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Consider $BV[a,b]$ the space of all bounded variation functions on a real interval $[a,b]$, endowed with the total variation norm $TV$. $AC[a,b]$, the space of absolutely continuous functions, is a subspace of $BV$. Is it closed?

MickG
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    The total variation "norm" is not a norm. – zhw. Jul 05 '16 at 22:35
  • I'm not sure of the relevance of $BV[a,b] \subset L^1[a,b].$ The equivalence classes do not match up. Changing a BV function at one point can change it's BV norm, but is irrelevant for the L^1 norm. – zhw. Jul 05 '16 at 23:10
  • @Ian By the way, your sequence $f_n$ is not Cauchy in BV. – zhw. Jul 05 '16 at 23:29

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The quantity $\|f\| = |f(a)| + \|f\|_{TV}$ is a natural norm on $BV[a,b].$ If $f\in AC,$ then this norm equals $|f(a)| + \int_a^b|f'|.$ Suppose $f_n$ is a sequence in $AC$ that is Cauchy in this norm. Then $f_n(a)$ is a Cauchy sequence in $\mathbb R,$ and $f_n'$ is Cauchy in $L^1[a,b].$ Thus $f_n(a) \to c$ in $ \mathbb R$ and $f_n'\to g$ in $L^1[a,b].$ We find then that $f_n(x) \to c + \int_a^x g$ in $BV.$ The last function is in $AC,$ proving $AC$ is closed in $BV.$

zhw.
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  • That seems to only work if $[a,b]\spse[0,1]$, But I guess I can just change the point from 0 to any point in the interval, and the limits of integration from $\int_0^1$ to \int_a^b$ to apply what you say in your answer to any interval. Right? Even infinite intervals perhaps? – MickG Jul 06 '16 at 07:37
  • Whoops, I am SO used to my LaTeX shortcuts that I typed \spse for \supseteq ($\supseteq$) above :). That is, $[a,b]\spse[0,1]$ meant $[a,b]\supseteq[0,1]$. – MickG Jul 06 '16 at 07:56
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    Yes, in my mind we were on $[0,1]$ all of a sudden. I'll edit. – zhw. Jul 06 '16 at 15:56
  • Is it important to choose $a$ as opposed to any point in [a,b]? I would guess not. – MickG Jul 06 '16 at 17:33
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    @MickG No, any point in that interval will do. – zhw. Jul 06 '16 at 18:16