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Let $\{f_{n}\}$ be a sequence of functions on $\mathbb{R}$ such that $f_{n} \rightarrow f$ uniformly and $|f_{n}| \leq B$ (uniformly) for all $n$. Is it true that $f$ is also bounded?

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Notice that $t\mapsto |t|$ is a continuous map. –  leo May 1 '12 at 4:30
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up vote 2 down vote accepted

You do not need uniform convergence. You can show that if $(f_n)$ is a sequence of functions uniformly bounded by $B>0$, and if $(f_n(x))$ converges to $f(x)$ for each $x$, then $f$ is bounded by $B$.

With uniform convergence, you don't need uniform boundedness. It is enough that each $f_n$ is bounded, possibly by different bounds.

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Yes. The limit function will be inside an $\varepsilon$ neighborhood of some $f_n$, for $n$ large enough. So, the limit function will differ from $f_n$ by at most $\epsilon$

A more complete argument is the following: given $\varepsilon>0$, find $n_0$ such that $|f_n(x)-f(x)|<\varepsilon$ for all $n>n_0$ and for all $x$. Then, by the triangular inequality:

$$|f(x)|=|f(x)-f_n(x)+f_n(x)|\leqslant|f_n(x)-f(x)|+|f_n(x)|\leqslant \varepsilon+B$$

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