其實是一個雙重極限對調的問題
題目如下
Let f be a continuous function on [0,1]. Prove that
1 n
∫ x f(x) dx
0
lim --------------- = f(1).
n→∞ 1 n
∫ x dx
0
我的作法是這樣:
By Weierstrass approximation theorem, there exist polynomials p_k(x) s.t.
p_k(x)→f(x) uniformly on [0,1].
n n n
∵|x p_k(x) - x f(x)| = |x ||p_k(x) - f(x)|≦|p_k(x) - f(x)| for x in [0,1]
n n
∴x p_k(x) → x f(x) uniformly on [0,1]
By uniform convergent theorem,
1 n 1 n
∫ x p_k(x) dx → ∫ x f(x) dx.
0 0
m_k i
Write p_k(x) = Σ a_(i_k) x .
i=0
1 n m_k a_(i_k)
Then ∫ x p_k(x) dx = Σ -----------.
0 i=0 i + n + 1
1 n 1 n
∫ x f(x) dx ∫ x p_k(x) dx
0 0
∴ lim --------------- = lim lim -----------------
n→∞ 1 n→∞ k→∞ 1 n
∫ x dx ∫ x dx
0 0
m_k a_(i_k) (n + 1)
= lim lim Σ -----------------
n→∞ k→∞ i=0 i + n + 1
如果這裡兩個極限順序可以對調，就會有下面的式子：
m_k a_(i_k) (n + 1)
= lim lim Σ ------------------
k→∞ n→∞ i=0 i + n + 1
m_k a_(i_k) (n + 1)
= lim Σ lim ------------------ (∵有限項取和)
k→∞ i=0 n→∞ i + n + 1
m_k
= lim Σ a_(i_k)
k→∞ i=0
= lim p_k(1) (∵p_k(x)→f(x) uniformly on [0,1])
k→∞
= f(1)
請問這裡極限順序可以對調嗎？
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※ 編輯: kemowu 來自: 140.115.221.107 (07/28 21:38)