Digital Signal Processing Reference
In-Depth Information
dg ( x )
dx
2.
0( g ( x ) monotone non increasing);
d 2 g
(
x
)
3.
0( g ( x ) convex).
dx 2
Then f ( x ) is Schur-convex on
D
(i.e., for x 0
x 1
...
x P− 1 ).
Proof. Throughout the proof remember that the condition x 0 ≥ x 1 ≥ ...≥
x P− 1 is assumed. The notation dg ( x k ) /dx denotes the derivative of g ( x )
evaluated at x k . Since d 2 g ( x ) /dx 2
0, we have
dg ( x 0 )
dx
dg ( x 1 )
dx ≥ ...≥
dg ( x P 1 )
dx
.
Using dg ( x ) /dx
0 we therefore get
dg ( x 0 )
dx ≤−
dg ( x 1 )
dx
dg ( x P 1 )
dx
0
≤−
...
≤−
.
We now combine this with the first condition of the corollary:
0
a 0
a 1
...
a P− 1 .
Since a k and
dg ( x k ) /dx are non-negative we conclude from the preceding
two sets of inequalities that
dg ( x 0 )
dx ≤−
dg ( x 1 )
dx
dg ( x P 1 )
dx
0
≤−
a 0
a 1
...
≤−
a P− 1
which is equivalent to
dg ( x 0 )
dx
dg ( x 1 )
dx
dg ( x P 1 )
dx
0
a 0
a 1
...
a P− 1
The function (21.36) therefore has the form (21.35) (just set g k ( x )= a k g ( x ))
and satisfies all the conditions of Theorem 21.4. It is therefore a Schur-
convex function in
D
.
For example, let
g ( x )= 1
x p ,
p> 0 .
Then, for x> 0 ,
< 0and d 2 g ( x )
dx 2
dg ( x )
dx
=
p
x p +1
= p ( p +1)
x p +2
> 0 .
Applying Corollay 21.1 it therefore follows that
P− 1
a k
x k
f ( x )=
,
p> 0
(21 . 37)
k =0
 
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