Seja
X
{\displaystyle \mathrm {X} }
normado e
C
{\displaystyle C}
convexo e fechado contido em
X
{\displaystyle \mathrm {X} }
x
0
∈
X
,
x
0
∉
C
{\displaystyle x_{0}\in X,x_{0}\not \in C}
, e seja
X
∗
{\displaystyle X^{*}}
Espaço Dual de
X
{\displaystyle \mathrm {X} }
, temos que existe
φ
∈
X
∗
{\displaystyle \varphi \in X^{*}}
tal que
φ
(
x
)
>
s
u
p
{
φ
(
x
)
:
x
∈
C
}
{\displaystyle \varphi (x)>sup\{\varphi (x):x\in C\}}
O Funcional de Minkowski
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Seja
X
{\displaystyle \mathrm {X} }
normado e
C
⊆
X
{\displaystyle C\subseteq X}
. O Funcional de Minkowski de
C
{\displaystyle C}
μ
C
:
X
→
R
∪
{
+
∞
}
{\displaystyle \mu _{C}:X\to \mathbb {R} \cup \{+\infty \}}
μ
C
(
x
)
=
i
n
f
{
λ
>
0
:
x
∈
λ
c
}
{\displaystyle \mu _{C}(x)=inf\{\lambda >0:x\in \lambda c\}}
i
n
f
(
∅
)
=
+
∞
{\displaystyle inf(\emptyset )=+\infty }
Um Lema importante antes da demonstração
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Se
C
{\displaystyle C}
0
∈
i
n
t
C
{\displaystyle 0\in intC}
μ
C
{\displaystyle \mu _{C}}
função sublinear e
{
x
∈
X
:
μ
C
(
x
)
<
1
}
⊆
C
⊆
{
x
∈
X
:
μ
C
(
x
)
≤
1
}
{\displaystyle \{x\in X:\mu _{C}(x)<1\}\subseteq C\subseteq \{x\in X:\mu _{C}(x)\leq 1\}}
Dicas para a demonstração do Lema
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Para provar que
μ
C
(
α
x
)
=
α
μ
C
(
x
)
{\displaystyle \mu _{C}(\alpha x)=\alpha \mu _{C}(x)}
B
(
x
,
r
)
=
{
y
∈
X
:
‖
x
−
y
‖
<
r
}
{\displaystyle B(x,r)=\{y\in X:\lVert x-y\rVert <r\}}
B
[
x
,
r
]
=
{
y
∈
X
:
‖
x
−
y
‖
≤
r
}
{\displaystyle B[x,r]=\{y\in X:\lVert x-y\rVert \leq r\}}
δ
>
0
{\displaystyle \delta >0}
0
∈
B
(
0
,
δ
)
⊆
C
{\displaystyle 0\in B(0,\delta )\subseteq C}
μ
C
(
0
)
=
0.
{\displaystyle \mu _{C}(0)=0.}
x
∈
X
∖
{
0
}
{\displaystyle x\in X\setminus \{0\}}
δ
‖
x
‖
⋅
x
∈
δ
B
x
=
B
[
0
,
δ
]
⊆
C
{\displaystyle {\frac {\delta }{\lVert x\rVert }}\cdot x\in \delta B_{x}=B[0,\delta ]\subseteq C}
Na prova de que
μ
C
(
x
+
y
)
≤
μ
C
(
x
)
+
μ
C
(
y
)
{\displaystyle \mu _{C}(x+y)\leq \mu _{C}(x)+\mu _{C}(y)}
C
{\displaystyle C}
Por fim, se
μ
C
(
x
)
<
1
{\displaystyle \mu _{C}(x)<1}
x
∈
λ
C
{\displaystyle x\in \lambda C}
λ
<
1
{\displaystyle \lambda <1}
x
∈
λ
C
⊆
C
{\displaystyle x\in \lambda C\subseteq C}
x
∈
C
{\displaystyle x\in C}
x
∈
1
⋅
C
{\displaystyle x\in 1\cdot C}
μ
C
(
x
)
≤
1
{\displaystyle \mu _{C}(x)\leq 1}
Demonstração da Versão Geométrica
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Sem perda de generalidade, podemos supor que
0
∈
C
{\displaystyle 0\in C}
δ
=
d
(
x
,
C
)
>
0
{\displaystyle \delta =d(x,C)>0}
C
{\displaystyle C}
Agora, seja
D
=
{
x
∈
X
:
d
(
x
,
C
)
≤
δ
/
2
}
{\displaystyle D=\{x\in X:d(x,C)\leq \delta /2\}}
0
∈
C
{\displaystyle 0\in C}
δ
4
B
x
⊆
D
{\displaystyle {\frac {\delta }{4}}B_{x}\subseteq D}
0
∈
i
n
t
D
{\displaystyle 0\in intD}
D
{\displaystyle D}
μ
D
{\displaystyle \mu _{D}}
μ
D
{\displaystyle \mu _{D}}
μ
D
>
1
{\displaystyle \mu _{D}>1}
x
0
∉
D
{\displaystyle x_{0}\not \in D}
Considere
φ
:
[
x
0
]
→
R
{\displaystyle \varphi :[x_{0}]\to \mathbb {R} }
φ
(
λ
x
0
)
=
λ
μ
D
(
x
0
)
{\displaystyle \varphi (\lambda x_{0})=\lambda {\displaystyle \mu _{D}}(x_{0})}
φ
{\displaystyle \varphi }
φ
(
λ
x
0
)
≤
μ
D
(
λ
x
0
)
{\displaystyle \varphi (\lambda x_{0})\leq \mu _{D}(\lambda x_{0})}
Teorema de Hahn-Banach , temos que existe
ϕ
~
∈
X
∗
{\displaystyle {\tilde {\phi }}\in X^{*}}
ϕ
~
∣
x
0
=
φ
{\displaystyle {\tilde {\phi }}\mid _{x_{0}}=\varphi }
ϕ
~
(
x
)
≤
μ
D
(
x
)
∀
x
∈
X
{\displaystyle {\tilde {\phi }}(x)\leq \mu _{D}(x)\forall x\in X}
ϕ
~
(
x
0
)
=
φ
(
x
0
)
=
μ
D
(
x
0
)
>
1
{\displaystyle {\tilde {\phi }}(x_{0})=\varphi (x_{0})=\mu _{D}(x_{0})>1}
Por outro lado, dado
x
∈
C
{\displaystyle x\in C}
x
∈
D
{\displaystyle x\in D}
ϕ
~
(
x
)
≤
μ
D
(
x
)
≤
1
{\displaystyle {\tilde {\phi }}(x)\leq \mu _{D}(x)\leq 1}
Assim,
ϕ
~
(
x
0
)
>
1
≥
s
u
p
{
ϕ
~
(
x
)
:
x
∈
C
}
{\displaystyle {\tilde {\phi }}(x_{0})>1\geq sup\{{\tilde {\phi }}(x):x\in C\}}
![{\displaystyle B[x,r]=\{y\in X:\lVert x-y\rVert \leq r\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f1d3ed0d8efbb04d25c48dd586b46d2bd586bccc)
![{\displaystyle {\frac {\delta }{\lVert x\rVert }}\cdot x\in \delta B_{x}=B[0,\delta ]\subseteq C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/37d64b4830a803370763af73f5cc414759335830)
![{\displaystyle \varphi :[x_{0}]\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/49678a41b8c95e085cb21d458c27b0275fbaa689)
