## Wavelets: Review I and Going Forward a Little

Let us recall what we have.

We have a collection of nested spaces…

$\dotsm\ V_{-3} \subset\ V_{-2} \subset\ V_{-1} \subset\ V_{0} \subset\ V_{1} \subset\ V_{2}\ \dotsm\ \$

… whose intersection is the trivial space and whose union is all square integrable functions on the real line:

$\cap\ V_i = \{0\}\$ and $\cup\ V_i = L^2(R)\$.

We assume that the space$V_0\$ is translation invariant and has the scaling property:

$f(x) \in V_0\ \text{ if and only if } f(x-k) \in V_0\$ for all integers k;

$f(x) \in V_j \text{ if and only if } f(2^{-j} x) \in V_0\$.

Finally, the only real theorem I have shown you says that if we also have an orthonormal basis for $V_0\$, then we can get an orthonormal basis for $L^2(R)\$:

$\psi_{j,k} (x) := 2^{j/2)} \psi(2^j\ x - k)\$

The proof is constructive. In addition to giving us that orthonormal basis, it gives us spaces $W_j\$, each of which is the orthogonal complement of $V_j\$ in $V_{j+1}\$. Furthermore, for fixed j, we have an orthonormal basis for $W_j\$:

$\psi_{j,k}(x)\$ will be an orthonormal basis of $W_j\$.

We have asserted two kinds of orthogonality:

• that we have orthonormal bases (for $V_0\$ and for $L^2(R)\$;
• that we have orthogonal direct sums.

Can we relax these two conditions separately? Yes.

I have shown you an explicit example — for the linear spline (semi-orthogonal wavelets — Strang & Nguyen call them that on the inside front cover, so I will too; they are also called pre-wavelets) — where the bases for $V_j\$ and $W_j\$ were not each orthogonal, but the spaces themselves were orthogonal.

I did not provide any existence theorem for that example, nor did I even tell you how I found the mother wavelet for that example. We’ll get there. But the example itself shows that we can have an orthogonal direct sum without having orthonormal bases for $V_j\$ and $W_j\$.

Let me empahsize that, at present, I have no theorem associated with the mother wavelet for the linear splines.

• The mother wavelet is supposed to be orthogonal to all the translates of the linear spline scaling function, and computation supports that.
• I infer that all the translates of the mother wavelet are also orthogonal to all the translates of the scaling function.
• Therefore all the translates of the mother wavelet are in W_0 (because it’s the orthogonal complement of $V_0\$).
• I say it seems clear that the same is true for scaled versions of the mother wavelet, so, for example, $\psi(2t-k) \in W_1\$.

But I don’t actually know some of the most important things I need:

• do the $W_j\$ spaces have the scaling property: $f(x) \in W_j \text{ if and only if }f(2^{-j}) \in W_0\$? (I believe it for the $\psi\$, but is it true for the spaces?)
• do the $\psi(t-k) \$ in $W_0\$ span $W_0\$?
• are they a basis for $W_0\$?

(Remember that $W_j\$ is defined as the orthogonal complement of $V_j\$, while the $V_j\$ are defied as the spans of the scaled and translated scalings functions. While we have finally gotten our hands on elements of $W_j\$, we don’t know very much about them. I do expect (but do not know) that once I can explain where the g coefficients came from, we’ll know that we do have bases and the scaling property.)

With the caveat that there are some some crucial concerns about the wavelets for the linear splines, let’s just keep using the linear splines as an example of a non-orthonormal basis. (They are, but are the wavelets?)

So, let me recast the dilation equation and its consequences in this framework. For a few moments let us forget that we have an orthonormal basis for $V_0\$.

Suppose we do have a basis, which may not be orthonormal. The translates of the linear splines are such a basis – for the space they span.

If we have a scaling function whose integer translates are a basis for $V_0\$, then I am quite sure (I didn’t say I had proven) that the translated and scaled functions are a basis for $V_1\$. Maybe I should say it the other way, that if the translated and scaled functions are a basis for $V_1\$, then we get the dilation equation:

$\varphi(t) = \sum_n {h(n)\ A\ \ \varphi(2\ t - n)}$.

That just says that the scaling function is an element of $V_1\$ and may be written in terms of this basis for $V_1\$. Of course, this equation serves as the definition of the h’s.

From the dilation equation, we concluded that the sum of the h’s = 2/A.

We also decided that the scaling function, as a solution of the dilation equation, was determined only to within a multiple. One way to characterize those various multiples is by the integral of the function:

$E = \int\ \varphi(t)\ dt \$.

I believe that people have shown that the sum of the even h’s is equal to the sum of the odd h’s. I do not yet know when this is true or how to prove it, so I treat it as a condition to be checked whenever I have a collection of h’s in my hand.

In particular, we saw that it was true for the linear splines – our example of a non-orthonormal basis – even though there were an odd number, specifically 3, nonzero h’s.

