2차원 유클리드 공간
R
2
{\displaystyle \mathbb {R} ^{2}}
(
0
,
0
)
{\displaystyle (0,0)}
원 의 방정식
x
2
+
y
2
−
1
=
0
{\displaystyle x^{2}+y^{2}-1=0}
을 생각하자. 이 방정식은
[
−
1
,
1
]
×
[
0
,
∞
)
{\displaystyle [-1,1]\times [0,\infty )}
y
=
1
−
x
2
∀
x
∈
[
−
1
,
1
]
{\displaystyle y={\sqrt {1-x^{2}}}\qquad \forall x\in [-1,1]}
또한 이 방정식은
[
−
1
,
1
]
×
(
−
∞
,
0
]
{\displaystyle [-1,1]\times (-\infty ,0]}
y
=
−
1
−
x
2
∀
x
∈
[
−
1
,
1
]
{\displaystyle y=-{\sqrt {1-x^{2}}}\qquad \forall x\in [-1,1]}
이 방정식을 만족시키는 연속 함수 는
[
−
1
,
1
]
×
R
{\displaystyle [-1,1]\times \mathbb {R} }
(
0
,
1
)
{\displaystyle (0,1)}
z
=
x
2
+
y
2
−
1
∀
x
,
y
∈
R
{\displaystyle z=x^{2}+y^{2}-1\qquad \forall x,y\in \mathbb {R} }
이렇게 정의한
F
{\displaystyle F}
R
3
{\displaystyle \mathbb {R} ^{3}}
z
=
0
{\displaystyle z=0}
(
0
,
1
)
{\displaystyle (0,1)}
x
′
2
+
y
′
2
−
1
=
0
{\displaystyle x'^{2}+y'^{2}-1=0}
(
x
′
,
y
′
)
{\displaystyle (x',y')}
(
x
′
,
y
′
)
=
(
0
,
1
)
{\displaystyle (x',y')=(0,1)}
x
=
x
′
{\displaystyle x=x'}
z
=
x
′
2
+
y
2
−
1
{\displaystyle z={x'}^{2}+y^{2}-1}
y
=
y
′
{\displaystyle y=y'}
y
{\displaystyle y}
y
<
y
′
{\displaystyle y<y'}
z
<
0
{\displaystyle z<0}
y
>
y
′
{\displaystyle y>y'}
z
>
0
{\displaystyle z>0}
z
=
x
2
+
y
2
−
1
{\displaystyle z=x^{2}+y^{2}-1}
(
0
,
1
)
{\displaystyle (0,1)}
y
=
y
(
x
)
{\displaystyle y=y(x)}
z
{\displaystyle z}
y
{\displaystyle y}
∂
z
/
∂
y
≠
0
{\displaystyle \partial z/\partial y\neq 0}
(
1
,
0
)
{\displaystyle (1,0)}
∂
z
/
∂
y
≠
0
{\displaystyle \partial z/\partial y\neq 0}
z
1
(
x
1
,
…
,
x
n
,
y
1
,
…
,
y
m
)
=
0
{\displaystyle z_{1}(x_{1},\dots ,x_{n},y_{1},\dots ,y_{m})=0}
⋮
{\displaystyle \vdots }
z
m
(
x
1
,
…
,
x
n
,
y
1
,
…
,
y
m
)
=
0
{\displaystyle z_{m}(x_{1},\dots ,x_{n},y_{1},\dots ,y_{m})=0}
에서
(
y
1
,
…
,
y
m
)
{\displaystyle (y_{1},\dots ,y_{m})}
(
x
1
,
…
,
x
n
)
{\displaystyle (x_{1},\dots ,x_{n})}
det
∂
(
z
1
,
…
,
z
m
)
∂
(
y
1
,
…
,
y
m
)
≠
0
{\displaystyle \det {\frac {\partial (z_{1},\dots ,z_{m})}{\partial (y_{1},\dots ,y_{m})}}\neq 0}
이 부등식의 좌변은 함수
z
1
,
…
,
z
m
{\displaystyle z_{1},\dots ,z_{m}}
y
1
,
…
,
y
m
{\displaystyle y_{1},\dots ,y_{m}}
야코비 행렬식 이다. 특히, 만약
z
1
=
a
11
x
1
+
⋯
a
1
n
x
n
+
b
11
y
1
+
⋯
+
b
1
m
y
m
{\displaystyle z_{1}=a_{11}x_{1}+\cdots a_{1n}x_{n}+b_{11}y_{1}+\cdots +b_{1m}y_{m}}
⋮
{\displaystyle \vdots }
z
m
=
a
m
1
x
1
+
⋯
+
a
m
n
x
n
+
b
m
1
y
1
+
⋯
+
b
m
m
y
m
{\displaystyle z_{m}=a_{m1}x_{1}+\cdots +a_{mn}x_{n}+b_{m1}y_{1}+\cdots +b_{mm}y_{m}}
일 경우, 위에서 정의한 야코비 행렬식은
x
1
,
…
,
x
n
{\displaystyle x_{1},\dots ,x_{n}}
det
(
b
i
j
)
m
×
m
{\displaystyle \det(b_{ij})_{m\times m}}
수학적 귀납법 을 사용하자.
