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.
Does $f$ monotone and $f\in L_{1}([a,\infty))$ imply $\lim_{t\to\infty} t f(t)=0$?
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real-analysis
measure-theory
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0If 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. – 2011-08-04
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0So: the way to find a counterexample will be to arrange that $f$ decreases to zero, but $t f(t)$ does not converge. – 2011-08-04
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0Careful: $f\to 0$ isn't strong enough to ensure $f$ is $L_1$. – 2011-08-04
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