推 ElvinN:我的印象中 第一題可以令g(t)=f(tx)/t^k 然後微分=0去証 01/31 12:24
會了，感謝！
g'(t) = [(tx)．f'(tx)t^(k-1) - kt^(k-1)f(tx)]/t^2k = 0
g(t) = c for every t≠0
f(x)=c as t=1
Thus f(tx)=(t^k)f(x).
※ 編輯: paperbattle 來自: 122.116.42.100 (01/31 13:28)
1.Let f:|R^n→|R be a C^1 function. Suppose that
x．▽f(x) = kf(x).
Prove that
k
f(λx) = λ f(x) for each x in |R^n and λ>0.
2.Let f_n be a sequence of continuously differentiable function on [a,b] such
that f_n(a) = f_n(b) = 0 and
b 2
∫ |f_n'(x)| dx ≦ M
a
for some fixed constant M. Prove that {f_n} contains a subsequence that
converges uniformly on [a,b].
看不出來f_n(x)或f_n'(x)是equicontinuous
3.Let f:(-1,2)→|R be a real analytic function. If f(1/k)=0 for all k, Show
that f is identically zero.
請給個hint
感謝
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