Supremum for a Sequence of Functions I understand that the supremum of a set is the least upper bound of the set. For example, $\sup \ (a,b) = b$. I understand that the supremum of a sequence is the least upper bound of the sequence. For example, if $(a_n) = \{1,2, \ldots\}$, then $\sup (a_n) = \infty$. 
I think I understand that the supremum of a function defined on an interval $I$ is just the least upper bound of the function on $I$. For example, if $f(x) = x^2$ and $I = [0,5)$, then $\sup f(x) = 25$. 
But what is the supremum of a sequence of functions supposed to be? If you have $f_j(x) = x/j$ defined on, say, $\mathbb{R}$, is the supremum supposed to be a function $f(x)$ that is an upper bound for $f_j(x)$ at each $j$ and each $x$, or is it a real number? What about the $\lim \sup f_j(x)$? I just can't wrap my head around it, and can't find a definition in any of my books. 
 A: It is normal to talk about supremum for sets of real numbers. So, for example, for $f(x) = x^2$ and $I = [0,5)$, we have
$$
\sup f(x) = \sup_{x \in I} f(x) = \sup  \{y \in \mathbb{R}: y = f(x), x \in I\}.
$$
Supremums for sequences of functions should be the same, so if $\{f_n\}$ is sequence of functions with $n \in \mathbb{N}$, one typically has 
$$
\sup f_n(x) = \sup_{n \in \mathbb{N}} f_n(x) = \sup \{ y \in \mathbb{R}: y = f_{n}(x), n \in \mathbb{N}\},
$$
where the set on the right hand side will depend on $x$. In other words, the supremum of a sequence of functions is still taking the supremum of a set of real numbers, but you will have a different set of real numbers for each $x$ value. In this sense, the supremum of a sequence of function is itself a function of $x$.
EDIT: Also, one can consider 
$$
\sup f_n(x) = \sup_{x \in I} f_n(x) = \sup \{y \in \mathbb{R}: y = f_n(x) , x \in I\},
$$
so that the set you are taking the supremum over depends on $n$ instead of $x$, but this is less commonly used I think. 