먼저
m
=
1
{\displaystyle m=1}
(
∂
f
/
∂
y
)
(
a
,
b
)
>
0
{\displaystyle (\partial f/\partial y)(a,b)>0}
∂
f
/
∂
y
{\displaystyle \partial f/\partial y}
δ
1
>
0
{\displaystyle \delta _{1}>0}
B
(
a
,
δ
1
)
⊆
U
{\displaystyle \operatorname {B} (\mathbf {a} ,\delta _{1})\subseteq U}
B
(
b
,
δ
1
)
⊆
V
{\displaystyle \operatorname {B} (b,\delta _{1})\subseteq V}
임의의
x
∈
B
(
a
,
δ
1
)
{\displaystyle \mathbf {x} \in \operatorname {B} (\mathbf {a} ,\delta _{1})}
y
∈
B
(
b
,
δ
1
)
{\displaystyle y\in \operatorname {B} (b,\delta _{1})}
(
∂
f
/
∂
y
)
(
x
,
y
)
>
0
{\displaystyle (\partial f/\partial y)(\mathbf {x} ,y)>0}
이에 따라
f
(
a
,
y
)
{\displaystyle f(\mathbf {a} ,y)}
B
(
b
,
δ
1
)
{\displaystyle \operatorname {B} (b,\delta _{1})}
f
(
a
,
b
)
=
0
{\displaystyle f(\mathbf {a} ,b)=0}
f
(
a
,
b
−
δ
1
)
<
0
<
f
(
a
,
b
+
δ
1
)
{\displaystyle f(\mathbf {a} ,b-\delta _{1})<0<f(\mathbf {a} ,b+\delta _{1})}
0
<
δ
2
<
δ
1
{\displaystyle 0<\delta _{2}<\delta _{1}}
임의의
x
∈
B
(
a
,
δ
2
)
{\displaystyle \mathbf {x} \in \operatorname {B} (\mathbf {a} ,\delta _{2})}
f
(
x
,
b
−
δ
1
)
<
0
<
f
(
x
,
b
+
δ
1
)
{\displaystyle f(\mathbf {x} ,b-\delta _{1})<0<f(\mathbf {x} ,b+\delta _{1})}
따라서, 임의의
x
′
∈
B
(
a
,
δ
2
)
{\displaystyle \mathbf {x} '\in \operatorname {B} (\mathbf {a} ,\delta _{2})}
f
(
x
′
,
y
)
{\displaystyle f(\mathbf {x} ',y)}
B
(
b
,
δ
1
)
{\displaystyle \operatorname {B} (b,\delta _{1})}
중간값 정리 에 따라
f
(
x
′
,
g
(
x
′
)
)
=
0
{\displaystyle f(\mathbf {x} ',g(\mathbf {x} '))=0}
g
(
x
′
)
∈
B
(
b
,
δ
1
)
{\displaystyle g(\mathbf {x} ')\in \operatorname {B} (b,\delta _{1})}
g
:
B
(
a
,
δ
2
)
→
R
{\displaystyle g\colon \operatorname {B} (\mathbf {a} ,\delta _{2})\to \mathbb {R} }
x
∈
B
(
a
,
δ
2
)
{\displaystyle \mathbf {x} \in \operatorname {B} (\mathbf {a} ,\delta _{2})}
f
(
x
,
g
(
x
)
)
=
0
{\displaystyle f(\mathbf {x} ,g(\mathbf {x} ))=0}
b
=
g
(
a
)
{\displaystyle b=g(\mathbf {a} )}
g
{\displaystyle g}
x
′
∈
B
(
a
,
δ
2
)
{\displaystyle \mathbf {x} '\in \operatorname {B} (\mathbf {a} ,\delta _{2})}
ϵ
>
0
{\displaystyle \epsilon >0}
f
(
x
′
,
g
(
x
′
)
−
ϵ
)
<
0
<
f
(
x
′
,
g
(
x
′
)
+
ϵ
)
{\displaystyle f(\mathbf {x} ',g(\mathbf {x} ')-\epsilon )<0<f(\mathbf {x} ',g(\mathbf {x} ')+\epsilon )}
δ
3
>
0
{\displaystyle \delta _{3}>0}
B
(
x
′
,
δ
3
)
⊆
B
(
a
,
δ
2
)
{\displaystyle \operatorname {B} (\mathbf {x} ',\delta _{3})\subseteq \operatorname {B} (\mathbf {a} ,\delta _{2})}
임의의
x
∈
B
(
x
′
,
δ
3
)
{\displaystyle \mathbf {x} \in \operatorname {B} (\mathbf {x} ',\delta _{3})}
f
(
x
,
g
(
x
′
)
−
ϵ
)
<
0
<
f
(
x
,
g
(
x
′
)
+
ϵ
)
{\displaystyle f(\mathbf {x} ,g(\mathbf {x} ')-\epsilon )<0<f(\mathbf {x} ,g(\mathbf {x} ')+\epsilon )}
따라서, 임의의
x
∈
B
(
x
′
,
δ
)
{\displaystyle \mathbf {x} \in \operatorname {B} (\mathbf {x} ',\delta )}
g
(
x
)
∈
B
(
g
(
x
′
)
,
ϵ
)
{\displaystyle g(\mathbf {x} )\in \operatorname {B} (g(\mathbf {x} '),\epsilon )}
g
{\displaystyle g}
h
:
B
(
a
,
δ
2
)
→
R
{\displaystyle h\colon \operatorname {B} (\mathbf {a} ,\delta _{2})\to \mathbb {R} }
h
(
B
(
a
,
δ
2
)
)
⊆
V
{\displaystyle h(\operatorname {B} (\mathbf {a} ,\delta _{2}))\subseteq V}
b
=
h
(
a
)
{\displaystyle b=h(\mathbf {a} )}
임의의
x
∈
B
(
a
,
δ
2
)
{\displaystyle \mathbf {x} \in \operatorname {B} (\mathbf {a} ,\delta _{2})}
f
(
x
,
h
(
x
)
)
=
0
{\displaystyle f(\mathbf {x} ,h(\mathbf {x} ))=0}
그렇다면, 다음과 같은 집합이
B
(
a
,
δ
2
)
{\displaystyle \operatorname {B} (\mathbf {a} ,\delta _{2})}
열린닫힌집합 임을 보이는 것으로 족하다.
