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--- 若$f''\in C[0,1],\,f(0)=f(1)=f'(0)=0,\,f'(1)=1,\,$则$\displaystyle{\int_0^1 (f'')^2 \ge 4}$ (http://mathchina.elinkage.net/cgi-bin/topic.cgi?forum=1&topic=709)


-- 作者: elim
-- 发布时间: 2016/08/23 02:59am

[b]题:[/b]若$f''\in C[0,1],\,f(0)=f(1)=f'(0)=0,\,f'(1)=1,\;$则$\displaystyle{\int_0^1 (f'')^2 \ge 4}\;$
$\qquad$找出$\displaystyle{\,\int_0^1 (f'')^2 = 4}\,$的条件. [url=http://www.mathchina.com/bbs/forum.php?mod=viewthread&tid=41269]src[/url]


-- 作者: elim
-- 发布时间: 2016/08/23 03:27am

[b]题:[/b]若$f''\in C,\,f(0)=f(1)=f'(0)=0,\,f'(1)=1,\;$则$\displaystyle{\int_0^1 (f'')^2 \ge 4}\;$
$\qquad$找出$\displaystyle{\int_0^1 (f'')^2 = 4}\,$的条件.
[b]解:[/b]令$h(x)=f(x)-x^2(x-1),\,$则$\underset{\,}{\;}(*)\;\small h''\in C[0,1],\,h(0)=h(1)=h'(0)=h'(1)=0$,
$\qquad (f'(x))^2 = (2(3x-1)+h''(x))^2=4(3x-1)^2+4(3x-1)h''(x)+(h''(x))^2\underset{\,}{\,}$
$\qquad\because\;\;{\small\displaystyle{\int_0^1}} 4(3x-1)h''(x)dx = 4[(3x-1)h'(x)-3h(x)]\big|_0^1=0,$
$\small\displaystyle{\qquad\therefore\;\;\int_0^1 (f'')^2 = 4+\int_0^1(h'')^2\ge 4.\;\;\big({\scriptsize\int_0^1 (f'')^2} = 4\iff (h''=0)\wedge (*)\iff (h=0)\big)}.\;\square$


-- 作者: elim
-- 发布时间: 2016/08/23 10:11am

[b]题:[/b]若$f''\in C,\,f(0)=f(1)=f'(0)=0,\,f'(1)=1,\;$则$\displaystyle{\int_0^1 (f'')^2 \ge 4}\;$
$\qquad$找出$\displaystyle{\int_0^1 (f'')^2 = 4}\,$的条件.
[b]解:[/b]考虑$\,\dfrac{d^2}{dx^2}(x^3-x^2)=6x-2,\;\dfrac{d^2}{dx^2}=f''\,$的关系. 由Schwarz不等式
$\qquad\small\displaystyle{\big(\int_0^1 (6x-2)f''(x)dx\big)^2\le \int_0^1(6x-2)^2dx\int_0^1(f''(x))dx}$ 及
$\qquad\small\displaystyle{\int_0^1(6x-1)f''(x)dx=[(6x-2)f'(x)-6f(x)]\big|_0^1 = 4 = \int_0^1(6x-2)^2dx}\;$即得
$\qquad\small\displaystyle{\int_0^1(f''(x))^2dx\ge 4.\;\;}$等式成立$\scriptsize\iff(f''=\lambda(6x-2))\wedge(f(0)=f(1)=f'(0)=0,\,f'(1)=1)$
$\qquad\iff (f=x^2(x-1))$


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