It's a problem in Bass's real analysis book. suppose $f:[1,\infty] \rightarrow R, f(1)=0, f'exist$ and is continuous and bounded and $f' \in L^2([1, \infty])$, let $g(x)=f(x)/x,$ show $g \in L^2([1, \infty])$ .
Progress
By the assumption, I know $f$ is absolutely continous, then I get the relationship between $f$ and $f'$ by property of AC, $f(x)=f(1)+ \int_{1}^{x} f'(t)dt=\int_{1}^{x} f'(t)dt$,then by H$\ddot{o}$lder inequality, $|f(x)|\leq (\int_{1}^{x} |f'(t)|^{2}dt)^{1/2}*(\int_{1}^{x} 1^{2}dt)^{1/2} $. Then the first part is bounded M, if the second part is finite, I can get the relationship between $f ~and~ f'$. But now the second part$(\int_{1}^{x} 1^{2}dt)^{1/2}$ which is not finite, then $\int |g|^2dx \leq \int M^2*(x-1)/x^2=M^2\int (1/x-1/x^2)dx $, $1/x $is not in $L^2$ and we can not get the result.what can I do next? thanks
From $$ f(x) = \int_{1}^{x}f'(y)\,dy $$ it follows that: $$ |f(x)| \leq \int_{1}^{x}|f'(y)|\,dy \leq \sqrt{\int_{1}^{x}f'(y)^2\,dy\cdot\int_{1}^{x}1\,dy}\leq \sqrt{x-1}\,\|f'\|_2\ll\sqrt{x}$$ hence: $$ I_N=\int_{1}^{N}\frac{f(x)^2}{x^2}\,dx=\left.-\frac{f(x)^2}{x}\right|_{1}^{N}+2\int_{1}^{N}\frac{f(x)\,f'(x)}{x}\,dx $$ always by the Cauchy-Schwarz inequality, fulfills the bound: $$ I_N \leq 2\|f'\|_2+2\|f'\|_2 \sqrt{I_N}\leq 2\|f'\|_2 \max(1,\sqrt{I_N}) $$ and assuming $I_N\geq 1$ it follows that: $$ I_N \leq 4\|f'\|_2^2 $$ from which $g\in L^2$ follows.