Seja
f
:
I
↦
R
n
{\displaystyle f:I\mapsto \mathbb {R} ^{n}}
a
∈
I
{\displaystyle a\in I\;}
f
−
1
(
v
)
{\displaystyle f^{-1}(v)\;}
v
∈
R
n
{\displaystyle v\in \mathbb {R} ^{n}\;}
f
′
(
a
)
=
0
{\displaystyle f'(a)=0\;}
A existência de
f
′
(
a
)
{\displaystyle f'(a)\;}
f
{\displaystyle f\;}
f
−
1
(
v
)
=
{
w
∈
I
/
f
(
w
)
=
v
}
{\displaystyle f^{-1}(v)=\{w\in I/f(w)=v\}}
a é um ponto de acumulação de
f
−
1
(
v
)
{\displaystyle f^{-1}(v)\;}
∀
ϵ
>
0
,
∃
w
∈
f
−
1
(
v
)
;
0
<
|
w
−
a
|
<
ϵ
{\displaystyle \forall \;\epsilon >0,\exists \;w\in f^{-1}(v);0<|w-a|<\epsilon }
∀
w
∈
I
,
∃
t
=
w
−
a
⇒
w
=
a
+
t
,
0
<
|
a
+
t
−
a
|
=
|
t
|
<
ϵ
⇒
f
(
a
+
t
)
=
v
,
0
<
|
t
|
<
ϵ
{\displaystyle \forall \;w\in I,\exists \;t=w-a\Rightarrow w=a+t,0<|a+t-a|=|t|<\epsilon \Rightarrow f(a+t)=v,0<|t|<\epsilon \;}
Como
f
(
a
)
=
v
{\displaystyle f(a)=v\;}
f
′
(
a
)
=
l
i
m
t
↦
0
f
(
a
+
t
)
−
f
(
a
)
t
{\displaystyle f'(a)=lim_{t\mapsto 0}{\frac {f(a+t)-f(a)}{t}}}
f
(
a
+
t
)
−
f
(
a
)
=
v
−
v
=
0
{\displaystyle f(a+t)-f(a)=v-v=0\;}
Portanto
f
′
(
a
)
=
0
{\displaystyle f'(a)=0\;}
Seja
f
:
I
↦
R
n
{\displaystyle f:I\mapsto \mathbb {R} ^{n}}
f
′
(
a
)
≠
0
{\displaystyle f'(a)\not =0\;}
a
∈
I
{\displaystyle a\in I\;}
L
⊂
R
n
{\displaystyle L\subset \mathbb {R} ^{n}\;}
t
k
↦
a
{\displaystyle t_{k}\mapsto a}
f
(
t
k
)
∈
L
{\displaystyle f(t_{k})\in L\;}
L
{\displaystyle L\;}
f
{\displaystyle f\;}
f
(
a
)
{\displaystyle f(a)\;}
Como
f
{\displaystyle f\;}
f
′
(
a
)
,
∀
a
∈
I
{\displaystyle f'(a),\forall \;a\in I}
f
′
(
a
)
≠
0
{\displaystyle f'(a)\not =0\;}
G
=
{
f
(
a
)
+
t
f
′
(
a
)
,
t
∈
R
}
{\displaystyle G=\{f(a)+tf'(a),t\in \mathbb {R} \}}
Devemos provar que
L
=
G
{\displaystyle L=G\;}
f
(
a
)
∈
L
,
f
′
(
a
)
{\displaystyle f(a)\in L,f'(a)\;}
Como
f
{\displaystyle f\;}
ϵ
>
0
,
∃
δ
>
0
;
x
∈
(
t
k
)
,
|
x
−
a
|
<
δ
⇒
|
f
(
x
)
−
f
(
a
)
|
<
ϵ
{\displaystyle \epsilon >0,\exists \;\delta >0;x\in (t_{k}),|x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon }
t
k
↦
a
,
∃
k
0
∈
N
;
k
>
k
0
⇒
|
t
k
−
a
|
<
δ
{\displaystyle t_{k}\mapsto a,\exists \;k_{0}\in \mathbb {N} ;k>k_{0}\Rightarrow |t_{k}-a|<\delta }
f
{\displaystyle f\;}
k
>
k
0
⇒
|
f
(
t
k
)
−
f
(
a
)
|
<
ϵ
{\displaystyle k>k_{0}\Rightarrow |f(t_{k})-f(a)|<\epsilon }
l
i
m
k
↦
∞
f
(
t
k
)
=
f
(
a
)
{\displaystyle lim_{k\mapsto \infty }f(t_{k})=f(a)}
Como
L
{\displaystyle L\;}
f
(
t
k
)
∈
L
{\displaystyle f(t_{k})\in L\;}
l
i
m
k
↦
∞
f
(
t
k
)
=
f
(
a
)
⇒
f
(
a
)
∈
L
{\displaystyle lim_{k\mapsto \infty }f(t_{k})=f(a)\Rightarrow f(a)\in L}
f
(
a
)
{\displaystyle f(a)\;}
L
{\displaystyle L\;}
L
{\displaystyle L\;}
f
(
a
)
{\displaystyle f(a)\;}
f
′
(
a
)
{\displaystyle f'(a)\;}
L
{\displaystyle L\;}
Seja
f
:
I
↦
R
n
{\displaystyle f:I\mapsto \mathbb {R} ^{n}}
a
∈
R
n
,
r
>
0
{\displaystyle a\in \mathbb {R} ^{n},r>0\;}
f
(
t
)
{\displaystyle f(t)\;}
t
∈
I
{\displaystyle t\in I\;}
a
{\displaystyle a\;}
r
{\displaystyle r\;}
t
0
∈
I
{\displaystyle t_{0}\in I\;}
f
′
(
t
)
{\displaystyle f'(t)\;}
f
(
t
)
−
a
,
∀
t
∈
I
{\displaystyle f(t)-a,\forall \;t\in I}
Por hipótese temos que
f
(
t
)
{\displaystyle f(t)\;}
t
∈
I
{\displaystyle t\in I\;}
a
{\displaystyle a\;}
r
{\displaystyle r\;}
t
0
∈
I
{\displaystyle t_{0}\in I\;}
f
′
(
t
)
{\displaystyle f'(t)\;}
f
(
t
)
−
a
,
∀
t
∈
I
{\displaystyle f(t)-a,\forall \;t\in I}
Seja a Esfera de centro
a
{\displaystyle a\;}
r
{\displaystyle r\;}
S
(
a
;
r
)
{\displaystyle S(a;r)\;}
Se
f
(
t
)
∈
S
(
a
,
r
)
,
∀
t
∈
I
{\displaystyle f(t)\in S(a,r),\forall t\in I}
f
(
t
0
)
∈
S
(
a
,
r
)
{\displaystyle f(t_{0})\in S(a,r)}
Para que o vetor velocidade
f
′
(
t
)
{\displaystyle f'(t)\;}
f
(
t
)
−
a
,
∀
t
∈
I
{\displaystyle f(t)-a,\forall \;t\in I}
⟨
f
(
t
)
−
a
,
f
′
(
t
)
⟩
=
0
,
∀
t
∈
I
{\displaystyle \left\langle f(t)-a,f'(t)\right\rangle =0,\forall \;t\in I}
Temos que
|
|
(
f
(
t
)
−
a
)
|
|
=
r
,
∀
t
∈
I
{\displaystyle ||(f(t)-a)||=r,\forall \;t\in I}
|
|
(
f
(
t
)
−
a
)
|
|
2
=
⟨
(
f
(
t
)
−
a
)
,
(
f
(
t
)
−
a
)
⟩
=
r
2
,
∀
t
∈
I
{\displaystyle ||(f(t)-a)||^{2}=\left\langle (f(t)-a),(f(t)-a)\right\rangle =r^{2},\forall \;t\in I}
d
d
t
[
⟨
f
(
t
)
,
f
(
t
)
⟩
−
2
⟨
f
(
t
)
,
a
⟩
+
⟨
a
,
a
⟩
]
=
2
⟨
f
(
t
)
,
f
′
(
t
)
⟩
−
2
⟨
a
,
f
′
(
t
)
⟩
=
2
⟨
f
(
t
)
−
a
,
f
′
(
t
)
⟩
=
0
,
∀
t
∈
I
{\displaystyle {\frac {d}{dt}}[\left\langle f(t),f(t)\right\rangle -2\left\langle f(t),a\right\rangle +\left\langle a,a\right\rangle ]=2\left\langle f(t),f'(t)\right\rangle -2\left\langle a,f'(t)\right\rangle =2\left\langle f(t)-a,f'(t)\right\rangle =0,\forall \;t\in I}
Por hipótese temos que
f
(
t
0
)
{\displaystyle f(t_{0})\;}
a
{\displaystyle a\;}
r
{\displaystyle r\;}
Logo
|
|
(
f
(
t
0
)
−
a
)
|
|
=
r
{\displaystyle ||(f(t_{0})-a)||=r\;}
Por hipótese temos que o vetor velocidade
f
′
(
t
)
{\displaystyle f'(t)\;}
f
(
t
)
−
a
,
∀
t
∈
I
{\displaystyle f(t)-a,\forall \;t\in I}
Logo
(
f
(
t
)
−
a
)
f
′
(
t
)
=
0
,
∀
t
∈
I
{\displaystyle (f(t)-a)f'(t)=0,\forall \;t\in I}
f
′
(
t
)
=
d
d
t
(
f
(
t
)
−
a
)
,
∀
t
∈
I
{\displaystyle f'(t)={\frac {d}{dt}}(f(t)-a),\forall \;t\in I}
Devemos mostrar que
f
(
t
)
∈
S
(
a
;
r
)
,
∀
t
∈
I
{\displaystyle f(t)\in S(a;r),\forall \;t\in I}
|
|
f
(
t
)
−
a
|
|
=
r
,
∀
t
∈
I
{\displaystyle ||f(t)-a||=r,\forall \;t\in I}
Pelo último resultado
(
f
(
t
)
−
a
)
f
′
(
t
)
=
0
⇒
(
f
(
t
)
−
a
)
d
d
t
(
f
(
t
)
−
a
)
=
0
,
∀
t
∈
I
{\displaystyle (f(t)-a)f'(t)=0\Rightarrow (f(t)-a){\frac {d}{dt}}(f(t)-a)=0,\forall \;t\in I}
Assim a primitiva é
(
f
(
t
)
−
a
)
(
f
(
t
)
−
a
)
=
k
⇒
|
|
(
f
(
t
)
−
a
)
|
|
=
k
,
∀
t
∈
I
{\displaystyle (f(t)-a)(f(t)-a)=k\Rightarrow ||(f(t)-a)||=k,\forall \;t\in I}
Como
|
|
(
f
(
t
0
)
−
a
)
|
|
=
r
{\displaystyle ||(f(t_{0})-a)||=r\;}
Seja
λ
:
[
a
,
b
]
↦
R
n
{\displaystyle \lambda :[a,b]\mapsto \mathbb {R} ^{n}}
t
∈
(
a
,
b
)
{\displaystyle t\in (a,b)\;}
⟨
λ
(
t
)
,
λ
′
(
t
)
⟩
=
0
{\displaystyle \left\langle \lambda (t),\lambda '(t)\right\rangle =0}
fechado aqui significa
λ
(
a
)
=
λ
(
b
)
{\displaystyle \lambda (a)=\lambda (b)\;}
λ
′
(
a
)
=
λ
′
(
b
)
{\displaystyle \lambda '(a)=\lambda '(b)\;}
Como
λ
:
[
a
,
b
]
↦
R
n
{\displaystyle \lambda :[a,b]\mapsto \mathbb {R} ^{n}}
∀
ϵ
>
0
,
∃
δ
>
0
;
t
∈
[
a
,
b
]
,
‖
t
−
t
0
‖
<
ϵ
⇒
‖
λ
(
t
)
−
λ
(
t
0
)
‖
<
δ
{\displaystyle \forall \;\epsilon >0,\exists \;\delta >0;t\in [a,b],\|t-t_{0}\|<\epsilon \Rightarrow \|\lambda (t)-\lambda (t_{0})\|<\delta }
Fixemos
t
0
∈
[
a
,
b
]
,
ϵ
=
max
{
‖
t
−
t
0
‖
;
t
∈
[
a
,
b
]
}
⇒
λ
(
t
)
∈
B
(
λ
(
t
0
)
;
δ
)
,
∀
t
∈
[
a
,
b
]
{\displaystyle t_{0}\in [a,b],\epsilon =\max\{\|t-t_{0}\|;t\in [a,b]\}\Rightarrow \lambda (t)\in B(\lambda (t_{0});\delta ),\forall \;t\in [a,b]}
Seja
H
=
{
λ
(
t
)
;
t
∈
[
a
,
b
]
}
{\displaystyle H=\{\lambda (t);t\in [a,b]\}}
λ
{\displaystyle \lambda \;}
∃
t
1
,
t
2
∈
[
a
,
b
]
;
λ
(
t
1
)
≤
λ
(
t
)
≤
λ
(
t
1
)
,
∀
t
∈
[
a
,
b
]
{\displaystyle \exists t_{1},t_{2}\in [a,b];\lambda (t_{1})\leq \lambda (t)\leq \lambda (t_{1}),\forall \;t\in [a,b]}
x
x
x
‖
λ
(
t
)
−
λ
(
t
0
)
‖
2
=
⟨
λ
(
t
)
−
λ
(
t
0
)
,
λ
(
t
)
−
λ
(
t
0
)
⟩
=
⟨
λ
(
t
)
,
λ
(
t
)
⟩
−
2
⟨
λ
(
t
)
,
λ
(
t
0
)
⟩
+
⟨
λ
(
t
0
)
,
λ
(
t
0
)
⟩
<
δ
2
{\displaystyle \|\lambda (t)-\lambda (t_{0})\|^{2}=\left\langle \lambda (t)-\lambda (t_{0}),\lambda (t)-\lambda (t_{0})\right\rangle =\left\langle \lambda (t),\lambda (t)\right\rangle -2\left\langle \lambda (t),\lambda (t_{0})\right\rangle +\left\langle \lambda (t_{0}),\lambda (t_{0})\right\rangle <\delta ^{2}}
2
⟨
λ
(
t
)
,
λ
′
(
t
)
⟩
−
2
⟨
λ
(
t
0
)
,
λ
′
(
t
)
⟩
=
2
⟨
λ
(
t
)
−
λ
(
t
0
)
,
λ
′
(
t
)
⟩
<
0
{\displaystyle 2\left\langle \lambda (t),\lambda '(t)\right\rangle -2\left\langle \lambda (t_{0}),\lambda '(t)\right\rangle =2\left\langle \lambda (t)-\lambda (t_{0}),\lambda '(t)\right\rangle <0}
R
⇒→⇐⇔↦
∖
⟨
⟩
‖
‖
∀
∈⊂
¯
1
2
∩
∪
e
{\displaystyle \mathbb {R} \Rightarrow \rightarrow \Leftarrow \Leftrightarrow \mapsto \setminus \left\langle \right\rangle \|\|\forall \in \subset {\bar {}}{1 \over 2}\cap \cup \;{\mbox{e}}}
Seja
f
:
(
a
,
b
)
→
R
n
{\displaystyle f:(a,b)\to \mathbb {R} ^{n}}
f
+
′
:
(
a
,
b
)
→
R
n
{\displaystyle f'_{+}:(a,b)\to \mathbb {R} ^{n}}
C
1
{\displaystyle C^{1}\;}
Para que f seja de classe
C
1
{\displaystyle C^{1}\;}
f
′
{\displaystyle f'\;}
f
′
{\displaystyle f'\;}
f
+
′
{\displaystyle f'_{+}\;}
f
−
′
{\displaystyle f'_{-}\;}
f
+
′
{\displaystyle f'_{+}\;}
f
−
′
{\displaystyle f'_{-}\;}
Seja
f
−
′
:
(
a
,
b
)
→
R
n
{\displaystyle f'_{-}:(a,b)\to \mathbb {R} ^{n}}
∀
y
∈
(
a
,
b
)
,
f
′
(
y
−
)
=
lim
t
→
0
−
f
(
y
+
t
)
−
f
(
y
)
t
=
lim
t
→
0
−
(
f
1
(
y
+
t
)
−
f
1
(
y
)
t
,
.
.
.
,
f
n
(
y
+
t
)
−
f
n
(
y
)
t
)
⇒
{\displaystyle \forall y\in (a,b),f'(y^{-})=\lim _{t\to 0^{-}}{f(y+t)-f(y) \over t}=\lim _{t\to 0^{-}}\left({f_{1}(y+t)-f_{1}(y) \over t},...,{f_{n}(y+t)-f_{n}(y) \over t}\right)\Rightarrow }
(
lim
t
→
0
−
f
1
(
y
+
t
)
−
f
1
(
y
)
t
,
.
.
.
,
lim
t
→
0
−
f
n
(
y
+
t
)
−
f
n
(
y
)
t
)
=
(
f
1
′
(
y
−
)
,
.
.
.
,
f
n
′
(
y
−
)
)
{\displaystyle \left(\lim _{t\to 0^{-}}{f_{1}(y+t)-f_{1}(y) \over t},...,\lim _{t\to 0^{-}}{f_{n}(y+t)-f_{n}(y) \over t}\right)=(f'_{1}(y^{-}),...,f'_{n}(y^{-}))}
Seja
f
+
′
:
(
a
,
b
)
→
R
n
{\displaystyle f'_{+}:(a,b)\to \mathbb {R} ^{n}}
∀
y
∈
(
a
,
b
)
,
f
′
(
y
+
)
=
lim
t
→
0
+
f
(
y
+
t
)
−
f
(
y
)
t
=
lim
t
→
0
+
(
f
1
(
y
+
t
)
−
f
1
(
y
)
t
,
.
