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Math 414 - Solutions to Chapter 4
Al Boggess
Spring 1999
- The given function
f (x) =
− 1 0 ≤ x < 1 / 4 4 1 / 4 ≤ x < 1 / 2 2 1 / 2 ≤ x < 3 / 4 − 3 3 / 4 ≤ x < 1
can be written as
f (x) = −φ(4x) + 4φ(4x − 1) + 2φ(4x − 2) − 3 φ(4x − 3)
Using the equations
φ(4x) = (ψ(2x) + φ(2x)) / 2 φ(4x − 1) = (φ(2x) − ψ(2x)) / 2 φ(4x − 2) = φ(4(x − 1 /2)) = (ψ(2(x − 1 /2)) + φ(2(x − 1 /2))) / 2 φ(4x − 3) = φ(4(x − 1 /2) − 1) = (φ(2(x − 1 /2) − ψ(2(x − 1 /2))) / 2
into the terms for f (x) and collecting terms, we have
f (x) = 3/ 2 φ(2x) − 1 / 2 φ(2x − 1) − 3 / 2 ψ(2x) + 5/ 2 ψ(2x − 1)
The terms involving ψ(2x) and ψ(2x − 1) belong to W 1. We need to further decompose the terms involving φ(2x) and φ(2x − 1). The key equations are
φ(2x) = (φ(x) + ψ(x))/ 2 φ(2x − 1) = (φ(x) − ψ(x))/ 2
Substituting these equations and collecting terms gives
f (x) = −φ(x) − (1/2)ψ(x) − 3 / 2 ψ(2x) + 5/ 2 ψ(2x − 1)
The term involving φ(x) is the V 0 component of f. The term involving ψ(x) is the W 0 component of f. The sum of the terms involving ψ(2x) and ψ(2x − 1) is the W 1 component of f.
- The dimension of a vector space is the number of basis elements. So suppose the dimension of A is n and the dimension of B is m. Let a 1 ,... an is a basis for A and b 1 ,... bm is a basis for B. We would like to show that the collection a 1 ,... an, b 1 ,... bm is a basis for A ⊕ B. First we show this collection spans. An arbitrary element of A ⊕ B is of the form a + b with a ∈ A and b ∈ B. Since a 1 ,... an is a basis for A, there are constants r 1 ,... , rn with
a = r 1 a 1 +... + rnan.
Likewise, there are constants s 1 ,... sm with
b = s 1 b 1 +... smbm.
Adding these two equations yields
a + b = r 1 a 1 +... + rnan + s 1 b 1 +... smbm
which shows this collection spans. To show linear independence, suppose that
r 1 a 1 +... + rnan + s 1 b 1 +... smbm = 0
We wish to show that all the coefficients (the rs and the ss) are zero. Let a = r 1 a 1 +... + rnan ∈ A and let b = s 1 b 1 +... smbm ∈ B. Thus the above equation becomes a + b = 0. Taking the inner product of both sides with this equation with b gives
〈(a + b), b〉 = 〈 0 , b〉 = 0.
Since 〈a, b〉 = 0 (by hypothesis), we conclude that 〈b, b〉 = 0, which means that s 1 b 1 +... smbm = b = 0. Since b 1 ,... , bm is a basis for B, they are linearly independent. So all the s-coefficients must be zero. In the same manner (taking the inner product with a), we can show that a = 0 and hence all the r-coefficients must be zero. If A and B are not orthogonal, they may have a nonempty intersection. In this case, the dimensions are related by
dim(A + B) = dim(A) + dim(B) − dim(A ∩ B).
