Exercise:
Let $f\in {{C}^{2}}\left[ 0,1 \right]\,\,\,\And \,\,f\left( 1 \right)=f\left( 0 \right)$
Show that, \(\int_{0}^{1}{{{\left( {{f}^{\left( 2 \right)}}\left( x \right) \right)}^{2}}dx\ge 120{{\left( \int_{0}^{1}{xf'\left( x \right)dx} \right)}^{2}}}\)
Solution: Take $u=f'\left( x \right)\,\,\And \,\,dv=x\Leftrightarrow du=f''\left( x \right)\,\,\,\,\And \,\,\,v=\frac{{{x}^{2}}}{2}$
So $\int_{0}^{1}{xf'\left( x \right)dx}=\left[ f'\left( x \right)\frac{{{x}^{2}}}{2} \right]_{0}^{1}-\int_{0}^{1}{\frac{{{x}^{2}}}{2}f''\left( x \right)dx}=\frac{f'\left( 1 \right)}{2}-\int_{0}^{1}{\frac{{{x}^{2}}}{2}f''\left( x \right)dx}$
by Fundamental theorem of calculus applied to $xf'\left( x \right)$ we get
$\int_{0}^{1}{\frac{d}{dx}\left( xf'\left( x \right) \right)dx=\int_{0}^{1}{d\left( xf'\left( x \right) \right)}}=\left[ xf'\left( x \right) \right]_{0}^{1}=f'\left( 1 \right)$ Also $\frac{d}{dx}\left( xf'\left( x \right) \right)=f'\left( x \right)+xf''\left( x \right)$
So $\int_{0}^{1}{\frac{d}{dx}\left( xf'\left( x \right) \right)dx}=\int_{0}^{1}{\left( f'\left( x \right)+xf''\left( x \right) \right)dx}=\int_{0}^{1}{f'\left( x \right)dx+\int_{0}^{1}{xf''\left( x \right)dx}}$
$=f\left( 1 \right)-f\left( 0 \right)+\int_{0}^{1}{xf''\left( x \right)dx}$ thus $f'\left( 1 \right)=\int_{0}^{1}{xf''\left( x \right)dx}$ (*)
Thus $\int_{0}^{1}{xf'\left( x \right)dx}=\int_{0}^{1}{\frac{x}{2}f''\left( x \right)dx}-\int_{0}^{1}{\frac{{{x}^{2}}}{2}f''\left( x \right)dx}=\int_{0}^{1}{\frac{x-{{x}^{2}}}{2}f''\left( x \right)dx}$ S.B.S
\[{{\left( \int_{0}^{1}{xf'\left( x \right)dx} \right)}^{2}}={{\left( \int_{0}^{1}{\frac{x-{{x}^{2}}}{2}f''\left( x \right)dx} \right)}^{2}}\underset{C.S}{\mathop{\le }}\,\int_{0}^{1}{{{\left( \frac{x-{{x}^{2}}}{2} \right)}^{2}}dx{{\int_{0}^{1}{\left( f''\left( x \right) \right)}}^{2}}dx}\]
But $\int_{0}^{1}{\frac{{{x}^{2}}+{{x}^{4}}-2{{x}^{3}}}{4}dx}=\frac{1}{4}\left( \frac{{{x}^{3}}}{3}+\frac{{{x}^{5}}}{5}-\frac{{{x}^{4}}}{2} \right)_{0}^{1}=\frac{1}{4}\left( \frac{1}{3}+\frac{1}{5}-\frac{1}{2} \right)=\frac{1}{120}$
Thus ${{\left( \int_{0}^{1}{xf'\left( x \right)dx} \right)}^{2}}\le \frac{1}{120}\int_{0}^{1}{{{\left( f''\left( x \right) \right)}^{2}}dx}$
Therefore $\int_{0}^{1}{{{\left( {{f}^{\left( 2 \right)}}\left( x \right) \right)}^{2}}dx\ge 120{{\left( \int_{0}^{1}{xf'\left( x \right)dx} \right)}^{2}}}$
So $\int_{0}^{1}{xf'\left( x \right)dx}=\left[ f'\left( x \right)\frac{{{x}^{2}}}{2} \right]_{0}^{1}-\int_{0}^{1}{\frac{{{x}^{2}}}{2}f''\left( x \right)dx}=\frac{f'\left( 1 \right)}{2}-\int_{0}^{1}{\frac{{{x}^{2}}}{2}f''\left( x \right)dx}$
by Fundamental theorem of calculus applied to $xf'\left( x \right)$ we get
$\int_{0}^{1}{\frac{d}{dx}\left( xf'\left( x \right) \right)dx=\int_{0}^{1}{d\left( xf'\left( x \right) \right)}}=\left[ xf'\left( x \right) \right]_{0}^{1}=f'\left( 1 \right)$ Also $\frac{d}{dx}\left( xf'\left( x \right) \right)=f'\left( x \right)+xf''\left( x \right)$
So $\int_{0}^{1}{\frac{d}{dx}\left( xf'\left( x \right) \right)dx}=\int_{0}^{1}{\left( f'\left( x \right)+xf''\left( x \right) \right)dx}=\int_{0}^{1}{f'\left( x \right)dx+\int_{0}^{1}{xf''\left( x \right)dx}}$
$=f\left( 1 \right)-f\left( 0 \right)+\int_{0}^{1}{xf''\left( x \right)dx}$ thus $f'\left( 1 \right)=\int_{0}^{1}{xf''\left( x \right)dx}$ (*)
Thus $\int_{0}^{1}{xf'\left( x \right)dx}=\int_{0}^{1}{\frac{x}{2}f''\left( x \right)dx}-\int_{0}^{1}{\frac{{{x}^{2}}}{2}f''\left( x \right)dx}=\int_{0}^{1}{\frac{x-{{x}^{2}}}{2}f''\left( x \right)dx}$ S.B.S
\[{{\left( \int_{0}^{1}{xf'\left( x \right)dx} \right)}^{2}}={{\left( \int_{0}^{1}{\frac{x-{{x}^{2}}}{2}f''\left( x \right)dx} \right)}^{2}}\underset{C.S}{\mathop{\le }}\,\int_{0}^{1}{{{\left( \frac{x-{{x}^{2}}}{2} \right)}^{2}}dx{{\int_{0}^{1}{\left( f''\left( x \right) \right)}}^{2}}dx}\]
But $\int_{0}^{1}{\frac{{{x}^{2}}+{{x}^{4}}-2{{x}^{3}}}{4}dx}=\frac{1}{4}\left( \frac{{{x}^{3}}}{3}+\frac{{{x}^{5}}}{5}-\frac{{{x}^{4}}}{2} \right)_{0}^{1}=\frac{1}{4}\left( \frac{1}{3}+\frac{1}{5}-\frac{1}{2} \right)=\frac{1}{120}$
Thus ${{\left( \int_{0}^{1}{xf'\left( x \right)dx} \right)}^{2}}\le \frac{1}{120}\int_{0}^{1}{{{\left( f''\left( x \right) \right)}^{2}}dx}$
Therefore $\int_{0}^{1}{{{\left( {{f}^{\left( 2 \right)}}\left( x \right) \right)}^{2}}dx\ge 120{{\left( \int_{0}^{1}{xf'\left( x \right)dx} \right)}^{2}}}$
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