Let $(f_n)_n$ be a sequence of continuous functions in $L^{\infty}(\mathbb{R}^N)$ and $(x_n)_n$ a sequence in $\mathbb{R}^N$. Then
$$
\limsup_n \left\|f_n\right\|_{\infty}=\sup_{(x_n)_n\subset \mathbb{R}^N}\left\{\limsup_n \left|f_n(x_n)\right|\right\}.
$$
My attempt:
to clarify the notation, let $\Omega$ be the set of all sequences ${\bf x}=(x_k)_k \subset \mathbb{R}^N$.
For $n$ fixed, we have $\left\|f_n\right\|_{\infty}=\sup_{\Omega} \left|f_n(x_n)\right|\geq \left|f_n(x_n)\right|$, for all ${\bf x}\in \Omega$. Taking the limsup on the inequality, we have
$$
\limsup_n \left\|f_n\right\|_{\infty}\geq\limsup_n \left|f_n(x_n)\right|.
$$
Since the left hand side does not depend on the sequence ${\bf x}$, we can take the supremum and get one of the inequalities we want. but for the other one, an argument like this doesn't seem to apply, as i think the best we have to start is "for all $\epsilon>0$, $\left|f_k(x_k)\right|\leq\limsup_n \left|f_n(x_n)\right|+\epsilon$, for all but finetely many (say, $m=m(\epsilon, {\bf x})$) indexes $k$ ". any tips will be appreciated.
Best Answer
For each $n=1,2,...$, let $x_{n}$ be such that $\|f_{n}\|_{\infty}-1/n<|f_{n}(x_{n})|$, then $\limsup_{n}(\|f_{n}\|_{\infty}-1/n)=\limsup_{n}\|f_{n}\|_{\infty}+\limsup_{n}(-1/n)=\limsup_{n}\|f_{n}\|_{\infty}$. But $\limsup_{n}\|f_{n}\|\leq\limsup_{n}|f_{n}(x_{n})|\leq\sup_{(x_{n})_{n}\subseteq\mathbb{R}^{n}}\limsup_{n}|f_{n}(x_{n})|$.
Note that $\limsup_{n}(a_{n}+b_{n})=\limsup_{n}a_{n}+\lim_{n}b_{n}$ for convergent $(b_{n})$.