Prove that $ \lim\limits_{x\to \infty}xf(x)=0 $

Question

I just need a hint not a whole solution please.

Problem: Let $f ∈ L_1(0, ∞)$ be monotone. Prove that $$ \lim\limits_{x\to \infty}xf(x)=0 $$

Isn't $f:x \mapsto x$ a trivial example? – Kenny Lau Sep 03 '17 at 15:18 — Kenny Lau, Sep 03 '17 at 15:18
That $f$ is not $L_1$ on the specified domain. – Ittay Weiss Sep 03 '17 at 15:19 — Ittay Weiss, Sep 03 '17 at 15:19

Rab · Answer 1 · 2017-09-03T15:27:08.207

0

If $f\in L_1(a,\infty)$, then it follows that it is absolutely convergence.

Then $\forall \epsilon >0, \mathbb{exists~}X_0~\mathbb{such~that~for~any~2x>}X_0, by~Cauchy:~ \int_{\frac{x}{2}}^{x}f(t)dt<\epsilon$

Try finding an upper bound for the integral.

An extra hint:

$f$ is monotonic and absolutely convergent, therefore $\lim_{x\to \infty} f(x)=0$ and $\int_{\frac{x}{2}}^{x}f(t)dt\le xf(x)$

edited Sep 03 '17 at 15:27

answered Sep 03 '17 at 15:22

Rab

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