Question: Let $ f $ be a real-valued function that differentiable on $ [a,+\infty) $. Suppose that $ f $ is monotonically decreasing, $ \lim_{x\to+\infty} f(x) = 0 $ and the integral $ \int_{a}^{+\infty} xf'(x) \mathrm{d}x $ converges, prove that the integral $ \int_{a}^{+\infty} f(x)\mathrm{d}x $ converges.
Could anyone solve this? Thanks!
Since $f$ is monotically decreasing to $f(+\infty) = 0$, $f$ is nonnegative. Hence, to prove that $\int_a^{+\infty} f$ converges, it is sufficient to prove that, $\int_a^b f$ is uniformly bounded above for every $b > a$.
Also, note that, since $f$ is monotically decreasing, $f' \leq 0$, so $\int_a^{+\infty} |x| |f'(x)| dx < +\infty$. In particular, this implies that $\int_a^{+\infty} |f'(x)| dx < +\infty$.
Let $b > a$. Then, integration by parts leads to $$\int_a^b f = b f(b) - a f(a) - \int_a^b x f'(x) dx$$
Concerning the first term, we write $f(b) = - \int_b^{+\infty} f'$, which is valid since $\int_a^{+\infty} |f'(x)| dx < +\infty$ and $f(+\infty) = 0$. Now $$ b f(b) = - b \int_b^{+\infty} f' \leq \int_b^{+\infty} |x f'(x)| dx$$
Eventually, this proves that $$ \int_a^b f \leq |af(a)| + 2 \int_a^{+\infty} | xf'(x)| dx$$ so $\int_a^{+\infty} f$ converges.
This is true even without assuming that $f$ is differentiable or even continuous.
Indeed, let $\mu=\mu_f$ be the Lebesgue--Stieltjes measure corresponding to a nonincreasing function $f$ on $[a,\infty)$ with $f(\infty-):=\lim_{x\to\infty}f(x)=0$, so that $$f(x+):=\lim_{y\downarrow x}f(y)=\int_{(x,\infty)}\mu(dt)$$ for all real $x\ge a$. The condition that $\int_a^\infty xf'(x)\,dx$ converges is now replaced by the more general condition $$\int_{(a,\infty)}\mu(dt)\,t<\infty.$$
Then $$\begin{aligned}\int_a^\infty dx\,f(x) &=\int_{(a,\infty)} dx\,f(x+) \\ &=\int_{(a,\infty)} dx\,\int_{(x,\infty)}\mu(dt) \\ &=\int_{(a,\infty)} \mu(dt)\int_{(a,t)}\,dx \\ &=\int_{(a,\infty)} \mu(dt)\,(t-a) \\ &=\int_{(a,\infty)} \mu(dt)\,t-af(a+)<\infty, \end{aligned}$$ as desired. The third inequality in the above display follows by Tonelli's theorem.
