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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?

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

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  • $\begingroup$ Ah, clear and concise. Thank you very much! $\endgroup$ – JohnDoe1234 Feb 26 '18 at 5:45
  • $\begingroup$ @JohnDoe1234: Nice question. Good try. $\endgroup$ – RRL Feb 26 '18 at 5:46
  • $\begingroup$ 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. $\endgroup$ – JohnDoe1234 Feb 26 '18 at 5:59
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    $\begingroup$ 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 $0$? $\endgroup$ – RRL Feb 26 '18 at 8:59
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    $\begingroup$ If $x^2f^2(x) \to L \neq 0$ as $x \to \infty$ then $xf^2(x) \sim L/x$ would not be integrable. $\endgroup$ – RRL Feb 26 '18 at 9:11

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