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Let $f:[0, \infty) \rightarrow \mathbb{R}$ be a continuous function such that $$ \lim_{x\rightarrow\infty} f(x)\int_0^x f^2(t) \ dt = 1 $$ I want to conclude that $$f(x) \sim \left(\frac{1}{3x}\right)^{1/3}$$ i.e. $$\lim_{x \rightarrow \infty} \frac{f(x)}{\left(\frac{1}{3x}\right)^{1/3}} = 1$$

I tried to argue by contradiction, but my attempt was terribly unsuccessful and now seems hopeless:

Suppose that the conclusion is false. Then, for some $\varepsilon > 0$ we have that

$$ \left|\frac{f(x)}{\left(\frac{1}{3x}\right)^{1/3}}-1\right|\ge \varepsilon $$ for arbitrarily large $x$. Equivalently, $$ \frac{f(x)}{\left(\frac{1}{3x}\right)^{1/3}} \ge 1 + \varepsilon $$ and OR $$ \frac{f(x)}{\left(\frac{1}{3x}\right)^{1/3}} \le 1 - \varepsilon $$

Any ideas to get started would be appreciated.

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    $\begingroup$ Your assumption to seek a contradiction does not make sense because $1-\varepsilon \geq 1+\varepsilon$ cannot hold for any positive $\varepsilon$. $\endgroup$
    – Gary
    Commented Mar 30, 2021 at 8:47
  • $\begingroup$ @Gary Right! I was going for $\left| \frac{f(x)}{(1/3x)^{1/3}} - 1\right| \ge \epsilon$. This is correct though right? $\endgroup$
    – fwd
    Commented Mar 30, 2021 at 8:53
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    $\begingroup$ Saying $|l-1|>\epsilon$ is equivalent to "$l>1+\epsilon$ or $l<1-\epsilon$", not "and". Makes a big difference. $\endgroup$ Commented Mar 30, 2021 at 11:03
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    $\begingroup$ one of the formulas says $\frac{f(x)}{\left(\frac{1}{3x}\right)^{1/3}}<\frac{f(x)}{\left(\frac{1}{3x}\right)^{1/3}}$. $\endgroup$ Commented Mar 30, 2021 at 11:09
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    $\begingroup$ I apologize. I must have clicked that "automatically move to chat button". I also think this idea won't work now that I've seen the grave errors I committed. $\endgroup$
    – fwd
    Commented Mar 30, 2021 at 11:29

1 Answer 1

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Let $g(x) = \int_{0}^{x} f(t)^2\mathrm{dt}$, we have by squaring the condition on $f$: $$\lim g^{\prime}(x) g(x)^2 = 1 $$

so for every $\varepsilon > 0$, there exists $A > 0$, s.t. $$1 - \varepsilon < g'(x)g(x)^2 < 1 + \varepsilon$$ by integrating from $A$ to $x$, we deduce: $$(1-\varepsilon)x + \frac1{3}g(A)^3 < \frac1{3}g(x)^3 < (1+\varepsilon)x + \frac1{3}g(A)^3$$ so: $$g(x) \sim \sqrt[3]{3x}$$ From $f(x)g(x) \sim 1$ we deduce: $$f(x) \sim \frac1{\sqrt[3]{3x}}$$

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