A
=
{
x
∈
B
(
a
,
δ
2
)
:
g
(
x
)
=
h
(
x
)
}
{\displaystyle A=\{\mathbf {x} \in \operatorname {B} (\mathbf {a} ,\delta _{2})\colon g(\mathbf {x} )=h(\mathbf {x} )\}}
우선 임의의
x
′
∈
A
{\displaystyle \mathbf {x} '\in A}
δ
4
>
0
{\displaystyle \delta _{4}>0}
B
(
x
′
,
δ
4
)
⊆
B
(
a
,
δ
2
)
{\displaystyle \operatorname {B} (\mathbf {x} ',\delta _{4})\subseteq \operatorname {B} (\mathbf {a} ,\delta _{2})}
임의의
x
∈
B
(
x
′
,
δ
4
)
{\displaystyle \mathbf {x} \in \operatorname {B} (\mathbf {x} ',\delta _{4})}
h
(
x
)
∈
B
(
b
,
δ
1
)
{\displaystyle h(\mathbf {x} )\in \operatorname {B} (b,\delta _{1})}
따라서, 임의의
x
∈
B
(
x
′
,
δ
4
)
{\displaystyle \mathbf {x} \in \operatorname {B} (\mathbf {x} ',\delta _{4})}
h
(
x
)
=
g
(
x
)
{\displaystyle h(\mathbf {x} )=g(\mathbf {x} )}
A
{\displaystyle A}
B
(
a
,
δ
2
)
{\displaystyle \operatorname {B} (\mathbf {a} ,\delta _{2})}
x
′
∈
A
′
∩
B
(
a
,
δ
2
)
{\displaystyle \mathbf {x} '\in A'\cap \operatorname {B} (\mathbf {a} ,\delta _{2})}
g
,
h
{\displaystyle g,h}
x
′
∈
A
{\displaystyle \mathbf {x} '\in A}
A
{\displaystyle A}
B
(
a
,
δ
2
)
{\displaystyle \operatorname {B} (\mathbf {a} ,\delta _{2})}
B
(
a
,
δ
2
)
{\displaystyle \operatorname {B} (\mathbf {a} ,\delta _{2})}
연결 집합 이며, 또한
A
≠
∅
{\displaystyle A\neq \varnothing }
A
=
B
(
a
,
δ
2
)
{\displaystyle A=\operatorname {B} (\mathbf {a} ,\delta _{2})}
x
∈
B
(
a
,
δ
2
)
{\displaystyle \mathbf {x} \in \operatorname {B} (\mathbf {a} ,\delta _{2})}
g
(
x
)
=
h
(
x
)
{\displaystyle g(\mathbf {x} )=h(\mathbf {x} )}
f
{\displaystyle f}
C
k
{\displaystyle {\mathcal {C}}^{k}}
g
{\displaystyle g}
C
k
{\displaystyle {\mathcal {C}}^{k}}
j
∈
{
1
,
…
,
n
}
{\displaystyle j\in \{1,\dots ,n\}}
x
′
,
x
′
+
Δ
x
j
e
j
∈
B
(
a
,
δ
2
)
{\displaystyle \mathbf {x} ',\mathbf {x} '+\Delta x_{j}\mathbf {e} _{j}\in \operatorname {B} (\mathbf {a} ,\delta _{2})}
Δ
y
(
x
′
,
x
′
+
Δ
x
j
e
j
)
=
g
(
x
′
+
Δ
x
j
e
j
)
−
g
(
x
′
)
{\displaystyle \Delta y(\mathbf {x} ',\mathbf {x} '+\Delta x_{j}\mathbf {e} _{j})=g(\mathbf {x} '+\Delta x_{j}\mathbf {e} _{j})-g(\mathbf {x} ')}
그러면 평균값 정리 에 따라 다음을 만족시키는
0
<
θ
(
x
′
,
x
′
+
Δ
x
j
e
j
)
<
1
{\displaystyle 0<\theta (\mathbf {x} ',\mathbf {x} '+\Delta x_{j}\mathbf {e} _{j})<1}
0
=
f
(
x
′
+
Δ
x
j
e
j
,
g
(
x
′
)
+
Δ
y
)
−
f
(
x
′
,
g
(
x
′
)
)
=
∂
f
∂
x
j
(
x
′
+
θ
Δ
x
j
e
j
,
g
(
x
′
)
+
θ
Δ
y
)
Δ
x
j
+
∂
f
∂
y
(
x
′
+
θ
Δ
x
j
e
j
,
g
(
x
′
)
+
θ
Δ
y
)
Δ
y
{\displaystyle {\begin{aligned}0&=f(\mathbf {x} '+\Delta x_{j}\mathbf {e} _{j},g(\mathbf {x} ')+\Delta y)-f(\mathbf {x} ',g(\mathbf {x} '))\\&={\frac {\partial f}{\partial x_{j}}}(\mathbf {x} '+\theta \Delta x_{j}\mathbf {e} _{j},\mathbf {g} (\mathbf {x} ')+\theta \Delta y)\Delta x_{j}+{\frac {\partial f}{\partial y}}(\mathbf {x} '+\theta \Delta x_{j}\mathbf {e} _{j},\mathbf {g} (\mathbf {x} ')+\theta \Delta y)\Delta y\end{aligned}}}
따라서, 다음이 성립한다.
