# Does $f$ monotone and $f\in L_{1}([a,\infty))$ imply $\lim_{t\to\infty} t f(t)=0$?

I want to show that if $f$ is non-increasing and $f\in L_{1}([a,\infty),m)$ where $m$ is Lebesgue measure then $\lim_{t\to\infty} t f(t)=0$. So far I've been able to show that $f\geq 0$ and that $\lim_{t\to\infty} f(t)=0$. Since monotone functions are differentiable a.e. I thought about using integration by parts but couldn't get anywhere with that. Any hints or suggestions would be greatly appreciated.

• If the limit exists, you can just say it's $x$ and use the delta-epsilon definition to bind $f$ inside $x/t\pm \epsilon$, then show it's not in $L_1$ unless $x=0$. I'm not sure how to demonstrate the limit exists thought. – anon Aug 4 '11 at 19:23
• So: the way to find a counterexample will be to arrange that $f$ decreases to zero, but $t f(t)$ does not converge. – GEdgar Aug 4 '11 at 19:28
• Careful: $f\to 0$ isn't strong enough to ensure $f$ is $L_1$. – anon Aug 4 '11 at 19:37
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## 2 Answers

Here's one way: $$2tf(2t)\leq2\int_t^{2t}f(x)\,dx\to0$$ as $t\to\infty$.

The result can be proved using integration by parts, as follows. Define the function $\alpha$ by $\alpha(x)=x$, $x \geq a$. Since $f$ is monotone, it is, in particular, of bounded variation. Since, moreover, $\alpha$ is continuous, the integration by parts theorem for Stieltjes integrals can be applied and gives $$\int_a^t {f(x)\,d\alpha (x)} = f(t)\alpha (t) - f(a)\alpha (a) - \int_a^t {\alpha (x)\,df(x)}$$ (see e.g. Example 6 here). Hence $$tf(t) = \int_a^t {f(x)\,dx} + \int_a^t {x\,df(x)} + af(a).$$ Assuming (without loss of generality) that $a > 0$, $\int_a^t {x\,df(x)}$ is non-increasing in $t$ (since $f$ is non-increasing). Since, moreover, $\lim _{t \to \infty } \int_a^t {f(x)\,dx} \in \mathbb{R}$ and $tf(t) \geq 0$ for any $t > a$, it thus follows that $\lim _{t \to \infty } \int_a^t {x\,df(x)} \in \mathbb{R}$ (for this limit cannot be $-\infty$). Therefore, letting $t \to \infty$ in the last equation, $$\exists \mathop {\lim }\limits_{t \to \infty } tf(t): = c \ge 0.$$ However, $c$ cannot be greater than $0$, for otherwise $f(t) \sim c/t$ as $t \to \infty$ yields a contradiction to the integrability of $f$. Hence $c=0$, and the result is proved.