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I have read that in order for a function to be a wavelet, it needs to fufull the L2 Norm property.
But I don't know what that is and there wasn't an explanation either. I know theres a L2 norm in relation to vectors, but that doesn't seam to be the same thing. So what does it mean for a function to be L2 conform?

Nick
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user2741831
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1 Answers1

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While some context is missing, like what the domain of your function is, for a function $f$ to have bounded $L^2$ norm means that $\int_D |f(x)|^2 dx<\infty$. The norm is induced by an inner product $\langle f, g \rangle = \int_D f(x)\overline{g(x)}dx$, and the collection of all functions with bounded $L^2$ norm on $D$ form a Hilbert space. There is a lot of nice theory of Hilbert spaces, and all of this is standard when studying measure theory and functional analysis, so a book on measure theory might be a good starting place.

Aaron
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  • thanks, but what is the line ontop of g(x) for, also why does f(x) need to be absolute when (-x)(-x)=xx? – user2741831 Jan 28 '20 at 17:35
  • Complex conjugation, in case you want to deal with complex valued functions instead of just real valued ones, in which case the absolute value becomes necessary because $|z|^2=z\overline z \neq z^2$ for complex numbers. – Aaron Jan 28 '20 at 17:36
  • ah, thanks makes sense – user2741831 Jan 28 '20 at 17:39