed version, and a difference version that records when the data is not smooth enough to look like its smooth version. This is the root to most practical discrete wavelet transforms. The Multi Resolution Analysis relates an (yet unknown) scaling function $\phi$ at two-scales, via a linear combination, with weights corresponding to the filter $H$ coefficients. This is explained for instance in Signal Processing: Fourier and Wavelet Representations, 6.1. Introduction, Scaling Function and its Properties, page 191. Finally, the spectrum of the scaling function is given by:
$$\Phi^{(\infty)}(\omega)= \prod_{k=1}^\infty \frac {1} {\sqrt 2} H\left( \frac {\omega} {2^k}\right) \Phi^{(\infty)}(0)$$
where iterated half-cut-off frequencies are apparent. The take-away message is: the spectrum of the scaling function is related to the that of the low-pass filter. And in the same manner, the spectrum of the wavelet function is related to the that of the high-pass filter. Some more pointers are given in:
The limit is that, if you start from continuous wavelets, very few of them can be faithfully implemented as discrete schemes as filters, and those filters, when they exist, don't look like wavelets. Because they don't have too. Their resemblance with wavelets only make sense when they are repeated over and over at different scales, across many dyadic sub-sampling.
And most discrete wavelets (like Daubechies') don't have an analytic formula. But if you look at the bottom line, from afar, the filters and the wavelets look similar in terms of envelope (red line below) and zero-crossings. So the global shape is not so different, as shown below with appropriate scaling of the wavelets and filters:
This similarity seems common for smooth-enough wavelets. Honestly, I have not investigated this into more details, because the actual ressemblance of filters and wavelets is not so important, their properties are way more of interest.
