TITLE: the upper bound of the supremums of the sequence
QUESTION [5 upvotes]: Let $f_{n}:S\to R$ be the sequence of uniformly continuous non-negative functions, $S$ is some normed space, such that for every $n$
$$
\underset{s\in C}{\sup} f_{n}(s) < \infty.
$$ 
Next, assume that for every $s$ in some subset $C \subset S$, with $S$ compact or only bounded) 
$$
\underset{n\to\infty}{\limsup}f_{n} (s) \leq g(s),
$$
where $g(s)$ is continuous and assume $\underset{s\in C}{\sup} g(s) < \infty$, if $S$ is not compact.
Is the following true
$$
\underset{n\to\infty}{\limsup} \underset{s\in C}{\sup} f_{n} (s) < \infty
$$
?

REPLY [3 votes]: This is not true. Take $S = C = [0, 1]  \subset \mathbb{R}$. Let $f_n$ be the function that is constant zero, except for a linear spike of height $n$ between $1/(n + 1)$ and $1/n$. Then note that $\limsup_n f_n(s) =  0$ for all $s \in [0,  1]$,so we may set $g(s)  = 0$.  Also, since each $f_n$ is continuous on a compact domain, each $f_n$ is uniformly continuous, and also  $\sup_{s \in  [0, 1]} f_n(s) = n <  \infty$. 
However, $\limsup_n \sup_{s \in [0, 1]}f_n(s) = \limsup_n n = \infty$.