∂
g
∂
x
j
(
x
′
)
=
lim
Δ
x
j
→
0
Δ
y
Δ
x
j
=
lim
Δ
x
j
→
0
(
−
(
∂
f
/
∂
x
j
)
(
x
′
+
θ
Δ
x
j
e
j
,
g
(
x
′
)
+
θ
Δ
y
)
(
∂
f
/
∂
y
)
(
x
′
+
θ
Δ
x
j
e
j
,
g
(
x
′
)
+
θ
Δ
y
)
)
=
−
(
∂
f
/
∂
x
j
)
(
x
′
,
g
(
x
′
)
)
(
∂
f
/
∂
y
)
(
x
′
,
g
(
x
′
)
)
{\displaystyle {\begin{aligned}{\frac {\partial g}{\partial x_{j}}}(\mathbf {x} ')&=\lim _{\Delta x_{j}\to \mathbf {0} }{\frac {\Delta y}{\Delta x_{j}}}\\&=\lim _{\Delta x_{j}\to \mathbf {0} }\left(-{\frac {(\partial f/\partial x_{j})(\mathbf {x} '+\theta \Delta x_{j}\mathbf {e} _{j},\mathbf {g} (\mathbf {x} ')+\theta \Delta y)}{(\partial f/\partial y)(\mathbf {x} '+\theta \Delta x_{j}\mathbf {e} _{j},\mathbf {g} (\mathbf {x} ')+\theta \Delta y)}}\right)\\&=-{\frac {(\partial f/\partial x_{j})(\mathbf {x} ',g(\mathbf {x} '))}{(\partial f/\partial y)(\mathbf {x} ',g(\mathbf {x} '))}}\end{aligned}}}
또한
(
∂
f
/
∂
x
j
)
(
x
′
,
g
(
x
′
)
)
{\displaystyle (\partial f/\partial x_{j})(\mathbf {x} ',g(\mathbf {x} '))}
(
∂
f
/
∂
y
)
(
x
′
,
g
(
x
′
)
)
{\displaystyle (\partial f/\partial y)(\mathbf {x} ',g(\mathbf {x} '))}
g
{\displaystyle g}
이제
m
>
1
{\displaystyle m>1}
V
{\displaystyle V}
y
=
(
y
1
,
y
~
)
{\displaystyle \mathbf {y} =(y_{1},{\tilde {\mathbf {y} }})}
f
=
(
f
1
,
f
~
)
{\displaystyle \mathbf {f} =(f_{1},{\tilde {\mathbf {f} }})}
det
(
D
y
~
f
~
(
a
,
b
)
)
≠
0
{\displaystyle \det(\mathrm {D} _{\tilde {\mathbf {y} }}{\tilde {\mathbf {f} }}\!(\mathbf {a} ,\mathbf {b} ))\neq 0}
g
~
:
B
(
a
,
δ
1
)
×
B
(
b
1
,
δ
1
)
→
R
m
−
1
{\displaystyle {\tilde {\mathbf {g} }}\colon \operatorname {B} (\mathbf {a} ,\delta _{1})\times \operatorname {B} (b_{1},\delta _{1})\to \mathbb {R} ^{m-1}}
δ
1
>
0
{\displaystyle \delta _{1}>0}
B
(
a
,
δ
1
)
⊆
U
{\displaystyle \operatorname {B} (\mathbf {a} ,\delta _{1})\subseteq U}
B
(
b
1
,
δ
1
)
×
g
~
(
B
(
a
,
δ
1
)
×
B
(
b
1
,
δ
1
)
)
⊆
V
{\displaystyle \operatorname {B} (b_{1},\delta _{1})\times {\tilde {\mathbf {g} }}(\operatorname {B} (\mathbf {a} ,\delta _{1})\times \operatorname {B} (b_{1},\delta _{1}))\subseteq V}
b
′
=
g
~
(
a
,
b
1
)
{\displaystyle \mathbf {b} '={\tilde {\mathbf {g} }}(\mathbf {a} ,b_{1})}
임의의
x
∈
B
(
a
,
δ
1
)
{\displaystyle \mathbf {x} \in \operatorname {B} (\mathbf {a} ,\delta _{1})}
y
1
∈
B
(
b
1
,
δ
1
)
{\displaystyle y_{1}\in \operatorname {B} (b_{1},\delta _{1})}
i
∈
{
2
,
…
,
m
}
{\displaystyle i\in \{2,\dots ,m\}}
f
i
(
x
,
y
1
,
g
~
(
x
,
y
1
)
)
=
0
{\displaystyle f_{i}(\mathbf {x} ,y_{1},{\tilde {\mathbf {g} }}(\mathbf {x} ,y_{1}))=0}
다음과 같은 함수
F
:
B
(
a
,
δ
1
)
×
B
(
b
1
,
δ
1
)
→
R
m
{\displaystyle F\colon \operatorname {B} (\mathbf {a} ,\delta _{1})\times \operatorname {B} (b_{1},\delta _{1})\to \mathbb {R} ^{m}}
F
(
x
,
y
1
)
=
f
1
(
x
,
y
1
,
g
~
(
x
,
y
1
)
)
∀
x
∈
B
(
a
,
δ
1
)
,
y
1
∈
B
(
b
1
,
δ
1
)
{\displaystyle F(\mathbf {x} ,y_{1})=f_{1}(\mathbf {x} ,y_{1},{\tilde {\mathbf {g} }}(\mathbf {x} ,y_{1}))\qquad \forall \mathbf {x} \in \operatorname {B} (\mathbf {a} ,\delta _{1}),\;y_{1}\in \operatorname {B} (b_{1},\delta _{1})}
그렇다면, 수학적 귀납법 가정에 의하여, 임의의
x
∈
B
(
a
,
δ
1
)
{\displaystyle \mathbf {x} \in \operatorname {B} (\mathbf {a} ,\delta _{1})}
y
1
∈
B
(
b
1
,
δ
1
)
{\displaystyle y_{1}\in \operatorname {B} (b_{1},\delta _{1})}
∂
F
∂
y
1
=
∂
f
1
∂
y
1
+
∑
j
=
2
m
∂
f
1
∂
y
j
∂
h
j
−
1
∂
y
1
=
∂
f
1
∂
y
1
+
D
y
~
f
1
(
D
y
~
f
~
)
−
1
D
y
1
f
~
=
∂
f
1
∂
y
1
+
∑
i
=
2
m
∑
j
=
2
m
(
−
1
)
i
+
j
1
det
D
y
~
f
