Interpretation of a limit inferior

Question

I was wondering if someone could provide me an explanation of the following liminf, because I am having a really hard time imagining what this really means.

Suppose that $u\in C([0,T]\times \Bbb{R}^m)\cap C^{1,2}((0,T]\times \Bbb{R}^m)$. Then, the following liminf $$ \liminf_{r\rightarrow \infty}\sup_{(t,x)\in [0,T]\times \Bbb{R}^m , |x|=r} u(t,x)\le 0 \tag{1} $$ implies that there exists a sequence of real numbers $\{X_n\}_{n\in\Bbb{N}}$ such that $X_n\rightarrow \infty$ as $n\rightarrow \infty$ and $$ \sup_{(t,x)\in [0,T]\times \Bbb{R}^m , |x|=X_n} u(t,x)\le \dfrac{1}{n}. $$

How can I interpret the left side of inequality (1). It looks like we are searching for the maximum value of $u$ on a "shell" for different $r$, then picking the smallest? And apparently (1) immediately implies that there is sequence, why is this true? Any hints are appreciated.

Answer 1

This is actually just some basic analysis and has nothing to do with PDEs, I guess you're overthinking it ;). By definition of $\liminf$, $\liminf_{k\rightarrow\infty}a_k=b$ in particular implies, that there exists a subsequence $a_{k_i}$, such that $\lim_{i\rightarrow\infty}a_{k_i}=b$. So in your example, $b=0$ and you can say e.g. $$a_k:=\sup_{(t,x)\in [0,T]\times \Bbb{R}^m , |x|=k} u(t,x),$$ and passing to a subsequence, you get $a_{k_i}\rightarrow 0$ for $i\rightarrow\infty$. By definition of convergence, for any $\epsilon=1/n$ you can find some $i(n)$, such that $a_{k_{i}}\leq 1/n$ for any $i\geq i(n)$. Wlog, you can choose $i(n)$ strictly increasing in $n$. You now call $X_n:=k_{i(n)}$, then you have $X_n\rightarrow\infty$ and $$ a_{X_n}=\sup_{(t,x)\in [0,T]\times \Bbb{R}^m , |x|=X_n} u(t,x)\leq\frac{1}{n} $$just as required.