Image Processing Reference
In-Depth Information
any 0 ≤i,j ≤n, the following hold: [38]
(i−j)c
u
≤ (iy
j
−jy
i
+ i)...........(a)
(i−j)c
l
≤ (iy
j
−jy
i
+ i)...........(b)
(i−j)m
u
≥ (y
i
−y
j
−i)...........(c)
(i−j)m
l
≥ (y
i
−y
j
−i)...........(d).
€
Lemma 3.4. The following hold:
∗
∗
(A) c
l
< c < c
u
if and only if m
l
(c) < m
u
(c)), where
∗
∗
m
l
(c) = max
i
(y
i
−c)/i and m
u
(c) = min
i
(y
i
+ 1−c)/i
∗
l
(m) < c
∗
u
(m), where
(B) m
l
< m < m
u
if and only if c
∗
l
(m) = max
i
∗
u
(m) = min
i
c
(y
i
−mi) and c
(y
i
+ 1−mi)
Proof: Part(A): Let c
l
< c < c
u
. Assume m
∗
l
(c) ≥ m
∗
u
(c). Clearly, there
exists p,q > 0 such that m
∗
l
(c) = (y
p
−c)/p and m
∗
u
(c) = (y
q
+ 1 −c)/q. So,
∗
∗
m
l
(c) ≥m
u
(c) implies (y
p
−c)/p ≥ (y
q
+ 1−c)/q. Rearranging terms we gets
(p−q)c ≥ (py
q
−qy
p
+ p).
Thus, from Lemma 3.3, (p−q)c ≥ (p−q)c
u
and (p−q)c ≥ (p−q)c
l
. If
p > q then c ≥ c
u
and if p < q, then c ≤ c
l
. p = q is ruled out as it implies
that 0 ≥ p > 0.
Hence c
l
< c < c
u
implies m
∗
l
(c) < m
∗
u
(c).
Conversely, let m
∗
l
(c) < m
∗
u
(c). Select any m such that m
∗
l
(c) ≤ m <
m
∗
u
(c). From the definition of m
∗
l
(c) and m
∗
u
(c) we get,
∀i,m
i
+ c−1 < y
i
≤m
i
+ c.
That is, y
i
= ⌊mi + c⌋. In other words, D(L(c,m)) = D
o
. Hence, from
Theorem 3.7, we get c
l
< c < c
u
.
Part(B) can be proved analogously.
€
Theorem 3.10. (The domain theorem). The domain Domain(D
o
) of a DSLS
D
o
is given by the following [38].
∗
l
(c),m
∗
u
(c)); or
Domain(D
o
)
=
[m
c
l
<c<c
u
∗
l
(m),c
∗
u
(m))
Domain(D
o
)
=
[c
m
l
<m<m
u
