# 1.2 Appendix a

This appendix contains outline proofs and derivations for the theorems and formulas given in early part of Chapter: The Scaling Function and Scaling Coefficients, Wavelet and Wavelet Coefficients . They are not intended to be completeor formal, but they should be sufficient to understand the ideas behind why a result is true and to give some insight into its interpretation aswell as to indicate assumptions and restrictions.

Proof 1 The conditions given by [link] and [link] can be derived by integrating both sides of

$\phi \left(x\right)=\sum _{n}h\left(n\right)\phantom{\rule{0.277778em}{0ex}}\sqrt{M}\phantom{\rule{0.166667em}{0ex}}\phi \left(M\phantom{\rule{0.166667em}{0ex}}x-n\right)$

and making the change of variables $y=Mx$

$\int \phi \left(x\right)\phantom{\rule{0.166667em}{0ex}}dx=\sum _{n}h\left(n\right)\int \sqrt{M}\phantom{\rule{0.166667em}{0ex}}\phi \left(Mx-n\right)\phantom{\rule{0.166667em}{0ex}}dx$

and noting the integral is independent of translation which gives

$=\sum _{n}h\left(n\right)\phantom{\rule{0.166667em}{0ex}}\sqrt{M}\phantom{\rule{0.166667em}{0ex}}\int \phi \left(y\right)\frac{1}{M}\phantom{\rule{0.166667em}{0ex}}dy.$

With no further requirements other than $\phi \in {L}^{1}$ to allow the sum and integral interchange and $\int \phi \left(x\right)\phantom{\rule{0.166667em}{0ex}}dx\ne 0$ , this gives [link] as

$\sum _{n}h\left(n\right)=\sqrt{M}$

and for $M=2$ gives [link] . Note this does not assume orthogonality nor any specific normalization of $\phi \left(t\right)$ and does not even assume $M$ is an integer.

This is the most basic necessary condition for the existence of $\phi \left(t\right)$ and it has the fewest assumptions or restrictions.

Proof 2 The conditions in [link] and [link] are a down-sampled orthogonality of translates by $M$ of the coefficients which results from the orthogonality of translates of the scaling function given by

$\int \phi \left(x\right)\phantom{\rule{0.166667em}{0ex}}\phi \left(x-m\right)\phantom{\rule{0.166667em}{0ex}}dx=E\phantom{\rule{0.166667em}{0ex}}\delta \left(m\right)$

$\int \left[\sum _{n},h,\left(n\right),\phantom{\rule{0.166667em}{0ex}},\sqrt{M},\phantom{\rule{0.166667em}{0ex}},\phi ,\left(Mx-n\right)\right]\left[\sum _{k},h,\left(k\right),\phantom{\rule{0.166667em}{0ex}},\sqrt{M},\phantom{\rule{0.166667em}{0ex}},\phi ,\left(Mx-Mm-k\right)\right]\phantom{\rule{0.166667em}{0ex}}dx=E\phantom{\rule{0.166667em}{0ex}}\delta \left(m\right)$

which, after reordering and a change of variable $y=M\phantom{\rule{0.166667em}{0ex}}x$ , gives

$\sum _{n}\sum _{k}h\left(n\right)\phantom{\rule{0.166667em}{0ex}}h\left(k\right)\int \phi \left(y-n\right)\phantom{\rule{0.166667em}{0ex}}\phi \left(y-Mm-k\right)\phantom{\rule{0.166667em}{0ex}}dy=E\phantom{\rule{0.166667em}{0ex}}\delta \left(m\right).$

Using the orthogonality in [link] gives our result

$\sum _{n}h\left(n\right)\phantom{\rule{0.166667em}{0ex}}h\left(n-Mm\right)\phantom{\rule{0.277778em}{0ex}}=\phantom{\rule{0.277778em}{0ex}}\delta \left(m\right)$

in [link] and [link] . This result requires the orthogonality condition [link] , $M$ must be an integer, and any non-zero normalization $E$ may be used.

