Riemann integrable

kingwinner

我暫時 prove 到咁多:

點 prove 個 hint?

thx!

Naozumi

1. Sup |f(x)g(x) - f(t)g(t)| = Sup [f(x)g(x) - f(t)g(t)]    (因為x,t是沒有次序的)
                                         = Sup f(x)g(x) - inf f(t)g(t)
2. 你已經證明hint果堆野係|f(x)g(x) - f(t)g(t)|的upper bounbf
3. Sup <= Upper bound

kingwinner

sorry, 唔係好明你既第1個 point wor...
點解 Sup |f(x)g(x) - f(t)g(t)| = Sup [f(x)g(x) - f(t)g(t)]  = Sup f(x)g(x) - inf f(t)g(t) ?? 點 prove?

我唔係好明個 hint 左面點可以整到 M_i (fg, P) - m_i (fg, P) 出黎

thx

Naozumi

f(1)g(1) - f(2)g(2)同 f(2)g(2) - f(1)g(1) 咪又係番個set {x,t in R:f(x)g(x) - f(t)g(t)} 度
冇話x要大過t 喎!!

Sup (A -B) = Sup A - inf B係基本properties,你要溫番書
(假設 Sup A, inf B exist)

kingwinner

如果 prove 到
Sup {f(x)g(x) - f(t)g(t): x,t E[x_(i-1),x_i]}
= Sup {f(x)g(x): x E[x_(i-1),x_i] } -  inf {f(t)g(t):  t E[x_(i-1),x_i]}
, 咁個 hint 就攪掂

我搵過幾本書都搵唔到點 prove 你個 property...

[ 本帖最後由 kingwinner 於 10-3-18 12:04 PM 編輯 ]

Naozumi

好簡單
Let a in A, b in B.
a - b <= Sup A - b
         <= Sup A - inf B ( b>= inf B)
So, Sup A - inf B is an upper bound of the set {a in A, b in B : a - b}

For all e > 0
Sup A - inf B - e = (Sup A - e/2) - (inf B + e/2)
By definition of Sup, there is an element a' in A s.t. a' > Sup A - e/2
By definition of inf, there is an element b' in B s.t. b' < inf B + e/2
hence, Sup A - inf B - e < a' - b'
i.e. Sup A - inf B is the least upper bound of the set {a in A, b in B : a - b}

pqrs853

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