I am a PhD student and during my research I was presented to the claim that


**For a positive definite function $f:\mathbb{R}\to \mathbb{R}$ continuous in $0$, with $0$ a stable point at $t=0$ for $x$, one has
$${\lim\inf}\_{t\to\infty} f(x(t))=0\Longrightarrow {\lim\inf}\_{t\to\infty} \|x(t)\|=0.$$**


In this context,



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> We say that a function $f:\mathbb{R}\to \mathbb{R}$ is positive definite if $f(x)\geq 0$ and $f(x)=0\iff x=0.$
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> We say that $p$ is a stable point at $t\_0$ if, for any neighborhood $H$ of $p$, there's a neighboorhood $V$ of $p$ s.t. if $x(t\_0)\in V$, then $x(t)\in H$ for all $t>t\_0.$
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I thought this is not true without more hypothesis. Could anyone have an ideia to (dis)prove this? If this is not true, any ideia about extra hypothesis? Thank you.



$\lim\inf_{t\to \infty} f(x(t))=0\Longrightarrow \lim\inf_{t\to\infty} \|x(t)\|=0$
Here is my take on your question. Let $f:\mathbb R\to\mathbb R\_+$ be a function such that $\{x:f(x)=0\}=\{0\}$ and $f$ is continuous at zero. Let $\phi:\mathbb R\to\mathbb R$ be locally Lipschitz continuous, so that the differential equation $x'=\phi(x)$ admits a unique local solution for any given initial condition $x\_0$. We denote the solution by $X:(t,x\_0)\mapsto X\_t(x\_0)$. Suppose also that $0$ is stable for this equation, in the sense that there exists a neighbourhood of zero from which all solutions are defined for all times, and $\lim\_{r\to0}\sup\_{|x\_0|\leq r}\sup\_{t\geq0}|X\_t(x\_0)|=0$. It is equivalent to your notion of stability, at least if you consider solutions to ODEs.


Consider the following property.
$$ (P)=(P)\_{f,\phi}:\text{For all }x\_0\in\mathbb R\text{ such that }\liminf\_{t\to\infty}|f(X\_t(x\_0))|=0,\text{ we have }\liminf\_{t\to\infty}|X\_t(x\_0)|=0. $$



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> **Question 1.**
> Do we always have $(P)$?
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Clearly this is not the case. Take any such $f$ that goes to zero at infinity (say $f:x\mapsto x^2/(1+x^4)$) and any such $\phi$ that admits solutions going to infinity (say $\phi:x\mapsto (x^3-x)/(1+x^4)$). Then any solution going to infinity satisfies the first but not the second condition (with the given $f$ and $\phi$, the set of $x\_0$ satisfying the first condition is $\mathbb R\setminus\{-1,1\}$, while for the the second it is $(-1,1)$).



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> **Question 2.**
> Does $(P)$ hold if we suppose that $f$ is [radially unbounded](https://en.wikipedia.org/wiki/Radially_unbounded_function)?
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No. It would be obvious that it fails if $f$ were allowed to have another point $x$ such that $f(x)=0$. But it also works with something like
$$f:x\mapsto \begin{cases}x^2(x-2)^2 & \text{for }x\neq1,\\1&\text{for }x=1.\end{cases}$$
Then just take $\phi:x\mapsto-x(x-1)(x-2)/(1+x^4)$.



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> **Question 3.**
> Does $(P)$ hold for $f$ lower semicontinuous and eventually bounded below\*?
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\* Say that we mean $\lim\_{R\to\infty}\inf\_{|x|\geq R}f(x)>0$, whereas radially bounded meant that this limit was $\infty$.


Yes. If the limit inferior of $t\mapsto f(X\_t(x\_0))$ is zero, then by definition there exists a sequence of times $t\_n\to\infty$ such that $f(X\_{t\_n}(x\_0))$ converges to zero. By the boundedness condition, there exists some $R,\varepsilon>0$ such that $|x|\geq R$ implies $f(x)>\varepsilon$; in particular we eventually have $|X\_{t\_n}(x\_0)|<R$. By compactness, we can extract a subsequence $n\mapsto\sigma(n)$ for which $X\_{t\_{\sigma(n)}}(x\_0)$ converges, say to some limit $x\_\infty\in[-R,R]$. Using the lower semicontinuity, we know that $\lim\_{n\to\infty}f(X\_{t\_{\sigma(n)}}(x\_0))\geq f(x\_\infty)$, so $f(x\_\infty)=0$ and $x\_\infty=0$. This means that $\liminf\_{t\to\infty}|X\_t(x\_0)|\leq\liminf\_{n\to\infty}|X\_{t\_{\sigma(n)}}(x\_0)|=|x\_\infty|=0$ as expected (and in fact we have $\liminf\_{t\to\infty}|X\_t(x\_0)|=0$ by stability).



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> **Question 4.**
> Does $(P)$ hold if $\phi$ admits no solution going to infinity?
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No; see question 2.



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> **Question 5.**
> Does $(P)$ hold if $\phi$ admits only one fixed point?
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Yes. 0 has to be a fixed point to be stable, so we can easily see by the intermediate value theorem that $\phi$ is non zero and has constant sign on $(-\infty,0)$ and $(0,+\infty)$. By stability, we see that its signs there are respectively positive and negative, so in fact we always have $\liminf\_{t\to\infty}|x(t)|=\lim\_{t\to\infty}|x(t)|=0$.