~
∂
f
i
∂
y
1
∂
f
1
∂
y
j
det
D
(
y
2
,
…
,
y
j
−
1
,
y
j
+
1
,
…
,
y
m
)
(
f
2
,
…
,
f
i
−
1
,
f
i
+
1
,
…
,
f
m
)
=
det
D
y
f
det
D
y
~
f
~
{\displaystyle {\begin{aligned}{\frac {\partial F}{\partial y_{1}}}&={\frac {\partial f_{1}}{\partial y_{1}}}+\sum _{j=2}^{m}{\frac {\partial f_{1}}{\partial y_{j}}}{\frac {\partial h_{j-1}}{\partial y_{1}}}\\&={\frac {\partial f_{1}}{\partial y_{1}}}+\mathrm {D} _{\tilde {\mathrm {y} }}f_{1}(\mathrm {D} _{\tilde {\mathrm {y} }}{\tilde {f}})^{-1}\mathrm {D} _{y_{1}}{\tilde {f}}\\&={\frac {\partial f_{1}}{\partial y_{1}}}+\sum _{i=2}^{m}\sum _{j=2}^{m}(-1)^{i+j}{\frac {1}{\det \mathrm {D} _{\tilde {\mathbf {y} }}{\tilde {\mathbf {f} }}}}{\frac {\partial f_{i}}{\partial y_{1}}}{\frac {\partial f_{1}}{\partial y_{j}}}\det \mathrm {D} _{(y_{2},\dots ,y_{j-1},y_{j+1},\dots ,y_{m})}(f_{2},\dots ,f_{i-1},f_{i+1},\dots ,f_{m})\\&={\frac {\det \mathrm {D} _{\mathbf {y} }\mathbf {f} }{\det \mathrm {D} _{\tilde {\mathbf {y} }}{\tilde {\mathbf {f} }}}}\end{aligned}}}
따라서
(
∂
F
/
∂
y
1
)
(
a
,
b
1
)
≠
0
{\displaystyle (\partial F/\partial y_{1})(\mathbf {a} ,b_{1})\neq 0}
g
1
:
B
(
a
,
δ
2
)
→
R
{\displaystyle g_{1}\colon \operatorname {B} (\mathbf {a} ,\delta _{2})\to \mathbb {R} }
0
<
δ
2
<
δ
1
{\displaystyle 0<\delta _{2}<\delta _{1}}
g
1
(
B
(
a
,
δ
2
)
)
⊆
B
(
b
1
,
δ
1
)
{\displaystyle g_{1}(\operatorname {B} (\mathbf {a} ,\delta _{2}))\subseteq \operatorname {B} (b_{1},\delta _{1})}
b
1
=
g
1
(
a
)
{\displaystyle b_{1}=g_{1}(\mathbf {a} )}
임의의
x
∈
B
(
a
,
δ
2
)
{\displaystyle \mathbf {x} \in \operatorname {B} (\mathbf {a} ,\delta _{2})}
F
(
x
,
g
1
(
x
)
)
=
0
{\displaystyle F(\mathbf {x} ,g_{1}(\mathbf {x} ))=0}
이제
g
:
B
(
a
,
δ
2
)
→
R
m
{\displaystyle \mathbf {g} \colon \operatorname {B} (\mathbf {a} ,\delta _{2})\to \mathbb {R} ^{m}}
g
(
x
)
=
(
g
1
(
x
)
,
g
~
(
x
,
g
1
(
x
)
)
)
∀
x
∈
B
(
a
,
δ
2
)
{\displaystyle \mathbf {g} (\mathbf {x} )=(g_{1}(\mathbf {x} ),{\tilde {\mathbf {g} }}(\mathbf {x} ,g_{1}(\mathbf {x} )))\qquad \forall \mathbf {x} \in \operatorname {B} (\mathbf {a} ,\delta _{2})}
그렇다면,
g
{\displaystyle \mathbf {g} }
g
(
B
(
a
,
δ
2
)
)
⊆
V
{\displaystyle \mathbf {g} (\operatorname {B} (\mathbf {a} ,\delta _{2}))\subseteq V}
f
1
(
x
,
g
(
x
)
)
=
F
(
x
,
g
1
(
x
)
)
=
0
∀
x
∈
B
(
a
,
δ
2
)
{\displaystyle f_{1}(\mathbf {x} ,\mathbf {g} (\mathbf {x} ))=F(\mathbf {x} ,g_{1}(\mathbf {x} ))=0\qquad \forall \mathbf {x} \in \operatorname {B} (\mathbf {a} ,\delta _{2})}
f
i
(
x
,
g
(
x
)
)
=
f
i
(
x
,
g
1
(
x
)
,
g
~
(
x
,
g
1
(
x
)
)
)
=
0
∀
x
∈
B
(
a
,
δ
2
)
,
i
∈
{
2
,
…
,
m
}
{\displaystyle f_{i}(\mathbf {x} ,\mathbf {g} (\mathbf {x} ))=f_{i}(\mathbf {x} ,g_{1}(\mathbf {x} ),{\tilde {\mathbf {g} }}(\mathbf {x} ,g_{1}(\mathbf {x} )))=0\qquad \forall \mathbf {x} \in \operatorname {B} (\mathbf {a} ,\delta _{2}),\;i\in \{2,\dots ,m\}}
g
(
a
)
=
(
g
1
(
a
)
,
g
~
(
a
,
b
1
)
)
=
b
{\displaystyle \mathbf {g} (\mathbf {a} )=(g_{1}(\mathbf {a} ),{\tilde {\mathbf {g} }}(\mathbf {a} ,b_{1}))=\mathbf {b} }
이러한
g
{\displaystyle \mathbf {g} }
m
=
1
{\displaystyle m=1}
f
{\displaystyle \mathbf {f} }
C
k
{\displaystyle {\mathcal {C}}^{k}}
i
∈
{
1
,
…
,
m
}
{\displaystyle i\in \{1,\dots ,m\}}
j
∈
{
1
,
…
,
n
}
{\displaystyle j\in \{1,\dots ,n\}}
f
i
(
x
,
g
(
x
)
)
=
0
{\displaystyle f_{i}(\mathbf {x} ,\mathbf {g} (\mathbf {x} ))=0}
∂
/
∂
x
j
{\displaystyle \partial /\partial x_{j}}
연쇄 법칙 에 따라 다음을 얻는다.
