文档介绍:: .
泛函分析<br****题解答
1、设(X,d)为一度量空间,令 U(Xo, ) {x|x X,d(x,Xo)
} S(x°, ) {x|x X,d(x,x°) },问
U(X0,)的闭包是否等于S(X0, ) o
解答:在一般度量空间中不成立 U(x°, ) S(x0,),例如:取
R1的度量子空间
X [0,1] U[2,3],贝U X 中
的开球 U(1,1) {x X;d(1,x) 1}的的闭包是[0,1],而 S(1,1) {x
X;d(1,x) 1}
[0,1] U{2}
2、设C [a,b]是区间[a,b]上无限次可微函数全体,
定义d(f,g)
1
r max
r 0 2r
a t b
| f(r)(t) g(r)(t)|
1 | f(r)(t)
g(r)(t)|,证明:
C [a, b]按d( f ,g)构成度量空间。
证明:(1 )
显然 d( f ,g) 0 且 d(f, g) 0
|f(r)(t) g(r)(t)| 0
tb 1 | f(r)(t) g(r)(t)|
1
r, - max
a
r, t [a,b]有
|f(r)(t) g(r)(t)|
0,特别当r 0,
t [a,b]时有 | f
(t)
g (t)| 0 t [a,b]有 f(t)
g(t) o
(2)由函数
f(t)占在[0,
)上单调增加,从而对
f ,g,h C [a,b]有
即三角不等式成立
d(f,g)
|f(r)(t) g(r)(t)|
1 |f(r)(t) g(r)(t)|
1
-max
2r
a t b
|f(r)(t) h⑴(t) h⑴(t) g⑴(t)|
丄
r 0 2^max1 | f(r)(t) h(r) (t) h(r)(t) g(r) (t)|
1 max 戸⑴⑴ h⑴(t)| |h⑴(t) g ⑴(t)|
r 0 2r a t b 1 | f (r)(t) h(r)(t)| |h(r)(t) g(r) (t)|
| f(r)(t) h(r)(t)|
(r)
(r)
(r)
(r)
1
— max
r 0 2r a t b1 | f(,)(t) h(,)(t)| |h(,)(t) g(,)(t)| 丄 |h(r)(t) g⑴(t)|
r 0 歹max 1 |f(r)(t) h(r)(t)|
1 max |f⑴(t) h(r)(t)|
r 02r a t b1 | f(r)(t) h(r)(t)|
d(f,h) d(h,g)
(r) (r)
(r)
(r)
(r)
(r)
| h(r) (t) g(r)(t)|
1 max |h(r)(t) g(r)(t)| 7 max
0 2r…
atb1 |h(r)(t) g(r)(t)|
d(f ,g) d(f,h) d(h,g) o
3、设B是度量空间X中的闭集,证明必有一列开集
01,。2 丄 On,L 包含 B,而且 | On B o n 1
1
证明:设B为度量空间X中的闭集,作集: On {x|d(x,