: 2.Assume that f_n: R→R (n=0.1.2...) is a sequence of differentiable
: functions s.t each f_n(x) is a solution of the equation
: y'(1 + x^2 + xy + y^2) = 1. If sup_n ∣f_n(1/n)∣＜∞,
: prove that there exists a subsequence f_n_k(x) s.t
: lim f_n_k(x) = f(x) exists for x in R.
: k→∞
: This limit f(x) must also be a solution of y'(1 + x^2 + xy + y^2) = 1.
: 謝謝~
早上起來想了個辦法，
不過小弟學藝不精，沒辦法很確定自己是不是對的。
所以想說PO上來給大家看看有沒有哪裡有問題@@
如果有錯誤不吝指正。也請各位鞭小力一點.. ..
==
Note that |f'_n(x)|=<1.
Let Q be the set of rational numbers. Since Q is countable and f_n(x) is
pointwise bounded
(Let S=sup|f_n(1/n)|, fix any x, |f_n(x)-f_n(1/n)|=|f'_n(t)(x-1/n)|
=<|x-1/n|<|x|+1
so |f_n(x)|-|f_n(1/n)|=<|x|+1, |f_n(x)|=< S+|x|+1)
, there is a subsequence of f_n, say f_n_k, such that {f_n_k}
converges pointwisely on R by the denseness of Q.
^^
此步要扣分，不過因為 f_n_k 原本就收斂 uniformly on every compact subset.
所以是連續函數。
We denote the limit by f(x).
By assumption, {f'_n_k} is uniformly bounded, and f_n_k converges on R.
Compute f''_n_k=<(2|x|+3|y|)/(1+x^2+xy+y^2)^2 < M for some M>0.
(Here y=f_n_k(x)) This means f'_n_k is equicontinuous on R.
There is a further subseq of f'_n_k, say f'_n_k_i, such that
f'_n_k_i converges pointwisely to g for some g.
Let z be any point on R, choose a compact set K such that z\in int(K).
Now f'_n_k_i converges uniformly on K. And lim f_n_k_i(x)=f(x),
therefore, lim f'_n_k_i(x)=f'(x) uniformly on K, specially at z
And the limit f(x) at x=z also satisfies that f'(z)(1+z^2+zf(z)+f^2(z))=1.
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◆ From: 140.112.50.253
※ 編輯: yusd24 來自: 140.112.50.253 (09/05 11:56)
※ 編輯: yusd24 來自: 140.112.50.253 (09/07 09:30)