Let $f_{n}:S\to R$ and $g_{n}: S\to R$ be two sequences of uniformly continuous non-negative functions, $S$ is some normed space and assume that $g_{n}(s)$ converges uniformly to $g$.
Next, assume that for every $s$ in some closed subset $\Omega \subset S$
$$
\underset{n\to\infty}{\limsup}f_{n} (s) \leq g(s).
$$
Is the following true $$ \underset{n\to\infty}{\limsup} \underset{s\in\Omega}{\sup} f_{n} (s) \leq \underset{s\in\Omega}{\sup} g(s) $$
No. As a counterexample, let $\Omega = S = \mathbb{R}$. Now, let $$f_n(x) = \begin{cases} 0,& x\leq n-1 \\ x-n+1,& n-1 < x\leq n \\ 1-(x-n),& n < x\leq n+1 \\ 0,& x > n+1 \end{cases}$$ and $g_n\equiv c$ for all $n$ for some $0 < c < 1$. We have that $g_n$ converges uniformly to $g\equiv c$. Furthermore, $\lim_{n\to \infty} f_n(s) = 0\leq g(s) = c$ for all $s\in \Omega$, but $\sup_{s\in \Omega} f_n(s) = 1$ for all $n$, so $\lim_{n\to \infty} \sup_{s\in \Omega} f_n(s) = 1 > \sup_{s\in \Omega} g(s) = c$.