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${f_n}$ is a sequence of continuous functions on R, and $f_n$ converges to f uniformly on R. If each of the functions $f_n$ is bounded, show that this does not imply that f is bounded.

Please help with the counterexample - I don't understand how to solve this.

EDIT: What if f was instead convergent on every finite interval [a,b] rather than on R? How does this change the result of the problem? I'm confused how to find a "loophole" in this case.

kiwifruit
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  • Hmm, if I take f(x)=x which is unbounded, I can't seem to think f anything converging to this.. – kiwifruit Mar 07 '14 at 04:53
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    The terms $x-1,\ x-1/2, \cdots$ aren't bounded. – coffeemath Mar 07 '14 at 05:16
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    There is no counterexample of this problem. See this. – Hanul Jeon Mar 07 '14 at 05:22
  • But this proves that ${f_n}$ is bounded, not f – kiwifruit Mar 07 '14 at 05:35
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    $f_n$ is uniformly bounded means $|f_n(x)|\le M$ for some $M$ regardless of $n$ and $x$, as uniform limit of $f_n$, we have $|f_n(x)-f(x)|<1$ for all $x$ once $n$ is large enough, then $|f(x)|\le 1+M$ for all $x$. – TTY Mar 07 '14 at 05:41
  • What if f was instead convergent on every finite interval [a,b] rather than on R? How does this change the result of the problem? I'm confused how to find a "loophole" in this case. – kiwifruit Mar 07 '14 at 05:56

1 Answers1

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If the sequence is uniformly convergent then we can find some $n$ such that $|f(t)-f_n(t)| < 1 $ for all $t$. Since $f_n$ is bounded by, say, $B$, then we must have $f$ be bounded by $B+1$.

This has nothing to do with continuity, just uniform convergence and boundedness.

copper.hat
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