## Introduction

I want to add one more property to the previous post. This is long enough that I do not want to insert it as an edit. It’s amazing that I ever considered even for a moment to insert it as an edit!

Just as we deduce what is called a partition of unity $\sum_k { \varphi(k)} = 1$ from the dyadic sum

$\sum_k { \varphi(\frac{k}{2^j})} = 2^j$

(by setting j = 0)

we can get a more general equation, if $\varphi$ is continuous. We can show that

$\sum_k { \varphi(t-k)} = 1\$,

where the sum, as before, is over all integer k. (The proof uses the previous dyadic sum and continuity to say something about non-integer values.) This may be called a generalized partition of unity.

That’s how Burrus et al. write it, but I find it useful to change the sign of k, and write

$\sum_k { \varphi(t+k)} = 1\$,

I may sometimes need negative values of k, but I like having t and k going in the same direction: increasing either of them is moving to the right on the number line.

For the Haar system, since the support is [0,1), the sum always reduces to one nonzero term, k = 0. For k ≠ 0, the value t+k is always outside the support interval. And since $\varphi(t) = 1$ for $t \in [0,1)\$, we understand that the equation is true.

Just in case I’ve never said it, or you don’t recognize it, the support of a function f is the part of its domain, the set of x, for which f(x) is nonzero. In particular, to say that a function defined on the real numbers has compact support means that it is zero outside of a compact (i.e., closed and bounded) interval. To say that the support of a function is the interval [0,3) means that the function is zero everywhere else outside that interval.

For the Daubechies D4, with support [0,3), the sum always reduces to two nonzero terms, but not necessarily the same 2. We always have a term for k = 0. For $t \in [0,1)\$ we have k = 1,2. For$t \in [1,2)\$, we can have $k = \pm 1\$. For $t \in [2,3)\$, we have k = -2,-1.

Computationally speaking, I might as well just write the sum for slightly more k’s than I need, because the computer will take care of the zero terms. But I better not omit k’s for which $\varphi(t+k) \ne 0\$.

Let me be explict about that. If the sum fails to be 1, it will be because we used too few values of k; we omitted a nonzero term. I will illustrate these calculations for the Daubechies D4.

I do not know if this condition is more useful directly or in reverse. That is, in reverse, if we find a scaling function for which the sum really isn’t 1, $\sum_k { \varphi(t+k)} \ne 1\$, then the scaling function $\varphi$ should not be continuous. Unless there are other hypotheses involved — and, I have to remind you, at my present level of understanding, there could be!

You might want to remember that, in general, $\varphi$ is a hypothetical solution to the dilation equation with some h’s. In particular, $\varphi$ may not even exist. And if it does exist, it may not be integrable. Hell, it may not even be a function, as we usually think of them, but a distribution (like the Dirac delta “function”).

That said, we are in the position of having entire families of scaling functions available to us. For them, we don’t need to worry about conditions which guarantee existence or integrability or continuity: we have the silly things right in our hands and can play with them, to verify whether properties hold or not.

To return to the D4 example, it’s easy enough to pick a few values and confirm that the sum is 1. That’s not a proof that the sum is always 1, but it was enough to send me to Daubechies herself (Daubechies, Ingrid; Ten Lectures on Wavelets. Society for Industrial & Applied Mathematics, 1992; ISBN 0 89871 274 2) to confirm there that D4 is continuous. (All the Dn are continuous.)

Having added this property, my checklist of 5 items grows to 6 items:

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

## Example: Daubechies D4

Let us look at the generalized partition of unity,

$\sum_k { \varphi(t+k)} = 1\$,

specifically for the Daubechies D4. The previous post verified the orginal five properties; we need to do this one.

First, let’s look at t = 1/2.

Recall the D4 scaling function. You might also recall that we compute it at rational points whose denominator is a power of 2.

Now I tell Mathematica® to list the values of $\{ \varphi(t+k)\}$ with t = 1/2, and k an integer from -3 to 3. That’s more than enough values of k).

Here’s the list, and the sum of the entries:

(Yes, $\varphi(3/2) = 0\$.)

Let’s think about that. We’re really trying to evaluate $\varphi$ at all the half-integer points on the real line. That infinite sum is really one representative of an equivalence class: we have, for example,

$\sum_k { \varphi(1/2+k)} = \sum_k { \varphi(-1/2+k)} = \sum_k { \varphi(3/2+k)}\$

etc. If I decide, for example, to check the sum for t = 13/2, but I still use the same limits for k…

the problem is not that the infinite series fails to be 1, but that I have chosen the wrong limits for k. The generalized partition of unity is a doubly-infinite series, in principle, but reduces to a finite sum if the scaling function has compact support (i.e. is zero outside a finite interval). I just didn’t get all the nonzero terms. My bad. Scaling function good.

Another way to say all that is, without loss of generality we can choose t in the interval [0,1) for checking a selection of values. That infinite series will include values in [1,2) and [2,3) (i.e. the other intervals where D4 is nonzero).

So let’s look at a few. I have chosen to let k range over -3 to 5 simply because it provides a reassuring number of zeroes on either end of the output list.

The following picture shows the results for t = 1/4, 3/8, 1/128, and 100/128.

## summary

The following six properties are consequences of the dilation equation, although we are used to having A = $\sqrt{2}\$ and two require E = 1. That means that some of them can be used to quickly determine that A ≠ $\sqrt{2}$ or that E ≠ 1. Alternatively, we could work out the generalized forms for E ≠ 1.

In particular, these properties do not require that integer translates be orthogonal. But we have yet to see an example of a scaling function that is not orthogonal to its integer translates. (Yes, of course, I have one just waiting for us.)

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