Suppose $f :[1, \infty)\to \mathbb{R}$, $f(1)=0$, $f^{'}$ exists , is continuous and bounded and $f^{'}\in L^{2}[1,\infty)$. Let $g(x)=\frac{f(x)}{x}$. Show that $g\in L^{2}[1.\infty)$. Also give a counterexample where the same hypothesis holds but $f^{'}\in L^{1}[1, \infty)$ instead and the same $g$ as defined is not in $L^{1}[1,\infty)$.
Now I have $$\int |g(x)|^{2}= \int |f(x)|^{2}. \frac{1}{|x|^{2}}dx \leq \int |f(x)|^{2}dx $$ Now If I can show somehow that $f\in L^{2}$ using the conditions given then I am done but I can't show that. Also I can't find a counterexample for the next part.
Using the Fundamental Theorem of Calculus, we have $$ ||g||_2=\Big[\int_1^{\infty}\Big|\frac{1}{x}\int_1^{x}f^{\prime}(t)\;dt\Big|^2\;dx\Big]^{\frac{1}{2}}=\Big[\int_1^{\infty}\Big|\int_1^{\infty}f^{\prime}(t)\frac{1_{t\leq x}}{x}\;dt\Big|^2\;dx\Big]^{\frac{1}{2}}$$ Now setting $u=\frac{t-1}{x-1}$, we get $$ \Big[\int_1^{\infty}\Big|\int_0^{1}f^{\prime}(1+(x-1)u)\frac{x-1}{x}\;du\Big|^2\;dx\Big]^{\frac{1}{2}}$$ and then by Minkowski's integral inequality, $$ ||g||_2\leq \int_0^{1}\Big[\int_1^{\infty}|f^{\prime}(1+(x-1)u)|^2\Big(\frac{x-1}{x}\Big)^2\;dx\Big]^{\frac{1}{2}}\;du$$ and setting $v=1+(x-1)u$ and using $0\leq\frac{x-1}{x}\leq 1$, it follows that $$ ||g||_2\leq \int_0^1u^{-\frac{1}{2}}\Big[\int_1^{\infty}|f^{\prime}(v)|^2\;dv\Big]^{\frac{1}{2}}\;du\leq 2||f^{\prime}||_2$$
For the counterexample when $p=1$, I suggest something along the lines of $\frac{1}{\log x}$, modified so that $f$ is differentiable on $[1,\infty)$ with $f(1)=0$.