Biomedical Engineering Reference
In-Depth Information
Fig. 2.11
Left panel
—Case 1: t
h. The graph of v
N
is shown in
solid line
, while the
graph of the shifted function
h
v
N
is shown in
dashed line
.The
shaded area
denotes the nonzero
contributions to the norm
k
h
v
N
v
N
k
2
L
2
.
Right panel
—Case2: t < h
D
t
C
s;0 < s < t.
The graph of v
N
is shown in
solid line
, while the graph of the shifted function
h
v
N
is shown in
the
dashed line
.The
shaded areas
denote nonzero contributions to the norm
k
h
v
N
v
N
k
2
L
2
.The
two colors
represent the contributions to the first and second integral in (
2.160
) separately
The last inequality follows from jhj <ı "=.2C/.
Case 2:
t < h
.
In this case we can write h D lt C s for some ł
2
N
,
0<s t. Similarly, as in the first case, we get (see Fig.
2.11
, right):
Z
.j
C
1/t
s
N
l
1
X
kv
N
v
j
C
l
2
2
L
2
.0;1/
k
h
v
N
v
N
k
L
2
.!
I
L
2
.0;1//
D
k
N
jt
j
D
1
!
: (2.160)
Z
.j
C
1/t
.j
C
1/t
s
kv
N
v
j
C
l
C
1
2
L
2
.0;1/
C
k
N
Now we use the triangle inequality to bound each term under the two integrals
from above by
P
l
C
1
i
D
1
kv
j
C
i
1
v
j
C
i
N
2
k
L
2
.0;1/
: After combining the two terms
N
together one obtains
N
l
1
l
C
1
X
X
i
D
1
kv
j
C
i
1
v
j
C
i
N
2
2
k
h
v
N
v
N
k
L
2
.!
I
L
2
.0;1//
t
k
L
2
.0;1/
: (2.161)
N
j
D
1
Lemma
2.1
now implies that the right-hand side of (
2.161
) is bounded by t.l C
1/C.Now,sinceh D lt C s we see that t h=l, and so the right-hand side
of (
2.161
) is bounded by
l
C
l
hC.Sincejhj <ıand from the form of ı we get
l
C
1
l
l
C
1
l
"
2
<":
2
L
2
.!
I
L
2
.0;1//
t.l C 1/C
k
h
v
N
v
N
k
hC
Thus, if we set ! D Œı=2;T ı=2 we have shown:
2
L
2
.ı=2;T
ı=2
I
L
2
.0;1//
<"; N2
N
k
ı=2
v
N
v
N
k
:
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