∂
f
i
∂
x
j
+
∑
k
=
1
m
∂
f
i
∂
y
k
∂
g
k
∂
y
j
=
0
{\displaystyle {\frac {\partial f_{i}}{\partial x_{j}}}+\sum _{k=1}^{m}{\frac {\partial f_{i}}{\partial y_{k}}}{\frac {\partial g_{k}}{\partial y_{j}}}=0}
이를 행렬로 표기하면
D
g
{\displaystyle \mathrm {D} \mathbf {g} }
g
{\displaystyle \mathbf {g} }
C
k
{\displaystyle {\mathcal {C}}^{k}}
다음과 같은 함수
F
:
U
×
V
→
R
m
{\displaystyle \mathbf {F} \colon U\times V\to \mathbb {R} ^{m}}
F
(
x
,
y
)
=
y
−
(
D
y
f
(
a
,
b
)
)
−
1
f
(
x
,
y
)
∀
x
∈
U
,
y
∈
V
{\displaystyle \mathbf {F} (\mathbf {x} ,\mathbf {y} )=\mathbf {y} -(\mathrm {D} _{\mathbf {y} }\mathbf {f} (\mathbf {a} ,\mathbf {b} ))^{-1}\mathbf {f} (\mathbf {x} ,\mathbf {y} )\qquad \forall \mathbf {x} \in U,\;\mathbf {y} \in V}
그러면 다음이 성립한다.
D
y
F
(
x
,
y
)
=
1
m
×
m
−
(
D
y
f
(
a
,
b
)
)
−
1
D
y
f
(
x
,
y
)
∀
x
∈
U
,
y
∈
V
{\displaystyle \mathrm {D} _{\mathbf {y} }\mathbf {F} (\mathbf {x} ,\mathbf {y} )=1_{m\times m}-(\mathrm {D} _{\mathbf {y} }\mathbf {f} (\mathbf {a} ,\mathbf {b} ))^{-1}\mathrm {D} _{\mathbf {y} }\mathbf {f} (\mathbf {x} ,\mathbf {y} )\qquad \forall \mathbf {x} \in U,\;\mathbf {y} \in V}
따라서
D
y
F
(
a
,
b
)
=
0
m
×
m
{\displaystyle \mathrm {D} _{\mathbf {y} }\mathbf {F} (\mathbf {a} ,\mathbf {b} )=0_{m\times m}}
det
D
y
f
(
a
,
b
)
≠
0
{\displaystyle \det \mathrm {D} _{\mathbf {y} }\mathbf {f} (\mathbf {a} ,\mathbf {b} )\neq 0}
0
<
c
<
1
{\displaystyle 0<c<1}
δ
1
>
0
{\displaystyle \delta _{1}>0}
B
¯
(
a
,
δ
1
)
⊆
U
{\displaystyle {\bar {\operatorname {B} }}(\mathbf {a} ,\delta _{1})\subseteq U}
B
¯
(
b
,
δ
1
)
⊆
V
{\displaystyle {\bar {\operatorname {B} }}(\mathbf {b} ,\delta _{1})\subseteq V}
임의의
x
∈
B
¯
(
a
,
δ
1
)
{\displaystyle \mathbf {x} \in {\bar {\operatorname {B} }}(\mathbf {a} ,\delta _{1})}
y
∈
B
¯
(
b
,
δ
1
)
{\displaystyle \mathbf {y} \in {\bar {\operatorname {B} }}(\mathbf {b} ,\delta _{1})}
D
y
F
(
x
,
y
)
<
c
{\displaystyle \mathrm {D} _{\mathbf {y} }\mathbf {F} (\mathbf {x} ,\mathbf {y} )<c}
det
D
y
f
(
x
,
y
)
≠
0
{\displaystyle \det \mathrm {D} _{\mathbf {y} }\mathbf {f} (\mathbf {x} ,\mathbf {y} )\neq 0}
또한,
F
(
a
,
b
)
=
b
{\displaystyle \mathbf {F} (\mathbf {a} ,\mathbf {b} )=\mathbf {b} }
0
<
δ
2
<
δ
1
{\displaystyle 0<\delta _{2}<\delta _{1}}
임의의
x
∈
B
(
a
,
δ
2
)
{\displaystyle \mathbf {x} \in \operatorname {B} (\mathbf {a} ,\delta _{2})}
‖
F
(
x
,
b
)
−
b
‖
<
(
1
−
c
)
δ
1
{\displaystyle \Vert \mathbf {F} (\mathbf {x} ,\mathbf {b} )-\mathbf {b} \Vert <(1-c)\delta _{1}}
이제 임의의
x
′
∈
B
(
a
,
δ
2
)
{\displaystyle \mathbf {x} '\in \operatorname {B} (\mathbf {a} ,\delta _{2})}
F
(
x
′
,
⋅
)
{\displaystyle \mathbf {F} (\mathbf {x} ',\cdot )}
B
¯
(
b
,
δ
1
)
{\displaystyle {\bar {\operatorname {B} }}(\mathbf {b} ,\delta _{1})}
c
{\displaystyle c}
립시츠 연속 함수 임을 증명하자. 우선 임의의
y
∈
B
¯
(
b
,
δ
1
)
{\displaystyle \mathbf {y} \in {\bar {\operatorname {B} }}(\mathbf {b} ,\delta _{1})}
0
<
θ
(
x
′
,
y
)
<
1
{\displaystyle 0<\theta (\mathbf {x} ',\mathbf {y} )<1}
‖
F
(
x
′
,
y
)
−
b
‖
≤
‖
F
(
x
′
,
y
)
−
F
(
x
′
,
b
)
‖
+
‖
F
(
x
′
,
b
)
−
b
‖
≤
‖
D
y
F
(
x
′
,
b
+
θ
(
y
−
b
)
)
‖
‖
y
−
b
‖
+
‖
F
(
x
′
,
b
)
−
b
‖
<
(
1
−
c
)
δ
1
+
c
δ
1
=
δ
1
{\displaystyle {\begin{aligned}\Vert \mathbf {F} (\mathbf {x} ',\mathbf {y} )-\mathbf {b} \Vert &\leq \Vert \mathbf {F} (\mathbf {x} ',\mathbf {y} )-\mathbf {F} (\mathbf {x} ',\mathbf {b} )\Vert +\Vert \mathbf {F} (\mathbf {x} ',\mathbf {b} )-\mathbf {b} \Vert \\&\leq \Vert \mathrm {D} _{\mathbf {y} }\mathbf {F} (\mathbf {x} ',\mathbf {b} +\theta (\mathbf {y} -\mathbf {b} ))\Vert \Vert \mathbf {y} -\mathbf {b} \Vert +\Vert \mathbf {F} (\mathbf {x} ',\mathbf {b} )-\mathbf {b} \Vert \\&<(1-c)\delta _{1}+c\delta _{1}\\&=\delta _{1}\end{aligned}}}
