4
$\begingroup$

Haar wavelets are defined as:

$$ \psi_{0,0}(t) = \begin{cases} 1, \text{ for } 0<t< 1/2\\ -1, \text{ for } 1/2<t<1 \\ 0, \text{ otherwise } \end{cases} $$ And for $n \geq 0$, $0 \leq k < 2^n$ $$\psi_{n,k} = 2^{n/2} \psi_{0,0}(2^n t -k).$$

I was able to prove the orthonormality of those functions, and tried to approximate polynomials and sin/cos functions, since I know that those are a basis, but both methods failed. This result has been proven by B.S. Kashin, A.A. Saakyan, "Orthogonal series" , Moscow (1984), but Russian isn't my forte. Full derivations and general pointers are appreciated.

I encountered these in the context of Brownian motion.

Edit: Wolfram alpha provides this graph of first few functionsSourc : WA

$\endgroup$
1
  • 1
    $\begingroup$ Note that $f\mapsto\langle1,f\rangle$ is continuous on $L^2[0,1]$. But for all the $\psi_{n,k}$ you have $\langle1,\psi_{n,k}\rangle=0$. So for the limit of any sum of these terms must also be mapped to zero and $1$ cannot lie in the closure of the span of the $\psi_{n,k}$ $\endgroup$
    – s.harp
    Jan 17, 2017 at 14:02

1 Answer 1

2
$\begingroup$

You need to include the constant function $1$ or it won't work. With this I want to show inductively that $$\chi_{[k2^{-n},(k+1)2^{-n}]}$$ lies in the span for all $0≤k<2^n$. This implies $\chi_{[a,b]}$ lies in the span (as an element of $L^2$, ie differing from that function on a null-set) whenever $a,b$ are of the form $2^n k$ for some $n,k$. That implies (since such points are dense in $[0,1]$) that the characteristic function of any sub-interval lies in the closure of the span, but the closure of that is $L^2[0,1]$ so the span of the wavelets was dense.

Start with $n=1$, $$\psi_{0,0}+1=2\chi_{[0,1/2]}\qquad 1-\psi_{0,0}=2\chi_{[1/2,1]}$$

Now $\psi_{n,k}=\chi_{[k2^{-n},(k+1/2)2^{-n}]}-\chi_{[(k+1/2)2^{-n},(k+1)2^{-n}]}$. From induction we are supposing $\chi_{[k2^{-n},(k+1)2^{-n}]}$ is in the span, so it follows that $$\chi_{[k2^{-n},(k+1/2)2^{-n}]}-\chi_{[(k+1/2)2^{-n},(k+1)2^{-n}]}+\chi_{[k2^{-n},(k+1)2^{-n}]}=2\chi_{[k2^{-n},(k+1/2)2^{-n}]}=2\chi_{[(2k)2^{-(n+1)},(2k+1)2^{-(n+1)}]}$$ Lies in the closure of the span for any $0≤2k<2^{n+1}$. The odd case follows by considering $$\chi_{[k2^{-n},(k+1)2^{-n}]}-(\chi_{[k2^{-n},(k+1/2)2^{-n}]}-\chi_{[(k+1/2)2^{-n},(k+1)2^{-n}]})=2\chi_{[(k+1/2)2^{-n},(k+1)2^{-n}]}=2\chi_{[(2k+1)2^{-(n+1)},(2k+2)2^{-(n+1)}]}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .