Funktio
f
:
R
→
R
{\displaystyle f:\mathbb {R} \rightarrow \mathbb {R} }
x
,
y
∈
R
{\displaystyle x,y\in \mathbb {R} }
0
≤
q
<
1
{\displaystyle 0\leq q<1}
|
f
(
x
)
−
f
(
y
)
|
≤
q
|
x
−
y
|
.
{\displaystyle |f(x)-f(y)|\leq q|x-y|.}
Yleisemmällä tasolla kontraktio määritellään kahden metrisen avaruuden välisenä funktiona. Tällöin yo. määritelmässä korvataan vain erotusten itseisarvot metriikoilla:[1]
f
:
X
→
Z
{\displaystyle f:X\rightarrow Z}
x
,
y
∈
X
{\displaystyle x,y\in X}
0
≤
q
<
1
{\displaystyle 0\leq q<1}
d
Z
(
f
(
x
)
,
f
(
y
)
)
≤
q
d
X
(
x
,
y
)
{\displaystyle d_{Z}(f(x),f(y))\leq qd_{X}(x,y)}
missä
d
Z
{\displaystyle d_{Z}}
d
X
{\displaystyle d_{X}}
Z
{\displaystyle Z}
X
{\displaystyle X}
Funktio
f
(
x
)
=
x
3
{\displaystyle f(x)={\tfrac {x}{3}}}
|
f
(
x
)
−
f
(
y
)
|
=
|
x
−
y
|
3
,
{\displaystyle |f(x)-f(y)|={\frac {|x-y|}{3}},}
eli
q
=
1
3
{\displaystyle q={\tfrac {1}{3}}}
Funktio
f
(
x
)
=
x
2
{\displaystyle f(x)=x^{2}}
|
f
(
0
)
−
f
(
2
)
|
=
|
0
−
4
|
=
4
>
2
=
1
⋅
|
0
−
2
|
.
{\displaystyle |f(0)-f(2)|=|0-4|=4>2=1\cdot |0-2|.}
Olkoon
f
:
I
→
R
{\displaystyle f:I\rightarrow \mathbb {R} }
I
⊂
R
{\displaystyle I\subset \mathbb {R} }
f
{\displaystyle f}
sup
x
∈
I
|
f
′
(
x
)
|
<
1
{\displaystyle \sup _{x\in I}|f'(x)|<1}
Todistus:
Olkoon
r
=
sup
x
∈
I
|
f
′
(
x
)
|
{\displaystyle r=\sup _{x\in I}|f'(x)|}
r
<
1
{\displaystyle r<1}
differentiaalilaskennan väliarvolauseen perusteella
|
f
(
x
)
−
f
(
y
)
|
/
|
x
−
y
|
=
|
f
′
(
c
)
|
≤
r
,
{\displaystyle |f(x)-f(y)|/|x-y|=|f'(c)|\leq r,}
kaikilla
x
,
y
∈
I
{\displaystyle x,y\in I}
f
{\displaystyle f}
I
{\displaystyle I}
toispuoleiset derivaatat päätepisteissä riittävät.)
Jos
q
<
r
{\displaystyle q<r}
x
∈
I
{\displaystyle x\in I}
δ
=
|
f
′
(
x
)
|
−
q
>
0
{\displaystyle \delta =|f'(x)|-q>0}
y
∈
I
∖
{
x
}
{\displaystyle y\in I\setminus \{x\}}
|
f
(
y
)
−
f
(
x
)
y
−
x
−
f
′
(
x
)
|
<
δ
,
{\displaystyle \left|{\frac {f(y)-f(x)}{y-x}}-f'(x)\right|<\delta ,}
jolloin
|
f
(
y
)
−
f
(
x
)
y
−
x
|
>
|
f
′
(
x
)
|
−
δ
>
q
{\displaystyle \left|{\frac {f(y)-f(x)}{y-x}}\right|>|f'(x)|-\delta >q}
q
<
r
{\displaystyle q<r}
r
≥
1
{\displaystyle r\geq 1}
f
{\displaystyle f}
Funktio
f
(
x
)
=
x
2
1
+
|
x
|
{\displaystyle f(x)={\frac {x^{2}}{1+|x|}}}
f
:
R
→
R
{\displaystyle f:\mathbb {R} \rightarrow \mathbb {R} }
f
:
[
a
,
b
]
→
R
{\displaystyle f:[a,b]\rightarrow \mathbb {R} }
0
≤
f
′
(
x
)
=
(
x
+
1
)
2
−
1
(
x
+
1
)
2
<
1
{\displaystyle 0\leq f'(x)={\frac {(x+1)^{2}-1}{(x+1)^{2}}}<1}
kaikilla
x
≥
0
{\displaystyle x\geq 0}
f
′
(
−
x
)
=
−
f
′
(
x
)
{\displaystyle f'(-x)=-f'(x)}
|
f
′
(
x
)
|
<
1
{\displaystyle |f'(x)|<1}
x
{\displaystyle x}
sup
x
∈
R
|
f
′
(
x
)
|
=
lim
x
→
∞
|
f
′
(
x
)
|
=
1
,
{\displaystyle \sup _{x\in \mathbb {R} }|f'(x)|=\lim _{x\rightarrow \infty }|f'(x)|=1,}
joten
sup
x
∈
I
|
f
′
(
x
)
|
<
1
{\displaystyle \sup _{x\in I}|f'(x)|<1}
jos ja vain jos väli
I
{\displaystyle I}
f
″
(
x
)
=
2
/
(
x
+
1
)
3
≥
0
{\displaystyle f''(x)=2/(x+1)^{3}\geq 0}
x
{\displaystyle x}
![{\displaystyle f:[a,b]\rightarrow \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc0d2d0b70573525d149ab82948308455d1853d0)