If the scaling function has been normalized so that its integral is one (E=1), then we had two further consequences:

• $\sum_k { \varphi(\frac{k}{2^j})} = 2^j \text{(for E = 1)}$
• $\sum_k { \varphi(t+k)} = 1 \text{(for E = 1)}\$

With the passage of time, I have decided (Hey, sometimes I’m slow! More like, overwhelmed.) that the general forms of those two equations are:

• $\sum_k { \varphi(\frac{k}{2^j})} = 2^j\ E\$
• $\sum_k { \varphi(t+k)} = E\$

We can confirm those by working thru the algebra in the general case; we can also just realize that going from E = 1 to E ≠ 1 effectively replaces the scaling function $\varphi\$ by $E\ \varphi\$. They appear to be independent of A.

(I have totally messed up the notation vis a vis Burrus et al., but I’m not going to change it now. If you are reading Burrus et al., you have every right to curse me out; my E is his $A_0\$, and his E is the integral of the square of the scaling function. I will remain consistent to the notation I have adopted, even though it is not that of Burrus et al. When I write “E”, its the integral of the scaling function, not of its square.)

These, then, are the six consequences I listed earlier. Actually, two of them really serve only to identify the constants A and E; but if someone just hands me a scaling function, I will compute A and E in order to see what conventions they’re using.

Let me write all 6 consequences together:

• The sum of the h’s = 2/A.
• $\varphi(t) = \sum_n {h(n)\ A\ \ \varphi(2\ t - n)}\$.
• $E = \int\ \varphi(t)\ dt \$ .
• The sum of the even h’s = the sum of the odd h’s.
• $\sum_k { \varphi(\frac{k}{2^j})} = 2^j\ E\$
• $\sum_k { \varphi(t+k)} = E\$

I was very happy to have an example of a non-orthogonal basis – the linear splines – for which I have already verified those properties.

So much for consequences of the dilation equation. What about the mother wavelet?

If we have singled out a particular function in $W_0\$, hence in $V_1\$, then we get an equation, similar to the dilation equation for the scaling function:

$\psi(t) = \sum_n {g(n)\ A\ \ \varphi(2\ t - n)}\$.

(I can’t imagine that that particular function is anything but the mother wavelet, and then this equation says that the coefficients called g are the ones associated with the mother wavelet.)

There is one particular consequence of having an orthonormal basis for $V_0\$ — the recipe for construction of the orthonormal basis of wavelets leads to the property that the sum of the g’s is zero.

That, in turn, implies that the integral of the mother wavelet — hence the integral of every wavelet — is zero.

But the proof that the integral of the mother wavelet is zero depends only on two properties:

• that the sum of the g’s is zero;
• that we may interchange a possibly infinite series and an integral.

The point I am trying to make is that:
if the sum of the g’s is zero — whether our bases are orthonormal or not — then we should expect the integral of the mother wavelet to be zero. If it is not, then there must have been a problem interchanging the integral and an infinite series (and I might be able to provide such an example).

To put that another way, I want to compute the sum of the g’s in any case.

And to be quite explicit, the proof goes by integrating both sides of the wavelet equation. From

$\psi(t) = \sum_n {g(n)\ A\ \ \varphi(2\ t - n)}\$.

we get

$\int \psi(t) \, dt = \int (\sum_n {g(n)\ A\ \ \varphi(2\ t - n)}) \, dt\$.

If we can interchange the integral and the summation – whic may be an infinite series rather than a finite sum – then we can extract the sum of the g’s:

$\int \psi(t) \, dt = \sum_n {g(n)\ \int (\ A\ \ \varphi(2\ t - n)}) \, dt\$,

and so long as the integral on the RHS is finite, we have that the integral of the mother wavelet is proportional to the sum of the g’s.

So if the sum of the g’s is zero, then the integral of the mother wavelet is zero, unless something went wrong interchanging the limit operations.

As I said before, if the sum of the g’s is zero, expect that the integral of the mother wavelet is zero – and if it isn’t, then realize that we’re looking at a pathological case.

To summarize, I have discussed only two properties of the mother wavelet: its sort-of dilation equation and that if the g’s add up to 0, then the integral of the mother wavelet ought to be zero:

• $\psi(t) = \sum_n {g(n)\ A\ \ \varphi(2\ t - n)}\$.
• $\sum_n {g(n)} = 0 \text{ implies } \int \psi(t) \, dt = 0$

I have not checked the sum of the g’s for the mother wavelet for our only non-orthonormal basis, the linear splines.

Why wait?

In the semi-orthogonal post, I handed you the g’s for the mother wavelet orthogonal to the translates of the linear spline scaling function:

$g = \left\{\frac{1}{24 \sqrt{2}},-\frac{1}{4 \sqrt{2}},\frac{5}{12 \sqrt{2}},-\frac{1}{4 \sqrt{2}},\frac{1}{24 \sqrt{2}}\right\}\$

They do indeed add up to zero. I expect that the integral of the mother wavelet is zero, and we can probably prove it by writing it in terms of the scaling function. Not now.

I should point out that even though we know so little about this mother wavelet, we know that it is in $V_1\$ because we were given the g’s that describe it wrt our basis $\{\varphi(2t-k)\}\$ of $V_1\$! And I was assured that it was, in fact, in $W_0\$.

I had intended to march right on into the consequences of our two kinds of orthogonality, but this post is probably long enough, and this is a good stopping point.