Теорема Риса о полноте — утверждение функционального анализа о полноте пространства Лебега
L
2
(
a
,
b
)
{\displaystyle L_{2}\left(a,b\right)}
Фридьеша Риса , установившего результат.
Каждая последовательность
{
f
n
(
x
)
}
{\displaystyle \left\{f_{n}(x)\right\}}
[
a
,
b
]
{\displaystyle \left[a,b\right]}
L
2
(
a
,
b
)
{\displaystyle L_{2}\left(a,b\right)}
Пусть задано произвольное
ϵ
>
0
{\displaystyle \epsilon >0}
n
ϵ
{\displaystyle n_{\epsilon }}
∫
a
b
{
f
n
+
p
(
x
)
−
f
n
(
x
)
}
2
d
x
<
ϵ
2
{\displaystyle \int _{a}^{b}\left\{f_{n+p}(x)-f_{n}(x)\right\}^{2}dx<\epsilon ^{2}}
n
⩾
n
ϵ
,
p
>
0
{\displaystyle n\geqslant n_{\epsilon },p>0}
ϵ
=
1
2
,
1
2
2
,
1
2
3
,
.
.
.
,
1
2
k
,
.
.
.
{\displaystyle \epsilon ={\frac {1}{2}},{\frac {1}{2^{2}}},{\frac {1}{2^{3}}},...,{\frac {1}{2^{k}}},...}
ϵ
=
1
2
k
{\displaystyle \epsilon ={\frac {1}{2^{k}}}}
n
k
{\displaystyle n_{k}}
n
1
<
n
2
<
.
.
.
.
<
n
k
<
.
.
.
{\displaystyle n_{1}<n_{2}<....<n_{k}<...}
∫
a
b
{
f
n
+
p
(
x
)
−
f
n
(
x
)
}
2
d
x
<
1
2
2
k
{\displaystyle \int _{a}^{b}\left\{f_{n+p}(x)-f_{n}(x)\right\}^{2}dx<{\frac {1}{2^{2k}}}}
n
=
n
k
,
n
+
p
=
n
k
+
1
{\displaystyle n=n_{k},n+p=n_{k+1}}
∫
a
b
{
f
n
k
+
1
(
x
)
−
f
n
k
(
x
)
}
2
d
x
<
1
2
2
k
{\displaystyle \int _{a}^{b}\left\{f_{n_{k+1}}(x)-f_{n_{k}}(x)\right\}^{2}dx<{\frac {1}{2^{2k}}}}
Неравенство Коши — Буняковского даст
∫
a
b
|
f
n
k
+
1
(
x
)
−
f
n
k
(
x
)
|
d
x
=
∫
a
b
|
f
n
k
+
1
(
x
)
−
f
n
k
(
x
)
|
∗
1
d
x
⩽
∫
a
b
d
x
∫
a
b
{
f
n
k
+
1
(
x
)
−
f
n
k
(
x
)
}
2
d
x
<
b
−
a
∗
1
2
k
{\displaystyle \int _{a}^{b}\left|f_{n_{k+1}}(x)-f_{n_{k}}(x)\right|dx=\int _{a}^{b}\left|f_{n_{k+1}}(x)-f_{n_{k}}(x)\right|*1dx\leqslant {\sqrt {\int _{a}^{b}dx}}{\sqrt {\int _{a}^{b}\left\{f_{n_{k+1}}(x)-f_{n_{k}}(x)\right\}^{2}dx}}<{\sqrt {b-a}}*{\frac {1}{2^{k}}}}
∫
a
b
|
f
n
1
(
x
)
|
d
x
+
∑
k
=
1
∞
|
f
n
k
+
1
(
x
)
−
f
n
k
(
x
)
|
2
d
x
{\displaystyle \int _{a}^{b}\left|f_{n_{1}}(x)\right|dx+\sum _{k=1}^{\infty }\left|f_{n_{k+1}}(x)-f_{n_{k}}(x)\right|^{2}dx}
f
(
x
)
∈
L
2
(
a
,
b
)
{\displaystyle f(x)\in L_{2}\left(a,b\right)}
∫
a
b
{
f
n
+
p
(
x
)
−
f
n
(
x
)
}
2
d
x
<
1
2
2
k
{\displaystyle \int _{a}^{b}\left\{f_{n+p}(x)-f_{n}(x)\right\}^{2}dx<{\frac {1}{2^{2k}}}}
n
=
n
k