- Suppose f =
∑ k akφ(2x^ −^ k) is orthogonal to^ V^0 which is spanned by the φ(x − l). We wish to show a 0 = −a 1 , a 2 = −a 3.. .. Now f is orthogonal to φ(x − l) for all l. Suppose l = 0, then since φ(x) is one for 0 ≤ x ≤ 1 and zero otherwise, we have
0 = 〈f, φ(x)〉 =
∫ (^1)
0
∑
k
akφ(2x − k)
The only values of k which contribute to this sum are k = 0 and 1. Any other translate of φ(2x) is zero on the interval 0 ≤ x ≤ 1. Therefore the above equation becomes
0 = a 0
∫ (^1)
0
φ(2x) + a 1
∫ (^1)
0
φ(2x − 1)
Since the graphs of φ(2x) and φ(2x − 1) are boxes of width 1/2 and height 1, both integrals are 1/2. Therefore the above equation becomes
0 = a 0 /2 + a 1 / 2
which implies that a 0 = −a 1 , as desired. In general, f is orthogonal to φ(x − l) for any l and so the above arguments can be repeated over the interval l ≤ x ≤ l + 1 and we conclude that a 2 l = −a 2 l+1.
- First we claim that if |f ′(x)| ≤ M then
|f (x) − f (y)| ≤ M |x − y|
This follows from the Mean Value Theorem, which states
f (x) − f (y) = f ′(c)(x − y)
for some c between x and y. Since |f ′(c)| ≤ M , the previous inequal- ity now follows. So on the interval k/ 2 n^ ≤ x ≤ (k + 1)/ 2 n^ (which has width 1/ 2 n), the function f varies by no more than M/ 2 n. The function fn(x) =
∑ k akφ(
nx − k) with ak = f (k/ 2 n) has constant value f (k/ 2 k) on the interval k/ 2 n^ ≤ x ≤ (k + 1)/ 2 n. Thus, fn and f differ by no more than M/ 2 n. To make this quantity less than , we must require M/ 2 n^ ≤
or M/ ≤ 2 n
or log 2 (M/) ≤ n
as claimed.
- Since Vj = Vj− 1 ⊕ Wj− 1 , the function φ(2j^ x − k) ∈ Vj can be written as:
φ(2j^ x − k) =
∑
l∈Z
αlφj− 1 ,l(x) + βlψj− 1 ,l(x) (1)
for some choice of constants αl and βl. Our goal is to compute βl. Since φj− 1 ,l and ψj− 1 ,l are orthonormal, we have
βl = 〈φ(2j^ x − k), ψj− 1 ,l〉L 2
= 2 (j−1)/^2
∫ (^) ∞
−∞
φ(2j^ x − k)ψ(2j−^1 x − l) dx.
Now we use the defining equation for ψ:
ψ(x) =
∑
k′
(−1)k
′ p 1 −k′ φ(2x − k′)
with x replaced by 2j−^1 x − l. We obtain
βl = 2(j−1)/^2
∫ (^) ∞
−∞
φ(2j^ x − k)
∑
k′
(−1)k
′ p 1 −k′ φ(2j^ x − 2 l − k′) dx.
Since { 2 j/^2 φ(2j^ x − k), k ∈ Z} is an orthonormal set, the only con- tributing term on the right occurs when 2l + k′^ = k or k′^ = k − 2 l and we obtain βl = 2−(j+1)/^2 (−1)kp 1 −k+2l ∗ (2)
So
βlψj− 1 ,l(x) = βl 2 (j−1)/^2 ψ(2j−^1 x − l) by definition of ψj− 1 ,l = 2 −^1 (−1)kp 1 −k+2lψ(2j−^1 x − l) by (∗).
Thus from (1), the projection of φ(2j^ x − k) onto Wj− 1 is ∑
l∈Z
βlψj− 1 ,l =
∑
l∈Z
2 −^1 p 1 −k+2lψ(2j−^1 x − l).
The function F is 2π - periodic because
F (ξ + 2π) = 2 π
∑
j∈Z
φ^ ˆ(ξ + 2π(j + 1)) ψˆ(ξ + 2π(j + 1))
= 2 π
∑
j′
φ^ ˆ(ξ + 2πj′) ψˆ(ξ + 2πj′)
where the last equality uses the change of index j′^ = j + 1. Since F is periodic, it has a Fourier series,
∑ αneinx, where the Fourier coefficients are given by αn = (1/ 2 π)
∫ (^2) π 0 F^ (ξ)e
−inξ (^) dξ. Thus, the or-
thonormality condition, (5), is equivalent to αn = 0, which in turn is equivalent to the statement F (ξ) = 0. This completes the proof.
