sparse image representation via combined transforms - Convex ...

3.7. PROOFS 77
the ijth component of matrix Σ is
Σ ij =
N−1
∑
k=0
λ k α1(k)α 2 2 (i)α 2 (j) 2 ( )
π(k +
N cos δ1 )(i + δ 2 )
cos
N
= α 2 (i)α 2 (j) 1 N
N−1
∑
k=0
( )
π(k + δ1 )(j + δ 2 )
[
λ k α1(k)
2 cos π(k + δ 1)(i − j)
+cos π(k + δ ]
1)(i + j +2δ 2 )
.
N
N
N
∑ N−1
In the above equation, let the first term 1 N k=0 λ kα1 2(k)cos π(k+δ 1)(i−j)
N
be the ijth
element of matrix Σ 1 , and let the second term 1 ∑ N−1
N k=0 λ kα1 2(k)cos π(k+δ 1)(i+j+2δ 2 )
N
be the
ijth element of matrix Σ 2 . It is easy to see that the matrix Σ 1 is Toeplitz, and the matrix
Σ 2 is Hankel. Inserting different values of δ 1 and δ 2 , we establish the Theorem 3.5.
Proof of Theorem 3.6
Consider the z-transform of sequence {h(n) :n ∈ Z} and sequence {g(n) :n ∈ Z}:
H(z) = ∑ n∈Z
h(n)z n ,
G(z) = ∑ n∈Z
g(n)z n .
From (3.27), note that all the even terms in H(z)H(z −1 ) have zero coefficient except the
constant term. Hence we have
H(z)H(z −1 )+H(−z)H(−z −1 )=2. (3.38)
Similarly from (3.28) we have
G(z)G(z −1 )+G(−z)G(−z −1 )=2. (3.39)
From (3.29), all the even terms in H(z)G(z −1 ) have zero coefficients, so we have
G(z)H(z −1 )+G(−z)H(−z −1 )=0. (3.40)