一題比較難的實變

Question

設 E_n =
{f in C[0,1] : 存在 x_0 in [0,1] 使得 |f(x)-f(x_0)| ≦ n|x-x_0|
for all x in [0,1]}

a) 證明 E_n 是 C[0,1] 內的一個 nowhere dense subset.

b) 證明所有在 [0,1] 上處處連續處處不可微的函數所成集合 E 是 Baire 第二類型.
(就是它沒辦法表示成 C[0,1] 內可數個 nowhere dense subset 的聯集)

問兩個地方 QQ 為什麼
|fk(j)(x) - fk(j)(x0,k(j))| → |f(x) - f(x0)| 和
|(f+g)(x) - (f+g)(x0)| + |f(x) - f(x0)| ≤ n|x - x0| + n|x - x0|
如果 f+g 和 f 都屬於 E_n, 那應該要假設兩個不同的 x0才合理吧 ?

|fk(j)(x) - fk(j)(x0,k(j))| → |f(x) - f(x0)| 這個我知道了
因為 f_n - f uniformly

因為 f_n 趨近於 f uniformly

Answer 1

The setting will be the uniform topology on C([0,1]) with norm ‖f‖ = supx |f(x)|.Lemma. En is closed for all n.Proof. Let n ∈ N, and let {fk} be a sequence in En converging to a function f in C([0,1]), with x0,k ∈ [0,1] such that for all x in [0,1], we have|fk(x) - fk(x0,k)| ≤ n|x - x0,k|.The Bolzano-Weierstraß theorem asserts the existence of a convergent subsequence x0,k(j) → x0, and hence for any x ∈ [0,1],|fk(j)(x) - fk(j)(x0,k(j))| → |f(x) - f(x0)| ≤ n|x - x0|, and so f ∈ En. Therefore, En is closed. ∎
Claim. En is nowhere dense in the uniform topology on C([0,1]). Proof. Let n ∈ N. Since we know that En is closed by the above lemma, it remains to show that En has empty interior. Let f ∈ En, ε > 0, and define g ∈ C([0,1]) by
g(x) = ε/2 - 3n|x-ε/6n|    if x ∈ [0,ε/3n)
periodically extended to the rest of [0,1] by setting
g(x+ε/3n) = g(x)   if x ∈ [ε/3n,1-ε/3n].

(Basically, the graph of g(x) looks like /\/\/\/\/..., with each line segment having slope 3n, with minimum 0 and maximum ε/2).
Then
‖(f+g) - f‖ = supx |(f+g)(x) - f(x)|
= supx |g(x)| = ε/2 < ε,
so f+g ∈ B(f,ε). I claim that f+g ∉ En; suppose otherwise, and let x0 ∈ [0,1]. There exists x close enough but not equal to x0 such that|g(x) - g(x0)| = 3n|x - x0|,but then|g(x) - g(x0)| ≤ |(f+g)(x) - (f+g)(x0)| + |f(x) - f(x0)| ≤ n|x - x0| + n|x - x0|= 2n|x - x0|,which leads to a contradiction. Hence, f+g ∉ En, and so B(f,ε) is not contained in En. Therefore, the interior of the closed set En is empty, and hence En is nowhere dense. ∎


Claim.  The class E of all C([0,1]) functions that are nowhere differentiable is of the second category.

Proof.  Suppose E is of the first category; by the Baire category theorem, C([0,1])∖E must be dense in C([0,1])∖E. Observe that C([0,1])∖E, the collection of C([0,1]) functions that is somewhere differentiable, is contained in ∪n En, the latter being nowhere dense as a consequence of the Baire category theorem. Therefore, C([0,1])∖E is also nowhere dense, leading to a contradiction. Hence, E is of the second category. ∎


2006-09-10 00:55:51 補充：
找到了！
http://www.math.uic.edu/~marker/math414/fs.pdf
Lemma 18，第5頁，設m=1即可。

Answer 2

你最後一段的 argument 是不是假設 E 是第一類型, 則 C[0,1] - E
會是第二類型, 但是只要在某點可微的函數一定屬於 ∪n En (這個我有證出來), 所以 C[0,1] - E 包含在 ∪n En, 但是 ∪n En 是第一類型
第一類型的子集也是第一類型, 矛盾
是這樣嗎 @@?