.
.
,
f
n
(
y
+
t
)
−
f
n
(
y
)
t
)
⇒
{\displaystyle \forall y\in (a,b),f'(y+)=\lim _{t\to 0^{+}}{f(y+t)-f(y) \over t}=\lim _{t\to 0^{+}}\left({f_{1}(y+t)-f_{1}(y) \over t},...,{f_{n}(y+t)-f_{n}(y) \over t}\right)\Rightarrow }
(
lim
t
→
0
+
f
1
(
y
+
t
)
−
f
1
(
y
)
t
,
.
.
.
,
lim
t
→
0
+
f
n
(
y
+
t
)
−
f
n
(
y
)
t
)
=
(
f
1
′
(
y
+
)
,
.
.
.
,
f
n
′
(
y
+
)
)
{\displaystyle \left(\lim _{t\to 0^{+}}{f_{1}(y+t)-f_{1}(y) \over t},...,\lim _{t\to 0^{+}}{f_{n}(y+t)-f_{n}(y) \over t}\right)=(f'_{1}(y^{+}),...,f'_{n}(y^{+}))}
Por ser
f
+
′
{\displaystyle f'_{+}\;}
f
1
+
′
,
.
.
.
,
f
n
+
′
{\displaystyle f'_{1^{+}},...,f'_{n^{+}}\;}
∀
ϵ
>
0
,
∃
δ
>
0
{\displaystyle \forall \epsilon >0,\exists \delta >0}
y
∈
(
a
,
b
)
,
‖
y
−
x
‖
≤
δ
⇒
‖
f
i
+
′
(
y
)
−
f
i
+
′
(
x
)
‖
≤
ϵ
,
1
≤
i
≤
n
{\displaystyle y\in (a,b),\left\|y-x\right\|\leq \delta \Rightarrow \left\|f'_{i+}(y)-f'_{i+}(x)\right\|\leq \epsilon ,1\leq i\leq n}
Seja
G
⊂
R
2
{\displaystyle G\subset \mathbb {R} ^{2}}
y
=
|
x
|
{\displaystyle y=|x|\;}
Mostre que existe um caminho
f
:
R
→
R
2
{\displaystyle f:\mathbb {R} \to \mathbb {R} ^{2}}
C
∞
{\displaystyle C^{\infty }}
f
(
R
)
=
G
{\displaystyle f(\mathbb {R} )=G}
f
(
0
)
=
(
0
,
0
)
{\displaystyle f(0)=(0,0)\;}
Prove que todo caminho diferenciável f nessas condições cumpre
f
′
(
0
)
=
0
{\displaystyle f'(0)=0\;}
Devemos mostrar que exist uma função f, nas condições acima.
Seja
g
:
R
→
R
2
{\displaystyle g:\mathbb {R} \to \mathbb {R} ^{2}}
g
(
x
)
=
(
x
,
|
x
|
)
=
{
(
x
,
x
)
,
se
x
≥
0
(
x
,
−
x
)
,
se
x
<
0
,
∀
x
∈
R
{\displaystyle g(x)=(x,|x|)={\begin{cases}(x,x),&{\mbox{se }}x\geq 0\\(x,-x),&{\mbox{se }}x<0\end{cases}},\forall x\in \mathbb {R} }
Assim
g
/
R
+
=
g
1
:
R
+
→
R
2
{\displaystyle g/\mathbb {R} _{+}=g_{1}:\mathbb {R} _{+}\to \mathbb {R} ^{2}}
g
1
(
x
)
=
(
x
,
x
)
{\displaystyle g_{1}(x)=(x,x)\;}
g
/
R
−
∗
=
g
2
:
R
−
∗
→
R
2
{\displaystyle g/\mathbb {R} _{-}^{*}=g_{2}:\mathbb {R} _{-}^{*}\to \mathbb {R} ^{2}}
g
2
(
x
)
=
(
x
,
−
x
)
{\displaystyle g_{2}(x)=(x,-x)\;}
Seja
f
:
R
→
R
2
{\displaystyle f:\mathbb {R} \to \mathbb {R} ^{2}}
f
(
t
)
=
(
t
k
+
1
,
|
t
|
t
k
)
=
{
(
t
k
+
1
,
t
k
+
1
)
,
se
t
≥
0
(
t
k
+
1
,
−
t
k
+
1
)
,
se
t
<
0
,
∀
t
∈
R
{\displaystyle f(t)=(t^{k+1},|t|t^{k})={\begin{cases}(t^{k+1},t^{k+1}),&{\mbox{se }}t\geq 0\\(t^{k+1},-t^{k+1}),&{\mbox{se }}t<0\end{cases}},\forall t\in \mathbb {R} }
Então
f
′
(
t
)
=
d
d
t
(
t
k
+
1
,
|
t
|
t
k
)
