# How to “explain in words” this expression involving liminf

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.

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.

• 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? – C. Bishop Feb 24 at 15:33
• @C.Bishop : Yes, "bounded away from and above zero". – Eric Towers Feb 24 at 21:00