※ 引述《pobm (妳會更快樂嗎?)》之銘言：
: p.114 2
: Suppose f'(x)>0 in (a,b), then f is strictly increasing in (a,b). let g be
: its inverse function. Prove that g is differentiable, and that
: 1
: g'(f(x))=------- x belong in (a,b)
: f'(x)
: 做起來有一點奇怪
: g(t)-g(f(x)) g(f(s))-g(f(x)) = s-x
: g'(f(x))= lim ---------------- = lim -----------------
: t->f(x) t-f(x) f(s)->f(x) f(s)-f(x)
: 1
: = lim ---------------- 可是要滿足所求lim下面不是應該是s->x嗎
: f(s)->f(x) f(s)-f(x)
: ------------
: s-x
: 但是f(s)->f(x)並沒有強制s->x阿
: 我是想過另外一個可能 就是如果f是f :[a,b]-> f([a,b])
: 那f conti 就保證了g conti (since [a,b] compact)
: 那這樣f(s)->f(x)就會imply s-x了
: 其實仔細看Rudin有時候會有點煩惱
: 常常要注意到底這裡的f是從哪映射到哪
: 謝謝^^
Since f'(x)>0 on (a, b), then f is injective on (a,b)
(i.e., the inverse of f exists.).
Since f is differentiable on (a, b), f is continuous on (a,b).
Let f(u)=s, f(v)=t for u, v on (a,b).
Estimate the magnitude:
g(s)-g(t) g(f(u))-g(f(v)) u-v f(u)-f(v)
----------- = ----------------- = -------------- = [-----------]^(-1)
s-t f(u)-f(v) f(u)-f(v) u-v
Since the last term tends to [f'(v)]^(-1) as u--->v, then g is differentiable
at t, i.e., g'(t) = [f'(v)]^(-1), so that g'(f(v)) = [f'(v)]^(-1)
note:因為當u趨近於v時，f(u)會趨近於f(v)，則s會趨近於t，
則可以從最後一個等式回推得到答案。
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有誤請指正
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