# Integrability condition implies boundedness and limit is zero

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 $$\left|\int_a^b x^2 f(x) f'(x) \, dx \right| \leqslant \left(\int_a^b x f^2(x) \, dx\right)^{1/2}\left(\int_a^b x^3 (f'(x))^2 \, dx\right)^{1/2}$$

• Ah, clear and concise. Thank you very much! – JohnDoe1234 Feb 26 '18 at 5:45
• @JohnDoe1234: Nice question. Good try. – RRL Feb 26 '18 at 5:46
• Actually, I'm not clear as to why the the above will force the limit to be 0? We know that it should be finite from the above. – JohnDoe1234 Feb 26 '18 at 5:59
• Since we know $xf^2$ is integrable and show $xff'$ is integrable the limit of $x^2|f(x)|^2$ as $x \to \infty$ exists. Could it not be $0$? – RRL Feb 26 '18 at 8:59
• If $x^2f^2(x) \to L \neq 0$ as $x \to \infty$ then $xf^2(x) \sim L/x$ would not be integrable. – RRL Feb 26 '18 at 9:11