7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – anon Aug 4 '11 at 19:23
  • $\begingroup$ So: the way to find a counterexample will be to arrange that $f$ decreases to zero, but $t f(t)$ does not converge. $\endgroup$ – GEdgar Aug 4 '11 at 19:28
  • $\begingroup$ Careful: $f\to 0$ isn't strong enough to ensure $f$ is $L_1$. $\endgroup$ – anon Aug 4 '11 at 19:37
  • $\begingroup$ After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?. $\endgroup$ – Clement C. Sep 3 '17 at 15:24
11
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.