Proving $\lim\limits_{x \to \infty} xf(x)=0$ if $\int_{0}^{\infty}f(x) dx$ converges.

Solution 1:

I assume that $f$ is non-negative (this is justified; see the end of the post for the somewhat tedious details). If $\int_0^{\infty}f(t)\,\mathrm dt$ converges, then, for each $\varepsilon>0$, there exists some $x_0>0$ such that $y>x_0$ implies that $$\int_y^{\infty}f(t)\,\mathrm dt<\frac{\varepsilon}{2}.\tag{1}$$ I will show that if $x>2x_0$, then $xf(x)<\varepsilon$, which will prove that claim that $\lim_{x\to\infty}xf(x)=0$.

Suppose that $x>2x_0$. Then, one has that \begin{align*} xf(x)=&\,2\frac{x}{2}f(x)=2\left(x-\frac{x}{2}\right)f(x)=2f(x)\int_{x/2}^x\,\mathrm dt=2\int_{x/2}^xf(x)\,\mathrm dt\\ \underset{\spadesuit}\leq&\,2\int_{x/2}^xf(t)\,\mathrm dt\leq2\int_{x/2}^{\infty}f(t)\,\mathrm dt\underset{\heartsuit}<2\frac{\varepsilon}{2}=\varepsilon. \end{align*} In this chain of inequalities,

• $\spadesuit$ follows from the fact that $f$ is non-increasing, so that $f(x)\leq f(t)$ for all $t\in[x/2,x]$; and

• $\heartsuit$ follows from (1), given that $x/2>x_0$.

The convergence of $\int_0^{\infty} f(t)\,\mathrm dt$ and the monotonicity of $f$ necessarily imply that $f(t)\geq0$ for all $t\geq0$. To see this, suppose, for the sake of contradiction, that $f(t_0)<0$ for some $t_0\geq 0$. Let \begin{align*} K\equiv&-f(t_0)>0. \end{align*} Let $M>0$ be an arbitrarily large positive number and choose some positive number $H$ sufficiently large so that $$H\geq\frac{t_0f(0)+M}{K}\tag{2}.$$ Finally, suppose that $x>t_0+H$. Then, one has that \begin{align*} \int_0^xf(t)\,\mathrm dt=&\,\int_0^{t_0}f(t)\,\mathrm dt+\int_{t_0}^xf(t)\,\mathrm dt\underset{\clubsuit}{\leq}t_0f(0)+(x-t_0)f(t_0)=t_0 f(0)-(x-t_0)K\\ <&\,t_0 f(0)-HK\leq t_0 f(0)-[t_0 f(0)+M]=-M, \end{align*} where $\clubsuit$ follows again from the fact that $f$ is non-increasing. Since $M$ can be arbitrarily large, it follows that $$\lim_{x\to\infty}\int_0^xf(t)\,\mathrm dt=-\infty,$$ which contradicts the convergence of $\int_0^{\infty} f(t)\,\mathrm dt$.

As the counterexamples in the comments reveal, monotonicity is indispensable. Indeed, I used them at two crucial steps: at $\spadesuit$ and $\clubsuit$.

Solution 2:

Let $F(x) =\int_0^x f(t) dt $. We are given that $\lim_{x \to \infty} F(x)$ exists. Call this limit $L$.

Suppose it is not true that $\lim_{x \to \infty} xf(x) = 0 $. Then there is a $c > 0$ such that $x f(x) > c$ for arbitrarily large $x$.

Suppose $x_0 f(x_0) > c$ for $x_0$ large enough that $\int_{x_0}^{\infty} f(t) dt < d $. Choose an $x_1 > x_0$ such that $x_1 f(x_1) > c$. Since $f$ is monotone decreasing, $\int_{x_0}^{x_1} f(t) dt > (x_1-x_0)f(x_1) > (x_1-x_0)(c/x_1) = c(1-x_0/x_1) $. If we choose $x_1 > 2 x_0$, then $\int_{x_0}^{x_1} f(t) dt > c/2 $.

So we have $d >\int_{x_0}^{\infty} f(t) dt >\int_{x_0}^{x_1} f(t) dt >c/2 $.

But since $\lim_{x \to \infty} \int_{x}^{\infty} f(t) dt =0 $, we can choose $x_0$ large enough so that $d$ is arbitrarily small. In particular, we can choose $d < c/2$, which contradicts the inequality above.

Therefore, $\lim_{x \to \infty} xf(x) = 0 $.

Solution 3:

There are counterexamples otherwise: $f(x)$ is $n$ if $n<x<n+1/(n^3)$, $f(x)$ is 0 otherwise. The integral exists, but $f(x)$ diverges.

Solution 4:

without loss of generality one can assume that $f(0)=a$. Then $\int_0^\infty f(x)dx=k=\int_0^af^{-1}(y)dy$ with first improper integral being type I and second type II. Since the second integral also converge, the assertion follows from $0<xf(x)<\int_0^{f(x)}f^{-1}(y)dy\to 0.$ (a picture might be helpful to see the simple geometric idea)