Fie
(
S
,
d
)
{\displaystyle (S,d)}
Aplicația
f
:
S
⟶
S
{\displaystyle f:S\longrightarrow S}
q
∈
(
0
,
1
)
{\displaystyle q\in (0,1)}
d
(
f
(
x
)
,
f
(
y
)
)
≤
q
d
(
x
,
y
)
(
∀
)
x
,
y
∈
S
(
1.1
)
{\displaystyle d(f(x),f(y))\leq qd(x,y)(\forall )x,y\in S\ (1.1)}
Punctul
c
∈
S
{\displaystyle c\in S}
f
:
S
⟶
S
{\displaystyle f:S\longrightarrow S}
f
(
c
)
=
c
.
(
1.2
)
{\displaystyle f(c)=c.\ (1.2)}
Fie
a
0
∈
S
{\displaystyle a_{0}\in S}
(
a
n
)
n
∈
N
{\displaystyle (a_{n})_{n\in \mathbb {N} }}
{
a
1
=
f
(
a
0
)
a
2
=
f
(
a
1
)
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
a
n
=
f
(
a
n
−
1
)
{\displaystyle \left\{{\begin{array}{rcl}a_{1}=f(a_{0})\\a_{2}=f(a_{1})\\...................\\a_{n}=f(a_{n-1})\end{array}}\right.\;}
Știind că S este un spațiu metric complet pentru a arăta că șirul
(
a
n
)
n
∈
N
{\displaystyle (a_{n})_{n\in \mathbb {N} }}
d
(
a
1
,
a
2
)
=
d
(
f
(
a
0
)
,
f
(
a
1
)
≤
q
d
(
a
0
,
a
1
)
{\displaystyle d(a_{1},a_{2})=d(f(a_{0}),f(a_{1})\leq qd(a_{0},a_{1})}
d
(
a
2
,
a
3
)
=
d
(
f
(
a
1
)
,
f
(
a
2
)
≤
q
d
(
a
1
,
a
2
)
≤
q
2
d
(
a
0
,
a
1
)
{\displaystyle d(a_{2},a_{3})=d(f(a_{1}),f(a_{2})\leq qd(a_{1},a_{2})\leq q^{2}d(a_{0},a_{1})}
d
(
a
3
,
a
4
)
=
d
(
f
(
a
2
)
,
f
(
a
3
)
≤
q
d
(
a
2
,
a
3
)
≤
q
3
d
(
a
0
,
a
1
)
{\displaystyle d(a_{3},a_{4})=d(f(a_{2}),f(a_{3})\leq qd(a_{2},a_{3})\leq q^{3}d(a_{0},a_{1})}
Prin inducție se obține:
d
(
a
n
,
a
n
+
1
)
≤
q
n
d
(
a
0
,
a
1
)
,
(
∀
)
n
∈
N
,
q
∈
(
0
,
1
)
.
{\displaystyle d(a_{n},a_{n+1})\leq q^{n}d(a_{0},a_{1}),(\forall )n\in \mathbb {N} ,q\in (0,1).}
Pe de altă parte, pentru orice
p
∈
N
{\displaystyle p\in \mathbb {N} }
d
(
a
n
,
a
n
+
p
)
≤
d
(
a
n
,
a
n
+
1
+
d
(
a
n
+
1
,
a
n
+
2
)
+
.
.
.
d
(
+
a
n
+
p
−
1
,
a
n
+
p
)
{\displaystyle d(a_{n},a_{n+p})\leq d(a_{n},a_{n+1}+d(a_{n+1},a_{n+2})+...d(+a_{n+p-1},a_{n+p})}
și folosind corespunzător inegalitatea (1.4) se obține:
d
(
a
n
,
a
n
+
p
)
≤
q
n
d
(
a
0
,
a
1
)
+
q
n
+
1
d
(
a
0
,
a
1
)
+
.
.
.
+
q
n
+
p
−
1
d
(
a
0
,
a
1
)
=
{\displaystyle d(a_{n},a_{n+p})\leq q^{n}d(a_{0},a_{1})+q^{n+1}d(a_{0},a_{1})+...+q^{n+p-1}d(a_{0},a_{1})=}
=
q
n
d
(
a
0
,
a
1
)
(
1
+
q
+
q
2
+
.
.
.
+
q
n
−
1
)
=
q
n
d
(
a
0
,
a
1
)
q
p
−
1
q
−
1
=
{\displaystyle =q^{n}d(a_{0},a_{1})(1+q+q^{2}+...+q^{n-1})=q^{n}d(a_{0},a_{1}){\frac {q^{p}-1}{q-1}}=}
=
q
n
d
(
a
0
,
a
1
)
1
−
q
p
1
−
q
≤
q
n
1
−
q
d
(
a
0
,
a
1
)
.
