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How to show that $BV[0,1]$, the set of all functions of bounded variation, is not complete under the supremum norm? Can one explicitly construct a Cauchy sequence which does not converge or find a sequence in $BV[0,1]$ which converges in the sup-norm in $B[0,1]$ (the space of all bounded functions) but outside $BV[0,1]$?

Is there any other way of showing $(BV[0,1],\|\cdot\|_{\infty})$ is not complete?

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Every polynomial has bounded variation. Not every continuous function on $[0,1]$ has bounded variation. By the Weierstrass theorem, the closure of the space of polynomials under the supremum norm is the space of all continuous functions.

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  • $\begingroup$ Uh huh , that is indeed a sophisticated solution . very nice +1) . Is there any explicit constructive argument ? $\endgroup$ – user228169 Mar 2 '16 at 14:03
  • $\begingroup$ Could a proof be given using the characterization of B.V. functions as the difference of two monotonic functions ? Ref. math.ubc.ca/~feldman/m321/variation.pdf $\endgroup$ – Jean Marie Mar 2 '16 at 14:22

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