Definição
M
(
f
,
D
)
{\displaystyle M(f,D)\;}
Definição Dizer que
#
M
(
f
,
D
)
=
1
{\displaystyle \#M(f,D)=1\;}
Definição Seja
v
¯
∈
]
−
∞
,
∞
[
;
v
¯
=
inf
x
∈
D
f
(
x
)
⇒
v
¯
∈
M
(
f
,
D
)
{\displaystyle {\bar {v}}\in \;]-\infty ,\infty [;{\bar {v}}=\inf _{x\in D}f(x)\Rightarrow {\bar {v}}\in M(f,D)}
v
¯
{\displaystyle {\bar {v}}\;}
Seja
f
:
D
→
R
{\displaystyle f:D\rightarrow \mathbb {R} }
contínua em D compacto.
Então
M
(
f
,
D
)
≠
∅
{\displaystyle M(f,D)\not =\emptyset }
editar
Suponha que f é ilimitada inferiormente, então
∀
k
∈
N
,
∃
x
k
∈
D
;
f
(
x
k
)
<
−
k
⇒
lim
k
→
∞
f
(
x
k
)
=
−
∞
{\displaystyle \forall \;k\in \mathbb {N} ,\exists \;x^{k}\in D;f(x^{k})<-k\Rightarrow \lim _{k\rightarrow \infty }f(x^{k})=-\infty }
{
x
k
}
⊂
D
{\displaystyle \{x^{k}\}\subset D}
{
x
k
}
{\displaystyle \{x^{k}\}\;}
Toda sequência limitada possui uma subsequência convergente . Assim
{
x
k
}
{\displaystyle \{x^{k}\}\;}
{
x
k
j
}
{\displaystyle \{x^{k_{j}}\}\;}
lim
j
→
∞
x
k
j
=
x
¯
{\displaystyle \;\lim _{j\rightarrow \infty }x^{k_{j}}={\bar {x}}}
f
(
x
¯
)
=
f
(
lim
j
→
∞
x
k
j
)
=
−
∞
{\displaystyle f({\bar {x}})=f(\lim _{j\rightarrow \infty }x^{k_{j}})=-\infty }
v
¯
=
inf
x
∈
D
f
(
x
)
>
−
∞
{\displaystyle {\bar {v}}=\inf _{x\in D}f(x)>-\infty }
k
∈
N
,
∃
x
k
∈
D
{\displaystyle k\in \mathbb {N} ,\;\exists \;x^{k}\in D}
v
¯
≤
f
(
x
k
)
<
v
¯
+
1
k
⇒
v
¯
≤
lim
k
→
∞
f
(
x
k
)
≤
v
¯
⇒
lim
k
→
∞
f
(
x
k
)
=
v
¯
{\displaystyle {\bar {v}}\leq f(x^{k})<{\bar {v}}+{1 \over k}\Rightarrow {\bar {v}}\leq \lim _{k\rightarrow \infty }f(x^{k})\leq {\bar {v}}\Rightarrow \lim _{k\rightarrow \infty }f(x^{k})={\bar {v}}}
Curva de nível
L
f
,
D
(
c
)
{\displaystyle L_{f,D}(c)}
editar
Definição Seja
f
:
D
→
R
,
c
∈
R
,
L
f
,
D
(
c
)
=
{
x
∈
D
;
f
(
x
)
≤
c
}
{\displaystyle f:D\rightarrow \mathbb {R} ,c\in \mathbb {R} ,L_{f,D}(c)=\{x\in D;f(x)\leq c\}}
Corolário da curva de nível compacta
editar
Sejam
D
∈
R
n
,
f
:
D
→
R
{\displaystyle D\in \mathbb {R} ^{n},f:D\rightarrow \mathbb {R} }
contínua em D. Se
∃
c
∈
R
,
∅
≠
{\displaystyle \exists \;c\in \mathbb {R} ,\emptyset \not =}
L
f
,
D
(
c
)
{\displaystyle L_{f,D}(c)\;}
é compacto.
Então
M
(
f
,
D
)
≠
∅
{\displaystyle M(f,D)\not =\emptyset }
editar
Prova: Pelo Teorema de Weierstrass
M
(
f
,
D
)
≠
∅
{\displaystyle M(f,D)\not =\emptyset }
∃
x
¯
∈
D
;
f
(
x
¯
)
≤
f
(
x
)
,
∀
x
∈
{\displaystyle \exists {\bar {x}}\in D;f({\bar {x}})\leq f(x),\forall \;x\in }
L
f
,
D
(
c
)
{\displaystyle L_{f,D}(c)\;}
).