따라서
F
(
x
′
,
y
)
∈
B
(
b
,
δ
1
)
⊆
B
¯
(
b
,
δ
1
)
{\displaystyle \mathbf {F} (\mathbf {x} ',\mathbf {y} )\in \operatorname {B} (\mathbf {b} ,\delta _{1})\subseteq {\bar {\operatorname {B} }}(\mathbf {b} ,\delta _{1})}
y
,
z
∈
B
¯
(
b
,
δ
1
)
{\displaystyle \mathbf {y} ,\mathbf {z} \in {\bar {\operatorname {B} }}(\mathbf {b} ,\delta _{1})}
0
<
θ
(
x
′
,
y
,
z
)
<
1
{\displaystyle 0<\theta (\mathbf {x} ',\mathbf {y} ,\mathbf {z} )<1}
‖
F
(
x
′
,
y
)
−
F
(
x
′
,
z
)
‖
≤
‖
D
y
F
(
x
′
,
z
+
θ
(
y
−
z
)
‖
‖
y
−
z
‖
≤
c
‖
y
−
z
‖
{\displaystyle {\begin{aligned}\Vert \mathbf {F} (\mathbf {x} ',\mathbf {y} )-\mathbf {F} (\mathbf {x} ',\mathbf {z} )\Vert &\leq \Vert \mathrm {D} _{\mathrm {y} }\mathbf {F} (\mathbf {x} ',\mathbf {z} +\theta (\mathbf {y} -\mathbf {z} )\Vert \Vert \mathbf {y} -\mathbf {z} \Vert \\&\leq c\Vert \mathbf {y} -\mathbf {z} \Vert \end{aligned}}}
즉,
F
(
x
′
,
⋅
)
{\displaystyle \mathbf {F} (\mathbf {x} ',\cdot )}
B
¯
(
b
,
δ
1
)
{\displaystyle {\bar {\operatorname {B} }}(\mathbf {b} ,\delta _{1})}
c
{\displaystyle c}
립시츠 연속 함수 이다. 또한
(
B
¯
(
b
,
δ
1
)
,
‖
⋅
‖
)
{\displaystyle ({\bar {\operatorname {B} }}(\mathbf {b} ,\delta _{1}),\Vert \cdot \Vert )}
완비 거리 공간 이므로, 바나흐 고정점 정리 에 따라
F
(
x
′
,
⋅
)
{\displaystyle \mathbf {F} (\mathbf {x} ',\cdot )}
B
¯
(
b
,
δ
1
)
{\displaystyle {\bar {\operatorname {B} }}(\mathbf {b} ,\delta _{1})}
g
(
x
′
)
∈
B
¯
(
b
,
δ
1
)
{\displaystyle \mathbf {g} (\mathbf {x} ')\in {\bar {\operatorname {B} }}(\mathbf {b} ,\delta _{1})}
f
(
x
′
,
g
(
x
′
)
)
=
0
{\displaystyle \mathbf {f} (\mathbf {x} ',\mathbf {g} (\mathbf {x} '))=\mathbf {0} }
이제 이렇게 정의한
g
:
B
¯
(
a
,
δ
2
)
→
R
m
{\displaystyle \mathbf {g} \colon {\bar {\operatorname {B} }}(\mathbf {a} ,\delta _{2})\to \mathbb {R} ^{m}}
x
′
,
x
′
+
Δ
x
∈
B
¯
(
a
,
δ
2
)
{\displaystyle \mathbf {x} ',\mathbf {x} '+\Delta \mathbf {x} \in {\bar {\operatorname {B} }}(\mathbf {a} ,\delta _{2})}
‖
g
(
x
′
+
Δ
x
)
−
g
(
x
′
)
‖
=
‖
F
(
x
′
+
Δ
x
,
g
(
x
′
+
Δ
x
)
)
−
F
(
x
′
,
g
(
x
′
)
)
‖
≤
‖
F
(
x
′
+
Δ
x
,
g
(
x
′
+
Δ
x
)
)
−
F
(
x
′
+
Δ
x
,
g
(
x
′
)
)
‖
+
‖
F
(
x
′
+
Δ
x
,
g
(
x
′
)
)
−
F
(
x
′
,
g
(
x
′
)
)
‖
≤
c
‖
g
(
x
′
+
Δ
x
)
−
g
(
x
′
)
‖
+
c
′
‖
Δ
x
‖
{\displaystyle {\begin{aligned}\Vert \mathbf {g} (\mathbf {x} '+\Delta \mathbf {x} )-\mathbf {g} (\mathbf {x} ')\Vert &=\Vert \mathbf {F} (\mathbf {x} '+\Delta \mathbf {x} ,\mathbf {g} (\mathbf {x} '+\Delta \mathbf {x} ))-\mathbf {F} (\mathbf {x} ',\mathbf {g} (\mathbf {x} '))\Vert \\&\leq \Vert \mathbf {F} (\mathbf {x} '+\Delta \mathbf {x} ,\mathbf {g} (\mathbf {x} '+\Delta \mathbf {x} ))-\mathbf {F} (\mathbf {x} '+\Delta \mathbf {x} ,\mathbf {g} (\mathbf {x} '))\Vert +\Vert \mathbf {F} (\mathbf {x} '+\Delta \mathbf {x} ,\mathbf {g} (\mathbf {x} '))-\mathbf {F} (\mathbf {x} ',\mathbf {g} (\mathbf {x} '))\Vert \\&\leq c\Vert \mathbf {g} (\mathbf {x} '+\Delta \mathbf {x} )-\mathbf {g} (\mathbf {x} ')\Vert +c'\Vert \Delta \mathbf {x} \Vert \end{aligned}}}
c
′
=
sup
x
∈
B
¯
(
a
,
δ
2
)
,
y
∈
B
¯
(
b
,
δ
1
)
‖
D
x
F
(
x
,
y
)
‖
{\displaystyle c'=\sup _{\mathbf {x} \in {\bar {\operatorname {B} }}(\mathbf {a} ,\delta _{2}),\;\mathbf {y} \in {\bar {\operatorname {B} }}(\mathbf {b} ,\delta _{1})}\Vert \mathrm {D} _{\mathbf {x} }\mathbf {F} (\mathbf {x} ,\mathbf {y} )\Vert }
즉, 다음이 성립한다.
