$\lim\inf_{t\to \infty} f(x(t))=0\Longrightarrow \lim\inf_{t\to\infty} \|x(t)\|=0$

Question

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,

We say that a function $f:\mathbb{R}\to \mathbb{R}$ is positive definite if $f(x)\geq 0$ and $f(x)=0\iff x=0.$

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 $t>t_0.$

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.

Answer 1

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. $$ (P)=(P)_{f,\phi}:\text{For all }x_0\in\mathbb R\text{ such that }\liminf_{t\to\infty}|f(X_t(x_0))|=0,\text{ we have }\liminf_{t\to\infty}|X_t(x_0)|=0. $$

Question 1. Do we always have $(P)$?

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 $(-1,1)$).

Question 2. Does $(P)$ hold if we suppose that $f$ is radially unbounded?

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 $\phi:x\mapsto-x(x-1)(x-2)/(1+x^4)$.

Question 3. Does $(P)$ hold for $f$ lower semicontinuous and eventually bounded below*?

* 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).

Question 4. Does $(P)$ hold if $\phi$ admits no solution going to infinity?

No; see question 2.

Question 5. Does $(P)$ hold if $\phi$ admits only one fixed point?

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 $\liminf_{t\to\infty}|x(t)|=\lim_{t\to\infty}|x(t)|=0$.

Answer 2

Suppose that $\liminf_{t\rightarrow\infty}\|x(t)\|\neq0$, so there exists a neighborhood $H$ of 0 and an increasing sequence $(t_n)$, tending to infinity, such that $x(t_n)\not\in H$. Let $V$ be a neighborhood that the stability suggests for $H$. By the definition, $x(t)\not\in V\cap f^{-1}(V)$, $t\geq0$. Hence, $f(x(t))\not\in V$ for $t\geq0$, whence $\liminf_{t\rightarrow\infty}f(x(t))\neq0$.