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Let $g$ be a non-negative function on $[0,+\infty[$ such that $g$ is $C^1$, non-increasing, positive, and $\lim_{t\to+\infty} g(t)=0$. Assume that $\limsup_{t\to+\infty} tg(t)>0$, then can it happen that $\liminf_{t\to+\infty} tg(t)=0$? When $g$ is not monotonous we can easily find an example where $\liminf_{t\to+\infty} tg(t)=0$ but can this really happen when $g$ is non-increasing?

What I tried:

  • Building a counter-example, but either the $\liminf$ was not $0$, e.g. $g(t)=\frac{1}{t+\sin(t)}$, there $\liminf_{t\to+\infty} tg(t)=1$, or I could not make $g$ non-increasing, e.g.: $g(t)=\frac{\sin(t)}{t}$.
  • Defining $h$ such that $g(t) = h(t)/t$ and then the conditions on $h$ are $$\begin{cases} h(t)\geq 0\\ \limsup_{t\to+\infty} h(t)>0 \\ \lim_{t\to+\infty} \frac{h(t)}{t} = 0 \\ \left(\frac{h(t)}{t}\right)' \leq 0 \iff h'(t)\leq \frac{h(t)}{t} \end{cases}$$ With these notations the question is: is it possible that $\liminf_{t\to+\infty} h(t)=0$?
  • Using the above notations, assume that $\liminf_{t\to+\infty} h(t)=0$ and denote $\limsup_{t\to+\infty} h(t)=C>0$. Then for any $\epsilon>0$ small enough, there exists $0<t_0\leq t_1$ such that $h(t_0)=\epsilon$ and $h(t_1)=C-\epsilon$. But then $C-2\epsilon = h(t_1)-h(t_0)= \int_{t_0}^{t_1} h'(s)\mathrm{d}s \leq \int_{t_0}^{t_1} \frac{h(s)}{s}\mathrm{d}s\leq (t_1-t_0)\frac{h(t_0)}{t_0}$ (because $h(s)/s$ is decreasing). Therefore $$t_1\geq t_0 + t_0\left(\frac{C}{\epsilon}-2\right) \xrightarrow[\epsilon\to 0]{} + \infty.$$ This tells us that the closer $h$ gets to zero at $t_0$, the longer time $t_1$ we need to wait until $h$ gets close to $C$ again, but I still cannot deduce that $\liminf_{t\to +\infty}\neq 0$.
  • I was close to finding a counter example with $h(t) = 1+\sin(\log(1+t))$ since $h(t)\geq 0$, $\liminf_{t\to+\infty} h(t)=0$, $\limsup_{t\to+\infty} h(t) = 2$. However $h'(t) = \frac{1}{t+1}\cos(\log(t+1))$ can sometimes be larger than $\frac{h(t)}{t} = \frac{1+\sin(\log(1+t))}{t}$.
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1 Answer 1

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A counter-example can be constructed by setting the value of $g$ (or $h$) along specific sequences of points $(t_n)_{n\in\mathbb{N}}$ and by interpolating between these points.

Set $t_0=1$ and let the sequence $(t_n)_{n\in\mathbb{N}}$ defined for all $n\in\mathbb N$ by $\begin{cases} t_{2n+1} = t_{2n}+1 \\ t_{2n+2} = nt_{2n+1} \end{cases}$.

Then consider a function $g:[0,+\infty[\to \mathbb{R}$ such that $\forall n\in\mathbb N$, $$\begin{cases} g(t_{2n+1}) = \frac{1}{nt_{2n+1}} \\ g(t_{2n+2}) = g(t_{2n+1}). \end{cases}$$ The sequence $(g(t_n))_{n\in\mathbb{N}}$ is non-increasing so we can construct a positive non-increasing and smooth function $g$ on $[0,+\infty[$ by smoothly interpolating between the points $(t_n,g(t_n))_{n\in\mathbb{N}}$.

Then observe that $\lim_{t\to+\infty} g(t)=0$ and by construction $t_{2n+2}g(t_{2n+2}) = t_{2n+2}g(t_{2n+1}) = \frac{t_{2n+2}}{nt_{2n+1}}=1$ and $t_{2n+1}g(t_{2n+1}) = \frac{1}{n}$, therefore: $$ \begin{cases} \limsup_{t\to+\infty} tg(t)\geq \limsup_{n\to+\infty} t_{2n}g(t_{2n})=1 \\ \liminf_{t\to+\infty} tg(t)\leq \liminf_{n\to+\infty} t_{2n+1}g(t_{2n+1})=\liminf_{n\to+\infty} \frac{1}{n} = 0. \end{cases}$$

In conclusion, there exists positive non-increasing smooth functions $g$ such that $\liminf_{t\to+\infty} tg(t)=0$ but $\limsup_{t\to+\infty} tg(t)>0$.

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