→ Vulpix : 其實i可改成檢查"D≦0"，ii就直接導致"不可能"。 11/28 18:50
推 OppOops : 推 11/28 22:24
→ Vulpix : 或者用兩方程的根應對稱於x=ξ得知ξ只能是0。 11/29 01:47
※ 編輯: Vulpix (111.243.103.5), 05/14/2017 13:39:20
※ 引述《TosakaRin (遠阪凜)》之銘言：
: 假設f是C^2([0,∞)), 且f' is not bounded. 那麼是否一定有
: f or f" is not bounded?
: 我覺得應該是對的 但不會說明
: 謝謝
這是Rudin的習題，而且習題後有Hint。
為了方便敘述，使用反證法。
pf):
Suppose that f and f" are bounded by M and N respectively,
where both M and N are positive.
By Taylor's Theorem, given ξ ≧ 0,
f(x) = f(ξ) + f'(ξ)(x-ξ) + (f"(η)/2)(x-ξ)^2
for some η between x and ξ.
=> f'(ξ)(x-ξ) = f(x) - f(ξ) - (f"(η)/2)(x-ξ)^2
=> |f'(ξ)(x-ξ)| ≦ 2M + (N/2)(x-ξ)^2
=> N(x-ξ)^2 + 2f'(ξ)(x-ξ) + 4M ≧ 0
& N(x-ξ)^2 - 2f'(ξ)(x-ξ) + 4M ≧ 0 for all x ≧ 0
i) D < 0
=> 4f'(ξ)^2 - 4N*4M < 0 => |f'(ξ)| < 2(MN)^0.5
ii) If D ≧ 0,
then the roots to N(x-ξ)^2 + 2f'(ξ)(x-ξ) + 4M = 0 are non-positive,
so are the roots to N(x-ξ)^2 - 2f'(ξ)(x-ξ) + 4M = 0.
By Vieta's formulae,
f'(ξ)/N + ξ ≦ 0
-f'(ξ)/N + ξ ≦ 0
Nξ^2 + f'(ξ)ξ + M ≧ 0 (will not be used)
Nξ^2 - f'(ξ)ξ + M ≧ 0 (will not be used)
=> ξ = 0, f'(0) = 0
Since ξ is arbitrary, f' is bounded by 2(MN)^0.5, which is a contradiction.
So f or f" must be unbounded.
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