Mas se
x
∈
D
∖
{\displaystyle x\in D\setminus }
L
f
,
D
(
c
)
{\displaystyle L_{f,D}(c)\;}
⇒
f
(
x
)
>
c
≥
f
(
x
¯
)
{\displaystyle \Rightarrow f(x)>c\geq f({\bar {x}})}
f
(
x
¯
)
≤
f
(
x
)
,
∀
x
∈
D
{\displaystyle f({\bar {x}})\leq f(x),\forall \;x\in D}
M
(
f
,
D
)
≠
∅
{\displaystyle M(f,D)\not =\emptyset }
Projeção de y sobre D
editar
Definição
P
D
(
y
)
=
x
¯
⇔
inf
x
∈
D
‖
x
−
y
‖
=
‖
x
¯
−
y
‖
{\displaystyle P_{D}(y)={\bar {x}}\Leftrightarrow \inf _{x\in D}\|x-y\|=\|{\bar {x}}-y\|}
Corolário da projeção de y sobre D
editar
∅
≠
D
∈
R
n
{\displaystyle \emptyset \not =D\in \mathbb {R} ^{n}}
Então
P
D
(
y
)
≠
∅
,
∀
y
∈
R
n
{\displaystyle P_{D}(y)\not =\emptyset ,\forall \;y\in \mathbb {R} ^{n}}
editar
Tome
y
∈
R
n
e
f
y
:
D
→
R
;
x
↦
f
y
(
x
)
=
‖
x
−
y
‖
{\displaystyle y\in \mathbb {R} ^{n}ef_{y}:D\rightarrow \mathbb {R} ;x\mapsto f_{y}(x)=\|x-y\|}
|
‖
x
‖
−
‖
y
‖
|
≤
‖
x
−
y
‖
{\displaystyle |\|x\|-\|y\||\leq \|x-y\|}
. Agora dado
ϵ
>
0
,
∃
δ
=
ϵ
,
‖
x
−
y
‖
<
δ
⇒
|
‖
x
‖
−
‖
y
‖
|
<
ϵ
{\displaystyle \epsilon >0,\exists \;\delta =\epsilon ,\|x-y\|<\delta \Rightarrow |\|x\|-\|y\||<\epsilon }
f
y
{\displaystyle f_{y}\;}
contínua .
Por outro lado,
L
f
y
,
D
(
c
)
=
{
x
∈
D
/
f
y
(
x
)
≤
c
}
=
{
x
∈
D
/
‖
x
−
y
‖
≤
c
}
=
D
∩
B
c
(
y
)
{\displaystyle L_{f_{y},D}(c)=\{x\in D/f_{y}(x)\leq c\}=\{x\in D/\|x-y\|\leq c\}=D\cap B_{c}(y)}
D
,
B
c
(
y
)
{\displaystyle D,B_{c}(y)\;}
D
∩
B
c
(
y
)
{\displaystyle D\cap B_{c}(y)}
B
c
(
y
)
{\displaystyle B_{c}(y)}
D
∩
B
c
(
y
)
⊂
B
c
(
y
)
{\displaystyle D\cap B_{c}(y)\subset B_{c}(y)}
L
f
y
,
D
(
c
)
=
D
∩
B
c
(
y
)
⇒
L
f
y
,
D
(
c
)
{\displaystyle L_{f_{y},D}(c)=D\cap B_{c}(y)\Rightarrow L_{f_{y},D}(c)}
Vimos que
f
y
{\displaystyle f_{y}\;}
contínua e
L
f
y
,
D
(
c
)
{\displaystyle L_{f_{y},D}(c)}
c
∈
R
{\displaystyle c\in \mathbb {R} }
L
f
y
,
D
(
c
)
≠
∅
{\displaystyle L_{f_{y},D}(c)\not =\emptyset }
M
(
f
y
,
D
)
{\displaystyle M(f_{y},D)\;}
≠
∅
{\displaystyle \not =\emptyset }
Mas
M
(
f
y
,
D
)
{\displaystyle M(f_{y},D)\;}
=
{
x
¯
∈
D
/
f
y
(
x
¯
)
≤
f
y
(
x
)
,
∀
x
∈
D
}
=
{
x
¯
∈
D
/
‖
x
¯
−
y
‖
≤
‖
x
−
y
‖
,
∀
x
∈
D
}
=
{\displaystyle =\{{\bar {x}}\in D/f_{y}({\bar {x}})\leq f_{y}(x),\forall \;x\in D\}=\{{\bar {x}}\in D/\|{\bar {x}}-y\|\leq \|x-y\|,\forall \;x\in D\}=}
=
{
x
¯
∈
D
/
inf
x
∈
D
‖
x
−
y
‖
=
‖
x
¯
−
y
‖
}
=
P
D
(
y
)
≠
∅
{\displaystyle =\{{\bar {x}}\in D/\inf _{x\in D}\|x-y\|=\|{\bar {x}}-y\|\}=P_{D}(y)\not =\emptyset }
Seja
D
=
{
x
∈
R
n
|
x
1
+
x
2
>
0
}
{\displaystyle D=\{x\in \mathbb {R} ^{n}|x_{1}+x_{2}>0\}}
f
:
D
→
R
,
f
(
x
)
=
x
1
2
+
x
2
+
1
x
1
+
x
2
{\displaystyle f:D\rightarrow \mathbb {R} ,f(x)=x_{1}^{2}+x_{2}+{1 \over x_{1}+x_{2}}}
Mostrar que
M
(
f
,
D
)
≠
∅
{\displaystyle M(f,D)\not =\emptyset }
editar
Suponhamos que
L
f
,
D
(
c
)
{\displaystyle L_{f,D}(c)\;}
c
∈
R
⇒
∃
{
x
k
}
⊂
L
f
,
D
(
c
)
{\displaystyle c\in \mathbb {R} \Rightarrow \exists \{x^{k}\}\subset L_{f,D}(c)}
lim
k
→
∞
‖
x
k
‖
{\displaystyle \lim _{k\rightarrow \infty }\|x^{k}\|}
lim
k
→
∞
‖
x
1
k
‖
{\displaystyle \lim _{k\rightarrow \infty }\|x_{1}^{k}\|}
lim
k
→
∞
‖
x
1
k
‖
{\displaystyle \lim _{k\rightarrow \infty }\|x_{1}^{k}\|}
![{\displaystyle {\bar {v}}\in \;]-\infty ,\infty [;{\bar {v}}=\inf _{x\in D}f(x)\Rightarrow {\bar {v}}\in M(f,D)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e7f58eaccff375d0eec52a7535282c289357da75)
