※ 引述《Eliphalet (真係廢到冇朋友)》之銘言:
: 有一個問題想請教大家
: 令 D 是 R^n 上的一個非空的開集合, f_k : D -> R 是 differentiable 的函數
: 假設 f_k -> f pointwise 且 ▽f_k -> g uniformly on D
: 那麼會不會有以下的結果
: f 是 differntiable ?
: ▽f_k -> ▽f uniformly on D ?
: 我知道 n=1 時是對的 那麼當 n >= 2 時呢 ?
答案是肯定的 ! f_k need not to be of class C^1.
參考的書來自
(Rudin , Principles Of Mathematical Analysis )
稍加改一下 n = 1 的證明 , 就可得到 n ≧ 2 情形.
只證open ball的情形就好了,在證明之前,先提一下下面定理7.11 .
Theorem 7.11:
Suppose φ_n —> φ uniformly on a set E in a metric space . Let
x be a limit point of E , and suppose that
lim φ_n(t) = An (n =1,2,3.........)
t->x
Then {An} converges , and
lim φ(t) = lim An
t->x n->∞
Theorem 7.17: ( 0<r<∞ )
{fn:Br—>|R} is a sequence of functions (Br is an open ball in |R^n),
differentiable on Br and such that {fn(x_0)} converges for some point
x_0 on Br , If {Dfn} converges uniformly on Br , then {fn} converges
uniformly on Br ,to a function f , and
Df(x) = lim Dfn(x) (for all x in Br)
n->∞
Proof:
( Notation: x,t are elements of Br , Dfn(x) = Ln(x) , Ln(x)->L(x) )
Let ε>0 be given . Choose N such that n≧N , m≧N ,implies
∣fn(x_0)-fm(x_0)∣< (ε/2)
and
∥Dfn(t)-Dfm(t)∥< (ε/(4r)) (for all t in Br)
∥ ∥
Ln(t) Lm(t)
we apply the mean value theorm (|R^n) to fn(t)-fm(t)
∣fn(t)-fm(t)-(fn(x)-fm(x))∣≦∣(Ln(c)- Lm(c))‧(t-x)∣
≦ ∥Ln(c)- Lm(c)∥∥t-x∥< (ε/(4r))∥t-x∥-------------(1)
where c is a ponit on the line segment joining x and t .
∴∣fn(x)-fm(x)∣≦∣fn(x)-fm(x)-(fn(x_0)-fm(x_0))∣+∣fn(x_0)-fm(x_0)∣
< (ε/(4r))*(2r)+(ε/2) =ε
{ fn(x) } converges uniformly on Br.
Let f be the limit function of { fn(x) } .
Now , fix a point x in Br .
∣fn(t)-fn(x)-Ln(x)‧(t-x)∣
Let φ_n(t) = ────────────────
∥t-x∥
∣f(t)-f(x)-L(x)‧(t-x)∣
Let φ(t) = ────────────────
∥t-x∥
∥φ_n(t) - φ_m(t)∥ ≦
∣fn(t)-fn(x)-Ln(x)‧(t-x)-(fm(t)-fm(x)-Lm(x)‧(t-x))∣
──────────────────────────── (..from (1))
∥t-x∥
< ε/(4r)+ε/(4r) =ε/(2r)
{φ_n} converges uniformly on Br -∣
∣=>φ_n->φ uniformly on Br
lim φ_n(t) = φ(t)----------------∣
n->∞
lim φ_n(t) = 0 = An
t->x
By Theorem 7.11
lim φ(t) = lim An = 0
t->x n->∞
that is
Df(x) = lim Dfn(x)
n->∞
Q.E.D
有打錯麻煩提醒我 !
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