TITLE: Integrable monotonic functions
QUESTION [7 upvotes]: Suppose that $f \in L^1(0,+\infty)$ is a monotonic function. Prove that $\lim_{x \to +\infty} x f(x)=0$.

REPLY [2 votes]: We can assume WLOG that $f$ is decreasing. Then $f$ is non-negative, otherwise, if $f(x_0)&lt0$ then $f(x)\leq f(x_0)&lt0$ for $x\geq x_0$ and $f$ wouldn't be integrable. We can write 
$$|xf(2x)|=\left|\int_x^{2x}f(2x)dt\right|=\int_x^{2x}f(2x)dt\leq \int_x^{2x}f(t)dt=\int_0^{2x}f(t)dt-\int_0^xf(t)dt.$$
Since $f$ is integrable, the RHS converges to $0$, so $\lim_{x\to 0}2xf(2x)=0$ and we are done.