Finite integral and limit

Question

Let $L:\mathbb{R} \to [0,\infty)$ a decreasing function such that

$\int_{0}^{\infty}x^{n-1}L(x)dx < \infty$

Prove that $\lim_{x \to \infty}x^{n}L(x) = 0$

Can anybody help me?

Answer 1

Suppose $x^{n}L(x)$ does not tend to $0$. Then there is a sequence $x_j$ increasing to $\infty$ and a $\delta >0$ such that $x_j^{n}L(x_j)\geq \delta$ for all $j$. We can find a subsequence $(y_k)$ of $(x_j)$ such that $y_{k+1}^{n} >2y_k^{n}$. [Use induction for this]. Now $\int_0^{\infty} x^{n-1} L(x)\, dx\geq \sum_k \int_{y_k}^{y_{k+1}} x^{n-1} L(x)\, dx \geq \sum_k L(y_{k+1})\frac {y_{k+1}^{n}-y_k^{n}} n \geq \delta \sum_k \frac {y_{k+1}^{n}-y_k^{n}} {ny_{k+1}^{n}} =\sum_k \frac {1-\frac {y_k^{n}} {y_{k+1}^{n}}} n=\infty$ because $y_{k+1}^{n} >2y_k^{n}$.