With $\lim \inf xf(x) = 0$ I need a sequence such that $f(x_n) \leq \frac{\epsilon}{x_n}$.

Question

Let continuous positive $f : [0, ∞) → \mathbb{R}$ satisfy $\int_{0}^{\infty}f(x)dx < \infty$. Show that for each ε > 0, there exists a sequence $\{x_n\}_n$ with $\lim_{n \rightarrow \infty} x_n = \infty$ so that $f(x_n) \leq \frac{\epsilon}{x_n}$.

Answer 1

Let $t_n$ be such that $t_n \rightarrow \infty $ with $ t_0=0,t_1 = 1, t_{n+1} = 2t_n$ for $n=1,2,3,....... $ then using mean value theorem we have $$\int_0^{\infty}f(x)dx = \sum_{n=0}^{\infty}\int_{t_n}^{t_{n+1}}f(x)dx= f(c_n)(t_{n+1}-t_n)=\sum_{n=0}^{\infty}t_nf(c_n) $$

where $t_n <c_n<t_{n+1}$

By hypothesis we have $\int_0^{\infty}f(x)dx < \infty$ hence we have $ \sum_{n=0}^{\infty}t_nf(c_n) < \infty$ hence given any $\epsilon >0$ we have an N such that $\forall n \geq N$ $t_nf(c_n) < \epsilon /2$

But $c_nf(c_n) < 2t_nf(c_n) < \epsilon$.

We relabel as $x_n = c_{N+n}$. Hence the result.

Answer 2

Claim: For any $\varepsilon>0,\exists(x_n)\to\infty$ such that $f(x_n)\le\frac{\varepsilon}{x_n}$

Proof:

• Assume, for a contradiction, that there is an $\varepsilon>0$ such that this is not possible.
• This would mean that ny sequence $(x_n)$ which satisfies $f(x_n)\le\frac{\varepsilon}{x_n}$ for all $n$ is bounded above, by, say, $M$.
• Then, for all $x>M$, $f(x)>\frac{\varepsilon}{x}$.
• But this contradicts the assumption that $f$ is integrable over $(0,\infty)$, as $$\int_M^{M'}f(x)\,dx \ge \int_M^{M'}\frac{\varepsilon}{x}\,dx=\varepsilon\log\frac{M'}{M}\to\infty \text{ as }M'\to\infty$$.
• Hence, no such $\varepsilon$ exists, and thus the claim holds.