I'm having trouble with the following problem.
Suppose that $f$ is a uniformly continuous function on $(0,\infty)$ with derivative $f^\prime(x)$ satisfying,
$$ \int_0^\infty xf^2(x)dx < \infty, \;\;\;\;\;\;\int_0^\infty x^3(f^\prime(x))^2dx < \infty$$
Prove that $\lim\limits_{x\rightarrow\infty}xf(x) = 0$.
I tried to prove by contradiction. I assumed that there exists $\epsilon > 0$ and a sequence $x_n \rightarrow \infty$ such that $|x_nf(x_n)| > \epsilon$ for all $n$. Using uniform continuity, I showed that,
$$ \int_{x_n-\delta}^{x_n+\delta}xf(x)dx > \epsilon\delta $$
where $\delta$ is the modulus of continuity. Therefore, $$ \int_0^\infty xf(x)dx \geq \sum\limits_{n=1}^\infty\int_{x_n-\delta}^{x_n+\delta}xf(x)dx > \infty $$
My initial reasoning for using this approach was to hopefully use the Cauchy-Schwarz inequality to arrive at the contradiction,
$$ \int_0^\infty xf(x)dx \leq \left(\int_0^\infty xf^2(x)dx\right)^{1/2}\left(\int_0^\infty xdx\right)^{1/2} $$
However, obviously the second integral is not finite so this approach probably will not work. Is there a slight modification I can use?
Hint:
Use integration by parts
$$\int_a^b x f^2(x) \, dx = \left.\frac{1}{2}(x f(x))^2\right|_a^b - \int_a^b x^2 f(x) f'(x) \, dx$$
and $$\left|\int_a^b x^2 f(x) f'(x) \, dx \right| \leqslant \left(\int_a^b x f^2(x) \, dx\right)^{1/2}\left(\int_a^b x^3 (f'(x))^2 \, dx\right)^{1/2}$$