请教两个积分不等式
设$f(x)$在$[a,b]$上连续可导，且$f(a) = 0$，求证1.\[(\int_{a}^{b}f(x)dx)^2 \le \frac{(b-a)^3}{3}\int_{a}^{b}(f'(x))^2dx\]
2.\[\int_{a}^{b}\abs{f(x)f'(x)}dx \le \frac{b-a}{2}\int_{a}^{b}(f'(x))^2dx\] [b]回复 [url=http://kuing.orzweb.net/redirect.php?goto=findpost&pid=19543&ptid=4340]1#[/url] [i]abababa[/i] [/b]
1、
\[(\int_a^bf(x)dx)^2=(\int_a^b\int_a^xf'(t)dtdx)^2=(\int_b^a(b-t)f'(t)dt)^2\]
由柯西不等式
\[\le \int_a^b(b-t)^2dt\int_b^a(f'(t))^2dt=\frac{(b-a)^3}{3}\int_b^a(f'(t))^2dt\] [b]回复 [url=http://kuing.orzweb.net/redirect.php?goto=findpost&pid=19543&ptid=4340]1#[/url] [i]abababa[/i] [/b]
2、
\[\int_a^b|f(x)f'(x)|dx=\int_a^b|\int_a^xf'(t)f'(x)dt|dx\le\int_a^b\int_a^x|f'(t)f'(x)|dtdx\]
\[=\int_a^b\int_t^b|f'(t)f'(x)|dxdt=\int_a^b\int_x^b|f'(t)f'(x)|dtdx\]
因此有
\[2\int_a^b\int_a^x|f'(t)f'(x)|dtdx=\int_a^b\int_a^x|f'(t)f'(x)|dtdx+\int_a^b\int_x^b|f'(t)f'(x)|dtdx\]
\[=(\int_a^b|f'(x)|dx)^2\le(b-a)\int_a^bf'(x)^2dx\]
因此
\[\int_a^b|f(x)f'(x)|dx\le\frac{b-a}{2}\int_a^bf'(x)^2dx\] [b]回复 [url=http://kuing.orzweb.net/redirect.php?goto=findpost&pid=19556&ptid=4340]3#[/url] [i]战巡[/i] [/b]
谢谢。第二题我通过
\[\int\abs{f'(x)}\int_{a}^{x}\abs{f'(t)}dtdx=\frac{1}{2}(\int_{a}^{x}\abs{f'(t)}dt)^2\]
然后得到
\[\int_{a}^{b}\abs{f(x)f'(x)}dx \le\int_{a}^{b}\abs{f'(x)}\int_{a}^{x}\abs{f'(t)}dtdx = \frac{1}{2}\left[(\int_{a}^{x}\abs{f'(t)}dt)^2\right]\mid_{a}^{b}=\frac{1}{2}(\int_{a}^{b}\abs{f'(t)}dt)^2\]
就是最后的柯西没想到，第一题也是，还是因为我不熟悉啊。
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