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Suppose that $f\in C^1([0,\infty),\mathbb{R})$,and $F(x)=\max_{x\leq y\leq 2x}|f(y)|$,then show that $$ \int_{0}^{\infty}F(x)dx\leq \int_{0}^{\infty}|f(x)|dx+\int_{0}^{\infty}x|f'(x)|dx $$

EDIT

Thanks to richard's counterexample,I realised now that the term $\dfrac{1}{x}$ should be replaced by $x$.

One try is to write $$ f(x)=\frac{1}{x}\int_{x}^{2x}(f(x)-f(y))dy+\frac{1}{x}\int_{x}^{2x}f(y)dy\\ =\frac{1}{x}\int_{x}^{2x}f(y)dy-\frac{1}{x}\int_{x}^{2x}\int_{x}^{y}f'(z)dz $$ then,we have the following $$ \dfrac{1}{x}\int_{x}^{2x}f(y)dy=2f(2x)-f(x)+\frac{1}{x}\int_{x}^{2x}yf'(y)dy $$and fact that $\left \vert \dfrac{1}{x} \displaystyle \int_{x}^{2x}f(y)dy \right \vert\leq F(x) $

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The inequality seems incorrect. Consider this example: $f(x)=\dfrac{1+\cos(2n\pi x)}{n}$, when $|x-n|\le\frac{1}{2n}$ for some $n\ge 1$; $f(x)=0$, otherwise. Then both integrals on the right hand side converge but the integral in the left hand side diverges. –  23rd Oct 31 '12 at 13:50
    
Dear richard.in your example,the integral in the LHS also converge,the function vanishes for large x,doesn't it ? –  sun Nov 1 '12 at 7:27
    
No. When $x\in[n,n+1]$ for some $n\ge 1$, $x\le n+1\le 2x$, so by definition, $F(x)\ge f(n+1)=\dfrac{2}{n+1}$. Therefore, $\int_0^\infty F(x)dx\ge\sum_{n=1}^\infty\dfrac{2}{n+1}=+\infty$. –  23rd Nov 1 '12 at 7:50
    
I think you mean for every n,but then the RHS diverges also. –  sun Nov 1 '12 at 7:59
    
No. Please check my construction carefully. Then you will find $\int_0^\infty|f(x)|d x\le \sum_{n=1}^\infty\dfrac{2}{n^2}<\infty$ and $\int_0^\infty\dfrac{|f'(x)|}{x}d x\le \sum_{n=1}^\infty\dfrac{8}{n(2n-1)}<\infty$ –  23rd Nov 1 '12 at 8:09
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