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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.

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  • $\begingroup$ For the example $f_j(x) = x/j$ on $\mathbb R$, there is no supremum since the $f_j$ are unbounded. $\endgroup$
    – Math1000
    Commented Oct 12, 2017 at 21:22

1 Answer 1

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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.

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  • $\begingroup$ Thanks. Your answer is very detailed. Would you mind also explaining $lim sup_{n\to \infty} f_n$? $\endgroup$
    – user398843
    Commented Feb 24, 2018 at 23:26
  • $\begingroup$ @user398843 Same reasoning as above, $limsupf_n(x)$ itself still a function where sup taking over infinite $f$ rather than finite $\endgroup$
    – LJNG
    Commented Aug 26, 2022 at 21:08

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