{\displaystyle =q^{n}d(a_{0},a_{1}){\frac {1-q^{p}}{1-q}}\leq {\frac {q^{n}}{1-q}}d(a_{0},a_{1}).}
Deci:
d
(
a
p
,
a
n
+
p
)
≤
q
n
1
−
q
d
(
a
0
,
a
1
)
,
(
∀
)
n
∈
N
,
(
∀
)
p
∈
N
,
q
∈
(
0
,
1
)
.
{\displaystyle d(a_{p},a_{n+p})\leq {\frac {q^{n}}{1-q}}d(a_{0},a_{1}),(\forall )n\in \mathbb {N} ,(\forall )p\in \mathbb {N} ,q\in (0,1).}
Presupunem că
d
(
a
0
,
a
1
)
≠
0
{\displaystyle d(a_{0},a_{1})\neq 0}
q
∈
(
0
,
1
)
⇒
q
n
1
−
q
d
(
a
0
,
a
1
)
→
0
∈
R
{\displaystyle q\in (0,1)\Rightarrow {\frac {q^{n}}{1-q}}d(a_{0},a_{1})\rightarrow 0\in \mathbb {R} }
(
(
∀
)
ε
>
0
,
(
∃
)
N
(
ε
)
{\displaystyle ((\forall )\varepsilon >0,(\exists )N(\varepsilon )}
astfel încât
(
∀
)
n
>
N
(
ε
)
{\displaystyle (\forall )n>N(\varepsilon )}
(
∀
)
p
∈
N
⇒
d
(
a
n
,
a
n
+
p
)
<
ε
{\displaystyle (\forall )p\in \mathbb {N} \Rightarrow d(a_{n},a_{n+p})<\varepsilon }
(
a
n
)
n
∈
N
{\displaystyle (a_{n})_{n\in \mathbb {N} }}
c
=
lim
n
→
∞
a
n
{\displaystyle c=\lim \limits _{n\rightarrow \infty }a_{n}}
c
∈
S
(
a
0
⟶
c
{\displaystyle c\in S(a_{0}\longrightarrow c}
În acest caz notăm:
c
=
lim
n
→
∞
a
n
{\displaystyle c=\lim \limits _{n\rightarrow \infty }a_{n}}
c
∈
S
{\displaystyle c\in S}
a
0
→
c
{\displaystyle a_{0}\rightarrow c}
a
n
→
c
{\displaystyle a_{n}\rightarrow c}
ε
>
0
{\displaystyle \varepsilon >0}
x
(
ε
)
>
o
{\displaystyle x(\varepsilon )>o}
n
>
N
(
ε
)
{\displaystyle n>N(\varepsilon )}
d
(
a
0
,
c
)
<
ε
{\displaystyle d(a_{0},c)<\varepsilon }
Observând și inegalitatea evidentă
d
(
f
(
a
0
)
,
f
(
c
)
)
≤
d
(
a
0
,
c
)
{\displaystyle d(f(a_{0}),f(c))\leq d(a_{0},c)}
d
(
f
(
a
)
,
f
(
c
)
)
<
ε
,
(
∀
)
n
>
N
(
ε
)
)
{\displaystyle d(f(a),f(c))<\varepsilon ,(\forall )n>N(\varepsilon ))}
f
(
a
0
)
→
f
(
c
)
{\displaystyle f(a_{0})\rightarrow f(c)}
a
n
+
1
→
f
(
c
)
{\displaystyle a_{n+1}\rightarrow f(c)}
a
n
+
1
→
c
{\displaystyle a_{n+1}\rightarrow c}
f
(
c
)
=
c
{\displaystyle f(c)=c}
c
∈
S
{\displaystyle c\in S}
Să arătăm acum unicitatea lui c.
Presupunem că mai există
c
1
∈
S
{\displaystyle c_{1}\in S}
f
(
c
1
)
=
c
1
{\displaystyle f(c_{1})=c_{1}}
d
(
c
,
c
1
)
=
d
(
f
(
c
)
,
f
(
c
1
)
)
≤
q
d
(
c
,
c
1
)
.
{\displaystyle d(c,c_{1})=d(f(c),f(c_{1}))\leq qd(c,c_{1}).}
Rezultă
(
1
−
q
)
d
(
c
,
c
1
)
≤
0
,
q
∈
(
0
,
1
)
{\displaystyle (1-q)d(c,c_{1})\leq 0,q\in (0,1)}
d
(
c
,
c
1
)
=
0
{\displaystyle d(c,c_{1})=0}
c
1
=
c
{\displaystyle c_{1}=c}
d
(
a
2
,
a
3
)
=
d
(
f
(
a
1
)
,
f
(
a
2
)
)
≤
q
d
(
a
1
,
a
2
)
≤
q
2
d
(
a
0
,
a
1
)
{\displaystyle d(a_{2},a_{3})=d(f(a_{1}),f(a_{2}))\leq qd(a_{1},a_{2})\leq q^{2}d(a_{0},a_{1})}
