Let $\{a_n\}\subset [0,\infty )$. It is well know that if $$\sum_{n \in \mathbb{N}}a_n < \infty$$ then we must have: $a_n \to 0$ as $n \to \infty$.
Now suppose that $a_n=a_n(u)$ with $\{a_n\}\subset [0,\infty )$ and $u \in U$ fixed in some metric space $U$ (possibly infinite dimensional).
Asssume the following condition: there exists $C>0$ such that $$\sum_{n \in \mathbb{N}}a_n(u) < C$$ for every $u \in U$. I remark that $C$ is independent of $u$.
Can I conclude that $\sup_{u} |a_n(u)| \to 0$?
Or do I need to assume that $U$ is compact?
The topology of $U$ makes no difference since $U$ is really just acting as an index set here. Take for example $U = \mathbb N$ and consider the sequences $$ a_n(u) = \begin{cases} 1, & \text{if $u = n$}, \\ 0, & \text{otherwise.} \end{cases} $$ (It's just the infinite grid with diagonal $1$.)
Then $$ \sum_{n \in \mathbb N} a_n(u) = 1 $$ for all $u$ (i.e., summing across the grid) and $$ \sup_{u} a_n(u) = 1 $$ for all $n$ (i.e., taking max/sup down the grid).
Change $U$ to be any other (infinite) set and you can play exactly the same game, it just won't look as neat if you can't easily compare the $u$ indices and the $n$ indices.