{\displaystyle n=n_{k}}
n
+
p
=
n
m
{\displaystyle n+p=n_{m}}
m
>
k
{\displaystyle m>k}
∫
a
b
{
f
n
m
(
x
)
−
f
n
k
(
x
)
}
2
d
x
<
1
2
2
k
{\displaystyle \int _{a}^{b}\left\{f_{n_{m}}(x)-f_{n_{k}}(x)\right\}^{2}dx<{\frac {1}{2^{2k}}}}
m
→
∞
{\displaystyle m\rightarrow \infty }
{
f
(
x
)
−
f
n
k
(
x
)
}
2
{\displaystyle \left\{f(x)-f_{n_{k}}(x)\right\}^{2}}
∫
a
b
{
f
(
x
)
−
f
n
k
(
x
)
}
2
d
x
⩽
sup
m
>
k
∫
a
b
{
f
n
m
(
x
)
−
f
n
k
(
x
)
}
2
d
x
<
1
2
2
k
{\displaystyle \int _{a}^{b}\left\{f(x)-f_{n_{k}}(x)\right\}^{2}dx\leqslant \sup _{m>k}\int _{a}^{b}\left\{f_{n_{m}}(x)-f_{n_{k}}(x)\right\}^{2}dx<{\frac {1}{2^{2k}}}}
f
(
x
)
∈
L
2
(
a
,
b
)
{\displaystyle f(x)\in L_{2}\left(a,b\right)}
∫
a
b
{
f
(
x
)
−
f
n
k
(
x
)
}
2
d
x
⩽
sup
m
>
k
∫
a
b
{
f
n
m
(
x
)
−
f
n
k
(
x
)
}
2
d
x
<
1
2
2
k
{\displaystyle \int _{a}^{b}\left\{f(x)-f_{n_{k}}(x)\right\}^{2}dx\leqslant \sup _{m>k}\int _{a}^{b}\left\{f_{n_{m}}(x)-f_{n_{k}}(x)\right\}^{2}dx<{\frac {1}{2^{2k}}}}
{
f
n
k
(
x
)
}
{\displaystyle \left\{f_{n_{k}}(x)\right\}}
f
(
x
)
{\displaystyle f(x)}
{
f
n
(
x
)
}
{\displaystyle \left\{f_{n}(x)\right\}}
неравенству треугольника имеем
‖
f
n
−
f
‖
⩽
‖
f
n
−
f
n
k
‖
+
‖
f
n
k
−
f
‖
{\displaystyle \left\|f_{n}-f\right\|\leqslant \left\|f_{n}-f_{n_{k}}\right\|+\left\|f_{n_{k}}-f\right\|}
ϵ
>
0
{\displaystyle \epsilon >0}
k
{\displaystyle k}
1
2
2
k
<
ϵ
2
{\displaystyle {\frac {1}{2^{2k}}}<{\frac {\epsilon }{2}}}
∫
a
b
{
f
(
x
)
−
f
n
k
(
x
)
}
2
d
x
⩽
sup
m
>
k
∫
a
b
{
f
n
m
(
x
)
−
f
n
k
(
x
)
}
2
d
x
<
1
2
2
k
{\displaystyle \int _{a}^{b}\left\{f(x)-f_{n_{k}}(x)\right\}^{2}dx\leqslant \sup _{m>k}\int _{a}^{b}\left\{f_{n_{m}}(x)-f_{n_{k}}(x)\right\}^{2}dx<{\frac {1}{2^{2k}}}}
‖
f
n
k
−
f
‖
<
ϵ
2
{\displaystyle \left\|f_{n_{k}}-f\right\|<{\frac {\epsilon }{2}}}
n
k
{\displaystyle n_{k}}
n
⩾
n
k
{\displaystyle n\geqslant n_{k}}
‖
f
n
−
f
n
k
‖
<
ϵ
2
{\displaystyle \left\|f_{n}-f_{n_{k}}\right\|<{\frac {\epsilon }{2}}}
{
f
n
(
x
)
}
{\displaystyle \left\{f_{n}(x)\right\}}
‖
f
n
−
f
‖
<
ϵ
{\displaystyle \left\|f_{n}-f\right\|<\epsilon }
n
⩾
n
k
{\displaystyle n\geqslant n_{k}}
Соболев В. И. 
![{\displaystyle \left[a,b\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f30926fb280a9fdf66fd931e14d4363cb824feaa)
