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Show that $BV[a, b]$ is not dense in $B[a, b]$ under the metric $||f||_\infty$.

I was wondering if I could get a hint.

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Weierstrass Approximation theorem is for continuous functions isn't it ? –  Dominic Michaelis Mar 1 '13 at 12:33
    
Ah, yes, my bad –  icaruss Mar 1 '13 at 12:38
    
I guess that $BV$ is closed in $B$. If this is true, and if you can prove it (of course! :-) ) then you are finished because clearly there exist bounded functions which are not of bounded variation such as $\sin(1/x)$ (in a neighborhood of $0$). –  Giuseppe Negro Mar 1 '13 at 12:41
    
@GiuseppeNegro But $BV$ is not closed in $B$, consider $x \mapsto x^2 \cdot \cos(1/x) \cdot \chi_{]1/n,1]}(x)$. –  Thomas Mar 1 '13 at 13:29
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up vote 5 down vote accepted

Suggestion: consider the function $1_\mathbb{Q}$ which takes the value 1 at every rational and 0 at every irrational. Show that if $\|f - 1_\mathbb{Q}\|_\infty < 1/2$ then $f$ has unbounded variation.

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Great example, +1. –  1015 Mar 1 '13 at 13:39
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