# $\liminf$ of product of increasing and decreasing functions

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 $$0 such that $$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}$$.

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 $$\limsup_{t\to+\infty} tg(t)>0$$.