= Decay of a nonnegative uniformly continuous integrable function
{title2=$f\geq0,\ \int_0^\infty f<\infty\Longrightarrow f(t)\to0$}
A <nonnegative function> that is a <uniformly continuous function> on the <interval> $[0,\infty)$ and has a finite <integral> tends to zero. Otherwise <uniform continuity> gives disjoint <intervals> of a fixed width on which the <function> exceeds a fixed positive value, contradicting <integrability>. This supplies a useful decay argument for <Lyapunov functions> with merely <continuous> <vector fields>.
To make the fixed width explicit, suppose some $\eta>0$ occurs as a lower bound $f(t_n)\geq\eta$ along a <sequence> with $t_n\to\infty$. Choose $\delta>0$ from <uniform continuity> for output tolerance $\eta/2$. On every <interval> $(t_n-\delta/2,t_n+\delta/2)$ lying in $[0,\infty)$, the <function> is greater than $\eta/2$. Extract a <subsequence> whose centers are more than $\delta$ apart. Each disjoint <interval> contributes at least $\eta\delta/2$ to the <integral>, so the first $N$ contribute at least $N\eta\delta/2$. This contradicts finiteness of the <integral>. The same input width at every center is the essential step; <continuity> alone would allow progressively narrower peaks.
Back to article page
