作者Dirichlet ( )
看板Math
标题Re: [分析] 初微(25)
时间Sun Aug 21 06:29:32 2005
※ 引述《PttFund (批踢踢基金只进不出)》之铭言：
: Suppose that a_ij > 0 for all i,j. Show that
: Σ Σ a_ij = Σ Σ a_ij.
: i j j i
设 Σ a_ij = b_i
j
case1. Σ Σ a_ij = +oo
i j
(a) b_(i_0) = +oo for some i_0 => for any M>0, there exists k in N s.t.
k k oo k
Σ a_(i_0)j > M, so Σ Σ a_ij > Σ Σ a_ij > Σ a_(i_0)j > M
j=1 j i j=1 i=1 j=1
故 Σ Σ a_ij = +oo
j i
(b) b_i < +oo for all i and Σ b_i = +oo => for 1/2^i, there exist n(i) in N
n(i)
s.t. b_i < (Σ a_ij) + 1/2^i (可调整 n(i) 使得 n(i) ≦ n(i+1) for all i)
j=1
then for any M>0, there exist k in N s.t.
k k n(i) k n(k)
M+1 < Σ b_i < (Σ Σ a_ij) + 1 < (Σ Σ a_ij) + 1
i=1 i=1 j=1 i=1 j=1
n(k) k
= (Σ Σ a_ij) + 1 < (Σ Σ a_ij) + 1
j=1 i=1 j i
故 Σ Σ a_ij > M, 即 Σ Σ a_ij = +oo
j i j i
case2. Σ b_i 收敛, 即 Σ Σ a_ij 收敛
i j
设 S = {x_0} U {x_n : n in N} 且 x_n -> x_0 as n -> oo
n
定义 f_i(x_n) = Σ a_ij, f_i(x_0) = b_i, g(x) = Σ f_i(x), x in S
j=1
f_i(x) ≦ b_i 且 Σ b_i 收敛, M-test => Σ f_i(x) = g(x) uniformly
又每个 f_i(x) 都在 x=x_0 连续, 故 g(x) 在 x=x_0 连续
oo n
Σ Σ a_ij = g(x_0) = lim g(x_n) = lim Σ Σ a_ij
i j n->oo n->oo i=1 j=1
oo
因为对每个 j = 1, 2, ... ,n 而言 Σ a_ij 都收敛, 故上式可拆开)
i=1
oo oo oo
= lim [(Σ a_i1) + (Σ a_i2) + ... + (Σ a_in)]
n->oo i=1 i=1 i=1
n oo
= lim Σ Σ a_ij = Σ Σ a_ij
n->oo j=1 i=1 j i
故 Σ Σ a_ij = Σ Σ a_ij ... Q.E.D.
i j j i
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