=
{
(
(
k
+
1
)
t
k
,
(
k
+
1
)
t
k
)
,
se
t
≥
0
(
(
k
+
1
)
t
k
,
−
(
k
+
1
)
t
k
)
,
se
t
<
0
,
∀
t
∈
R
{\displaystyle f'(t)={d \over dt}(t^{k+1},|t|t^{k})={\begin{cases}((k+1)t^{k},(k+1)t^{k}),&{\mbox{se }}t\geq 0\\((k+1)t^{k},-(k+1)t^{k}),&{\mbox{se }}t<0\end{cases}},\forall t\in \mathbb {R} }
f é de classe
C
∞
{\displaystyle C^{\infty }\;}
f
(
R
)
=
G
{\displaystyle f(\mathbb {R} )=G}
f
(
0
)
=
(
0
,
0
)
{\displaystyle f(0)=(0,0)\;}
Devemos mostrar que
f
′
(
0
)
=
0
{\displaystyle f'(0)=0\;}
f
′
(
0
)
=
lim
t
→
0
+
f
(
0
+
t
)
−
f
(
0
)
t
=
lim
t
→
0
−
f
(
0
+
t
)
−
f
(
0
)
t
⇒
lim
t
→
0
+
(
t
k
+
1
,
t
k
+
1
)
−
(
0
,
0
)
t
=
lim
t
→
0
−
(
t
k
+
1
,
−
t
k
+
1
)
−
(
0
,
0
)
t
⇒
{\displaystyle f'(0)=\lim _{t\to 0^{+}}{f(0+t)-f(0) \over t}=\lim _{t\to 0^{-}}{f(0+t)-f(0) \over t}\Rightarrow \lim _{t\to 0^{+}}{(t^{k+1},t^{k+1})-(0,0) \over t}=\lim _{t\to 0^{-}}{(t^{k+1},-t^{k+1})-(0,0) \over t}\Rightarrow }
lim
t
→
0
+
(
t
k
,
t
k
)
=
lim
t
→
0
−
(
t
k
,
−
t
k
)
=
(
0
,
0
)
{\displaystyle \lim _{t\to 0^{+}}(t^{k},t^{k})=\lim _{t\to 0^{-}}(t^{k},-t^{k})=(0,0)}
![{\displaystyle {\frac {d}{dt}}[\left\langle f(t),f(t)\right\rangle -2\left\langle f(t),a\right\rangle +\left\langle a,a\right\rangle ]=2\left\langle f(t),f'(t)\right\rangle -2\left\langle a,f'(t)\right\rangle =2\left\langle f(t)-a,f'(t)\right\rangle =0,\forall \;t\in I}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f897a86c77664f619fa01fc965f50c2c482bd064)
![{\displaystyle \lambda :[a,b]\mapsto \mathbb {R} ^{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6bc5eaa760b28df50addeccb669697bfb708d9fd)
![{\displaystyle \forall \;\epsilon >0,\exists \;\delta >0;t\in [a,b],\|t-t_{0}\|<\epsilon \Rightarrow \|\lambda (t)-\lambda (t_{0})\|<\delta }](https://wikimedia.org/api/rest_v1/media/math/render/svg/67eb1d2f35fe1ab473674e51fc01895959221b32)
![{\displaystyle t_{0}\in [a,b],\epsilon =\max\{\|t-t_{0}\|;t\in [a,b]\}\Rightarrow \lambda (t)\in B(\lambda (t_{0});\delta ),\forall \;t\in [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cc430d56a2fd398a7ab10230fca03eaed38077b8)
![{\displaystyle H=\{\lambda (t);t\in [a,b]\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/39523ef07870e223cc13bafa434a4ce6d218e01e)
![{\displaystyle \exists t_{1},t_{2}\in [a,b];\lambda (t_{1})\leq \lambda (t)\leq \lambda (t_{1}),\forall \;t\in [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7cc7f61bb08971066793f66e1cce29f4292a91a9)
