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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