3.1 Hal Space V0

A grayscale image is made up of multiple pixel points, and again, these pixels are made of a non-negative number from 0 (black) to 255 (white). Suppose we now have a small grayscale image. The grayscale value in the first line is 110,100,120,140,130,100,100. These grayscale figures are drawn in Figure 3.1

Figure 3.1 Grayscale

Naturally, we ask, can we use a function f (t) to represent this line of data? We can use the function φ (t) introduced from example 1.2. Φ (t) is

So we're going to start using φ (t) to build f (t).

Φ (T) and F (t) 3.2

Notice that all f (t) are continuous for all t∈r except in t=1,2,3,4,5.

• Definition of V0 space

We can use the example above to create a vector space, and all of the breakpoints are in integers. We might as well try to define that this space is made up of φ (t) and its transformation function φ (t-k) spanned, where k∈z, so the element in this space can be expressed as

In Equation 3.3 We deliberately blur the range of sums. Are we really going to have to do some binding on the sum? In fact, in the image processing, k as long as the coverage of all the ranks can be. In this case, we call these pixels a tightly braced element.

However, if K is set to a finite value, we will not be able to model some useful functions. This is like a function of this infinitely long support set, such as the sin and cos linear combinations. Of course, all the functions that are actually used are limited. So, we have to discuss the function in L2 (R) space, and we don't need to discuss infinity, but we need to discuss the value of the function is approximate to 0, so we have the following definition

Definition 3.1 (HAL space V0 and Hal function φ (t))φ (t) already written in (3.1), we define V0 space below

(The author notes: This name is to commemorate the Hungarian mathematician N Alfred Haar (1885-1933), the mathematician founded the theory of orthogonal function)

This space is all the piecewise constant functions where the breakpoint is positive in L2 (R). In exercise 3.4 You need to prove that this space is a L2 (R) subspace.

By definition, you can know that φ (t) and φ (t-k) Form the entire space. In Exercise 3.1 you will know that they are linearly unrelated. Therefore, they are a group of bases in this space

• The orthogonality of Φ (T) and its transformation

Let's choose subscript j!. The =k function φ (t-j) and φ (t-k). φ (t-j) in the interval [j,j+1) is not 0, likewise φ (t-k) in [k,k+1] is not 0. If j! =k, their product is 0, and 3.3 shows

So the inner product is

So they are orthogonal. Again, we've noticed that Φ (t-k) 2=φ (t-k), so no matter how we pan, it's always 1 of the area around the horizontal axis.

Note that our discussion in this chapter is a real function. So we've modified the conjugate in the definition of 1.8 to calculate the inner product.

The subject 3.1 (orthogonal base of V0) is given in definition 3.1 for V0, and its orthogonal base is {Φ (t-k)}k∈z.

proof: by definition as well as exercise 3.1 you will know that {Φ (t-k)}k∈z is linearly independent, and the previous calculation is also very clear that the inner product of V0 is

Wherein Δj,k in the formula (2.16) gives

Let's look at some examples of V0

Example 3.1 (element in V0) determines whether the following function is located in V0

(a) where

(b) Step function

(c) function

(d) function

Solution:

For (a), we make the following simplification

It is not difficult to see that the break point is 2, and the place of the segment is positive and

So f (t) ∈v0

for (b), when t≥1 and k=1,2,3,4 ...., g (t) ≥1. However, we are practicing 1.13 (b) know that the function

does not belong to L2 (R). Also because on [0,∞] g (t) ≥r (t). So. Although the discontinuity of this function is also positive, because it is not absolutely integrable, it does not belong to the space V0

for (c), we can see that this is a linear combination of the function φ (t-k). But we still have to take some time to see it. So let's start with a little bit of simplicity

These rigorous proofs should be designed to sum the infinite series and to exchange the order of indefinite integrals. These should be taught in analytical math classes, and interested readers can read Rudin's books. The simplification of H (t) 2 is as follows

We're integrating it on R.

From (3.5) We know that the integral of φ (t-k) and φ (t-j) is 0, only when j=k is 1, so the formula can be simplified to

From the point of integration, we know that the P-order converges. (Author note: We can try to prove the sum of this thing by Fourier series as ∏2/6, see Kammler's book). So we know that H (t) belongs to the V0 space.

(d) The crux of the problem is that the coefficients of each are related to the function of φ (4t). This is a scaled HAL function.

We draw φ (4t) in Figure 3.4. Now φ (4t) =φ (4 (T-1/4)), so we can think of this as moving the 1/4 units to the right, the rest and so on, we draw this function in Figure 3.4

Figure 3.4 function φ (4t) and L (t)

• Extended from L2 (R) to V0

3.1 Hal Space V0