→ Dirichlet:(c) 就如 math1209 所說的 03/10 13:37
推 ikikiki:謝謝，我了解了 03/11 16:53
※ 引述《ikikiki (小優)》之銘言：
: Assume that f : R^n → R^n is continuously differentiable, and that there is
: a positive number c such that
: || f(x) - f(y) ||≧ c || x-y || for all x,y in R^n.
: (a) Prove that f(R^n) is open.
: (b) Prove that f(R^n) is close.
: (c) What can you conclude from (b) and (c)
: 攤手了…..╮(﹋﹏﹌)╭..
: 和今年台大考題好像…可惜分數也飛了！
: 請教大師
(a) Fix x, and take y =/= x we have
c <= || f(x)-f(y)|| / ||x-y||
= (|| f(x)-f(y)|| - ||Df(x)(x-y)|| + ||Df(x)(x-y)||) / ||x-y||
= (|| f(x)-f(y)|| - ||Df(x)(x-y)||) / ||x-y|| + ||Df(x)(x-y)||) / ||x-y||
<= (|| f(x)-f(y)-Df(x)(x-y)||) / ||x-y|| + ||Df(x)||
Let y -> x we get c <= ||Df(x)||
以上對所有 |R^n 內之 x 皆成立，故有 ||Df(x)|| 不為 0 對所有 |R^n 內之 x
成立，這又導致 |Df(x)| 不為 0 對所有 |R^n 內之 x 成立。因 |R^n 是 open，
由開映射定理知 f(|R^n) 是 open.
(b) 取 f(|R^n) 內一數列 f(x_n), 其中 x_n 屬於 |R^n,
且滿足 f(x_n) -> y in |R^n.
=> ||f(x_n) - f(x_m)|| >= c||x_n - x_m||
因 f(x_n) 是柯西數列, 由上式易知 x_n 也是柯西數列,
又 |R^n 是完備空間, 有 x_n -> x in |R^n. 再由連續性
=> f(x_n) -> f(x) as n -> oo
極限唯一性得 y = f(x), 故 y 屬於 f(|R^n). 因此 f(|R^n) 是閉集.
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