Proof 3 (Corollary 2) The result that

$\sum _{n}h\left(2n\right)=\sum _{n}h\left(2n+1\right)=1/\sqrt{2}$

$\sum _{n}h\left(M\phantom{\rule{0.166667em}{0ex}}n\right)=\sum _{n}h\left(M\phantom{\rule{0.166667em}{0ex}}n+k\right)=1/\sqrt{M}$

is obtained by breaking [link] for $M=2$ into the sum of the even and odd coefficients.

$\sum _{n}h\left(n\right)=\sum _{k}h\left(2k\right)+\sum _{k}h\left(2k+1\right)={K}_{0}+{K}_{1}=\sqrt{2}.$

Next we use [link] and sum over $n$ to give

$\sum _{n}\sum _{k}h\left(k+2n\right)h\left(k\right)=1$

which we then split into even and odd sums and reorder to give:

$\begin{array}{}\sum _{n}\left[\sum _{k},h,\left(2k+2n\right),h,\left(2k\right),+,\sum _{k},h,\left(2k+1+2n\right),h,\left(2k+1\right)\right]& =\sum _{k}\left[\sum _{n},h,\left(2k+2n\right)\right]h\left(2k\right)+\sum _{k}\left[\sum _{n},h,\left(2k+1+2n\right)\right]h\left(2k+1\right)& =\sum _{k}{K}_{0}h\left(2k\right)+\sum _{k}{K}_{1}h\left(2k+1\right)={K}_{0}^{2}+{K}_{1}^{2}=1.\end{array}$

Solving [link] and [link] simultaneously gives ${K}_{0}={K}_{1}=1/\sqrt{2}$ and our result [link] or [link] for $M=2$ .

If the same approach is taken with [link] and [link] for $M=3$ , we have

$\sum _{n}x\left(n\right)=\sum _{n}x\left(3n\right)+\sum _{n}x\left(3n+1\right)+\sum _{n}x\left(3n+2\right)=\sqrt{3}$

which, in terms of the partial sums ${K}_{i}$ , is

$\sum _{n}x\left(n\right)={K}_{0}+{K}_{1}+{K}_{2}=\sqrt{3}.$

${K}_{0}^{2}+{K}_{1}^{2}+{K}_{2}^{2}=1.$

Equation [link] and [link] are simultaneously true if and only if ${K}_{0}={K}_{1}={K}_{2}=1/\sqrt{3}$ . This process is valid for any integer $M$ and any non-zero normalization.

Proof 3 If the support of $\phi \left(x\right)$ is $\left[0,N-1\right]$ , from the basic recursion equation with support of $h\left(n\right)$ assumed as $\left[{N}_{1},{N}_{2}\right]$ we have

$\phi \left(x\right)=\sum _{n={N}_{1}}^{{N}_{2}}h\left(n\right)\phantom{\rule{0.277778em}{0ex}}\sqrt{2}\phantom{\rule{0.166667em}{0ex}}\phi \left(2x-n\right)$

where the support of the right hand side of [link] is $\left[{N}_{1}/2,\left(N-1+{N}_{2}\right)/2\right)$ . Since the support of both sides of [link] must be the same, the limits on the sum, or, the limits on the indices of the non zero $h\left(n\right)$ are such that ${N}_{1}=0$ and ${N}_{2}=N$ , therefore, the support of $h\left(n\right)$ is $\left[0,N-1\right]$ .

Proof 4 First define the autocorrelation function

$a\left(t\right)=\int \phi \left(x\right)\phantom{\rule{0.166667em}{0ex}}\phi \left(x-t\right)\phantom{\rule{0.166667em}{0ex}}dx$

and the power spectrum

$A\left(\omega \right)=\int a\left(t\right)\phantom{\rule{0.166667em}{0ex}}{e}^{-j\omega t}\phantom{\rule{0.166667em}{0ex}}dt=\int \int \phi \left(x\right)\phantom{\rule{0.166667em}{0ex}}\phi \left(x-t\right)\phantom{\rule{0.166667em}{0ex}}dx\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{e}^{-j\omega t}\phantom{\rule{0.166667em}{0ex}}dt$

which after changing variables, $y=x-t$ , and reordering operations gives

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