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

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.

• Your assumption to seek a contradiction does not make sense because $1-\varepsilon \geq 1+\varepsilon$ cannot hold for any positive $\varepsilon$.
– Gary
Commented Mar 30, 2021 at 8:47
• @Gary Right! I was going for $\left| \frac{f(x)}{(1/3x)^{1/3}} - 1\right| \ge \epsilon$. This is correct though right?
– fwd
Commented Mar 30, 2021 at 8:53
• Saying $|l-1|>\epsilon$ is equivalent to "$l>1+\epsilon$ or $l<1-\epsilon$", not "and". Makes a big difference. Commented Mar 30, 2021 at 11:03
• one of the formulas says $\frac{f(x)}{\left(\frac{1}{3x}\right)^{1/3}}<\frac{f(x)}{\left(\frac{1}{3x}\right)^{1/3}}$. Commented Mar 30, 2021 at 11:09
• 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.
– fwd
Commented Mar 30, 2021 at 11:29

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