Prove that $\int_{0}^{+\infty}f(x) dx < +\infty$ then $\displaystyle \lim_{x \to +\infty}xf(x) =0$.

Question

Suppose that $$ \int_{0}^{+\infty}f(x) \,\mathrm dx< +\infty, $$ the limit of $x\,f(x)$ when $x$ goes to $+\infty$ exists and $\forall x >0$, $f(x) \ge 0$. Can we conclude that $$ \lim_{x \to +\infty}xf(x)=0 ? $$ I guess that if $\displaystyle \lim_{x \to +\infty} x f(x) = c \ne 0$, then $U(x)$ will be equivalent to $f(x) = \frac{c}{x}$ when $x$ goes to $+\infty$. Then $f$ will not be integrable. It is a contradiction. Did I make any mistake?
Thank you for your time

Answer 1

Yes, the conjecture is true.

Suppose we have a function $f: \{ x\in \mathbb{R} : x \geq 0\} \to \mathbb{R}$, integrable on every finite interval, with

$$ \lim_{x \to +\infty} x f(x) = L \neq 0 $$

I don't require that $f(x) \geq 0$.

If $L<0$, we can replace $L$ with $-L$ and $f$ with $-f$ to get a case with $L>0$, so assume without loss of generality $L>0$.

Define $$ I(x) = \int_0^x f(t)\, dt $$

and statements about the improper integral $\int_0^{+\infty} f(x)\, dx$ actually say things about the behavior of $I(x)$ as $x \to +\infty$.

From the first limit, we have that there exists $N>0$ such that $x > N$ implies $|x f(x) - L| < \frac{L}{2}$. This then implies that $x f(x) > \frac{L}{2}$, and $f(x) > \frac{L}{2x}$. So when $x>N$,

$$ \begin{align*} I(x) &= \int_0^N f(t)\, dt + \int_N^x f(t)\, dt \\ I(x) &\geq \int_0^N f(t)\, dt + \int_N^x \frac{L}{2t}\, dt \\ I(x) &\geq \int_0^N f(t)\, dt + \frac{L}{2t}(\ln x - \ln N) \end{align*} $$

The first integral is a constant with respect to $x$, so $\lim_{x \to +\infty} I(x) = +\infty$. We also write this conclusion as

$$ \int_0^{+\infty} f(x)\, dx = +\infty $$