It is assumed that the sequence of functions converges and that each function is bounded, then can the limiting function be unbounded?
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You should be clear about what "converges" means. I assume you mean "converges point wise". – zhw. Jul 27 '16 at 18:16
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You're right,I should've mentioned that. I guess I'm looking for uniform convergence, since the sequence of functions is supposed to be Cauchy wrto the infinite norm. – DpS Jul 27 '16 at 18:21
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The uniform limit of bounded functions is bounded. Just choose $\epsilon =1$ in the definition. – zhw. Jul 27 '16 at 18:23
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If the upper bound is unbounded, you have it. Take $f_n(x)=n$. – Dec 14 '18 at 20:19
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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 $\lVert f\rVert_\infty \leq \lVert f_N\rVert_\infty+1$.
Clement C.
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You can even make the $f_n$'s continuous if you want, setting $$ f_n(x) = x^2 \mathbb{1}{[-n,n]}(x) + n^2\mathbb{1}{\mathbb{R}\setminus[-n,n]}(x), x\in\mathbb{R}. $$ – Clement C. Jul 27 '16 at 18:10
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What if I also add the condition that the sequence of functions should be Cauchy with respect to the infinite norm? – DpS Jul 27 '16 at 18:16
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Then you are good: the space of bounded-real valued functions is complete for the sup norm. (see e.g. these lecture notes or this answer). – Clement C. Jul 27 '16 at 18:21
