TITLE: Can the limit of a sequence of bounded functions be unbounded?
QUESTION [9 upvotes]: It is assumed that the sequence of functions converges and that each function is bounded, then can the limiting function be unbounded?

REPLY [8 votes]: Yes, if you only have pointwise convergence. Take $(f_n)_n$ defined by
$$
f_n(x) = x^2 \mathbb{1}_{[-n,n]}(x), x\in\mathbb{R}.
$$
This converges pointwise to the function $f\colon x\in\mathbb{R}\mapsto x^2$, which is not bounded. But each $f_n$ is itself bounded (namely, $\lVert f_n\rVert_\infty = n^2$).
Following a comment: however, if it exists, the uniform limit (i.e., with regard to the supremum norm $\lVert\cdot\rVert_\infty$) of a sequence of real-valued bounded functions will be bounded. (Follows e.g. from the fact that the space of bounded-real valued functions is complete for the sup norm, see this); or from a direct proof$^{(\dagger)}$).

$(\dagger)$ Taking $\varepsilon = 1$, there exists $N\geq 0$ such that $\lVert f-f_n\rVert_\infty \leq 1$ for all $n\geq N$. In particular, for this specific, fixed $N$, by the triangle inequality
$\lVert f\rVert_\infty \leq \lVert f_N\rVert_\infty+1$.

REPLY [4 votes]: Let $f_n(x)=|x|$ for $|x|\le n$, and let $f(x)=n$ for $|x|\gt n$.