Application of the extremum principle to investigating certain extremal problems
Authors: Mikołajczyk L. , Walczak S. 
Published: June 2010

Denote by C, K, X, respectively, a complex plane, the disc

{
z

∈

C:

|
z
|

>

1
}

\{ z\, \in \,{\textbf {C:}}\,\left | z \right |\, > \,1\}

and any compact Hausdorff space. Denote by P a set of probabilistic measures defined on Borel subsets of the space X. For

μ

∈

P

\mu \, \in \,P

, let

f
(
z
)

=

∫
X

q
(
z
,

t
)

d
μ

,

z

∈

K

f(z)\, = \,\int _X {q(z,\,t)\,d\mu } ,\,z\, \in \,K

, and

F

=

{
f
:

μ

∈

P
}

{\mathcal {F}}\, = \,\{ f:\,\mu \, \in \,P\}

. Consider a finite sequence of real functions

F
0

,

F

1
,

…
,

F
m

{F_0},\,{F_{1,}}\, \ldots ,\,{F_m}

defined in the space

R

2
n

{R^{2n}}

. Let

ζ
1

,

…
,

ζ
k

{\zeta _1},\, \ldots ,\,{\zeta _k}

be fixed points of the disc K and
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" alttext="eta left-parenthesis f right-parenthesis equals left-bracket re f Superscript left-parenthesis 0 right-parenthesis Baseline left-parenthesis zeta right-parenthesis comma im f Superscript left-parenthesis 0 right-parenthesis Baseline left-parenthesis zeta 1 right-parenthesis comma ellipsis comma re f Superscript left-parenthesis n 1 right-parenthesis Baseline left-parenthesis zeta 1 right-parenthesis comma im f Superscript left-parenthesis n 1 right-parenthesis Baseline left-parenthesis zeta 1 right-parenthesis semicolon ellipsis comma re f Superscript left-parenthesis 0 right-parenthesis Baseline left-parenthesis zeta Subscript k Baseline right-parenthesis comma im f Superscript left-parenthesis 0 right-parenthesis Baseline left-parenthesis zeta Subscript k Baseline right-parenthesis comma ellipsis comma re f Superscript left-parenthesis n Super Subscript k Superscript right-parenthesis Baseline left-parenthesis zeta Subscript k Baseline right-parenthesis comma im f Superscript left-parenthesis n Super Subscript k Superscript right-parenthesis Baseline left-parenthesis zeta Subscript k Baseline right-parenthesis right-bracket">

η
(
f
)

=

[
re

f

(
0
)

(
ζ
)
,

im

f

(
0
)

(

ζ
1

)
,

…
,

re

f

(

n
1

)

(

ζ
1

)
,

im

f

(

n
1

)

(

ζ
1

)
;

…
,

re

f

(
0
)

(

ζ
k

)
,

im

f

(
0
)

(

ζ
k

)
,

…
,

re

f

(

n
k

)

(

ζ
k

)
,

im

f

(

n
k

)

(

ζ
k

)
]

\eta (f)\, = \,[\operatorname {re} \,{f^{(0)}}(\zeta ),\,\operatorname {im} \,{f^{(0)}}({\zeta _1}),\, \ldots ,\,\operatorname {re} \,{f^{({n_1})}}({\zeta _1}),\,\operatorname {im} \,{f^{({n_1})}}({\zeta _1});\, \ldots ,\,\operatorname {re} \,{f^{(0)}}({\zeta _k}),\,\operatorname {im} \,{f^{(0)}}({\zeta _k}),\, \ldots ,\,\operatorname {re} \,{f^{({n_k})}}({\zeta _k}),\,\operatorname {im} \,{f^{({n_k})}}({\zeta _k})]

, where

f

∈

F

,

n

=

n
1

+

⋯

+

n
k

+

k

f\, \in \,{\mathcal {F}},\,n\, = \,{n_1}\, + \, \cdots \, + \,{n_k}\, + \,k

. Let

F
j

(
f
)

=

F
j

(
η
(
f
)
)
,

j

=

0
,

1
,

…
,

m

{F_j}(f)\, = \,{F_j}(\eta (f)),\,j\, = \,0,\,1,\, \ldots ,\,m

. We consider the following extremal problem. Determine a minimum of the functional

F
0

(
f
)

{F_0}(f)

under the conditions

F
j

(
f
)

⩽

0
,

j

=

1
,

2
,

…
,

m
,

f

∈

F

{F_j}(f)\, \leqslant \,0,\,j\, = \,1,\,2,\, \ldots ,\,m,\,f\, \in \,{\mathcal {F}}

. We apply the extremum principle to solve this problem. In the linear case this problem was investigated in [11].