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标题: [求助]谁能帮我翻译一下这篇文章?~ [打印本页]
作者: angel_lys 时间: 2007-11-23 22:16
标题: [求助]谁能帮我翻译一下这篇文章?~
文章是讲一致收敛和微分的~我英语不好看不明白~请大家帮帮忙啊~~~~
Theorem 9.13. Assume that each term of is a real-valued function having a finite derivative at each point of an open interval
) e7 T" @9 {' P: _" n& X( _9 L . Assume that for at least one point* A8 C D; S5 l2 X' y
in. l1 H a4 d( X! U, y1 r7 j
the sequence converges. Assume further that there exists a function g such that0 F" b5 X d# t( @; H, H
uniformly on2 V4 J6 @* R$ G* `1 K5 y8 j2 s F
. Then:
4 ?3 `2 X- t$ a" L" b5 P ^6 M
a) There exists a function f such that5 c& N+ z0 v7 Q
uniformly on; |9 ]: q* P) }8 {6 z
.
b) For each x in# f( y/ H% F/ S
the derivative$ Y- V- e. V. M/ l0 Q, v! v
exists and equal
, T* S- z G7 x& L .
Proof. Assume that% }7 U3 x& X, u4 N% l8 a: `
and define a new sequence
3 m+ F: }3 h: r+ L) L as follows:
8 n/ Z) T1 R2 g; G% Q
- e4 b6 z% B% Z; y5 @% o (8)
0 w: k' \% r7 x/ X' ?" J
The sequence
5 `; x% n u6 M2 d so formed depends on the choice of c. Convergence of follows from the hypothesis, since" ~, }6 |/ l, d7 `0 n
. We will prove next that* k! j. d/ h2 k4 M( V7 E- q
converges uniformly on$ R6 _; L: ` i# \
. If , we have
* Z8 e9 w6 r. d& a; o* V! Y
,
- [1 K$ a1 t. f; w' s7 f (9)
5 a7 g0 D7 m- H4 j9 X
where0 m( a( t+ U/ z6 _: ]
. Now
$ t! |+ h# Q K: H0 a+ g exists for each x in
1 v# F% @ y, n( g0 c' m2 m9 Y and has the value1 m( e! Z+ W" u. l' c" F9 w& ]
. Applying the Mean-Value Theorem in (9), we get
,
, ~: `8 {$ _4 L, B5 v4 K7 g8 A
% K& v4 F& m* p3 X- W (10)
where
( q( H4 [" V) y+ a1 ?1 u( @. P lies between x and c. Since5 \6 Y" N" x* B4 S
converges uniformly on0 k& V5 ^; e+ l _3 y
(by hypothesis), we can use (10), together with the Cauthy condition, to deduce that8 z! q5 z% a& R/ v9 K, x# T* E0 j
converges uniformly on$ Z7 |. r4 P+ E
.
Now we can show that
4 `9 {0 v7 t+ @6 l I converges uniformly on
& [& O% N, B, D0 q* g: y . Let us form the particular sequence" T8 {3 \* P# b8 R1 ^
corresponding to the special point
. G6 \7 S$ M; [ T7 J1 o3 `% e for which. V7 Y* h& K# B5 e A( n/ a
is assumed to converge. Form (8) we can write
an equation which holds for every x in9 G) c7 w+ d' F8 _5 N( n/ @
. Hence we have
This equation, with the help of the Cauthy condition, establishes the uniform convergence of on$ R# P' b- Z0 t: g- l1 }& o
. This proves (a).
To prove (b), return to the sequence
( E% p' A# p" F$ G4 D G" s, Q defined by (8) for an arbitrary point c in, L7 [9 S( K1 K E+ Q
and let
4 d0 c, y9 x0 ~$ e- `1 f7 D6 c . The hypothesis that; J1 O, V5 ^ }$ f
exists means that . In other words, each
4 Z- j$ K5 p1 r2 Y" ~3 s is continuous at c. Since% t; B: h ^" o0 T
uniformly on
6 p: p% l' ?) f2 G+ g: X8 @ , the limit function G is also continuous at c. This means that
. }" \; v) K: `2 l (11)
the existence of the limit being part of the conclusion. But for
% W$ b2 U/ Q! H7 x, O- E/ [ , we have
Hence, (11) states that the derivative9 z8 m8 J5 p6 z3 P: w8 _2 J- r
exists and equals h4 y7 v8 Y; d8 [6 L% f; t
. But
hence! s( P1 z2 d8 v& G% o
. Since c is an arbitrary point of# T% B2 X3 N$ z; q P6 k1 F
, this proves (b).
When we reformulate Theorem 9.13 in terms of series, we obtain
2 ~ [$ K, J$ x8 K; P
Theorem 9.14. Assume that each( e) l/ A# H# ~$ {
is a real-valued function defined on
7 m3 Q+ G& W: T* {9 ^0 a such that the derivative
6 P( R: d# g+ F+ l+ D2 j exists for each x in0 Q- y7 L- y: e* |; G6 Y2 Z
. Assume that, for at least one point
0 O* K; Q* x9 }( C% p; R in b/ ` F. c+ J5 m4 s5 l
, the series
5 i, n0 a7 w2 R* n7 L/ k converges. Assume further that there exists a function g such that (uniformly on
; j, q2 p {+ c U- ` ). Then:
a)7 e% a+ ~( r+ u" l7 K& ]
There exists a function f such that
f% r2 j' D+ ]5 r" m4 e8 { (uniformly on
3 R, N; n0 u5 ]' k( g ).
b)
+ g& p) F9 w0 Q0 `. E If , the derivative
2 Q9 s* n5 c+ M8 U7 h# I exists and equals
8 T% A- C( Y5 t! q) G( a% a8 { .
作者: lzh0601 时间: 2008-4-19 13:22
我怎么只看到脚本文件呢?
作者: tinkertinker 时间: 2008-7-28 14:22
“Theorem 9.13. Assume that each term of is a。。。。。” 这里面有没有漏了字?
`/ H0 I4 g) s. s( p
[此贴子已经被作者于2008-7-28 14:24:38编辑过]
作者: rao 时间: 2009-1-13 21:47
看不清楚。。。。。
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