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Every $f \in L^\infty([0,1])$ induces a continuous linear functional on BV via $g \mapsto \int f g \mathrm{d}x$. I believe $L^\infty([0,1])$ is also separable in BV$^\ast$, while BV$^\ast$ is not separable, so it cannot be the case the BV$^\ast$ completion of $L^\infty([0,1])$ is all of BV$^\ast$.

Is there a nice characterization of the BV$^\ast$ completion of $L^\infty([0,1])$?

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    $\begingroup$ I have not thought through the details, but have you tried integrating by parts, since IIRC g is in BV if and only if it is an indefinite integral of a (Borel) measure on [0,1]? $\endgroup$
    – Yemon Choi
    Commented Sep 7, 2020 at 14:14

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