Let $\Omega\subset\mathbb{R}^N$ and $F:\Omega\times\mathbb{R}\to\mathbb{R}$ be a function and consider the expression $$ \liminf_{| u|\to+\infty} \frac{F(x, u(x))}{|u|^p +| u|^q} <0,$$ for some $p, q >1$. I am trying to "explain in words" this expression in a easy-to-understand way.

Could anyone please suggest me something?

I was thinking something like

"No matter how far you look toward infinity, there will be $|u|$ such that $F(x, u)$ is a negative quantity"

but I am not totally convinced.

Could anyone please help? Thank you in advance!


Your sentence is weaker than what the limit is saying. As an illustration, $-1/x$ on $x \in (0,\infty)$ has all its values negative, but its limit infimum is zero.

It is generally useful to speak in terms of "tail"s when talking about limits infimum and supremum. Here, a "tail" is: for each choice of $u_0$, restrict the limit as $|u| \rightarrow \infty$ to $|u| > u_0$. Taking the limit on the "complement of a bounded ball" might be more descriptive than "tail", depending on your setting.

For the limit infimum to be negative...

There is a quantity, $m < 0$ such that for each tail of the limit there are infinitely many choices of $(x, u(x))$ in that tail having $F(x, u(x)) < m$.

Note that the limit requires you to sort by magnitude(s) of $u$, but there is no restriction on the choice of $x$ among those $u(x)$ having the same magnitude. So $x$ is part of your choice for points giving the various infima.

We can probably take this a step further...

All tails of the limit have infinitely many $(x,u(x))$ where $F$ is bounded away from and below zero.

"Bounded away from" captures the existence of $m$ in the prior version. It means there is a gap between zero and all the values of $F$ from the called out $(x,u(x))$ points.

  • $\begingroup$ Thank you for your answer. I really like your input "All tails of the limit have infinitely many (x,u(x)) where F is bounded away from and below zero", I think it is an easy way to describe the situation. Unfortunately, there is a mistake in my question, "<0" should be replaced with ">0" (I'm gonna edit), but I think your reasoning still works just replacing "below zero" with "above zero", isn't it? $\endgroup$ – C. Bishop Feb 24 at 15:33
  • 1
    $\begingroup$ @C.Bishop : Yes, "bounded away from and above zero". $\endgroup$ – Eric Towers Feb 24 at 21:00

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