If $f \in C([0, \infty))$ and $\lim_{x\rightarrow\infty} f(x)\int_0^x f^2(t) \ dt = 1$, then $f(x) \sim \left(\frac{1}{3x}\right)^{1/3}$

Question

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.

Answer 1

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: $$f(x) \sim \frac1{\sqrt[3]{3x}}$$