I'm looking into wavelets to approximate a known square-integrable function $$ f(x) = \sum_{j,k} a_{j,k} \times 2^{j/2}\psi(2^j x - k), \qquad a_{j,k}=2^{j/2}\int f(x) \psi(2^jx-k)dx$$ and I'm happy with Haar wavelets $\psi$ to begin with, mostly because I can calculate the coefficients $a_{j,k}$ in closed form (see below). My 2 questions are:

  1. How do I truncate the double sum over $j,k$? I naively tried $-N\leq j,k\leq N$ for some values of $N$ but that doesn't work for my displaced ramp input function $f(x) = (x - c)^+$ when $c>1$.
  2. For the Haar wavelets, integration by parts yields $$a_{j,k} = -2^{j/2}[F(2^{-j}k) - F(2^{-j}(k + 1/2)) + F(2^{-j}(k+1))]$$ where $F$ is antiderivative of $f$. I did not see this result in any of my online readings, so I am curious why this might not be a useful result?

Thanks! p.


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