‖
g
(
x
′
+
Δ
x
)
−
g
(
x
′
)
‖
≤
c
′
1
−
c
‖
Δ
x
‖
{\displaystyle \Vert \mathbf {g} (\mathbf {x} '+\Delta \mathbf {x} )-\mathbf {g} (\mathbf {x} ')\Vert \leq {\frac {c'}{1-c}}\Vert \Delta \mathbf {x} \Vert }
따라서
g
{\displaystyle \mathbf {g} }
g
{\displaystyle \mathbf {g} }
f
{\displaystyle \mathbf {f} }
C
k
{\displaystyle {\mathcal {C}}^{k}}
g
{\displaystyle \mathbf {g} }
C
k
{\displaystyle {\mathcal {C}}^{k}}
x
′
,
x
′
+
Δ
x
∈
B
¯
(
a
,
δ
2
)
{\displaystyle \mathbf {x} ',\mathbf {x} '+\Delta \mathbf {x} \in {\bar {\operatorname {B} }}(\mathbf {a} ,\delta _{2})}
Δ
y
(
x
′
,
x
′
+
Δ
x
)
=
g
(
x
+
Δ
x
)
−
g
(
x
)
{\displaystyle \Delta \mathbf {y} (\mathbf {x} ',\mathbf {x} '+\Delta \mathbf {x} )=\mathbf {g} (\mathbf {x} +\Delta \mathbf {x} )-\mathbf {g} (\mathbf {x} )}
그러면 다음이 성립한다.
‖
Δ
y
+
(
D
y
f
(
x
′
,
g
(
x
′
)
)
)
−
1
D
x
f
(
x
′
,
g
(
x
′
)
)
Δ
x
‖
≤
‖
(
D
y
f
(
x
′
,
g
(
x
′
)
)
)
−
1
‖
‖
(
D
y
f
(
x
′
,
g
(
x
′
)
)
)
Δ
y
+
D
x
f
(
x
′
,
g
(
x
′
)
)
Δ
x
‖
=
‖
(
D
y
f
(
x
′
,
g
(
x
′
)
)
)
−
1
‖
‖
f
(
x
′
+
Δ
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{\displaystyle {\begin{aligned}\Vert \Delta \mathbf {y} +(\mathrm {D} _{\mathbf {y} }\mathbf {f} (\mathbf {x} ',\mathbf {g} (\mathbf {x} ')))^{-1}\mathrm {D} _{\mathbf {x} }\mathbf {f} (\mathbf {x} ',\mathbf {g} (\mathbf {x} '))\Delta \mathbf {x} \Vert &\leq \Vert (\mathrm {D} _{\mathbf {y} }\mathbf {f} (\mathbf {x} ',\mathbf {g} (\mathbf {x} ')))^{-1}\Vert \Vert (\mathrm {D} _{\mathbf {y} }\mathbf {f} (\mathbf {x} ',\mathbf {g} (\mathbf {x} ')))\Delta \mathbf {y} +\mathrm {D} _{\mathbf {x} }\mathbf {f} (\mathbf {x} ',\mathbf {g} (\mathbf {x} '))\Delta \mathbf {x} \Vert \\&=\Vert (\mathrm {D} _{\mathbf {y} }\mathbf {f} (\mathbf {x} ',\mathbf {g} (\mathbf {x} ')))^{-1}\Vert \Vert \mathbf {f} (\mathbf {x} '+\Delta \mathbf {x} ,\mathbf {g} (\mathbf {x} ')+\Delta \mathbf {y} )-\mathbf {f} (\mathbf {x} ',\mathbf {g} (\mathbf {x} '))-(\mathrm {D} _{\mathbf {y} }\mathbf {f} (\mathbf {x} ',\mathbf {g} (\mathbf {x} ')))\Delta \mathbf {y} -\mathrm {D} _{\mathbf {x} }\mathbf {f} (\mathbf {x} ',\mathbf {g} (\mathbf {x} '))\Delta \mathbf {x} \Vert \\&=o(\Vert \Delta \mathbf {x} \Vert )\qquad (\Delta \mathbf {x} \to \mathbf {0} )\end{aligned}}}
마지막 등호는
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{\displaystyle \Vert \Delta \mathbf {y} \Vert \leq (c'/(1-c))\Vert \Delta \mathbf {x} \Vert }
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{\displaystyle \mathrm {D} \mathbf {g} (\mathbf {x} ')=-(\mathrm {D} _{\mathbf {y} }\mathbf {f} (\mathbf {x} ',\mathbf {g} (\mathbf {x} ')))^{-1}\mathrm {D} _{\mathbf {x} }\mathbf {f} (\mathbf {x} ',\mathbf {g} (\mathbf {x} '))}
가 성립하며,
g
{\displaystyle \mathbf {g} }
C
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{\displaystyle {\mathcal {C}}^{k}}
![{\displaystyle [-1,1]\times [0,\infty )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/653be9b4122feb85fa600cfab3bb236badff3988)
![{\displaystyle y={\sqrt {1-x^{2}}}\qquad \forall x\in [-1,1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f33d428e67dddbb1e30517f259ed0d688e91f67b)
![{\displaystyle [-1,1]\times (-\infty ,0]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a055b0d4d37c026ed9647a1913cceb245e55272)
![{\displaystyle y=-{\sqrt {1-x^{2}}}\qquad \forall x\in [-1,1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f857a5cbefbf32d0c920817c238cc6b3d1aa5570)
![{\displaystyle [-1,1]\times \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/7dd5e0a7c17676c2788b6033e5a2b0eb58613d82)
